The work done during expansion is 2.73×10³J and the heat transferred to the surrounding is -2.73×10³ J
What is isothermal process ?The thermodynamic process which takes place in constant temperature of the system is called isothermal process.
What is work done in isothermal process?Due to change in volume of the system at constant temperature, some work is done .As, the temperature of gas is fixed. This process is isothermal.Now, work done (W) = nRT ln ( V2/V1)where W = work done by the systemn= number of moles
R= universal gas constant
V2= volume of the system after expansion
V1= volume of the system before expansion
W= (1)(8.31)(273)ln ( 10/3)= 2.73×10³ J.
Also in isothermal process the amount of work done by the system = -( the amount of heat transferred to the surrounding)So the amount of heat transferred is -2.73×10³ JThus, we can conclude that a) the amount of work done is 2.73×10³J and b) the amount of heat transferred to the surrounding is -(2.73×10³J).
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This technique allowed multiple peoples DNA to be compared to look for
similarities and differences in order to solve crime
DNA Examination
DNA Elimination
ODNA Sequencing
ODNA Fingerprinting
*2
DNA sequencing is the technique which allows multiple peoples DNA to be compared to look for similarities and differences in order to solve crime and is denoted as option C.
What is DNA?This is referred to as the deoxyribonucleic acid and contains the genetic components of organisms and is also located in the nucleus. On the other hand, DNA sequencing refers to the process of determining the nucleic acid sequence.
Each organisms has a unique DNA sequence which is why it is used to identify individuals through the use of bodily fluids such as saliva, blood etc.
It is used to compare the similarities and differences in order to solve crime thereby making it the most appropriate choice.
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An astronaut on a spacewalk 200km above Earth drops a hammer (mass 2kg), which goes into orbit about the Earth (radius 6,400km) How long does it take the hammer to orbit the Earth?
Answer:
See below
Explanation:
T = period of orbit = sqrt ( 4 pi^2 r^3 / (G Me) )
G = gravitational constant 6.6743 x 10^-11 m^3 / (kg-s^2)
Me = mass of earth = 6 x 10^24 kg
r = radius = 6600 km = 6,600,000 m
plug in the values to find T = 5323.75 seconds
(check my math)
A flat circular coil carrying a current of 8.80 A has a magnetic dipole moment of 0.194 A⋅m2 to the left. Its area vector A⃗ is 4.0 cm2 to the left.
a) How many turns does the coil have?
b) An observer is on the coil's axis to the left of the coil and is looking toward the coil. Does the observer see a clockwise or counterclockwise current?
c) If a huge 45.0 T external magnetic field directed out of the paper is applied to the coil, what magnitude of torque results?
d) What direction of torque results?
Hi there!
a)
We can use the equation for the magnetic dipole moment to solve for the number of turns:
[tex]\mu_m = NIA\vec{n}[/tex]
[tex]\mu_m[/tex] = Magnetic dipole moment (0.194 Am²)
N = Number of loops (?)
A = Area of loop (4.0 cm²)
[tex]\vec{n}[/tex] denotes the area vector, or the normal line perpendicular to the area.
We first need to convert cm² to m² using dimensional analysis.
[tex]4.0 cm^2 * \frac{0.01m}{1 cm} * \frac{0.01 m}{1cm} = 0.0004 m^2[/tex]
Rearranging the equation to solve for 'N':
[tex]N = \frac{\mu_m}{IA}\\\\N = \frac{0.194}{(8.8)(0.0004)} = \boxed{55.11 \text{ turns}}[/tex]
**Since we cannot have part of a turn, the coil has about 55 turns.
b)
For this, we can use the Right-Hand-Rule for current. Looking at the coil from the left with your curled fingers going around the coil with the fingertips pointing through and to the left in the direction of the magnetic moment, your thumb points in the COUNTERCLOCKWISE direction.
c)
Now, let's use the equation for the torque produced by a magnetic field:
[tex]\tau = \mu_m \times B[/tex]
This is a cross-product, but since our magnetic field is perpendicular to the magnetic moment, we can disregard it.
Plugging in the values for the magnetic moment and the magnetic field:
[tex]\tau = 0.194 * 45 = \boxed{8.73 Nm}[/tex]
d)
Using the other RHR (current, field, force), the coil will spin about its vertical axis in the field. In more detail, if you look at the coil from the left-hand side with its opening towards you, from this perspective, the left of the coil will come towards you, and the right side of the coil will move away.
State the newton's law of motion and give
application of each law.
Answer:
In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
Explanation:
The four tires of an automobile are inflated to a gauge pressure of 2.02×10^5 Pa. Each tire has an area of 217 cm^2 in contact with the ground. Determine the weight of the automobile.
The weight of the automobile is 17,533.6 N.
Weight of the automobile
The weight of the automobile is calculated as follows;
P = F / A
F = (W/4)
P = (W/4) / A
P = W/4A
W = 4AP
where;
P is pressure A is areaW = (4)(217 x 10⁻⁴ m²)(2.02 x 10⁵ Pa)
W = 17,533.6 N
Thus, the weight of the automobile is 17,533.6 N.
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Two spheres with uniform surface charge density, one with a radius of 7.4 cm and the
other with a radius of 5.0 cm, are separated by a center-to-center distance of 38 cm. The
spheres have a combined charge of +55μC and repel one another with a force of 0.62 N.
Assume that the charge of the first sphere is greater than the charge of the second sphere.
Question 1.
What is the surface charge density in the sphere of radius 7.4?
Question 2
What is the surface density on the 2nd sphere?
EXPRESS ANSWER USING TWO SIGNIFICANT FIGURES.
(a) The surface charge density in the sphere of radius 7.4 is 0.0322 C/m³.
(b) The surface density on the 2nd sphere is 3.48 x 10⁻⁴ C/m³.
Total charge of the spheresF = kq₁q₂/r²
Fr² = kq₁q₂
q₁q₂ = Fr²/k
where;
r is distance between the chargesk is Coulomb's constantq₁q₂ = (0.62 x 0.38²) / (9 x 10⁹)
q₁q₂ = 9.95 x 10⁻¹² C
q₂ = 9.95 x 10⁻¹² C/q₁
From the question;
q₁ + q₂ = 55 x 10⁻⁶
q₁ + 9.95 x 10⁻¹² /q₁ = 55 x 10⁻⁶
q₁² + 9.95 x 10⁻¹² = 55 x 10⁻⁶q₁
q₁² - 55 x 10⁻⁶q₁ + 9.95 x 10⁻¹² = 0
solve the quadratic equation using formula method;
q₁ = 5.48 x 10⁻⁵ C
q₂ = 55 x 10⁻⁶ C - 5.48 x 10⁻⁵ = 1.82 x 10⁻⁷ C
Volume of the first sphereV1 = ⁴/₃πr³
V1 = (⁴/₃ π)(0.074)³ = 1.7 x 10⁻³ m³
Surface charge density = (5.48 x 10⁻⁵ C) / (1.7 x 10⁻³ m³) = 0.0322 C/m³
Volume of the second sphereV2 = (⁴/₃ π)(0.05)³ = 5.236 x 10⁻⁴ m³
Surface charge density = ( 1.82 x 10⁻⁷ C) / (5.236 x 10⁻⁴ m³) = 3.48 x 10⁻⁴ C/m³
Thus, the surface charge density in the sphere of radius 7.4 is 0.0322 C/m³.
The surface density on the 2nd sphere is 3.48 x 10⁻⁴ C/m³.
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do you think an Electromagnet can be used for separating plastic bags from a garbage heap? explain
Answer:
No
Explanation:
Plastic bags are not magnetic materials, only magnetic materials (such as iron) can be attracted by the magnet.
Hope this helps.
A proton moving along the positive x-axis enters a uniform magnetic field which is directed along the
positive y-axis. If the magnitude of the field is 0.7 T, and the charge and speed of the proton are 1.5 x10-19 C
and 3.0 x 106 m/s respectively. Calculate the force acting on the proton
Answer: The force acting on the proton of charge 1.5 x10-19 C moving with velocity 1.5 x10-19 C under the influence of a magnetic field of 0.7 T will be 3.15×10^-13 N.
Explanation: To find the answer we need to know more about the Lorentz magnetic force.
What is the Lorentz magnetic force acting on the proton?Consider a proton of charge q moving with a velocity v in a magnetic field, then the Lorentz magnetic force exerted on the proton can be expressed as,F= q (v× B)
[tex]F= qvBsin\alpha[/tex] where, [tex]\alpha[/tex] is the angle between v and B.
In the question, it is given that,[tex]B=0.7 T\\q=1.5*10^{-19}C\\v= 3*10^{6}m/s.\\\alpha =90 degree.\\[/tex] because, from the question it is clear that the proton is moving along x axis and the magnetic field is along the y axis.
Thus, we can find the force acting on the proton as,[tex]F=qvBsin\alpha =1.5*10^{-19}C*3*10^6 m/s*0.7T*sin (90)\\F=3.15*10^{-13}N[/tex]
Thus, we can conclude that the Lorentz force acting on the proton will be 3.15×10^-13 N.
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In response to a magnetic field of 0.7 T, a proton with a charge of 1.5 x 10-19 C travelling at a speed of 1.5 x 10-19 C will experience a force of 3.15 x 10^-13 N.
We need to learn more about the Lorentz magnetic force in order to locate the solution.
What does the proton experience as the Lorentz magnetic force?If you imagine a proton with charge q travelling at speed v in a magnetic field, you can write down the Lorentz magnetic force acting on the proton as,F= Q (v× B)
[tex]F=QvBsin\alpha[/tex]
[tex]\alpha[/tex] the angle between v and B is, where.
It is stated in the query that, the magnetic field is along the y axis and the proton is travelling along the x axis. Thus, the angle will be 90 degrees.As a result, we can identify the proton's driving force as,[tex]F=3.15*10^{-13}N[/tex]
Thus, we can infer that the proton will be subject to a 3.15 10^-13 N Lorentz force.
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At t=0s a small "upward" (positive y) pulse centered at x = 4.0 m is moving to the right on a string with fixed ends at x=0.0m and x = 13.0 m . The wave speed on the string is 3.5 m/s . At what time will the string next have the same appearance that it did at t=0s?
The next time the string will have the same appearance that it did at t=0s is 2.29 s.
Frequency of the wave
v = fλ
f = v/λ
where;
λ is wavelengthhalf of the upward pulse is a quarter of wavelength = ¹/₄ x 4 m = 1 m
f = 3.5/1
f = 3.5 Hz
Time of motion when the pulse is at 4 mt1 = 4/3.5 = 1.143 s
The next time the string will have the same appearance that it did at t=0s.
d = 4 m x 2 = 8 m
t2 = 8/3.5
t2 = 2.29 s
Thus, the next time the string will have the same appearance that it did at t=0s is 2.29 s.
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emily is standing 150 feet from a circular target with radius 3 inches. will she hit the target if her aim is off by 0.2 degrees?
Answer:
no
Explanation:
The angle subtended by the radius of the target at Emily's distance can be found using the tangent relation.
tan(α) = opposite/adjacent = (1/4 ft)/(150 ft) = 1/600
The angle is found using the inverse relation -
α = arctan(1/600) ≈ 0.095°
If Emily's aim is off by 0.2°, she will miss the target by several inches.
Emily's projectile will miss her aiming point by ... (150 ft)tan(0.2°) ≈ 0.524 ft ≈ 6.28 in
a body has a mass of 2kg.it accelerats from 20m/s to 40m/s in 4 seconds.the resultant force is
The resultant force is 8N
Given that mass is 2kg , v= 40m/s, u =20m/s and we need to calculate resultant force
F=ma
m is given
so for a
v-u/t=a { first equation of motion }
40-20/4= 4
so a=4
F = ma =2*4 = 8N
The difference between the forces that are acting on an object as part of a system is known as the resultant force.
v = u + at is the first equation of motion. Here, v denotes the end speed, u the starting speed, an acceleration, and t the passage of time. The first equation of motion is provided by the velocity-time relation, which may be used to calculate acceleration.
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A toroidal solenoid has 600 turns, cross-sectional area 6.90 cm2, and mean radius 4.30 cm.
a) Calculate the coil's self-inductance.
b) If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms , calculate the self-induced emf in the coil.
c) The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a?
(a) The coil's self-inductance is 7.26 mH.
(b) The self-induced emf in the coil is 7.26 V
(c) The direction of the induced emf is from b to a.
Coil's self-inductance
L = N²μA/I
L = (600² x 4π x 10⁻⁷ x 6.9 x 10⁻⁴)/(0.043)
L = 7.26 x 10⁻³ H
L = 7.26 mH
Self-induced emf in the coilemf = N(ΔBA)/t
where;
B is magnetic fieldA is area N is number of turnst is timeB = μNI/L
B1 = (4π x 10⁻⁷ x 600 x 5)/0.043
B1 = 0.0876 T
B2 = (4π x 10⁻⁷ x 600 x 2)/0.043
B2 = 0.035 T
emf = NΔBA/t
emf = (600)(0.0876 - 0.035)(6.9 x 10⁻⁴) / (3 x 10⁻³)
emf = 7.26 V
The direction of the induced emf is always opposite to the direction of the applied current.
Thus, the direction of the induced emf is from b to a.
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A fisherman and his young son are in a boat on a small pond. Both are wearing life jackets. The son is holding a large helium filled balloon by a string. Consider each action below independently and indicate whether the level of the water in the pond, Rises, Falls, is Unchanged or Cannot tell.
The son pops the helium balloon.
The fisherman knocks the tackle box overboard and it sinks to the bottom.
The son finds a cup and bails some water out of the bottom of the boat
The fisherman lowers himself in the water and floats on his back.
The fisherman lowers the anchor and it hangs one foot above the bottom of the pond.
The son gets in the water, looses his grip on the string, letting the balloon escape upwards.
Based on the weight of the objects in the water, the water level in the given scenario is as follows;
unchangedrisesfallsunchangedrisesrisesWhat changes will be observed in each of the given scenario?The rise or fall of a fluid when an object is placed in it is determined by the density, mass, and volume of the object.
According to Archimedes' principle, the upthrust or upward force acting on a body fully or partially immersed in a fluid, is equal to the weight of the fluid displaced.
Considering the given situations:
The son pops the helium balloon - water level is unchanged because the weight of the balloon is negligible.Fisherman knocks the tackle box overboard and it sinks to the bottom - water level will rise since the tackle box has significant weightSon finds a cup and bails some water out of the bottom of the boat - water level will fall since some volume of water is being removed and added to the boat.Fisherman lowers himself in the water and floats on his back - water level is unchanged because no additional weight is added to the waterFisherman lowers the anchor and it hangs one foot above the bottom of the pond - water level will rise since the anchor will displace some volume of water equal to its weight in the pondSon gets in the water, looses his grip on the string, letting the balloon escape upwards- water level will rise since the weight of the balloon and string are significant.In conclusion, the rise in water level depends on the weight of the objects.
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Technetium -99 (half-life = 6.01) is used in the medical imaging. How many half-lives would go by in 44.0h?
The number of half-lives that would go by in 44 h is 7 half-lives
What is half life?This is the time taken for half a substance to decay
How to determine the number of half lifeThe number of half-lives that will elaspe after 44 h can be obtained as illustrated below:
Half-life (t½) = 6.0 hTime (t) = 44 hNumber of half-lives (n) =?n = t / t½
n = 44 / 6.01
n = 7
Thus, 7 half-lives will elaspe after 44 h
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A 45-kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.
_______m
The altitude or height of the pole vaulter as she crosses the bar is 4.04 m.
What is the height of the pole vaulter?The height of the pole vaulter is determined from the change in kinetic energy which is equal to the potential energy at that height.
Potential energy = Change in kinetic energymgh = m(v - u)²/2h = (v - u)²/2g
h = (10 - 1.1)²/2 * 9.8
h = 4.04 m.
In conclusion, the height is determined from the potential energy at that height.
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if f=30 m=4 calculate value of m in equation
m=f/a
Answer:
a = 7.5
Explanation:
m = f/a
f = 30
m = 4
Thus;
a = f/m
a = 30/4
a = 7.5
A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg.
The tension in Rope A is 7.42x10^2 N, and the tension Rope B carries is 7.52x10^2 N.
What is the tension in rope C?
Answer: A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg. The tension in Rope A is 7.42x10^2 N, and the tension Rope B carries is 7.52x10^2 N. Then, the tension in rope C is 376N
Explanation: To find the correct answer, we have to know more about the Basic forces that acts upon a body.
What is force and which are the basic forces that acts upon a body?A push or a pull which changes or tends to change the state or rest, or motion of a body is called Force.Force is a polar vector as it has a point of application.Positive force represents repulsion and the negative force represented attraction.There are 3 main forces acting on a body, such as, weight mg, normal reaction N, and the Tension or pulling force.How to solve the problem?Given that,
[tex]T_A=7.42*10^2N\\T_B=7.54*10^2N[/tex]
From the free body diagram, we get Tension in the rope C as,[tex]T_C=T_B sin30\\thus,\\T_C=7.52*10^2*0.5=376N[/tex].
Thus, we can conclude that the tension in the rope c will be 376N.
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376N is the tension in rope C.
In order to determine the right response, we must have a better understanding of the fundamental forces that affect a body.
What exactly is force, and what are the fundamental forces that affect a body?Force is a push or a pull that modifies or tends to modify the condition, rest, or motion of a body.Given that it has a point of application, force is a polar vector.Repulsion is represented by positive force, and attraction by negative force.A body is subject to three main forces: weight mg, normal response N, and tension or pulling force.How can the issue be resolved?Tension in the rope C is determined from the free body diagram as,[tex]T_c=T_Bsin30\\T_c=7.52*10*2*0.5=376N[/tex]
Thus, we can infer that the rope's c tension will be 376N.
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Please I need help! This is the last question I need for this assignment!
Part A
Compare the temperature change for cold sand and cold water when the same amount of hot water was added. What do you discover?
Answer:
When the same amount of heat is added to cold sand and cold water, the temperature change of sand will be higher because of its lower specific heat capacity.
What is specific heat capacity?
Specific heat capacity is the quantity
of heat required to raise a unit mass of
a substance by 1 kelvin.
Specific heat capacity of water and sand
{refer to the above attachment}
Δθ = Q/mc
Thus, for an equal mass of water and sand, when the same amount of heat is added to cold sand and cold water, the temperature change of sand will be higher because of its lower specific heat capacity.
10. When the force applied on an object is doubled, how does the pressure exert on the object change?
Answer:
It doubles as well.
Explanation:
Assuming that the area on which the force acts remains constant, and knowing that by definition:
P = F/A
We see that if we have a double force:
p = 2 F/A
Then the pressure is doubled.
Answer:
pressure doubles if area is constant
Explanation:
we know,
p=F/A
when force is double by keeping area constant
F=2F
Then,
Change in pressure (p') will be
P'=2F/A
or,p'=2×F/A
or,p'=2p(since, P=F/A)
therefore, when force is double by keeping area constant the pressure will increase by 2 times
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A person has a reasonable chance of surviving a car crash if the deceleration is no more than 30 'g's (g=9.8m/s2). Calculate the force on a 70 kg person undergoing this acceleration. What distance is travelled if the person is brought to rest at this rate from 100 km/h?
Thus, the distance traveled by automobile is [tex]=1.31m[/tex]
How do you find the distance traveled by automobile?Newton's second law can be used to compute the force being applied to the subject. The force is calculated as the product of the vehicle's acceleration and the person's mass. Newton's third law can be used to calculate the distance driven by a car.The mass of the person is [tex]m=65kg[/tex]
The deceleration of the automobile is [tex]a=30g[/tex]
The acceleration due to gravity is [tex]g=9.81m/s.[/tex]
The initial velocity of the automobile is [tex]v_{i}=100 km/h*\frac{1m/s}{3.6km/h} =27.78m/s[/tex]
Determination of the force exerted on the person:
The force applied to the person can be calculated using Newton's second law as follows:
[tex]F=ma[/tex]
[tex]30g[/tex]Here, m denotes the person's mass, while a denotes the car's acceleration. Because the driver is moving at 30g, the car's acceleration is assumed to be positive.
Substitute the values as 70 kg for [tex]m,(30g)[/tex] for a, and[tex]9.81m/s^{2}[/tex]
for g in the above equation.
[tex]F=70kg*30g[/tex]
[tex]=70kg*30g*9.81m/s^{2} (\frac{1N}{1kg*m/s^{2} })[/tex]
[tex]=20,601N[/tex]
Thus, the force acting on the person is [tex]=20,601N[/tex]
Determination of the distance traveled by automobile:
The final velocity of the car can be calculated using Newton's third law and represented as follows:
[tex]v_{f} ^{2}=v_{i} ^{2}+2ad[/tex]
[tex]d=\frac{v_{f} ^{2}=v_{i} ^{2}}{2a}[/tex]
Here [tex]v_{f}[/tex] is the final velocity of the automobile. The final velocity of the automobile becomes 0 m/s as the automobile comes to rest. The acceleration is taken to be negative because the automobile is decelerating at [tex]30g[/tex]
Substitute the values as [tex]0 m/s[/tex] for[tex]v_{f} =27.78m/s[/tex], for[tex]v_{i} ,(-30g)[/tex], for a, and [tex]9.81m/s^{2}[/tex]for g in the above equation.
[tex]d=\frac{(0 m/s)^{2} - (27.78m/s)^{2} }{2*(-30)*9.81m/s^{2} }[/tex]
[tex]=1.31m[/tex]
Thus, the distance traveled by automobile is[tex]=1.31m[/tex]
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An action force is 50 N to the left. The reaction force must be:
A. 50 N right
B. 50 N down
C. 50 N left
D. 50 N up
A rectangle measuring 30.0 cm by 40.0 cm is located inside a region of a spatially uniform magnetic field of 1.65 T, with the field perpendicular to the plane of the coil (Figure 1). The coil is pulled out at a steady rate of 2.00 cm/s traveling perpendicular to the field lines. The region of the field ends abruptly as shown.
a) Find the emf induced in this coil when it is all inside the field.
b) Find the emf induced in this coil when it is partly inside the field.
c) Find the emf induced in this coil when it is all outside the field.
The solution for the questions below is mathematically given as
induced emf=0induced emf=0.0132Vinduced emf=0What is the emf induced in this coil when it is all inside the field.?(A)
Generally, the equation for the flux is mathematically given as
[tex]\Phi _{B}=B.A[/tex]
[tex]\Phi _{B}=B.A\\\\\Phi _{B}=B(40*30*10^{-4}) ,[/tex]
So, the magnetic flux through the coil is constant.
From faradays law,
[tex]\varepsilon =-\frac{\mathrm{d} \Phi _{B}}{\mathrm{d} t}[/tex]
---(1) for the induced emf. Since magnetic flux is constant, LHS. of (1) =0
induced emf=0
(B)
Let x be the length of the coil's magnetic field area.
[tex]then, \Phi _{B}=B.A=B(40*x*10^{-4}) \\\\\frac{\mathrm{d}\Phi _{B} }{\mathrm{d} t}=40\\B*10^{-4}*\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
induced emf=0.0132V
(C)
In conclusion, Therefore, there is no variation in the magnetic flux across the coil when magnetic flux=0.
induced emf=0
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A 0.80-kg mass attached to the end of a string swings in a vertical circle (radius = 2.0 m). When the mass is at the highest point of the circle, the speed of the mass is 9.0 m/s. What is the magnitude of the force of the string on the mass at this position?
Answer:
Approximately [tex]25\; {\rm N}[/tex] assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
This mass is in a circular motion of radius [tex]r[/tex]. Hence, when the velocity of the mass is [tex]v[/tex], the acceleration of this mass should be [tex](v^{2} / r)[/tex]. The net force on this mass should be [tex](\text{net force}) = (m\, v^{2}) / r[/tex] towards the center of the circle.
When this [tex]m = 0.80\; {\rm kg}[/tex] mass is at the top of this circle, both gravitational pull and the force of the string (tension) point downwards. Hence, the net force on this mass would be:
[tex](\text{net force}) = (\text{weight}) + (\text{tension})[/tex].
Thus:
[tex]\begin{aligned} (\text{tension}) &= (\text{net force}) -(\text{weight})\\ &= \frac{m\, v^{2}}{r} - m\, g \\ &= m\, \left(\frac{v^{2}}{r} - g\right) \\ &= 0.80\; {\rm kg}\times \left(\frac{(9.0\; {\rm m\cdot s^{-1}})^{2}}{2.0\; {\rm m}} - 9.81\; {\rm m\cdot s^{-2}}\right) \\ &\approx 25\; {\rm kg \cdot m\cdot s^{-2}} \\ &= 25\; {\rm N}\end{aligned}[/tex].
If Earth were a perfect sphere, would you weigh more or less at the equator than at the poles? Explain
Answer:
You would weigh the same.
Explanation:
At the moment, since Earth is not a perfect sphere, the Earth "bulges out" at the equator, so you're further from the centre of the Earth. Since gravity acts through a body's center of mass, the further you are from the centre the weaker the gravitational acceleration you will feel, because gravity weakens over distance.
So, you're actually lighter at the equator than you'd be at the poles.
However, if the Earth was a perfect sphere, this "bulge" at the equator would not happen, and so you would weigh the same at the poles and at the equator.
Hope this makes sense.
Select the correct answer. Which graphs show the correct relationship between kinetic energy and mass? A. Graph representing a relationship between mass on the x-axis and kinetic energy on the y-axis. the line starts at the origin and increases upwards B. Graph representing the relationship between mass on the x-axis and kinetic energy on the y-axis. the curve starts at the origin and keeps on increasing C. Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis D. Graph representing a relationship between mass on the x-axis and kinetic energy on the y-axis. the line starts at the origin and increases as it goes with a small bend Reset Next
C. Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis.
What is kinetic energy?
Kinetic energy is the energy possessed by a body due to its motion.
K.E = ¹/₂mv²
where;
m is mass of the objectThe kinetic energy of a body is directly proportional to the mass of the object.
Thus, the correct relationship between kinetic energy and mass is Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis.
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an electron is moving with a speed of 0.85c in a direction opposite to that of photon. calculate the relative velocity of the electron and photon.
Answer:
1.85c
Explanation:
a photon moves at c, the electron is moving at 0.85c, and since they are moving in opposing directions, the relative speed would be 1.85c
The relative velocity of the electron and photon is equal to the speed of light in a vacuum, which is 3 × 10⁸ m/s The negative sign indicates that they are moving in opposite directions.
The formula for adding velocities in special relativity is given by:
V(relative) = (V₁ - V₂) ÷ (1 - (V₁ × V₂ ÷ c²))
where:
V₁ = velocity of the electron
V₂ = velocity of the photon
c = speed of light in a vacuum
The relative velocity is:
V(relative) = (0.85c - c) ÷ (1 - (0.85c × c ÷ c²))
V(relative) = (0.85c - c) ÷ (1 - 0.85)
V(relative) = (-0.15c) ÷ (0.15)
V(relative) = -c
Therefore, The relative velocity of the electron and photon is equal to the speed of light in a vacuum, which is 3 × 10⁸ m/s The negative sign indicates that they are moving in opposite directions.
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A baseball player holds a 1.42 N baseball in his hand, a distance of 34.0 cm from the elbow joint as shown in the figure below. The biceps attached at a distance of 2.75 cm from the elbow exerts an upward force of 12.6 N on the forearm. Consider the forearm and hand to be a uniform rod with a mass of 1.20 kg. Calculate the net torque acting on the forearm and hand.
Answer:
90.3N
Explanation:
⊥mg = (0.170 m)(1.20 kg) 9.81 m/s
τ ball = r⊥Wball = (0.340 m)(1.42 N) = − 0.483 N ⋅m
F − 2.001− 0.483 N ⋅m = 0
F = 2.484 N ⋅m
0.0275 m = 90.3 N
The net torque acting on the forearm and hand is 90.3N
What is torque?
Torque is a measure of the pressure that can motivate an object to rotate about an axis. simply as pressure is what causes an object to accelerate in linear kinematics, torque is what reasons an item to collect angular acceleration. Torque is a vector amount.
⇒mg = (0.170 m)(1.20 kg) 9.81 m/s
⇒torque = r⊥weight of the ball
⇒ (0.340 m)(1.42 N) = − 0.483 N ⋅m
⇒F = − 2.001− 0.483 N ⋅m = 0
⇒F = 2.484 N ⋅m
⇒0.0275 m = 90.3 N
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Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers.
a) For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas.
b) What magnetic field does the line produce at ground level as a percent of the earth's magnetic field, which is 0.50 G .
c) Is this value of magnetic field cause for worry?
Yes. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
No. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
Yes. Since this field is much greater than the earth's magnetic field, it would be expected to have more effect than the earth's field.
No. Since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.
(a) The magnetic field the line produced at ground level is 3.75 x 10⁻⁶ T.
(b) The lines magnetic field as a percent of the earth's magnetic field is 7.5%.
(c) No, since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.
Magnetic field the line produced at ground levelB = (μ x I) / (2πr)
B = (4π x 10⁻⁷ x 150) / (2π x 8)
B = 3.75 x 10⁻⁶ T
Percent of the earth's magnetic fieldx = 3.75 x 10⁻⁶ /0.5G
x = (3.75 x 10⁻⁶ ) / (0.5 x 10⁻⁴)
x = 0.075
x = 0.075 x 100% = 7.5%
Thus, we can conclude that, since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.
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a. A light wave moves through glass (n = 1.5) at an angle of 15°. What angle will it have when it moves from the glass into water (n = 1.33)? (4 points)
b. Draw a ray diagram to locate the image of the arrow, as refracted through the lens shown. Write 2 - 3 sentences describing the type of image and its size relative to the object. What type of mirror could be used to form an image of the same type and size? (8 points)
c. An object is located 65 cm from a concave mirror with a focal length of 45 cm. What is the image distance? Is the image real or virtual? (6 points)
Answer in detail with the correct units and steps to solve. Will mark brainliest.
A wave is a phenomenon that does not cause a permanent displacement in the particles of the medium through which it passes. And it transfers energy from one end of the medium to the other. Examples of waves include light waves, sound waves, water waves, x-rays, radiowaves, etc. Thus the required answers for each part of the question are:
a. The angle that the light wave would have is [tex]7.5^{o}[/tex].
b. The type of mirror that can be used is a plane mirror.
c. The image distance is 146.3 cm.
ii. The image formed by the mirror is a real image.
a. When a ray of light passes from one medium to another, then refraction occurs. The refraction depends on the refractive index of the medium considered.
Thus from Snell's law, we have:
refractive index, n, = [tex]\frac{Sin i}{Sin r}[/tex]
where: i is the angle of incidence, and r is the refracted angle.
Now given that n = 1.5, and i = 15.
Then;
n = [tex]\frac{Sin i}{Sin r}[/tex]
1.5 = [tex]\frac{Sin 15}{Sin r}[/tex]
Sin r = [tex]\frac{0.2588}{1.5}[/tex]
= 0.17253
r = [tex]Sin^{-1}[/tex] 0.17253
= 9.936
r ≅ [tex]10^{o}[/tex]
Since the light wave now moves from the glass into water, the determined refracted angle now becomes its angle of incidence in water. So that;
n = [tex]\frac{Sin i}{Sin r}[/tex]
1.33 = [tex]\frac{Sin 10}{Sin r}[/tex]
Sin r = [tex]\frac{0.17365}{1.33}[/tex]
= 0.1306
r = [tex]Sin^{-1}[/tex] 0.1306
= 7.504
r = [tex]7.5^{o}[/tex]
Therefore, the angle that the light wave would have is [tex]7.5^{o}[/tex].
b. The image formed would be the same size as that of the object. And also the same distance as that of the object to the pole of the lens.
The type of mirror that can be used is a plane mirror.
The ray diagram is attached to this answer.
c. From the mirror formula;
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
where; f is the focal length of the mirror, u is the object's distance to the mirror, and v is the image's distance to the mirror.
Given; u = 65 cm, and f = 45 cm, then:
[tex]\frac{1}{45}[/tex] = [tex]\frac{1}{65}[/tex] + [tex]\frac{1}{v}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{1}{45}[/tex] - [tex]\frac{1}{65}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{20}{2925}[/tex]
v = [tex]\frac{2925}{20}[/tex]
v = 146.25 cm
The image distance is 146.3 cm.
ii. The image formed by the mirror is a real image.
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A golf ball is driven from a point A with a speed of 40 m/s at an angle of elevation of 30°. On its downward flight, the ball hits an advertising hoarding at a height 15.1 m above the level of A, as shown in the diagram below. Find A. The time taken by the ball to reach its greatest height above A, B. The time taken by the ball to travel from A to B, C. The speed with which the ball hits the hoarding. 20
(a) The time taken by the ball to reach its greatest height is 2 seconds.
(b) The time taken by the ball to travel from A to B, C is 4.08 m/s.
(c) The speed with which the ball hits the hoarding is 69.3 m/s.
Maximum height of the projectile
H = u²sin²θ/2g
H = (40² x (sin 30)²) / (2 x 9.8)
H = 20.4 m
Time for the ball to travel 20.4 mh = ut - ¹/₂gt²
20.4 = (40 x sin 30)t - (0.5)(9.8)t²
20.4 = 20t - 4.9t²
4.9t² - 20t + 20.4 = 0
solve the quadratic equation using formula method;
t = 2 seconds
Time taken for the ball to travel A, B , CThis is the time of motion of the ball;
T = 2usinθ/g
T = (2 x 40 sin 30)/9.8
T = 4.08 s
Speed of the ball when it hits the hoardingv = u + gt
vy = 40 sin30 + (9.8 x 4.08)
vy = 59.98 ms
vx = 40 x cos30
vx = 34.64 m/s
vf = √(vy² + vx²)
vf = √(59.98² + 34.64²)
vf = 69.3 m/s
Thus, the time taken by the ball to reach its greatest height is 2 seconds.
The time taken by the ball to travel from A to B, C is 4.08 m/s.
The speed with which the ball hits the hoarding is 69.3 m/s.
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