The weight of a 100-kg person decreases by 4 N when he goes from sea level to mountain top at an altitude of 5000 m.
What determines the weight of a person?The weight of a person is determined by the mass of the body and the acceleration due to gravity.
The acceleration due to gravity, g is dependent on the mass of the earth, M the radius of the earth and the gravitational force constant , G.
Mathematically, the acceleration due to gravity at the mountain top is determined using the formula:
g = GM/r²where:
G = 6.67 × 10⁻¹¹ Nm²/kg²
M = 5.9736 x 10²⁴ kg
r = 638000 + 5000 = 6385000
g = (6.67 × 10⁻¹¹ * 5.9736 x 10²⁴ )(6385000)²
g = 9.77 m/s²
His weight at the mountain top will be:
weight = 100 * 9.77
weight = 977 N
Weight at sea level = 100 * 9.81 = 981 N
Decrease in weight = 981 - 977
Decrease in weight = 4 N
In conclusion, the weight of the man varies according to his distance from the earth.
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Find the orbital speed of an ice cube in the rings of Saturn. The mass of Saturn is 5.68 x 1026 kg, and use an orbital radius of 3.00 x 105 km. (G = 6.67 × 10-11 N ∙ m2/kg2)
The orbital speed of an ice cube in the rings of Saturn is determined as 11,237.7 m/s.
What is orbital speed?The orbital speed of an astronomical body or object is the speed at which it orbits around the center of mass of the most massive body.
Orbital speed of ice cube in the rings of SaturnThe orbital speed of ice cube in the rings of Saturn is calculated as follows;
v = √GM/r
where;
G is universal gravitation constantM is mass of Saturnr is the distance of the ice cube = 3 x 10⁸ mv = √(6.67 x 10⁻¹¹ x 5.68 x 10²⁶)/(3 x 10⁸)
v = 11,237.7 m/s
Thus, the orbital speed of an ice cube in the rings of Saturn is determined as 11,237.7 m/s.
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For the circuit shown below with a 12.0 V battery, find the total current through the battery and the current through each resistor with the following resistances: R1=5.0 Ohms, R2=10.0 Ohms, R3=15.0 Ohms, R4=33.0 Ohms. Show all your work!
The total current through the battery is 4.77 A.
The current through resistor R1 is 2.4 A.
The current through resistor R2 is 1.2 A
The current through resistor R3 is 0.8 A
The current through resistor R4 is 0.36 A.
Total resistance of the circuitThe total resistance of the circuit is calculated as follows;
1/Rt = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄
1/Rt = 1/5 + 1/10 + 1/15 + 1/33
1/Rt = 0.397
Rt = 1/0.397
Rt = 2.518 ohms
Total current flowing in the circuitI = V/Rt
I = 12/2.518
I = 4.77 A
Current in resistor R₁I₁ = V/R₁
I₁ = 12/5
I₁ = 2.4 A
Current in resistor R₂I₂ = 12/10
I₂ = 1.2 A
Current in resistor R₃I₃ = 12/15
I₃ = 0.8 A
Current in resistor R₄I₄ = 12/33
I₄ = 0.36 A
Thus, the total current through the battery is 4.77 A.
The current through resistor R1 is 2.4 A.
The current through resistor R2 is 1.2 A
The current through resistor R3 is 0.8 A
The current through resistor R4 is 0.36 A.
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In our solar system, which celestial object is known as the dwarf planet?
Answer:
unfournatletly
Explanation:
i have no clue sorry to waste ur time ill rather not say a answer that will be incorrect.
Answer:
Pluto
Explanation:
Pluto was part of our solar system till 2006In 2006 international scientific committee removed it from planets listIt's known as dwarf planet nowHow does an atom of rubidium-85 a rubidium ion with a +1 charge?
A. The atom loses 1 electron to have a total of 47.
B. The atom gains proton to have a total of 38.
C. The atom loses 1 electron to have a total of 36,
D. The atom gains 1 proton to have a total of 86.
Answer:
C. The atom loses 1 electron to have a total of 36
Explanation:
The number of electrons in a Rubidium atom is 37. Since the atom loses 1 electron, it has 36 left.
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.17 times a second. A tack is stuck in the tire at a distance of 0.351 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed
The tangential speed of the wheel is determined as 4.786 m/s.
Tangential speed of the wheel
The tangential speed of the wheel is calculated as follows;
v = ωr
where;
ω is angular speed in rad/sr is radius of the circular pathv = (2.17 x 2π rad)/s x 0.351 m
v = 4.786 m/s
Thus, the tangential speed of the wheel is determined as 4.786 m/s.
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Calculate the magnitude of the electric field at one corner of a square 1.10 m on a side if the other three corners are occupied by 4.05×10−6 C charges. What is the direction of the electric field?
The direction of magnetic field is south-east and the magnitude is
[tex]23.66[/tex] × [tex]10^{3} N/C[/tex].
Here, the magnitude of CD and BC will be cancelled, as they both are in the opposite direction and equal to each other.
the magnitude, towards the diagonal AC will result in CP, which is the direction of the electric field.
magnitude of electric field can be defined as :- The force per charge on the test charge is a straight forward way to define the size of the electric field.
To find the magnitude of the electric field use the formula
[tex]E = kq/ r^{2}[/tex]
inserting the values,
[tex]E = 9. 10^{9}[/tex] × [tex]4.05[/tex] × [tex]10^{-6} / 1.1 \sqrt{2}[/tex]
[tex]E= 36.45[/tex] × [tex]10^{3}[/tex] / [tex]1.54[/tex]
[tex]E = 23.66[/tex] × [tex]10^{3}[/tex] N/C
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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.
Calculate the magnitude of the tension in the string marked A. (You'll need to get the various positions from the graph. The ends of the strings are exactly on one of the tic marks.)
The magnitude of the tension in the string marked A is 52.5N
Generally, the equation for is mathematically given as
Let's take θ be an angle at A
So, tanθ = 3/8
Let's take α be an angle at B (Below X)
tanα = 5/4
Let's take β be an angle at C (Below x)
tanβ = 1/6
First we take the Horizontal Components
74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)
By solving the equation, we get
A = 78.9 - 0.668B … (1)
Now, we take the vertical components
74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)
By solving the equation, we get
40.07 = 1.015B
B = 39.5N
By substituting the value of B in equation (1)
A = 78.9 - 0.6668× 39.5
A = 52.5N
Hence, the magnitude of the tension in the string marked A is 52.5N
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K donates, or transfers, one electron to bromine, which has 7 electrons. Both K and Br are now stable with 8 electrons. K becomes a positive ion and Br becomes a negative ion. The positive K ion and the negative Br ion attract each other to form an ionic bond.
K is cation by losing of electron whereas Br is anion due to accepting of electrons.
Is charge appears when an atom lose or accept electron?Yes, the positive ion appears on K and become cation whereas the negative ion bears on Br which make it anion because of losing and gaining of electron by these atoms. This transferring of electrons leads to formation of ionic bonds between them.
So we can conclude that K is cation by losing of electron whereas Br is anion due to accepting of electrons.
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. What is the potential energy of a 0.40 kg ball at a height of 9.2 m?
Explanation:
gravitational potential energy = mgh (must be in S.I. unit)
m= 0.4 kg ; g= 10m/s (gravitational acceleration occurs); h=9.2 m
hence mgh=0.4×10×9.2= 36.8J
unit for energy is joules and since the variables are in S.I. unit, we can use Joules as the final unit for measurement
Answer:
36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s²
Explanation:
Potential Energy = Mass x Gravitational Acceleration x Height
It can be expressed as [tex]\displaystyle{PE = mgh}[/tex] where PE is potential energy, m is mass, g is gravitational acceleration and h is height.
In this case, we know mass is 0.40 kg and height of 9.2 m as well as gravitational acceleration is defined to be 9.8 m/s² (You can also define g = 10 m/s²)
Therefore, substitute given information in formula:
[tex]\displaystyle{PE=0.40 \ \times \ 9.8 \ \times 9.2 }\\\\\displaystyle{PE=36.06 \ J}[/tex]
With g = 10 m/s², you'll get:
[tex]\displaystyle{PE = 0.40 \ \times 10 \ \times 9.2}\\\\\displaystyle{PE = 36.8 \ J}[/tex]
Note that J is for joule unit.
Therefore, the answer is 36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s² - both work.
1. A block of mass 0.4kg resting on the top of an inclined plane of height 20m starts to slide down on the surface of the incline to its foot, and then continues its slide horizontally. At a distance of 5m from the foot of the incline there is another block of the same mass resting on the horizontal surface to undergo an elastic collision. Next to the second block, there is a light spring of constant k = 4000N/m fixed freely against a wall. The spring is supposed to make a head-on collision with the second block. See the arrangements as in 1. Assuming all surfaces being frictionless, (a) calculate the kinetic energy of the firs block just at the foot of the incline; (b) calculate the kinetic and gravitational potential energies of the first block halfway down the incline; (c) calculate the speeds of the two blocks just after their collision; (d) compute the maximum compression of the spring resulted from its collision with the second block; (e) determine the maximum work done by the spring on the second block.
The kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.
To find the answer, we need to know about the concept of collision and kinetic energy.
How to find the kinetic energy of the first block just at the foot of the incline?Given that, the block of mass 0.4kg resting on the top of an inclined plane of height 20m.Thus, at the top of the incline it has a potential energy, and the kinetic energy will be equal to zero, or we can say that the total energy of the system is equal to the potential energy at topmost point.[tex]TE=KE+PE\\T=PE=m_1gh=(0.4*9.8*20)=78.4 J[/tex]
We have to find the kinetic energy of the first block just at the foot of the incline, and at the bottom point the PE=0, or we can say that the total energy or the potential energy is converted into kinetic energy.[tex]TE=KE=78.4J[/tex]
What is the kinetic and gravitational potential energies of the first block halfway down the incline?At the halfway, the PE will be,[tex]U'=m_1gh'=mg\frac{h}{2} \\U'=39.2J[/tex]
As we know that, the energy is conserved at each point of the motion.[tex]TE=78.4 J\\KE'+PE'=78.4J\\KE'=78-U'=78.4-39.2=39.2J[/tex]
How to find the speeds of the two blocks just after their collision?We have the KE at bottom point as, 78.4J. Thus, the velocity of first block at the bottom before collision will be,[tex]KE=\frac{1}{2} mv^2=78.4J\\v=\sqrt{\frac{2KE}{m} } =4m/s[/tex]
This is the velocity of the block 1 of mass m1 before collision, we can say, u1.As we know that, the 2 nd block of mass m2 is at rest, thus, u2=0.Given that, the collision is elastic. Thus, both the KE and the momentum will be conserved.[tex]\frac{1}{2}m_1u_1^2+ \frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
We have,[tex]m_1=m_2=m\\u_1=4m/s\\u_2=0\\v_1=?\\v_2=?[/tex]
Substituting this in both the equations, we get,[tex]\frac{1}{2}m*4^2=\frac{1}{2}m(v_1^2+v_2^2)\\(v_1^2+v_2^2)=16[/tex] from resolving KE equation.
[tex]4m=m(v_2+v_1)\\4=v_2+v_1\\v_1=4-v_2[/tex] From resolving momentum conservation.
solving both, we get,[tex]v_2=4m/s\\v_1=0[/tex]
Thus, we can conclude that, the kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.
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A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in Figure 4.29(b). The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
Figure 4.29
(a) Calculate the minimum force F he must exert to get the block moving.
40.873
(b) What is its acceleration once it starts to move, if that force is maintained?
m/s2
(a) The minimum force F he must exert to get the block moving is 38.9 N.
(b) The acceleration of the block is 0.79 m/s².
Minimum force to be applied
The minimum force F he must exert to get the block moving is calculated as follows;
Fcosθ = μ(s)Fₙ
Fcosθ = μ(s)mg
where;
μ(s) is coefficient of static frictionm is mass of the blockg is acceleration due to gravityF = [0.1(36)(9.8)] / [(cos(25)]
F = 38.9 N
Acceleration of the blockF(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32
a = F(net)/m
a = 28.32/36
a = 0.79 m/s²
Thus, the minimum force F he must exert to get the block moving is 38.9 N.
The acceleration of the block is 0.79 m/s².
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A proton is moved from the negative to the positive plate of a parallel-plate arrangement. The plates are 1.50 cm apart, and the electric field is uniform with a magnitude of 1 500 N/C.
What is the proton’s potential energy change?
What is the potential difference between the plates?
What is the potential difference between the negative plate and a point midway between the plates?
If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate?
(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
(b) The potential difference between the negative plate and a point midway between the plates is 11.25 V.
(c) The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
Potential energy of the proton
U = qΔV
where;
q is charge of the protonΔV is potential differenceU = q(Ed)
U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)
U = 3.6 x 10⁻¹⁸ J
Potential difference between the negative plate and a point midwayΔV = E(0.5d)
ΔV = 0.5Ed
ΔV = 0.5 (1500)(1.5 x 10⁻²)
ΔV = 11.25 V
Speed of the protonU = ¹/₂mv²
U = mv²
v² = 2U/m
where;
m is mass of proton = 1.67 x 10⁻²⁷ kgv² = (2 x 3.6 x 10⁻¹⁸) / ( 1.67 x 10⁻²⁷)
v² = 4.311 x 10⁹
v = √(4.311 x 10⁹)
v = 6.57 x 10⁴ m/s
Thus, the proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
The potential difference between the negative plate and a point midway between the plates is 11.25 V.
The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
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A baseball (m=145g) traveling 39 m/s moves a fielder's glove backward 23 cm when the ball is caught.
What was the average force exerted by the ball on the glove?
Express your answer to two significant figures and include the appropriate units.
The average force exerted by the ball on the glove is 480 N.
What is the force exerted by the ball on the glove?
The average force exerted on the glove by the ball is equal in magnitude to the force on the ball.
Force = mass * accelerationmass = 145 g = 0.145 kg
Acceleration of the baseball, a = (v² - u²)/2s
where:
v is final velocity = 0
u is initial velocity = 39 m/s
s is distance = -23 cm = 0.23 m
a = (0 - 39²)/2(-0.23)
a = 3306.52 m/s²
Force = 0.145 * 3306.52
Force = 479.4 N
Average force = 480 N
In conclusion, force is derived from the product of mass and acceleration.
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Your friend's 10.8 g graduation tassel hangs on a string from his rearview mirror. When he accelerates from a stoplight, the tassel deflects backward toward the rear of the car at an angle of 5.23 ∘ relative to the vertical.
A) Find the tension in the string holding the tassel.
B) At what angle to the vertical will the tension in the string be twice the weight of the tassel?
The tension in the string holding the tassel and the vertical will the tension in the string
T = 0.1953 NФ = 34.4 °What is the tension in the string holding the tassel. ?Generally, the equation for Tension is mathematically given as
[tex]TCos\theta = mg[/tex]
Therefore
[tex]TCos6.58^{o} = 19.8*10^{-3}*9.8[/tex]
T = 0.1953 N
b).
Where
[tex]T* sin \theta = ma[/tex]
[tex]0.1953*Sin6.58 \textdegree = 19.8*10^{-3}*a[/tex]
a = 1.13 m/s^2
In conclusion
T* sinФ = ma
2msinФ = ma
2sinФ = a
[tex]sin\theta = \frac{a}{2}[/tex]
[tex]\theta = sin^{-1}\frac{a}{2} \\\\\theta= sin^{-1}\frac{1.13}{2}[/tex]
Ф = 34.4 °
In conclusion, The tension in the string holding the tassel and the vertical will the tension in the string
T = 0.1953 N
Ф = 34.4 °
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A truck is carrying a refrigerator as shown in the figure. The height of the refrigerator is 158.0 cm, the width is 60.0 cm. The center of gravity of the refrigerator is 67.0 cm above the midpoint of the base.
What is the maximum acceleration the truck can have so that the refrigerator doesn't tip over? (The static friction between the truck and the refrigerator is very large.)
The maximum acceleration the truck can have so that the refrigerator does not tip over is 1.86m/s².
What is acceleration?Acceleration is the change in velocity with time.
The maximum acceleration is obtained by taking moments about the tipping point of rotation.
Moment = force * perpendicular distanceClockwise moment = Anticlockwise momentF₂ * 1.58 m = F₁ * 0.67 m
where
F₂ is tipping force = mass * acceleration, a
F₁ is weight = mass * acceleration due to gravity, g
The weight acts at a distance half the width of the refrigerator = 30 cm or 0.3 m
Height of refrigerator is 158 cm 0r 1.58 cm
m * a * 1.58 = m * 9.81 * 0.30
a = 1.86 m/s²
The maximum acceleration the truck can have so that the refrigerator does not tip over is 1.86m/s².
In conclusion, if the maximum acceleration is exceeded, the refrigerator will tip over.
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A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car moves 30.0 m. Neglecting friction between car and road, find v and the horizontal force exerted on the car.
(a) the speed v
______m/s
(b) the horizontal force exerted on the car (Enter the magnitude.)
_______N
The horizontal force applied is 160 N while the velocity is 2.03 m/s.
What is the speed of the car?The work done by the car is obtained as the product of the force and the distance;
W = F x
F = ?
x = 30.0 m
W = 4,800 J
F = 4,800 J/30.0 m
F = 160 N
But F = ma
a = F/m
a = 160 N/2.30 ✕ 10^3-kg
a= 0.069 m/s
Now;
v^2 = u^2 + 2as
u = 0/ms because the car started from rest
v = √2as
v = √2 * 0.069 * 30
v = 2.03 m/s
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A certain force gives mass m1, an acceleration of 12 m/s² and mass m2 an acceleration of 3.3. m/s². What acceleration will this force give to the combined mass?
Acceleration is the rate of change of velocity. Usually, acceleration means the speed is transforming, but not always. When an object moves in a circular path at a steady speed, it is still accelerating, because the focus of its velocity is changing.
a = 4,552 m / s², b) a = 2,588 m / s²
Newton's second law is
F = ma
a = F / m
in this case the force remains constant
indicate us
* for a mass m₁
a₁ = F/m₁
a₁ = 12, m/ s²
* for a mass m₂
a₂= 3.3 m / s²
acceleration m = m₂-m₁
substitute
[tex]$\begin{aligned}&a=\frac{F}{m_{2}-m_{1}} \\&1 / a=\frac{m_{2}}{F}-\frac{m_{1}}{F}\end{aligned}$[/tex]
let's calculate
1/a=1/3.3 - 1/12 = 0.21969
a = 4,552 m / s²
a= 4,552m/s²
What is speed and acceleration?Speed estimates the rate of movement of an object, that is, the distance traveled per unit of time. Acceleration calculates the rate of change of velocity, that is, the change in velocity between two different moments
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A 149-g baseball is dropped from a tree 15.0 m above the ground.
With what speed would it hit the ground if air resistance could be ignored?
Express your answer to three significant figures and include the appropriate units.
If it actually hits the ground with a speed of 9.00 m/s , what is the magnitude of the average force of air resistance exerted on it?
Express your answer to three significant figures and include the appropriate units.
1. The speed with which the ball hits the ground is 17.1 m/s
2. The magnitude of the average force of air resistance exerted on it is 0.77 N
1. How to determine the velocity with which the ball hits the groundInitial velocity (u) = 0 m/s Acceleration due to gravity (g) = 9.8 m/s²Height (h) = 15 m Final velocity (v) =?v² = u² + 2gh
v² = 2gh
Take the square root of both side
v = √(2 × 9.8 × 15)
v = 17.1 m/s
2. How to determine the forceWe'll begin by calculating the time to reach the ground. This is illustrated below:
Acceleration due to gravity (g) = 9.8 m/s²Height (h) = 15 m Time (t) =?h = ½gt²
15 = ½ × 9.8 × t²
15 = 4.9 × t²
Divide both side by 4.9
t² = 15 / 4.9
Take the square root of both side
t = √(15 / 4.9)
t = 1.75 s
Now we can determine the force. This can be obtained as illustrated below:
Mass (m) = 149 g = 149 / 1000 = 0.149 KgInitial velocity (u) = 0 m/sFinal velocity (v) = 9 m/sTime (t) = 5 ms = 1.75 sForce (F) = ?F = m(v –u) / t
F = 0.149(9 – 0) / 1.75
F = 0.77 N
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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
1. What is the First Cosmic Speed i.e. the speed of a satellite on a low lying circular orbit around this planet? (Planet-X doesn't have any atmosphere.)
2. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
3. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
(a) The speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.
(b) The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.
(c) The radius of the synchronous orbit of a satellite is 69,801 km .
Speed of the satellitev = √GM/r
where;
M is mass of the planetr is radius of the planetv = √[(6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 7,338.93 m/s
Escape velocity of the satellitev = √2GM/r
v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 10,378.82 m/s
Speed of the satellite at the given periodv = 2πr/T
r = vT/2π
r = (7,338.93 x 16.6 x 3600 s) / (2π)
r = 69,801 km
Thus, the speed of a satellite on a low lying circular orbit around this planet is 7,338.93 m/s.
The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.
The radius of the synchronous orbit of a satellite is 69,801 km .
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A 72.5-kg hiker starts at an elevation of 1230 m and climbs to the top of a peak 2660 m high.
What is the hiker's change in potential energy?
Express your answer to three significant figures and include the appropriate units.
What is the minimum work required of the hiker?
Express your answer to three significant figures and include the appropriate units.
Change in potential energy = 1.02 * 10⁶ J
Minimum work required of the hiker = 1.89 * 10⁶ J
What is the change in potential energy of the hiker?The potential energy of a body is calculated as follows:
Potential energy = mghChange in potential energy = Final PE - Initial PE
Change in potential energy = mg(H - h)
Change in potential energy = 72.5 * 9.81 * (2660 - 1230)
Change in potential energy = 1.02 * 10⁶ J
The minimum work required of the hiker is the potential energy at the highest point.
Minimum work = mgH
Minimum work required of the hiker = 1.89 * 10⁶ J
In conclusion, potential energy is energy due to state or position of a body.
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For a projectile launched horizontally at 650 m/s and travels for 5.75 seconds, What is the range?
The range of horizontal projectile motion is 3737.5 m.
A projectile's horizontal velocity is constant (a never changing value), Gravity causes a downward vertical acceleration with a magnitude of 9.8 m/s/s. A projectile's vertical velocity fluctuates by 9.8 m/s per second, A projectile's vertical motion is unrelated to its horizontal motion. we know that sinθ is maximum at 90°. Therefore horizontal range will maximum at 45°.
To calculate the range of horizontal projectile motion we use;
Δx = vₓ t
where , Δx = Range
vₓ = Velocity
t = Time
Initial horizontal velocity, vₓ = 650 m/s
Time = 5.75 sec
Δx = 650 × 5.75
= 3737.5 m
Therefore, the range of horizontal projectile motion is 3737.5 m.
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The motor of a weed trimmer spins
at 9,000 rpm. The amount of time
required for the motor to reach this
speed would NOT be affected by...
A. the distribution of
trimmer line inside the
spool
B. the mass of the spool
containing the trimmer
line
C. the direction in which
the torque is applied
The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool. Option A
What is a weed trimmer?The weed trimmer is a device that is used to trim the grasses on a lawn or a field. This device has a rotating shaft that does the actual trimming of the grasses.
The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool thus it does not affect the amount of time required to reach 9,000 rpm speed.
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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
The vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
Tension in the cable
Apply the principle of moment and calculate the tension in the cable;
Clockwise torque = TL sinθ
Anticlockwise torque = ¹/₂WL
TL sinθ = ¹/₂WL
T sinθ = ¹/₂W
T = (W)/(2 sinθ)
T = (29 x 9.8)/(2 x sin57)
T = 169.43 N
Vertical component of the forceT + F = W
F = W - T
F = (9.8 x 29) - 169.43
F = 114.77 N
Thus, the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
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Calculate the amount of air in a room 6m long, 5m wide and 3mm high.
Answer:
0.09kg of air
Explanation:
The dimensions of the room are given
change the height to meters by dividing it by thousand.
For the volume multiplying the length,width and height (all should be in the same unit most suitable being meters).
Volume refers to the amount of space inside a box or a object.
The amount of air is equal to the volume.
Answer:
90 m^3
Explanation:
Volume of the room:
6 m * 5 m * 3 m = 90 m^3 <=====( I changed 3mm to 3 m)
if 3mm is not a typo mistake
volume becomes ( 3 mm = .003 m)
6 m * 5 m * .003 m = .09 m^3 ( though unlikely )
What is the speed of a giraffe that has a
mass of
75 kg and a kinetic energy of 600 J?
O 2 m/s
O
4 m/s
O 8 m/s
O 16 m/s
A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal.
The horizontal component of the force exerted by the hi.nge on the beam is 8.662x10^1 N.
What is the magnitude of the force that the beam exerts on the hi.nge?
288.51 N is the magnitude of the force that the beam exerts on the hi.nge.
Given
Mass 0f beam = 40 Kg
The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N
Angle between the beam and cable is = 90°
Angle between beam and the horizontal component = 31°
As the system of the beam, hi_nge and cable are in equilibrium.
The magnitude of the force that the beam exerts on the hi_nge can be calculated by -
F =The horizontal component of force + the vertical component of force
F = 86.62 N + 40 × 9.8 × sin 31°
F =86.62 N + 201.89 N
F = 288.51 N
Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51 N.
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The three displacement vectors have magnitude of A=5.00m,B=5.00m and C=4.00m.Find the resultant (magnitude and directional angle) of the three vectors?
3.00 m is the magnitude of the resultant vector
137.1° is the directional angle of the resultant vector.
1) To find the magnitude of the resultant
Discover each vector's individual components first.
We must consider the signs of the components because the angles in the figure are measured in various ways.
In this case, both the y component of vector C and the x component of vector A are negative.
The vectors' components are as follows, using a little trigonometry:
Magnitude of A = 5 m
[tex]A_{x}[/tex] = - (5.00m) cos20° = -5 ×0.408 = -4.698 m
[tex]A_{y}[/tex] = + (5.00m)sin20° = +5 × 0.342 = +1.710 m
Magnitude of B = 5m
[tex]B_{x}[/tex] = +(5.00m)cos60° = 5 × 0.5 = +2.5m
[tex]B_{y}[/tex] = +(5.00m)sin60° = 5 × √3/2 = +4.33 m
Cx = 0
Cy = -4.00m
The sum of all three vectors, which we refer to as R, produces components.
Rₓ = Aₓ + Bₓ + Cₓ
= -4.698 + 2.5 + 0
= -2.198 m
[tex]R_{y}[/tex] = [tex]A_{y} + B_{y} + C_{y}[/tex]
= +1.71 +4.33 - 4.00
= 2.040 m
R = [tex]\sqrt{R_{x^{2} }+R_{y^{2} }[/tex]
= [tex]\sqrt{(-2.198)^{2 } + (2.040)^{2} }[/tex]
= 3.00 m
2) To find the directional angle of resultant
tan θ = 2.040/-2.198 = -0.928
θ = -42.9°
Such a vector would be in the so-called "fourth quadrant," as it is well known. However, we discovered that R has a negative x component and a positive y component, indicating that such a vector must reside in the "second quadrant.
The calculator accidentally returned an angle that is 180 degrees off, thus we must add 180 degrees to the naïve angle in order to get the right angle.
Therefore, R's actual direction is determined by-
θ = -42.9° + 180° = 137.1°
Hence, the magnitude of resultant vector is 3.00 m and directional angle is 137.1°
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I need help with my homework
C. The center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.
What is center of mass?The center of mass of an object is the unique point where the weighted relative position of the distributed mass sums to zero.
What is center of gravity?Center of gravity is the point from which the weight of a body or system may be considered to act.
Thus, the center of mass and center of gravity will coincide if the acceleration due to gravity is constant or uniform.
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Define the term work and state its unit. An ant is dragging a house-fly and the elephant is pushing a big tree which is not moving. Who is doing work, the ant or the elephant? Justify your answer. 922.5 205
Ant is performing a work
what is work?
Work is the force applied on an individual with respect to displacement.
Work = Force × displacement
Unit is Nm
Elephant is pushing bt there is no displacement occurred so the work of elephant is zero.
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The ant works, but the elephant does not.
Who works, how do find the ant and the elephant?Work done = Force × Displacement.
If there are ants and houseflies,
Ants drag the house bug, so they use specific force to move the house bug from one point to another, so we can say they work.
In the case of the elephant and the tree,
When the elephant pushes the tree (applying a force), the tree does not move, i.e., there is no displacement, so there is work.
Work done = Force × Displacement
= Force × 0
= 0
Therefore,
The ant works, but the elephant does not.
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Discuss the aspects of either the Gemini Program or the Soyuz Program.
Answer:
The Gemini program
Explanation:
The Gemini Program was the second human spaceflight program hosted by Nasa in the year 1961. Taking place between mission Mercury and Apollo, the Gemini spacecraft carried two people to space and marked the foundation to the upcoming Apollo mission to Moon. It was a series of missions into the outer orbital which took place between 1965 and 1966. Prior to the Gemini missions, NASA had little to no information about space and space traveling. It was crucial for them to get acquainted with life outside before establishing successful Moon landings. And the series of Gemini missions helped them do just that.