Answer:
pictures please
Explanation:
I need a picture so I can tell you
What affects fuel consumption in automobiles?
A. Drag
B. Nothing
C. Air resistance
D. Time of day
Answer:
A and C
Explanation:
drag (the area of lower air pressure behind the car when moving) and mostly air resistance (the work to push the air in front of us away to move through - the faster we go, the stronger the air resists to move aside).
D. A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weight to be 7.84 N. She then weighs the crown while it is immersed in water, and now the scale reads 6.86 N. Is the crown made of pure gold? Find the density of the crown and compare it to the it to density of the gold.
Answer:
Wc = 7.84 weight of crown
Ww = 7.84 - 6.86 = .98 weight of water displaced
Density = 7.84 / .98 = 8 crown is 8 X that of water
Since gold has a density of 19.3 that of water the crown is certainly not 100 percent (if any) gold
An object of height 5 cm is kept in front of a
lens. An inverted image of height 5 is formed. identity the lens and position of the
object and image.
Answer:
The type of lens involved is a convex lens and the object is positioned at the center of the curvature of the convex lens.
According to the question, a real, inverted, and same-sized image of the object is formed.
A concave lens is a diverging lens and always forms virtual images. But convex lenses are converging lenses and are the only type of lens that produce real and inverted images of the corresponding objects.
When an object is placed at the center of the curvature of a convex lens, its corresponding image is formed on the opposite side of the convex lens. The image formed is real and inverted.
The distance of the image from the lens is equal to the distance of the object from the lens.
For a convex lens, the distance of the center of curvature from the lens is double the focal length of the lens.
That's why the convex lens and center of curvature are the correct answer to this question.
Explanation:
A car is being tested on a track. The driver approaches the test section at a speed of 28 m s−1. He then accelerates at a uniform rate between two markers separated by 100 m. The car reaches a speed of 41 m s−1.
The uniform acceleration of the car is 4.485 m/s².
Acceleration of the car
The uniform acceleration of the car is calculated as follows;
v² = u² + 2as
a = (v² - u²)/2s
where;
v is final velocity = 41 m/su is initial velocity = 28 m/ss is distance = 100 ma is acceleration = ?a = (41² - 28²)/(2 x 100)
a = 4.485 m/s²
Thus, the uniform acceleration of the car is 4.485 m/s².
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A block with a mass of 33.0 kg is pushed with a horizontal force of 150 N. The block moves at a constant speed across a level, rough surface a distance of 4.95 m.
(a) What is the work done (in J) by the 150 N force?
_________J
(b) What is the coefficient of kinetic friction between the block and the surface?
________
The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.
What is the work done?The work done is given by the use of the formula;
W = F * x
Where;
F = force applied
x = distance covered
W = 150 N * 4.95 m = 742.5 J
Now;
The coefficient of kinetic friction is given by;
μ = F/mg
μ = 150/ 33 * 9.8
μ = 0.46
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Part B
A roller coaster ride starts with the roller coaster car being pulled to the top of the first hill with pulley system. The car is
released from the top with an initial velocity close to zero, then accelerates downward. From that first hill, the roller coaster just
coasts; there is no driving force, other than gravity, to keep It going. Assuming no friction, what can you say about the height of
the other hills in the roller coaster ride?
The highest point of a roller coaster is almost always the first hill. In the majority of roller coasters, the hills get smaller as the train travels down the track.
To find the answer, we have to know more about the mechanical energy of a system.
How to find the answer?Since it influences the mechanical energy of the system, the first hill must be the highest.One of the fundamental tenets of physics is that, in the absence of friction, mechanical energy must be conserved. Mechanical energy is the product of kinetic energy and potential energy.When the vehicles ascend the first hill on the roller coaster, mechanical energy is provided to the system because the speed is zero at this point.Mechanical energy = U = mgh
Where m represents the car mass, g represents gravity, and h represents height
If the system is to continue moving, the other hills on the mountain must be lower than the first hill. When the vehicles are released, this energy is converted into kinetic and potential energy when it lowers and ascends, but the sum of these two cannot be larger than the starting energy.Finally, by applying the principle of energy conservation, we may determine that, the initial hill must be the highest.
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A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale.
Evaluate the magnitude of the force on the left hand pole.
The magnitude of the force on the left-hand pole is mathematically given as
f'=0.167N
Q=45 degrees
What is the magnitude of the force on the left-hand pole.?Generally, the equation for Force is mathematically given as
F=mg/sinФ
Therefore
F(17.1*10^{-3})*9.8/sin 45
F=0.237N
Considering horizontal axis or plane
f'-fcos=0
Therefore
f'=0.237*cos45
f'=0.167N
In conclusion, the slope
tanФ=30/30
tanФ=1
Ф=45
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Derive the following equations of motion
1. v = at
2. s = ut + at²
3. v² = u² + 2as
according to definition of acceleration
a=v-u/t
t=v-u/a(equation 1)
according to the formula of average velocity
v+u/2*s/t
s=v+u/2*t(equation 2)
now putting the value of t in equation 2
s=v+u/2*v-u/a
s=v^2-u^2/2a
v^2=u^2+2as
please make me brainlist
Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×10^3 km, mass of Earth = 5.98×10^24 kg, G = 6.67×10^-11 Nm^2/kg^2.)
Answer: The speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth is 4188.11 m/s.
Explanation: To find the answer, we need to know about the equation of motion of a satellite around earth.
What is the equation of motion of a satellite around earth?We have gravitational force of attraction between the satellite of mass m and earth of mass M as,[tex]F_g=\frac{GMm}{r^2}[/tex]
The expression for centripetal force of,[tex]F_c=\frac{mv^2}{r} \\[/tex]
These two forces are equal for a satellite around earth.[tex]\frac{GMm}{r^2} =\frac{mv^2}{r} \\thus,\\v=\sqrt{\frac{GM}{r} }[/tex]
How to solve the problem?Given that,[tex]r=3.57 R_E=3.57*6.37*10^3=22.74*10^3 km\\M=5.98*10^24kg\\G=6.67*10^{-11}Nm^2/kg^2[/tex]
Thus, the speed of the satellite will be,[tex]v=\sqrt{\frac{6.67*10^{-11}*5.98*10^{24}}{22.74*10^6m} } =4188.11 m/s[/tex]
Thus, we can conclude that the speed of satellite will be 4188.11 m/s.
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In a circular orbit with a radius that is 3.57 times the mean radius of the Earth, a satellite moves at a speed of 132.43km/s.
In order to get the solution, we must understand the satellite's planetary motion equation.
What is the satellite's orbital motion equation?The earth's mass M and the satellite's mass M are attracted to one another by gravity.[tex]F_g=\frac{GMm}{r^2}[/tex]
The term used to describe centripetal force of,[tex]F_c=\frac{MV^2}{r}[/tex]
When a satellite orbits the earth, these two forces are equivalent. Thus, the velocity will be,[tex]\frac{GMm}{r^2}=\frac{mV^2}{r}\\V=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10{-11}*5.98*10^{24}}{22.74*10^3} } \\V=132.43*10^3m/s[/tex]
As a result, we may estimate that the satellite will move at a speed of 132.43 km/s.
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The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×10^3 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10^-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×10^5 m around the planet Glob. Calculate the speed of the satellite.
(a) The initial speed of the rock as it left the astronaut's hand is 19.35 m/s.
(b) The speed of the satellite is 50.24 m/s.
Acceleration due to gravity of the planet
g = GM/R²
where;
M is mass of the planetR is radius of the planetg = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(63200)²
g = 0.13 m/s²
Initial speed of the rockv² = u² - 2gh
where;
v is final velocityu is initial velocityat maximum height, v = 0
u² = 2gh
u = √2gh
u = √(2 x 0.13 x 1,440)
u = 19.35 m/s
Speed of the satellitev = √GM/r
M is mass of the planet Globr is the total distance from the center of the planet Globr = radius of planet Glob + radius of the satellite
r = 63200 m + 145,000 m = 208,200 m
v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(208,200)]
v = 50.24 m/s
Thus, the initial speed of the rock as it left the astronaut's hand is 19.35 m/s.
The speed of the satellite is 50.24 m/s.
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A yellow train of mass 100 kg is moving at 8 m/s toward an orange train of mass 200 kg traveling in the opposite direction on the same track at a speed of 1 m/s. What is the initial momentum of the yellow and orange trains combined?
A. 200 kgm/s
B. 1000 kgm/s
C. 800 kgm/s
D. 600 kgm/s
The initial momentum of the yellow and the orange train is 1000kgm/s.
Momentum is the product of the mass and velocity of any object.
Momentum is denoted by P.
Momentum P = mv , where m = mass and v = velocity.
Given:Mass of the orange train = 200kg
Velocity of the orange train = 1m/s
So, the momentum of the orange train will be,
∴ P = mv
P = 200 x 1
P = 200 kgm/s
∴ The initial momentum of the orange train is 200kgm/s.
Mass of the yellow train = 100kg
Velocity of the yellow train = 8m/s
So, the momentum of the yellow train will be,
∴ P = mv
P = 100 x 8
P = 800 kgm/s
∴ The initial momentum of the yellow train is 800kgm/s.
Therefore, the initial momentum of the yellow and the orange train is 1000kgm/s.
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A step down transformer with an input voltage of 220V decreases the voltage to half of the input .There is a current flowing of 15A in the primary coil. Find the output current
The output current for the given step down transformer is 30A.
What is the step down transformer?
A step down transformer is a passive device that converts high voltage power to low voltage power, while the output current is higher than the input current. They are used in power adaptors and rectifiers to decrease the voltage to the desired level. It works according to Faraday's law of Electromagnetic induction.
The current in the windings of a step down transformer is inversely proportional to the voltage in windings as:
Input voltage / Output voltage = Output current / Input current
220 / 110 = Outout current / 15
Output current = 30A
Hence, the output current (30A) obtained is higher than the given input current (15A).
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A man pushing a crate of mass
m = 92.0 kg
at a speed of
v = 0.845 m/s
encounters a rough horizontal surface of length
ℓ = 0.65 m
as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.351 and he exerts a constant horizontal force of 280 N on the crate.
(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
magnitude_____N
What is the direction?
1. Opposite as the motion of the crate
2. Same as the motion of the crate
(b) Find the net work done on the crate while it is on the rough surface.
______J
(c) Find the speed of the crate when it reaches the end of the rough surface.
_______m/s
(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is 23.7 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
Magnitude of net force on the crateF(net) = F - μFf
F(net) = 280 - 0.351(92 x 9.8)
F(net) = -36.46 N
Net work done on the crateW = F(net) x L
W = -36.46 x 0.65
W = - 23.7 J
Acceleration of the cratea = F(net)/m
a = -36.46/92
a = - 0.396 m/s²
Speed of the cratev² = u² + 2as
v² = 0.845² + 2(-0.396)(0.65)
v² = 0.199
v = √0.199
v = 0.45 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is 23.7 J.
The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
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describe one similarity and one difference between the velocity on the reference circle and the velocity on the pendulum
SHM can be acquired by perpendicular projection of uniform circular motion of a particle on its diameter such a particle is called reference particle and its circular path is called reference circle.
A pendulum reaches its maximum velocity when the block is at its lowest point (the pendulum is vertical and pointing straight down). We can then use the term for conservation of energy to determine the maximum height of the block.
What is the velocity at the bottom of a pendulum?
As the pendulum swings downward, gravity converts this potential energy into kinetic energy, so that at the bottom of the swing, the pendulum bob has zero potential energy, and its kinetic power, (1/2)mv2, equals the initial potential energy (mgh). (So the velocity, v, equals √(2gh).)
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Two stretched copper wires both experience the same stress. The first wire has a radius of 3.9×10-3 m and is subject to a stretching force of 450 N. The radius of the second wire is 5.1×10-3 m. Determine the stretching force acting on the second wire.
The stretching force acting on the second wire, given the data is 588 N
Data obtained from the questionRadius of fist wire (r₁) = 3.9×10⁻³ mForce of first wire (F₁) = 450 NRadius of second wire (r₂) = 5.1×10⁻³ mForce of second wire (F₂) =?How to determine the force of the second wireF₁ / r₁ = F₂ / r₂
450 / 3.9×10⁻³ = F₂ / 5.1×10⁻³
cross multiply
3.9×10⁻³ × F₂ = 450 × 5.1×10⁻³
Divide both side by 3.9×10⁻³
F₂ = (450 × 5.1×10⁻³) / 3.9×10⁻³
F₂ = 588 N
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an 8 kilogram ball is fired horizontally form a 1 times 10^3 kilogram cannon initially at rest after having been fired the momentum of the ball is 2.4 time 10^3 kilogram meters per second east calculate the magnitude of the cannons velocity after the ball is fired
The magnitude of the cannons velocity after the ball is fired will be 2.4m/s.
To find the answer, we have to know more about the law of conservation of linear momentum.
How to find the magnitude of the cannons velocity after the ball is fired?The law of conservation of linear momentum states that, the initial momentum of a system will be equal to the final momentum.Here, then initial momentum of the system will be equal to zero, since the initial velocity of the cannon is equal to zero.Given that,[tex]M_B=8kg\\M_C=1*10^3 kg\\P_f=2.4*10^3 kgm/s.\\[/tex]
When the ball is fired from the canon, then the cannon will have a recoil velocity in the opposite direction of motion of the ball.Since it has a final momentum towards east, the recoil momentum will be in the west.Thus, the velocity of the cannon after when the ball is fired will be,[tex]P_f=M_CV_C\\V_C=\frac{P_f}{M_C}=\frac{2.4*10^3}{1*10^3} =2.4m/s \\west[/tex]
Thus, we can conclude that, the magnitude of the cannons velocity after the ball is fired is 2.4m/s.
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After the ball is shot, the cannon will move at a speed of 2.4 m/s.
We must learn more about the rule of conservation of linear momentum in order to locate the solution.
How can I determine the cannon's post-ball velocity magnitude?
According to the law of conservation of linear momentum, a system's starting and ultimate velocities are equal.Since the cannon's initial velocity is 0 in this case, the system's initial momentum will be equal to zero as well.We have,[tex]m=8kg\\M=1000kg\\P_f=2.4*10^3kgm/s[/tex]
When a cannon fires a ball, the cannon will recoil quickly and in the opposite direction of the ball's motion.Given that it is now moving eastward, the recoil momentum is towards the west.As a result, when the ball is fired, the cannon's velocity will be,[tex]P_f=MV\\V=\frac{P_f}{M} =2.4m/s[/tex]
Thus, we can infer that the cannon's maximum speed once the ball is fired is 2.4 m/s.
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john has 4 apples , is train is 7 minutes early calculate te mass of the sun
Answer:
The mass of Sun doesn't change with respective to the conditions.
Michael has 4 Apples, which may increase his own mass or weight but not the Sun's .
His train is 7 minutes, but this doesn't mean the Sun has been made to change. The train coming late affects the time management and delays work.
So, As the per the question, It is evident that Sun's Mass is still the same irrespective of conditions .
Hence, The required answer Sun's Mass is 2*10^30 kg
Explanation:
Determine the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/60 of its value at the Earth's surface.
The distance from the Earth's center to the point outside the Earth is 55800 Km
How to determine the distance from the surface of the EarthAcceleration due to gravity of Earth = 9.8 m/s²Acceleration due to gravity of the poin (g) = 1/60 × 9.8 = 0.163 m/s²Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²Mass of the Earth (M) = 5.97×10²⁴ KgDistance from the surface of the Earth (r) =?g = GM / r²
Cross multiply
GM = gr²
Divide both sides by g
r² = GM / g
Take the square root of both sides
r = √(GM / g)
r = √[(6.67×10¯¹¹ × 5.97×10²⁴) / 0.163)]
r = 4.94×10⁷ m
Divide by 1000 to express in Km
r = 4.94×10⁷ / 1000
r = 4.94×10⁴ Km
How to determine the distance from the center of the EarthDistance from the surface of the Earth (r) = 4.94×10⁴ KmRadius of the Earth (R) = 6400 Km Distance from the centre of the Earth =?Distance from the centre of the Earth = R + r
Distance from the centre of the Earth = 6400 + 4.94×10⁴
Distance from the centre of the Earth = 55800 Km
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two particles P and Q are shot vertically up. Particle P is first up with a velocity of 40m/s. After 4 seconds, particle Q is also shot up. Find a)where the two particles meet,if at the point of meeting,P has a velocity 15m/s. b ) the velocity with which Q is shot. take g=10m/s^2
The velocity with which the particle Q is fired is 15m/s upwards.
What is velocity?The vector quantity velocity (v), represented by the equation v = s/t, quantifies displacement (or change in position, s) over the change in time (t). Speed (or rate, r) is a scalar number, denoted by the equation r = d/t, that quantifies the distance traveled (d) over the change in time (t).
Assume the upward direction is positive and the downward direction is negative.
Velocity P, = 40m/s
Distance traveled by P,
Using the first equation of motion for particle P,
v = u + at
⇒ 0 = 40 + (-10)t
⇒ t = 4s
This is the time it takes for P to rise
Now, the maximum height(s) reached by the particle P is,
Using the second equation of motion,
s = ut + 1/2at²
⇒ s = 40×4 + 1/2 × (-10) × 4²
⇒ s = 80m
a) A particle P falls when Q is shot after 4 seconds from the initial time:
V² = U² + 2aH₁
⇒ H₁ = 15² - 0/ 2(-10)
⇒ H₁ = 11.25m
Particles P and Q meet at a distance from the ground (H₂).
Height, H₂ = s - H
⇒ H₂ = 80 - 11.25
= 68.75m
Particles P and Q meet at a distance of 68.75m from the ground.
v = u + at₁
⇒ 15 = 0 + (-10) t₁
⇒ t₁ = 1.5s
It is equal to P's fall time and Q's rise time.
b) For particle Q
H₂ = u₂t₁ + 1/2at₁
⇒ 11.25 = u₂ × 1.5 + 1/2 (-10) × 1.5
⇒ u₂ = 15 m/s
Therefore,
The velocity with which the particle Q is fired is 15m/s upwards.
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Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b). By what angle has planet Y rotated through during this time?
According to mathematics, the planet's angle is stated as
dY=704 degrees.
What is the current rotational angle of planet Y?We may demonstrate this by using Kepler's third law, which asserts that a planet's orbit squared is a function of cubed radius.
The equation for the period is often expressed numerically as
[tex](periodX / periodY)^2 = (radius X / radius Y)^3[/tex]
Therefore
(pX / pY)^2 = 4^3
(pX / pY)^2 = 64
[tex]\sqrt{(pX / pY )^2}= \sqrt{64}[/tex]
pX / pY=8
In conclusion, planet Y travels 8 times further than planet X does in the same amount of time since one orbit on planet X takes 8 times longer to complete.
planet Y travels ;
dY=8 * 88.0
dY= 704 degrees
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Why is air resistance friction not useful for an airplane?
A. Causes turbulence
B. Speeds it up
C. Slows it down
D. Creates heat
The correct option is C. Air resistance friction is not useful for an airplane because it slows it down.
What is air resistance?Air resistance is the opposition to motion of an object caused by air flow.
High air resistance may cause turbulence during the motion of an air plane. This can lead to engine failure or some mishap during flight.
Air resistance is also a type of friction between air and another material such as airplane.
Effect of air resistance on airplaneFriction between the air and the plane slows the airplane down. This is known as air resistance, or drag.
The faster an object travels through the air, the more it has to fight against drag.
Thus, air resistance friction is not useful for an airplane because it slows it down.
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The decibel level of the sound from a certain hair dryer is measured at 60 dB. Find the intensity of the sound.
Based on the calculations, the sound intensity level is equal to 1.0 × 10⁻⁵ W/m².
How to determine intensity of the sound?Mathematically, sound intensity level can be calculated by using this formula:
[tex]\beta = 10 log(\frac{I}{I_{ref}} )[/tex]
Where:
I is the intensity of the sound.
Note: The reference value of sound intensity is equal to 1.0 × 10⁻¹² W/m².
Rewriting the formula, we have:
β/10 = logI - logIo
Substituting the parameters into the formula, we have;
60/10 = logI - log(1.0 × 10⁻¹²)
6 = logI + 12
logI = 6 - 12
logI = -6
I = 1.0 × 10⁻⁵ W/m².
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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 being applied to the beam by the high-tension cable is 36.37 N.
Calculation of the cable's tension-
Utilizing the moment principle, determine the cable's tension:
Torque clockwise = TL sin∅
Counterclockwise torque = 1/2WL
TL sin∅ = 1/2WL
⇒T sin∅ = 1/2W
⇒T = W/2sin∅
⇒T = (29* 9.8)/ (2*sin57)
⇒T = 169.43 N
Calculation of the vertical component of the force-
Forces that operate perpendicular to the surface vertical plane are called the vertical force. Gravity always pulls objects straight down to the earth's core.
T+F = W
⇒F = W-T
⇒F = (21*9.8)-169.43
⇒F = 36.37 N
So, the force on the beam has a vertical component of 36.37 N.
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A circular loop of wire with radius 0.0410 m and resistance 0.169 Ω is in a region of spatially uniform magnetic field, as shown in the following figure (Figure 1). The magnetic field is directed out of the plane of the figure. The magnetic field has an initial value of 7.78 T and is decreasing at a rate of -0.605 T/s.
a) Is the induced current in the loop clockwise or counterclockwise?
b) What is the rate at which electrical energy is being dissipated by the resistance of the loop?
(a) The induced current in the loop will be counterclockwise.
(b) The rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.
Direction of the currentThe induced current in the loop will be counterclockwise to the direction of magnetic field.
Emf induced in the loopemf = -NdФ/dt
emf = -NBA/dt
where;
A is area of the loop
A = πr² = π(0.041)² = 5.28 x 10⁻³ m²
emf = -(-0.605 - 7.78) x 5.28 x 10⁻³
emf = 8.385 x 5.28 x 10⁻³
emf = 0.0442 V
Rate of energy dissipationP = emf²/R
P = (0.0442)²/0.169
P = 0.012 W
Thus, the rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.
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P = Patm + pgh is which law
The law that relates the absolute pressure to atmospheric pressure and gauge pressure is Pascal law.
What is Pascal law?
Pascal's law states that when an object is immersed in a fluid, it experiences equal pressure on all surfaces.
P = Patm + pgh
where;
P is absolute pressurepgh is gauge pressurePatm is atmospheric pressureThus, the law that relates the absolute pressure to atmospheric pressure and gauge pressure is Pascal law.
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A tennis player tosses a tennis ball straight up and then catches it after 1.77 s at the same height as the point of release.
(a) What is the acceleration of the ball while it is in flight?
magnitude ______ m/s2
Which direction?
1. Upward
2. Downward
3. The magnitude is zero
(b) What is the velocity of the ball when it reaches its maximum height?
magnitude _______________ m/s
Which direction?
1. Upward
2. Downward
3. The magnitude is zero
(c) Find the initial velocity of the ball.
______m/s upward
(d) Find the maximum height it reaches.
___________m
(a) The acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.
(b) The velocity of the ball when it reaches its maximum height is zero.
(c) The initial velocity of the ball is 17.36 m/s.
(d) The maximum height it reaches is 15.36 m.
Acceleration of the ball
The acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.
Velocity of the ball at maximum heightThe velocity of the ball decreases as the ball moves upwards and eventually becomes zero at maximum height.
Initial velocity of the ballv = u - gt
at maximum height, final velocity, v = 0
0 = u - gt
u = gt
u = 9.81 x 1.77
u = 17.36 m/s
Maximum height reached by the projectileh = ut - ¹/₂gt
h = 17.36(1.77) - ¹/₂(9.81)(1.77²)
h = 15.36 m
Thus, the acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.
The velocity of the ball when it reaches its maximum height is zero.
The initial velocity of the ball is 17.36 m/s.
The maximum height it reaches is 15.36 m.
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Two long, straight, parallel wires, 10.0 cm apart carry equal 4.00-A currents in the same direction, as shown in (Figure 1).
a) Find the magnitude of the magnetic field at point P1 , midway between the wires.
b) What is its direction?
c) Find the magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 .
d) What is its direction?
e) Find the magnitude of the magnetic field at point P3 , 20.0 cm directly above P1 .
f) What is its direction?
(a) The magnitude of the magnetic field at point P1 , midway between the wires is 1.005 x 10⁻⁴ T and the direction will be out of the page.
(b) The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.
(c) The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1 is 3.88 x 10⁻⁶ T and the direction is downwards.
Magnetic field midway between the wiresB = μ/2π[I₁/0.5r + I₂/0.5r]
B = (μ/2π) x (I/0.5r + I/0.5r)
B = (μ/2π) x (2I/0.5r)
B = μI/0.5r
B = 2μI/r
where;
I is current in the wiresr is the distance between the wiresB = (2 x 4π x 10⁻⁷ x 4)/(0.1)
B = 1.005 x 10⁻⁴ T
The direction of the magnetic field is out of the page.
Magnetic field at 25 cm right of P1B = μI/2πd
d = 5 cm + 25 cm = 30 cm
B = (4π x 10⁻⁷ x 4)/(2π x 0.3)
B = 2.67 x 10⁻⁶ T
The direction of the magnetic field is into the page towards P1.
Magnetic field at 20 above P1B = μI/2πd
d = √(20² + 5²)
d = 20.62 cm
B = (4π x 10⁻⁷ x 4)/(2π x 0.2062)
B = 3.88 x 10⁻⁶ T
The direction of the magnetic field is downwards towards P1.
Thus, the magnitude of the magnetic field at point P1 , midway between the wires is 1.005 x 10⁻⁴ T and the direction will be out of the page.
The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.
The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1 is 3.88 x 10⁻⁶ T and the direction is downwards.
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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?
Solutions for the three problems are is mathematically given as
v=7338.9349[tex]V=14677.86986 \mathrm{~m} / \mathrm{s}[/tex][tex]&r=30581248.06 \times 10^{4} \mathrm{~m} / \mathrm{s}[/tex]What is the First Cosmic Speed i.e. the speed of a satellite on a low-lying circular orbit around this planet?(a) First cosmic speed (arbitral velocily)
[tex]v=\sqrt{\frac{G M}{r}}\\v=\sqrt{\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24}}{5870 \times 10^{3}}}[/tex]
v=7338.9349
(b) Second cosmic speed (escape velo.)
$$
\begin{gathered}
[tex]V=\sqrt{\frac{2 G M}{r}}\\\\V=\sqrt{2} \sqrt{\frac{G M}{r}}\\\\=V\sqrt{2} \times 7338.9349 \\[/tex]
[tex]V=14677.86986 \mathrm{~m} / \mathrm{s}[/tex]
(c) In conclusion, in a circular orbit, the gravitational force is gets balanced by centripetal force
[tex]&m_{q \omega \omega^{2}}=\frac{G M M}{r^{2}} \\&r^{3}=\frac{G M}{\omega^{2}}=\frac{G M}{4 \pi^{2}} T^{2} \\&r^{3}=\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24} \times(16.6 \times 3600)^{7}}{4 \pi^{2}} \\[/tex]
[tex]&r=30581248.06 \times 10^{4} \mathrm{~m} / \mathrm{s}[/tex]
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ATTA-KAY PHYSICS 121 3. A ball A is left to roll down an inclined plane of inclination 30º. Just at the moment a second ball B is pushed up the plane with a velocity of 45ms¹. The balls met at a point where the velocity of B is 1.8 times the velocity of A. Calculate (a) the velocities of A and B when they meet. (b) when the two balls meet. (c) where the two ball meet. (g = 10 ms²)
vr>vs because the rolling ball acquires rotational as well as translational kinetic energy.
The accelerating force acting on the ball as it goes along a smooth plane is mgsin. Its acceleration is therefore equal to gsin. The mgsin acts down the plane as the ball travels down the rough inclined plane, but friction develops that acts up the plane.
Since both balls' potential energy is lost at the same rate, their KEs are actually equal at the base of the planes. However, a ball sliding down a smooth plane has only translational kinetic energy, but a ball rolling down a rough plane contains both translational and rotational kinetic energy at the bottom of the plane. As a result, the ball's translational KE will be lower than its translational Kinetic energy.
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A satellite is in a circular orbit very close to the surface of a spherical planet. The period of the orbit is 2.35 hours. What is density of the planet? Assume that the planet has a uniform density.
The density of the planet is determined as 1,974.26 kg/m³.
Density of the planet
√(⁴/₃πGρ) = 2π/(2.35 x 3600)
where;
ρ is density of the planetG is universal gravitation constant√(⁴/₃πGρ) = 2π/(2.35 x 3600)
√(⁴/₃πGρ) = 2π/8460
(⁴/₃πGρ) = (2π/8460)²
⁴/₃πGρ = 4π²/(8460)²
ρ = 12π/(8460² x 4G)
ρ = (12π) / (8460² x 4 x 6.67 x 10⁻¹¹)
ρ = 1,974.26 kg/m³
Thus, the density of the planet is determined as 1,974.26 kg/m³.
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