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sertanlavr [38]
2 years ago
6

A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air

borne. The ramp makes an angle of 22.0º to the ground, and the ball travels a distance of 5.00 m on the ramp. What is the maximum height the ball reaches, above the point where it was kicked, if (a) the ramp is frictionless and (b) there is a coefficient of friction of 0.150 between the ramp and the ball? Assume the ball slides, rather than rolls, up the ramp.
Physics
1 answer:
Bess [88]2 years ago
7 0

Answer:

Explanation:

given,

initial velocity of the ball = 20 m/s

angle of ramp = 22°

ball travel at a distance  = 5 m

a) for friction less  

  \dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh

             v^2 = u^2 - 2gh

             v = \sqrt{u^2- 2 g h cos 22^0}

             v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0 }

                    v = 17.58 m/s

b) considering the friction

  \dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh-\mu_kmgl

             v^2 = u^2 - 2gh-2\mu_kmgl

             v = \sqrt{u^2- 2 g h cos 22^0-2\mu_kgl}

             v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0-2\times 0.15\times 9.8 \times 5 }

                   v = 17.16 m/s

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3 years ago
A fixed coil of wire with 10 turns and an area of 0.055 m2 is placed in a perpendicular magnetic field. This field oscillates in
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Answer:

Part a)

Average EMF for half cycle is

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Part b)

For one complete cycle we will have

E_{avg} = 0

Part c)

Maximum induced EMF will be at

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minimum induced EMF is at

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Explanation:

As we know that magnetic field is oscillating in direction as well as magnitude

so induced EMF is given as

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Part a)

For average value of EMF from positive maximum to negative maximum which is equal to half cycle

so we have

E_{avg} = NBA\omega \frac{2}{T}\int_0^{T/2} sin(\omega t) dt

E_{avg} = \frac{2NBA\omega}{\pi}

E_{avg} = \frac{2(10)(0.12)(0.055)(2\pi (10))}{\pi}

E_{avg} = 2.64 V

Part b)

For one complete cycle we will have

E_{avg} = NBA\omega \frac{1}{T}\int_0^T sin(\omega t) dt

E_{avg} = 0

Part c)

Maximum induced EMF will be at

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here we know

T = \frac{1}{f} = 0.1 s

t = 0.025 s and 0.075 s

minimum induced EMF is at

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t = 0.05s and 0.1 s

8 0
3 years ago
7. When will an object's displacement and distance traveled be different?
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If an object changes direction while travelling will an object's displacement and distance travelled be different.

Some people believe that distance and displacement are simply different names for the same quantity. However, distance and displacement are not the same thing. If an object changes direction while travelling, the total distance travelled is greater than the displacement between those two points.

The magnitude of the displacement is always less than or equal to the distance because it is measured along the shortest path between two points.

When the direction of displacement does not change, the magnitude of the displacement and distance are the same. When a body travels in a straight line, for example, its displacement and distance are the same.

Learn more about displacement and distance brainly.com/question/3243551

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8 0
1 year ago
Suppose there is a uniform magnetic field, B, pointing into the page (so your index finger will point into the page). If the vel
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Answer:

Explanation:

We shall show all given data in vector form and calculate the direction of force with the help of following formula

force F = q ( v x B )

q is charge , v is velocity and B is magnetic field.

Given B = - Bk ( i is  right  , j is  upwards  and k is straight up the page  )

v = v j

F = q ( vj x - Bk )

= -Bqvi

The direction is towards left .

a ) If velocity is down

v = - v j

F = q ( - vj x - bk )

= qvB i

Direction is right .

b ) v = v i

F = q ( vi x - Bk )

= qvB j

force is upwards

c ) v = - vi

F = q ( -vi x - Bk )

= -qvBj

force is downwards

d ) v = - v k

F = q( - vk x -Bk  )

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No force will be created

e ) v =  v k

F = q(  vk x -Bk  )

= 0

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An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the
katrin [286]
<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

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