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mamaluj [8]
3 years ago
7

A student produces a power of p = 0.87 kw while pushing a block of mass m = 75 kg on an inclined surface making an angle of θ =

8.5 degrees with respect to the horizontal. the coefficient of kinetic friction between the block and the incline is μk = 0.16. randomized variables p = 0.87 kw m = 75 kg θ = 8.5 degrees μk = 0.16 50% part (a) write an expression for the maximum constant speed, vm, the block travels at under the power applied by the student.
Physics
1 answer:
Paul [167]3 years ago
6 0

When block is pushed upwards along the inclined plane

the net force applied on the block will be given as

F_{net} = mg sin\theta + \mu_k mg cos\theta

here we know that

m = 75 kg

\theta = 8.5 degree

\mu_k = 0.16

now plug in all values into this

F_{net} = 75\times 9.8 sin8.5 + 0.16 \times 75\times 9.8 cos8.5

F = 225 N

now for finding the power is given as

P = Fv

0.87 \times 10^3 = 225 \time v

v = \frac{870}{225} = 3.87 m/s

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An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much
cupoosta [38]

Answer:

time will elapse before it return to  its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × 10^{7} m/s

uniform electric field E = 1.18 × 10^{4} N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × 10^{-31} kg ×a = 1.18 × 10^{4} × 1.602 × 10^{-19}

a = 20.75 × 10^{14} m/s²

so acceleration is 20.75 × 10^{14} m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × 10^{7} = 0 + 20.75 × 10^{14} (t)

t = 11.80 × 10^{-9} s

so time will elapse before it return to  its staring point is

time = 2t

time = 2 ×11.80 × 10^{-9}

time is 23.6 × 10^{-9} s

time will elapse before it return to  its staring point is 23.6 ns

7 0
3 years ago
a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r
dimaraw [331]

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

spring constant k= 2250 Nm

now according to principal of conservation of energy we observe,

the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.

mgh= μ (mgl) +1/2 kx²

10 x 10 x 3= μ(600) +(1125) (0.09)

μ(600) =300 - 101.25

μ = 198.75÷600

μ =0.33125

The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

Learn more about kinetic friction here-

brainly.com/question/13754413

#SPJ4

4 0
2 years ago
Which arrow represents the change of state described<br> above?<br> M<br> N<br> P<br><br> Q
Anon25 [30]

Answer: The Q arrow

Explanation: when the solid is heated it changes into a liquid state first this action represented the Q arrow

5 0
2 years ago
. A book is moved once around the perimeter of a table of dimensions 2.00 m by 3.00 m. If the book ends up at its initial positi
Lynna [10]

Answer:

10

Explanation:

displacement would be 10 because knowledge

5 0
3 years ago
A 0.80 kg basketball traveling upward at 5.0 m/s impacts an 8.0 10 kg tennis ball traveling downward at 5.0 m/s. The basketball’
vlabodo [156]

To solve this problem we will apply the concepts related to the conservation of momentum. This can be defined as the product between the mass and the velocity of each object, and by conservation it will be understood that the amount of the initial momentum is equal to the amount of the final momentum. By the law of conservation of momentum,

m_1u_1+m_2u_2 = m_1v_1+m_2v_2

Here,

m_1 = Mass of Basketball

m_2 = Mass of Tennis ball

u_1 = Initial velocity of Basketball

u_2 = Initial Velocity of Tennis ball

v_1 = Final velocity of Basketball

v_2 = Final velocity of the tennis ball

Replacing,

(0.8)(0.5)+(0.1)(-5.0)=(0.8)(0.3)+(0.1)v_2

Solving for the final velocity of the tennis ball

v_2 = 11m/s^2

Therefore the velocity of the tennis ball after collision is 11 m/s

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3 years ago
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