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vlada-n [284]
3 years ago
8

A horizontal applied force of magnitude 25 N acts on a block sliding on a horizontal surface. The force of friction between the

block and the surface has magnitude 8 N. If the direction of the applied force is reversed which of the following is true?
A. the magnitude of the velocity of the block will remain constant.
B. the magnitude of the acceleration of the block decreases
C. The magnitude of the net force on the block increases as compared to before.
D. the magnitude of the net force on the block will remain the same as before​
Physics
1 answer:
kompoz [17]3 years ago
3 0

' A ' and ' D ' are both correct statements.

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One similarity between work and power is that in order to calculate both you must know
svetoff [14.1K]
D.) In order to calculate both of them, we must know the "FORCE" on the system.
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Which isotope is least likely to be stable?<br> A) Ca-40 <br> B) Cl-38 <br> C) U-235 <br> D) Xe-136
aksik [14]

Answer:

C) U-235

Explanation:

On Isotope stability, an isotrope will be least stable when Proton and newtron are both odd.

8 0
3 years ago
Need some help please
11Alexandr11 [23.1K]

Answer:

x^2\neq -\frac{5}{13}

First option

Explanation:

<u>Operations with functions </u>

Given two functions f, g, we can perform a number of operations with them including addition, subtraction, product, division, composition, and many others .

We have

f(x)=20x^3-7x^2+3x-7

g(x)=-13x^2-5

We are required to find

\displaystyle \frac{f}{g}(x)

We simply divide f by g as follows

\displaystyle \frac{f}{g}(x)=\frac{20x^3-7x^2+3x-7}{-13x^2-5}

We know rational functions may have problems if the denominator can be zero for some values of x. We must find out if there are such values and exclude them from the domain of the new-found function. We must ensure

-13x^2-5\neq 0

or equivalently

x^2\neq -\frac{5}{13}

Thus the first option is correct

Note: Since x^2 is always a positive number (for x real), our function does not really have any restriction in its domain

8 0
3 years ago
A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. W
Karo-lina-s [1.5K]
Before we find impulse, we need to find the initial and final momentum of the ball.

To find the momentum of the ball before it hit the floor, we need to figure out its final velocity using kinematics.

Values we know:
acceleration(a) - 9.81m/s^2 [down]
initial velocity(vi) - 0m/s
distance(d) - 1.25m [down]

This equation can be used to find final velocity:

Vf^2 = Vi^2 + 2ad

Vf^2 = (0)^2 + (2)(-9.81)(-1.25)

Vf^2 = 24.525

Vf = 4.95m/s [down]

Now we need to find the velocity the ball leaves the floor at using the same kinematics concept.

What we know:
a = 9.81m/s^2 [down]
d = 0.600m [up]
vf = 0m/s

Vf^2 = Vi^2 + 2ad

0^2 = Vi^2 + 2(-9.81)(0.6)

0 = Vi^2 + -11.772

Vi^2 = 11.772

Vi = 3.43m/s [up]

Now to find impulse given to the ball by the floor we find the change in momentum.

Impulse = Momentum final - momentum initial

Impulse = (0.120)(3.43) - (0.120)(-4.95)

Impulse = 1.01kgm/s [up]
8 0
3 years ago
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A 375-g stone hangs from a thin light string that is wrapped around the circumference of a pulley with a moment of inertia of 0.
Monica [59]

Answer:

The magnitude of the acceleration of the stone is 19.87 m/s²

Explanation:

Given;

mass of stone, m = 375 g = 0.375 kg

moment of inertia, I = 0.0125 kg.m²

radius of the pulley, r = 26 cm = 0.26 m

Torque generated by the pulley on the stone is given as;

τ = F x r = Iα

where;

F is applied force on the stone due to its weight

r is the radius of the pulley

I is moment of inertia

α is angular acceleration (rad/s²)

Force, F = mg = 0.375 x 9.8 = 3.675 N

Torque, τ = F x r

τ = 3.675 x 0.26

τ = 0.9555 N.m

τ = Iα

Angular acceleration, α = τ / I

α = 0.9555 / 0.0125

α = 76.44 rad/s²

Finally, determine linear acceleration, a,  in m/s²

a = αr

a = 76.44 x 0.26

a = 19.87 m/s²

Therefore, the magnitude of the acceleration of the stone is 19.87 m/s²

4 0
3 years ago
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