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Levart [38]
2 years ago
6

In an experiment, a variable, position-dependent force FC) is exerted on a block of mass 1.0 kg that is moving on a horizontal s

urface. The frictional force between the block and the surface has a constant magnitude of Ff. In addition to the final velocity of the block, which of the following information would students need to test the hypothesis that the work done by the net force on the block is equal to the change in kinetic energy of the block as the block moves from 2 = 0 to z = 5 m?
A The function F(2) for 0 < x < 5 and the value of Fr.

B) The function a(t) for the time interval of travel and the value of F.

C The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F.

D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of Ft.

E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F.
Physics
1 answer:
marshall27 [118]2 years ago
8 0

Answer:

C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of Fr.

Explanation:

Yo want to prove the following equation:

W_N=\Delta K\\\\

That is, the net force exerted on an object is equal to the change in the kinetic energy of the object.

The previous equation is also equal to:

F(x)x-F_f=\frac{1}{2}m(v_f^2-v_o^2)    (1)

m: mass of the block

vf: final velocity

v_o: initial velocity

Ff: friction force

F(x): Force

x: distance

You know the values of vf, m and x.

In order to prove the equation (1) it is necessary that you have C The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F.  Thus you can calculate experimentally both sides of the equation.

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This time particle A starts from rest and accelerates to the right at 65.5 cm/s
FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

d = v_i t + \frac{1}{2}at^2

d = 0 + \frac{1}{2}(65.5)t^2

d_1 = 32.75 t^2 cm

Now we know that B moves with constant speed so in the same time B will move to another distance

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now we know that B is already 349 cm down the track

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then in that case

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32.75 t^2 = 349 + 44 t

on solving above kinematics equation we have

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4 0
3 years ago
After striking both mirrors, at what angle relative to the incoming ray does the outgoing ray emerge?
PIT_PIT [208]
The appropriate response is Zero degrees. The beam will leave the two mirrors along a way parallel to the one it came in on. This is the guideline of the corner reflector, which is frequently utilized as a radar target. Take note of that the corner reflector utilizes three reflecting surfaces (that are set up at 90o from each other) rather than the two like are being utilized here. Wikipedia has a truly awesome drawing that shows this two-dimentional issue pleasantly. A moment connection is given to the article on the corner reflector and the 3-D angles.
4 0
2 years ago
Please help Need good grade
ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the  range

Using formula of range

R=\dfrac{v^2\sin(2\theta)}{g}...(I)

The range for the other angle is

R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}....(II)

Here, distance and speed are same

On comparing both range

\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}

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\sin120=\sin2(\alpha-60)

120=2\alpha-120

\alpha=\dfrac{120+120}{2}

\alpha=120^{\circ}

Hence, The other angle is 120°

6 0
3 years ago
Two 100 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Bat -10 m/s when th
tatiyna

Answer:

b) -10 m/s

Explanation:

In perfectly elastic head on collisions of identical masses, the velocities are exchanged with one another.

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2 years ago
In the reaction 2Ca + O2 → 2CaO for every 2 Ca you will need how much O2?
Serjik [45]

Answer:

1

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