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ICE Princess25 [194]
4 years ago
11

A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above

. If FP remains constant but the angle θ is increased, which of the following is true at some later time?
A.The force of friction between the block and surface will increase.

B.The force of friction between the block and surface will decrease.

C.The weight of the block will increase.

D.The weight of the block will decrease.

Physics
2 answers:
gizmo_the_mogwai [7]4 years ago
6 0

Answer:

B.The force of friction between the block and surface will decrease.

Explanation:

The force of friction is given by

F_s = \mu N

where \mu is the coefficient of friction and N is the normal force.

When the student pulls on the block with force F_p at an angle \theta, the normal force on the block becomes

N  = Mg- F_psin(\theta)

and hence the frictional force becomes

F_s = \mu (Mg- F_psin(\theta)).

Now, as we increase \theta, sin(\theta) increases which as a result decreases the normal force Mg- F_psin(\theta), which also means the frictional force decreases; Hence choice B stands true.

<em>P.S: Choice D is tempting but incorrect since the weight </em>W=mg<em> is independent of the external forces on the block. </em>

Virty [35]4 years ago
3 0

Answer:

The force of friction between the block and surface will decrease.

Explanation:

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4 0
3 years ago
You serve a tennis ball from a height of 1.80 m above the ground. The ball leaves your racket with a velocity of 18.0 m/s at an
Delicious77 [7]

Answer:

Yes, ball will clear the net

Explanation:

First we have to find the range of projectile motion.

Data given,

Ф = 7°

Initial velocity = 18 m/s

R = (V)^2.sin2Ф/g

Now by putting values

R = 7.99 m

Now for height

h = v^2.(sinФ)^2/2g

by putting values

h = 0.245 m

Since range is less than our distance (11.83 m) from net, so still it is not clear that ball will clear the net or not.

So, now from the maximum height, we have to calculate the horizontal distance of ball to net.

Now velocity in projectile motion is in two dimensions.

V(x) = 18 m/s

V(y) = 0 m/s (because at maximum height, ball will stop and then start again, so y-component of velocity will be 0 but since there will be no acceleration along x-axis, so V(x) will be 18 m/s)

Now, by formula S = V(y)t + (1/2)gt^2

we can calculate time which is required by the ball to reach net from the maximum height it has achieved.

Now, tricky part is to calculate S, because without it we can not calculate t.

So, by data given in question, we know that the ball is served at height of 1.8 m and it achieved the height of 0.245 m. But net is at height of 1.07 m.

So, the vertical distance downward, which ball will travel from maximum height to net will be

S = 1.8 + 0.245 - 1.07

S = 0.975 m

Since we know V(y) = 0 m/s

S = (1/2)gt^2

t = (2S/g)^(1/2)

t = 0.44 s

Now time for both vertical and horizontal distance are same,

So, for horizontal distance "D(x)"

D(x) = V(x) x t (Since, no acceleration along x axis, so we can use simple formula to calculate distance)

D(x) = 18 x 0.44

D(x) = 8.029 m

Now please notice that at maximum height, range was half, so at that point ball covered distance "a"

a = 3.99 m

From maximum height to net, as we calculated, ball covered

D(x) = 8.029 m

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a + D(x) = 3.99 + 8.029

a + D(x) = 12.024 m

which is more than your total distance from net which is 11.83 m. So, the ball will clear the net.

7 0
3 years ago
Which of the following is an example of the effect of a natural force ?
Ulleksa [173]
A. Rain falling from the sky.
8 0
4 years ago
Read 2 more answers
Conduction
sashaice [31]

conducted

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7 0
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laila [671]
False because if it already has speed it means it was already accelerated and is now getting slower
8 0
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