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melamori03 [73]

3 years ago

10

A ball thrown horizontally from the top of a building hits the ground in 0.600 s. If it had been thrown with twice the speed in

the same direction, it would have hit the ground in:________.
a. 4.0 s.
b. 1.0 s.
c. 0.50 s.
d. 0.25 s.
e. 0.125 s.

1 answer:

ludmilkaskok [199]3 years ago

5 0

Answer:

none of the answers is correct, the time is the same t₁ = t₂ = 0.600 s

Explanation:

This is a kinematics exercise, analyze the situation a bit. The vertical speed in both cases is the same is zero, the horizontal speed in the second case is double (vₓ₂ = 2 vₓ₁)

let's find the time to hit the ground

y = y₀ + I go t - ½ g t²

0 = y₀ - ½ g t²

t = √ 2y₀ / g

with the data from the first launch

y₀i = ½ g t²

y₀ = ½ 9.8 0.6²

y₀ = 1,764 m

with this is the same height the time to descend in the second case is the same

t₂ = 0.600 s

this is because the horizontal velocity change changes the offset on the x axis, but does not affect the offset on the y axis

Therefore, none of the answers is correct, the time is the same

t₁ = t₂ = 0.600 s

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The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

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4 years ago

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daser333 [38]

I beleve the answer is D).

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4 years ago

Stella [2.4K]

The weight of the plane is (mass) x (acceleration of gravity).
Neither of these changes when the plane rises from the ground.
Its weight on the ground is equal to (the same as) its weight in the air.

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3 years ago

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Which tool do we use to measure wind

wariber [46]

Answer:

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Explanation:

windmill anemometer is commonly used by news stations to measure winds speeds

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3 years ago

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Mary applies a force of 73 N to push a box with an acceleration of 0.48 m/s^2. When she increases the pushing force to 84 N, the

nikklg [1K]

Answer: $68.75\ kg, 0.06$

Explanation:

Mary applies a force of 73 N to create an acceleration of $0.48\ m/s^2$

When She increases force to 84 N, it creates an acceleration of $0.64\ m/s^2$

Friction opposes the motion of box

$\Rightarrow 73-f=m\times 0.48\quad \ldots(i)\\\Rightarrow 84-f=m\times 0.64\quad \ldots(ii)$

Subtract (i) from (ii)

$\Rightarrow 11=m(0.64-0.48)\\\Rightarrow m=68.75\ kg$

Therefore friction is

$\Rightarrow f=73-68.75\times 0.48\\\Rightarrow f=73-33\\\Rightarrow f=40\ N$

Here, friction is kinetic friction which is given by

$\Rightarrow f=\mu_kmg\\\Rightarrow 40=\mu_k 68.75\times 9.8\\\Rightarrow \mu_k=0.061$

5 0

3 years ago