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Aneli [31]
3 years ago
11

4. ""Giving in"" during a catching action in baseball More than one answer is correct A. reduces the magnitude of the force requ

ired to stop the ball. B. increases the magnitude of the force required to stop the ball. C. increases the contact time of the hand with the ball. D. does neither increase nor decrease the impulse required to stop the ball.
Physics
1 answer:
blondinia [14]3 years ago
3 0

Answer:

A. reduces the magnitude of the force required to stop the ball.

C. increases the contact time of the hand with the ball.

D. does neither increase nor decrease the impulse required to stop the ball.

Explanation:

As we know that the force required to stop the ball is given as

F = \frac{\Delta P}{\Delta t}

here we know that

impulse = \Delta P

so we have

impulse = m(v_f - v_i)

now we know that time to stop the ball is increased due to which the force to stop the ball is decreased

so we have correct answer will be

A. reduces the magnitude of the force required to stop the ball.

C. increases the contact time of the hand with the ball.

D. does neither increase nor decrease the impulse required to stop the ball.

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A car slows down at -5.00 m/s^2 until it comes to a stop after travelling 15.0 m. How much time did it take to stop?
Xelga [282]
In physics, there are already derived equation that are based on Newton's Law of Motions. The rectilinear motions at constant acceleration have the following equations:

x = v₁t + 1/2 at²
a = (v₂-v₁)/t

where
x is the distance travelled
v₁ is the initial velocity
v₂ is the final velocity
a is the acceleration
t is the time

Now, we solve first the second equation. Since it mentions that the car comes eventually to a stop, v₂ = 0. Then,

-5 = (0-v₁)/t
-5t = -v₁
v₁ = 5t

We use this new equation to substitute to the first one:
x = v₁t + 1/2 at²
15 = 5t(t) + 1/2(-5)t²
15 = 5t² - 5/2 t²
15 = 5/2 t²
5t² = 30
t² = 30/5 = 6
t = √6 = 2.45

Therefore, the time it took to travel 15 m at a deceleration of -5 m/s² is 2.45 seconds.

5 0
3 years ago
calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end
sdas [7]

Answer:0.8\ mm

Explanation:

Given

length of wire l=75\ cm

change in length \Delta l=1.85\ mm

mass of wire m=10\ kg

Young's modulus for silver E=7.9\times 10^{10}\ N/m^2

load on wire F=mg

F=10\times 9.8=98\ kg

change in length is given by

\Delta l=\dfrac{Pl}{AE}

Where A=area of cross-section

A=\dfrac{Pl}{\Delta lE}

A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}

A=\dfrac{73.5}{14.615\times 10^{7}}

A=5.029\times 10^{-7}\ m^2

also wire is the shape of cylinder so cross-section is given by

A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2

\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }

\Rightarrow d^2=64.02\times 10^{-8}

\Rightarrow d=8\times 10^{-4}\ m

\Rightarrow d=0.8\ mm

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