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Maslowich
3 years ago
15

An arrow strikes a target moving at 75 m/s and embeds itself 15 cm into the target. If the arrow stopped with constant accelerat

ion, how long (in seconds) after striking the target did it take for the arrow to come to rest?
Physics
1 answer:
zhenek [66]3 years ago
5 0

Answer:

Answer:

4 ms

Explanation:

initial velocity, u = 75 m/s

final velocity, v = 0

distance, s = 15 cm = 0.15 m

Let the acceleration is a and the time taken is t.

Use third equation of motion

v² = u² + 2 a s

0 = 75 x 75 - 2 a x 0.15

a = - 18750 m/s^2

Use first equation of motion

v = u + at

0 = 75 - 18750 x t

t = 4 x 10^-3 s

t = 4 ms

thus, the time taken is 4 ms.

Explanation:

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A 1.75 kg stone falls from the top of a cliff and strikes the ground at a speed of 58.8 m/s. What is the height of the cliff??
Reil [10]

Answer:

172.9m

Explanation:

h = 1/2 gt^

First calculat for t using

v = gt

t = v/g = 58.8/10

= 5.88secs

now h = 1/2 x 10 x 5.88^2

h =1/2 x 10 x 34.57

= 345.74/2

= 172.9m

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3 years ago
Acceleration of 1.5 ms expressed in km /hr2? ​
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You’re answer is 5 because !! :)
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A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the grou
True [87]

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

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At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

6 0
3 years ago
HELP!!!
aniked [119]

Answer:

p = m .v momentum = mass • velocity. [kg • m/s] [kg] [m/s]. Kinetic Energy. KE = 12 • m • v ... 1. A 1500 kg car traveling at 15 m/s to the south collides with a 4500 kg truck that is ... What is the final velocity of the two-vehicle mass? ... m/s. What is the velocity of the joined cars after the collision? ... 5) = (1.5x104+1.5x604) VELVE.

Explanation:

4 0
2 years ago
Consider the following waves representing electromagnetic radiation: An illustration shows two waves representing electromagneti
Murljashka [212]

Answer:

a) red wave hs a longer wavelength than the green wave

b)f = 1.875 10¹¹ Hz ,  f_green = 5.45 10¹⁴Hz

c)   E = 1.24 10⁻²² J , E_green = 3.6 10⁻¹⁹ J

d) The red wave is in the infrared range, heat waves

The green wave is in the visible wavelength

Explanation:

a) The green wave are on the left in the electromagnetic spectrum so the red wave has a longer wavelength than the green wave

The green wavelength is in the range of 550 10⁻⁹ m

The speed of the wave is

            c = λ f

            f = c /λ

b) The frequency of the red wave is

            f = 3 10⁸ / 1.6 10⁻³

            f = 1.875 10¹¹ Hz

For the green wave

           f_green = 3 10⁸/550 10⁻⁹

           f_green = 5.45 10¹⁴Hz

c) The photon energy is given by the Planck equation

             E = h f

             E = 6.63 10⁻³⁴ 1.875 10¹¹

             E = 1.24 10⁻²² J

For the green wave

              E_green = 6.63 10⁻³⁴ 5.45 10¹⁴

              E_green = 3.6 10⁻¹⁹ J

d) The speed of electromagnetic waves is constant and has a value of 3 108 m / s

e)  

The red wave is in the infrared range, heat waves

The green wave is in the visible wavelength

6 0
3 years ago
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