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just olya [345]
3 years ago
13

The manufacturer of a bulletproof vest wants the vest to be able to stop a bullet with a mass of 0.4 kg and a velocity of 1800 m

/s. what is the momentum of this bullet?
a. 720 kg×m/s
b. 740 kg×m/s
c. 820 kg×m/s
d. 4500 kg×m/s
Physics
1 answer:
11111nata11111 [884]3 years ago
7 0
We know that momentum = mass times velocity
So a. 720 kgm/s
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A ball is thrown vertically upwards from the edge of the cliff and hits the ground at the base of the cliff with a speed of 30 m
olya-2409 [2.1K]

To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.

v = v_0 -gt

Here,

v = Final velocity

v_0 = Initial velocity

g = Acceleration due to gravity

t = Time

At t = 4s, v = -30m/s (Downward)

Therefore the initial velocity will be

-30 = v_0 -9.8(4)

v_0 = 9.2m/s

Now the position can be calculated as,

y = h +v_0t -\frac{1}{2}gt^2

When it has the ground, y=0 and the time is t=4s,

0 = h+(9.2)(4)-\frac{1}{2} (9.8)(4)^2

h = 41.6m

Therefore the cliff was initially to 41.6m from the ground

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3 years ago
An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f
sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

f=\frac{v_s}{2L}         (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

f=\frac{343m/s}{2(0.47228m)}=362.36Hz      

Closed tube:

f'=\frac{v_s}{4L'}

L': length of the closed tube = 0.702821m

f'=\frac{343m/s}{4(0.702821m)}=122.00Hz

Next, you use the following formula for the beat frequency:

f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

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