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QveST [7]
3 years ago
9

A water balloon was dropped from a high window and struck its target 1.1 seconds later. If the balloon left the person's hand at

-5.0 m/sec, what was its velocity on impact?
Physics
1 answer:
GrogVix [38]3 years ago
5 0
The acceleration of gravity on Earth is 9.8 meters per second every second.  So
in 1.1 second of falling, an object gains (9.8 x 1.1) = 10.78 meters per second of
downward speed.

The balloon was already falling at 5 m/sec when the time began, and it gained
another 10.78 m/sec on the way.  So at the end of 1.1 second, its downward
speed was (5.0 + 10.78) = 15.78 m/sec.

Its velocity was 15.78 m/sec downward. 
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A sinusoidal transverse wave travels along a long, stretched, string. the amplitude of this wave is 0.0885 m, it's frequency is
timofeeve [1]

Answer:

(a) 0.177 m

(b) 16.491 s

(c) 25 cycles

Explanation:

(a)

Distance between the maximum and the minimum of the  wave = 2A ............ Equation 1

Where A = amplitude of the wave.

Given: A = 0.0885 m,

Distance between the maximum and the minimum of the wave = (2×0.0885) m

Distance between the maximum and the minimum of the wave = 0.177 m.

(b)

T = 1/f ...................... Equation 2.

Where T = period, f = frequency.

Given: f = 4.31 Hz

T = 1/4.31

T = 0.23 s.

If 1 cycle pass through the stationary observer for 0.23 s.

Then, 71.7 cycles will pass through the stationary observer for (0.23×71.7) s.

= 16.491 s.

(c)

If  1.21 m contains  1 cycle,

Then, 30.7 m will contain (30.7×1)/1.21

= 25.37 cycles

Approximately 25 cycles.

6 0
2 years ago
A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height
AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

E_i=\frac{1}{2}m_1v_1^2

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

E_f = (m_1 +m_2)gh_{max}

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m

4 0
3 years ago
A 12-kg hammer strikes a nail at a velocity of 8.5 m/s and comes to rest in a time interval of 8.0 ms. (A) What is the impulse g
dimulka [17.4K]

Answer:

102 kg.m/s

Explanation:

m = Mass of hammer = 12 kg

v = Final velocity = 8.5 m/s

u = Initial velocity = 0

t = Time taken = 8 ms

Force acting over a given amount of time or change in momentum is known as impulse.

Impulse

J=Ft\\\Rightarrow J=p_2-p_1\\\Rightarrow J=m(v-u)\\\Rightarrow J=12(8.5-0)\\\Rightarrow J=102\kg.m/s

Impulse given to the nail is 102 kg.m/s

6 0
2 years ago
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What happens when a proton is placed directly in the path of the proton cannon?
blondinia [14]

Answer:coherent light

Explanation:

5 0
3 years ago
Whats the net force​
zavuch27 [327]

Answer:

13n pushing left

Explanation:

1563 - 1550

3 0
3 years ago
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