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ANEK [815]
2 years ago
8

Tyler stands at rest on a skateboard. He has a mass of 120 kg. His friend (m = 60 kg) jumps into his arms at a speed of 2 m/s. I

f Tyler catches his friend, how fast will they move on the skateboard?
Physics
1 answer:
Andrews [41]2 years ago
5 0
Momentum question. This is an inelastic collision, so 

m1v1+m2v2=Vf(m1+m2)
Vf=(m1v1+m2v2)/(m1+m2)=[(120kg)(0m/s)+(60kg)(2m/s)] / (120kg+60kg)
Vf=120kg m/s  /   180kg
Vf=0.67m/s

0.67m/s
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What mass needs to be attached to a spring with a force constant of 7N/m in order to make a simple harmonic oscillator oscillate
Alex_Xolod [135]

Answer:

Mass will be 4.437 kg

Explanation:

We have given force constant k = 7 N/m

Time period of oscillation T = 5 sec

So angular frequency \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{5}=1.256rad/sec

We know that angular frequency is given by

\omega =\sqrt{\frac{k}{m}}

1.256 =\sqrt{\frac{7}{m}}

Squaring both side

1.577 =\frac{7}{m}

m = 4.437 kg

6 0
2 years ago
A car with an initial velocity of 16.0 meters per second east slows uniformly to 6.0 meters per second east in 4.0 seconds. What
pychu [463]
(6-16)/4.0=-2.5 m/s²
Acceleration of the car is -2.5 m/s²
5 0
2 years ago
A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t
sertanlavr [38]

Explanation:

The given data is as follows.

       C = 20 \times 10^{-6} F

        R = 100 \times 10^{3} ohm

        Q_{o} = 100 \times 10^{-6} C

          Q = 13.5 \times 10^{-6} C

Formula to calculate the time is as follows.

          Q_{t}  = Q_{o} [e^{\frac{-t}{\tau}]

       13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]

               0.135 = e^{\frac{-t}{2}}

         e^{\frac{t}{2}} = \frac{1}{0.135}

                         = 7.407

           \frac{t}{2} = ln (7.407)

                      t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

4 0
2 years ago
Before leaving the house in the morning, you plop some stew in your slow cooker and turn it on Low. The slow cooker has a 160 Oh
guajiro [1.7K]

Answer:

Total charge flow through the cooker is 21600 C

Explanation:

As we know that the current flow through the cooker is given by Ohm's law

here it is given as

V = i R

i = \frac{V}{R}

i = \frac{120}{160}

i = \frac{3}{4} A

now the charge flow through it is given as

Q = i t

total time is t = 8 hours

Q = \frac{3}{4}(8 \times 60 \times 60)

Q = 21600 C

7 0
2 years ago
The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is c
puteri [66]

Answer:

64.59kpa

Explanation:

See attachment

6 0
2 years ago
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