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Artist 52 [7]
3 years ago
13

A 0.041-kg bullet has a kinetic energy of 600j. What is the velocity of the bullet?

Physics
1 answer:
Paladinen [302]3 years ago
5 0

171.0798 M/S

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

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A movie stunt performer is filming a scene where he swings across a river on a vine. The safety crew must use a vine with enough
Julli [10]

Answer:

1125.66956 N

Explanation:

m = Mass of stunt performer

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of the swing = 7 m/s

T = Tension

r = Radius of the swing = Length of vine = 11.5 m

From the free body diagram

T-mg-m\frac{v^2}{r}=0\\\Rightarrow T=mg+m\frac{v^2}{r}\\\Rightarrow T=m(g+\frac{v^2}{r})\\\Rightarrow T=80(9.81+\frac{7^2}{11.5})\\\Rightarrow T=1125.66956\ N

The minimum tension force the vine must be able to support without breaking is 1125.66956 N

7 0
3 years ago
A generator consists of a rectangular loop with turns of wire spinning at in a uniform magnetic field. The generator output is c
iren [92.7K]
<h3><u>Full question:</u></h3>

A generator consists of a 18-cm by 12-cm rectangular loop with 500 turns of wire spinning at 60 Hz in a 25 mT uniform magnetic field. The generator output is connected to a series RC circuit consisting of a 150 Ω and a 35 μF capacitor.What is the average power delivered to the circuit?

<h3><u>Answer:</u></h3>

The average power delivered to the circiut is 27.5 W

<h3><u>Solution:</u></h3>

Induced emf amplitude = N A B w  

N- Number of turns of the coil

B- Magnetic field

=500 \times(0.18 \times 0.12) \times 0.025 \times(2 \pi 60)

Induced emf amplitude = 101.8 Volt

\begin{aligned}&X_{c}=\frac{1}{(2 p i f C)}=\frac{1}{2 \times \pi \times 60 \times 35 \times 10^{-6}}\\&X_{c}=75.8\ \mathrm{ohm}\end{aligned}

R = 150 ohm

\begin{aligned}&z=\sqrt{\left[R^{2}+X_{c}^{2}\right]}=168 \text { ohm }\\&I_{\text {peak}}=\frac{101.8}{168}=0.606\ \mathrm{A}\\&P_{a v g}=\frac{l^{2} R}{2}=27.5\ \mathrm{W}\end{aligned}

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