If all the energy she put into bending the bow is completely
transmitted to the arrow, then the arrow has the 100 joules
of kinetic energy when it leaves the bow.
Kinetic energy = (1/2) (mass) (speed)²
100 J = (1/2) (0.5 kg) (speed²)
Divide each side by 0.25 kg: 100 J / 0.25 kg = speed²
[ joule ] = [ newton-meter ] = kg-m²/sec²
100 kg-m²/sec² / 0.25 kg = speed²
400 m²/sec² = speed²
Take the square root of each side: speed = √400 m/s
20 m/s
(about 44.7 mph)
Answer:
Force exerted = 25.41 kN
Explanation:
We have equation of motion
v² = u²+2as
u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s
0² = 345²+2 x a x 0.089
a = -668679.78 m/s²
Force exerted = Mass x Acceleration
Mass of bullet = 38 g = 0.038 kg
Acceleration = 668679.78 m/s²
Force exerted = 25409.83 N = 25.41 kN
Answer:
The potential energy of the ball is 784 joules. And the kinetic energy of it is 392 while falling halfway down.
Explanation:
PE = mass (2kg) * Gravitational acceleration (9.8 m/s^2)* height (40 meters)
KE = 1/2 mass (1 kg) * velocity^2 (19.8)
Multiple point perspective is a system of perspective in which: A. there are a number of vanishing points, usually created by multiple objects.
<h3>What is
multiple point perspective?</h3>
Multiple point perspective can be defined as a system of perspective in which each items (objects) has a number of interacting lines with different vanishing points.
This ultimately implies that, a multiple point perspective is a system of perspective which is created with a number of vanishing points, that are usually created by multiple objects.
Read more on multiple point perspective here: brainly.com/question/5398474
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(a) The lowest frequency (called fundamental frequency) of a wire stretched under a tension T is given by

where
L is the wire length
T is the tension
m is the wire mass
In our problem, L=10.9 m, m=55.8 g=0.0558 kg and T=253 N, therefore the fundamental frequency of the wire is

b) The frequency of the nth-harmonic for a standing wave in a wire is given by

where n is the order of the harmonic and f1 is the fundamental frequency. If we use n=2, we find the second lowest frequency of the wire:

c) Similarly, the third lowest frequency (third harmonic) is given by