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MaRussiya [10]
3 years ago
5

A 3.0 g bullet traveling at a speed of 400 m/s enters a tree and exits the other side with a speed of 200 m/s. Where did the bul

let's lost KE go, and what was the energy transferred?
Physics
1 answer:
cricket20 [7]3 years ago
6 0

Answer:

1800J

Explanation:

Step one:

given data

mass of bullet m= 3g= 0.03kg

initial velocity u = 400m/s

final velocity v= 200 m/s

Step two:

1.The bullet's lost kinetic energy went inside the tree.

2. The energy transferred is computed as

= initial KE- KE final

Initial KE= 1/2mu^2

Initial KE= 1/2*0.03*400^2

Initial KE= 1/2*0.03*160000

Initial KE= 1/2*4800

Initial KE= 2400J

KE final= 1/2mv^2

KE final= 1/2*0.03*200^2

KE final= 1/2*0.03*40000

KE final= 1/2*1200

KE final= 600J

KE transferred = 2400-600

KE transferred= 1800J

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Answer:

option a

Explanation:

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On the other hand, sound waves are mechanical waves which can only travel in a medium as they can only move forward by oscillating material particles.

Thus, the student should conclude option a.

8 0
3 years ago
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An electric heater carries a current of 13.5 A when operating at a voltage of 120V . What is the resistance of the heater?
Aloiza [94]

An electric heater that carries a current of 13.5 A when operating at a voltage of 120V , will have  the resistance of the heater as 8.89 ohm.

<h3>What is Ohm's law?</h3>

Ohm's law  can be described as the law in physics that   focus on how voltage or potential difference between two points is directly proportional to the current or electricity that is moving in the direction of the resistance.

An this is also  directly proportional to the resistance of the circuit, hence it can be calculated using the Ohm's law is V=IR.

from V=IR

R=V/I

= 120/13.55

=8.89 ohm.

Therefore,  electric heater that carries a current of 13.5 A when operating at a voltage of 120V , will have  the resistance of the heater as 8.89 ohm.

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4 0
1 year ago
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a

Explanation:

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3 years ago
Is vibrating considered work
allsm [11]

Answer: no it shouldn't be

Explanation:

8 0
3 years ago
A sled with no initial velocity accelerates at rate of 5.0 m/s^2 down a hill . How long does it take the sled to go 45 m to the
Leya [2.2K]

The time taken by the sled to reach 45 m to the bottom is 4.24 s.

The sled's initial velocity <em>u</em> is zero. It has an acceleration <em>a</em> of a value 5 m/s² down the hill and it travels a distance <em>s</em> equal to 45 m down the hill.

Use the equation of motion,

s=ut+\frac{1}{2} at^2

Substitute 0 m/s for u and rewrite the equation for t.

t=\sqrt{\frac{2s}{a}}  \\ =\sqrt{\frac{2(45 m)}{5.0m/s^2} } \\ =4.24s

The sled takes a time of 4.24 s to reach the bottom of the hill

6 0
4 years ago
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