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kifflom [539]
3 years ago
11

In a video game, a flying coconut moves at a constant velocity of 20 meters/second. The coconut hits an obstacle and moves in th

e opposite direction with a constant velocity of 10 meters/second. What’s the change in velocity of the coconut?
A. -30 m/s
B. -10 m/s
C. 10 m/s
D. 30 m/s
Physics
1 answer:
qwelly [4]3 years ago
4 0
Change in velocity = Final velocity - Initial velocity

Initial velocity = 20 m/s
Final velocity = 10 m/s

Change in velocity = 10 - 20 = -10m/s

Correct option B

Hope This Helps You!
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Firdavs [7]
C is the answer to the question
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A) explain how the data matches each line on the graph (give name and line letter for each)
stepladder [879]
The answer you have is right good job 12
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2 years ago
A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance
seropon [69]

Answer: Option B. R = (1/2)gt^2

Explanation:

S = R (horizontal distance)

V^2 = 2gS

V^2 = 2gR

R = V^2 / 2g

But V = gt

R = (gt)^2 / 2g

R = (g^2 x t^2) / 2g

R = gt^2 / 2

But t^2 = 2h/g

R = ( g x 2h/g) / 2

R = h

But h = (1/2)gt^2

R = h = (1/2)gt^2

4 0
3 years ago
Practical steam engines utilize 450ºC steam, which is later exhausted at 270ºC.
Naily [24]

(a) 0.249 (24.9 %)

The maximum efficiency of a heat engine is given by

\eta = 1-\frac{T_C}{T_H}

where

Tc is the low-temperature reservoir

Th is the high-temperature reservoir

For the engine in this problem,

T_C = 270^{\circ}C+273=543 K

T_H = 450^{\circ}C+273=723 K

Therefore the maximum efficiency is

\eta = 1-\frac{T_C}{T_H}=1-\frac{543}{723}=0.249

(b-c) 0.221 (22.1 %)

The second steam engine operates using the exhaust of the first. So we have:

T_H = 270^{\circ}C+273=543 K is the high-temperature reservoir

T_C = 150^{\circ}C+273=423 K is the low-temperature reservoir

If we apply again the formula of the efficiency

\eta = 1-\frac{T_C}{T_H}

The maximum efficiency of the second engine is

\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221

8 0
3 years ago
Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statemen
muminat

Answer:

Option A

Explanation:

From the question we are told that:

Mass m=0.20kg

Velocity v=4m/s

Generally the equation for momentum for Ball A is mathematically given by

Initial Momentum

 M_{a1}=mV

 M_{a1}=0.2*4

 M_{a1}=0.8

Final Momentum

 M_{a2}=-0.8kgm/s

Therefore

 \triangle M_a=-1.6kgm/s

Generally the equation for momentum for Ball B is mathematically given by

Initial Momentum

 M_{b1}=mV

 M_{b1}=0.2*4

 M_{b1}=0.8

Final Momentum

 M_{b2}=-0 kgm/s

Therefore

 |\triangle M_a|>|\triangle Mb|

Option A

4 0
3 years ago
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