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kotykmax [81]
4 years ago
9

What is the average rate of change of f(x), represented by the graph, over the interval [-2, 2)?

Physics
1 answer:
Sedbober [7]4 years ago
8 0

Answer:

...

Explanation:

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4 years ago
A ball has m=10 kg and v=10 m/s. It has the same momentum as which of the following?
crimeas [40]

Answer:

Choice d. All three balls listed in choice a, b, and c have a momentum of 100\; \rm kg \cdot m \cdot s^{-1} each, same as that of the 10\; \rm kg ball moving at 10\; \rm m \cdot s^{-1}. Assumption: all four balls are moving in the same direction.

Explanation:

The momentum p of an object is equal to the product of its mass m and its velocity v.

Momentum of a 10\; \rm kg ball moving at 10\; \rm m \cdot s^{-1}:

p = m \cdot v = 10\; \rm kg \times 10\; \rm m \cdot s^{-1} = 100\; \rm kg \cdot m \cdot s^{-1}.

Similarly:

  • Momentum of the ball in choice a: 100 \; \rm kg \times 1\; \rm m \cdot s^{-1} = 100\; \rm kg \cdot m\cdot s^{-1}.
  • Momentum of the ball in choice b: 5 \; \rm kg \times 20\; \rm m \cdot s^{-1} = 100\; \rm kg \cdot m\cdot s^{-1}.
  • Momentum of the ball in choice c: 2 \; \rm kg \times 50\; \rm m \cdot s^{-1} = 100\; \rm kg \cdot m\cdot s^{-1}.

Hence the conclusion that these balls have the same momentum.

7 0
3 years ago
A 90-kg astronaut is stranded in space at a point 6.0 m from his spaceship, and he needs to get back in 4.0 min to control the s
Stella [2.4K]

Momentum = 0.5 * 4 = 2 
to conclude the man’s velocity after he throws the piece of equipment, divide this number by the man’s mass. 

v = 2/90 

This is about 0.0222 m/s. To know if he can move 6 meters at velocity in 4minutes, use the following equation. 

d = v * t, t = 4 * 60 = 240 s 
d = 2/90 * 240 = 5⅓ meters. 

This is ⅔ of a meter from the spaceship. To know the velocity that he must have to move 6 meter, use the same equation. 

6 = v * 240 
v = 6/240 
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His final momentum = 90 * 6/240 = 2.25 


To know the velocity of the package, divide this number by the mass of the package. 
v = 2.25/0.5 = 4.5 m/s

8 0
3 years ago
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