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shutvik [7]
3 years ago
15

The main goal of science is to A. Learn how to use the laws of nature to benefit humans. B. Learn about the universe and the law

s of nature. C. Replace old theories with new, more interesting theories. D. Develop new inventions and processes.
Physics
1 answer:
marishachu [46]3 years ago
7 0

The main goal of science is to learn about the universe and the laws of nature. <em>(B)</em>

The goal of Applied Science, Technology, and Engineering is to develop new inventions and processes (D), that can use the laws of nature to benefit humans. (A)

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zysi [14]

Answer:

yes it doesn't matter

Explanation:

it doesn't matter because troughs and crests are the same and either can be used

7 0
2 years ago
What is the velocity of a car that travels -50 meters in 5 seconds then 10 meters in 5 seconds
diamong [38]

Answer:

2 m/s

Explanation:

The first part of the question the car is going in reverse or negative along the x axis. Then the second part the car is moving forward along the x axis. So the car would only have velocity in the current direction of movement. So our equation for velocity is as follows.

v = d/t

v = 10 m/5 s

v = 2 m/s

3 0
2 years ago
Sometimes the tar of a newly paved road gets sticky on a very hot summer day. What causes this?
Ksivusya [100]
I’m pretty sure it is caused by the heat of the sun warming it up back into its original state of tar
6 0
3 years ago
A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the
Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2= 85.75 J

The speed of the ball after leaving the bat is:

K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

V_f^2-V^2=-2gh

V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m

V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2

V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}

V_f=42.5m/s

3 0
2 years ago
How do I solve for horizontal force? And how can I find the acceleration to find the force? And how do I find the power during s
Snezhnost [94]

Answer:hhmm c

Explanation:

5 0
3 years ago
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