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omeli [17]
3 years ago
12

Potential energy and kinetic energy are forms of what kind of energy? 1. chemical 2. nuclear 3. electromagnetic 4. heat 5. mecha

nical
Physics
1 answer:
blsea [12.9K]3 years ago
4 0
Heat and mechanical energy.
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This speed is measured with respect to the space station the spacecraft was originally launched from. In interstellar space the
valentinak56 [21]

Answer:

15193.62 m/s

Explanation:

t = Time taken = 6.5 hours

u = Initial velocity = 0 (Assumed)

m = Mass of rocket = 1380 kg

F = Thrust force = 896 N

v = Final velocity

a = Acceleration of the rocket

Force

F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{896}{1380}\\\Rightarrow a=0.6493\ m/s^2

Equation of motion

v=u+at\\\Rightarrow v=0+0.6493\times 6.5\times 60\times 60\\\Rightarrow v=15193.62\ m/s

The velocity of the rocket after 6.5 hours of thrust is 15193.62 m/s

5 0
3 years ago
What statement is true according to newton’s first law of motion?
Airida [17]

Answer:

  c.  in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.

Explanation:

First law: things keep doing what they are doing, unless force is applied.

5 0
2 years ago
Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction
Norma-Jean [14]

Answer:

The ball would hit the floor approximately 0.55\; \rm s after leaving the table.

The ball would travel approximately 5.5\; \rm m horizontally after leaving the table.

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

Let \Delta h denote the change to the height of the ball. Let t denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let v_0(\text{vertical}) denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible:\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t.

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}.

The height of the table was 1.5\; \rm m. Therefore, after hitting the floor, the ball would be 1.5\; \rm m \! below where it was before leaving the table. Hence, \Delta h = -1.5\;\rm m.

The equation becomes:

\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}.

Solve for t:

\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55.

In other words, it would take approximately 0.55\; \rm s for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at v(\text{horizontal}) =10\; \rm m \cdot s^{-1}) until the ball hits the floor.

The ball was in the air for approximately t = 0.55\; \rm s and would have travelled approximately v(\text{horizontal})\cdot t \approx 5.5\;\rm m horizontally during the flight.

4 0
2 years ago
How much heat is absorbed by 60g of copper when it is heated from 20°C to 80°C
nexus9112 [7]

Answer:

I HOPE THIS IS CORRECT

Explanation:

It is heated from 20°C to 80°C. We need to find the heat absorbed. It can be given by the formula as follows : So, 1386 J of heat is absorbed.

7 0
2 years ago
HELP NOWWWWW PLEASE
Finger [1]

Answer:it is A

Explanation:

ok

7 0
2 years ago
Read 2 more answers
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