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mash [69]
3 years ago
13

When a firecracker explodes, what types of energy does it give off?

Physics
1 answer:
fomenos3 years ago
7 0
The answer is c. 
Sound, light and heat energy.

Hope this helped :)
You might be interested in
How long would it take a plane to fly 1,054 km with an average speed of 886 km/h?
Leviafan [203]

Answer:

1.19 hours

Explanation:

divide distance by speed. hope this helps

8 0
2 years ago
a ball is thrown striaght up in the air and then falls back to earth. if the downward fall takes 2.2s, how fast is the ball trav
lapo4ka [179]

The velocity of the ball when it strikes the ground, given the data is 21.56 m/s

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • Time to reach ground from maximum height (t) = 2.2 s
  • Initial velocity (u) = 0 m/s
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Final velocity (v) =?

<h3>How to determine the velocity when the ball strikes the ground</h3>

The velocity of the ball when it strikes the ground can be obtained as illustrated below:

v = u + gt

v = 0 + (9.8 × 2.2)

v = 0 + 21.56

v = 21.56 m/s

Thus, the velocity of the ball when it strikes the ground is 21.56 m/s

Learn more about motion under gravity:

brainly.com/question/22719691

#SPJ1

5 0
2 years ago
According to Newton's first law, an object at rest will _____.
Vesnalui [34]
Stay at rest unless moved my force! :)
8 0
2 years ago
Please help!!~
lisov135 [29]

Answer:

C. Waves transfer energy, but not matter.

Explanation: hope this helps :)

7 0
3 years ago
A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o
iVinArrow [24]
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s
- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, m \frac{v^2}{r}, so we can write:
mg-N=m \frac{v^2}{r} (1)
By rearranging the equation and substituting the numbers, we find N:
N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
mg=m \frac{v^2}{r}
from which we find
v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s
7 0
3 years ago
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