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Svetlanka [38]
2 years ago
15

You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand at 30 m/s. What is the kinetic energy (KE) of the volley

ball?
Physics
2 answers:
FinnZ [79.3K]2 years ago
8 0

awnser :

Explanation:

Lena [83]2 years ago
3 0

Answer:

I believe the answer is KE = 945 J

Explanation:

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Type O negative blood is considered the universal red blood cell donor because it _____.
marissa [1.9K]

Answer:

Honestly i think the answer is B

Explanation:

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3 years ago
Costal residents must do many things to prepare for hurricanes
tangare [24]
The answer is A it’s more safer that way
3 0
2 years ago
A cylindrical piece of aluminum is 6.00cm y’all and 2.00cm in radius how much does it weigh
Harlamova29_29 [7]

Answer:

2.00 N

Explanation:

Weight is mass times gravity:

W = mg

Mass is density times volume:

m = ρV

Volume of a cylinder is:

V = πr²h

Finding the volume:

V = π (2.00 cm)² (6.00 cm)

V = 75.4 cm³

The density of aluminum is 2.7 g/cm³.  Finding the mass:

m = (2.7 g/cm³) (75.4 cm³)

m = 204 g

Finding the weight:

W = (0.204 kg) (9.8 m/s²)

W = 2.00 N

The aluminum cylinder weighs 2.00 N.

3 0
2 years ago
How did this formula come about? Please show me the formula derivation process
kirill [66]

Answer:

so simple it is a square formula

6 0
3 years ago
Estimate the number of Ping-Pong balls that can be packed into an average size room (without crushing them). Given that Ping-Pon
scoray [572]

Answer: 3,893,845.918 Ping-Pong balls

Explanation:

The volume of an average room is:

V_{room}=(length)(width)(height) (1)

V_{room}=(12 ft)(18 ft)(9 ft)=1944 ft^{3} (2)

Now let’s transform this V_{room} to units of cm^{3}, knowing 1 ft=30.48 cm:

V_{room}=1944 ft^{3}\frac{{(30.48 cm)}^{3}}{1ft^{3}}=55,047,949.78 cm^{3} (3)

On the other hand, we have Ping-Pong balls with a radius r=1.5 cm, and their volume is given by:

V_{balls}=\frac{4}{3} \pi r^{3} (4)

V_{balls}=\frac{4}{3} \pi (1.5 cm)^{3} (5)

V_{balls}=14.137 cm^{3} (6)

Now, the number n of Ping-Pong balls that can be packed into the room is:

n=\frac{V_{room}}{V_{balls}} (7)

n=\frac{55,047,949.78 cm^{3}}{14.137 cm^{3}}  (8)

n=3,893,845.918 This is the number of Ping-Pong balls that can be packed into an average size room

7 0
2 years ago
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