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Sati [7]
2 years ago
14

A football has a mass of 2 kg and it accelerates at 20 m/s2. What is its force? (You can use a calculator)

Physics
2 answers:
attashe74 [19]2 years ago
8 0
10 mps is the answer
kap26 [50]2 years ago
6 0

Answer:

The answer is 40mps.

Explanation:

1. Force = (mass of an object) x (acceleration)

2. Force = ( "has a mass of 2 kg" ) x ( "accelerates at 20 m/s^2" )

3. Plug the numbers into the formula

4. F = 2 x 20

5. F = 40 mps

<em>Hope I helped!! Please mark me branliest :)</em>

- sophi

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Vitamins and iron are nutrients that provide energy to the human body.<br> A true<br> B false
SIZIF [17.4K]
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State the Pascal's principle <br>​
krok68 [10]

Answer:

Pascal's principle, also called Pascal's law, in fluid (gas or liquid) mechanics, statement that, in a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container

8 0
2 years ago
An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at
andrew-mc [135]

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

6 0
3 years ago
Glycerin is poured into an open U-shaped tube until the height in both sides is 20cm. Ethyl alcohol is then poured into one arm
Liula [17]

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining \rho_A: Ethyl alcohol density and \rho_G: Glycerin density , we can write:

\rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3

Simplifying:

\rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1)

On the other hand:

h_1 + h_2 = \Delta h + h_3

Rearranging:

h_1 - \Delta h = h_3 - h_2 (2)

Replacing  (2) in (1):

\rho_A\times h_1 = \rho_G \times (h_1 - \Delta h)

Rearranging:

\frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h

Data:

h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3}

\frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}}  = \Delta h

7.5 cm =  \Delta h

7 0
3 years ago
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