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skelet666 [1.2K]
3 years ago
15

What kind of energy does a football flying through the air have? Explain.

Physics
1 answer:
Nadusha1986 [10]3 years ago
8 0

1# The ball flying through the air has two types of energy, kinetic from its velocity and potential from its height off the ground.

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Newton’s first law because the airplane needs space to fully stop due to its inertia
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I need help with these Physics problems​
adoni [48]

Answer:

1. 3 m

2. 27 s

Explanation:

1. "A car traveling at +33 m/s sees a red light and has to stop.  If the driver can accelerate at -5.5 m/s², how far does it travel?"

Given:

v₀ = 33 m/s

v = 0 m/s

a = -5.5 m/s²

Unknown: Δx

To determine the equation you need, look for which variable you don't have and aren't solving for.  In this case, we aren't given time and aren't solving for time.  So look for an equation that doesn't have t in it.

Equation: v² = v₀² + 2aΔx

Substitute and solve:

(0 m/s)² = (33 m/s)² + 2(-5.5 m/s²) Δx

Δx = 3 m

2. "A plane starting from rest at one end of a runway accelerates at 4.8 m/s² for 1800 m.  How long did it take to accelerate?"

Given:

v₀ = 0 m/s

a = 4.8 m/s²

Δx = 1800 m

Unknown: t

Equation: Δx = v₀ t + ½ a t²

Substitute and solve:

1800 m = (0 m/s) t + ½ (4.8 m/s²) t²

t ≈ 27 s

4 0
3 years ago
The action force is the balloon pushing the air out. What is the magnitude of the reaction force of the air pushing on the ballo
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Understanding Newtons second law that everything has an equal and opposite reaction. The reaction force from a balloons air being pushed out is the preasurized air it had to push out into the open air.
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You are attracting electricity<span />
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(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i
PilotLPTM [1.2K]

Answer:

Part a)

a_c = 1.67 \times 10^{-10} m/s^2

Part b)

v = 2.18 \times 10^5 m/s

Explanation:

Time period of sun is given as

T = 2.60 \times 10^8 years

T = 2.60 \times 10^8 (365 \times 24 \times 3600) s

T = 8.2 \times 10^{15} s

Now the radius of the orbit of sun is given as

R = 3.00 \times 10^4 Ly

R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m

R = 2.84 \times 10^20 m

Part a)

centripetal acceleration is given as

a_c = \omega^2 R

a_c = \frac{4\pi^2}{T^2} R

a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})

a_c = 1.67 \times 10^{-10} m/s^2

Part b)

orbital speed is given as

v = \frac{2\pi R}{T}

v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}

v = 2.18 \times 10^5 m/s

5 0
3 years ago
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