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Pie
3 years ago
13

A gymnast jumps directly up onto a beam that is 1.5 m high. From the beam, she jumps 0.5 m straight up and lands back on the bea

m in the same spot. What is the total distance she traveled?
A. 0.5 m
B. 1.0 m
C. 1.5 m
D. 2.5 m
Physics
1 answer:
tatiyna3 years ago
5 0

Answer:

D. 2.5 m

Explanation:

1.5 m up to the beam + 0.5 m up + 0.5 m down = 2.5 m total

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Two Jupiter-size planets are released from rest 1.40×10^11 m apart.. What are their speeds as they crash together?
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M₁ = mass of planet #1 
M₂ = mass of planet #2 
M = total mass 
R₁ = radius of planet #1 
R₂ = radius of planet #2 
d₁ = initial distance between planet centers 
d₂ = final distance between planet centers 
a = semimajor axis of plunge orbit 
v₁ = relative speed of approach at distance d₁ 
v₂ = relative speed of approach at distance d₂ 

M₁ = M₂ = 1.8986e27 kilograms 
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G = 6.6742e-11 m³ kg⁻¹ sec⁻² 
GM = 2.5343e17 m³ sec⁻² 
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d₂ = R₁ + R₂ = 1.42984e8 meters 
v₁ = 0 
v₂ = √[GM(2/d₂−1/a)] 
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v₂ = 59508.4 m/s </span>
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The time to fall is 1337.7 days

.</span>I hope my answer has come to your help. Thank you for posting your question here in Brainly.
7 0
3 years ago
1. What is the kinetic energy of a frog that has mass of 13kg and can hop<br> 9m/s? *
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Answer:526.6J

Explanation:I did a test

8 0
3 years ago
You're standing in a large room when a loud noise is made and you hear a series of echoes. Which characteristi
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4 0
2 years ago
Water exits a garden hose at a speed of 1.2 m/s. If the end of the garden hose is 1.5 cm in diameter and you want to make the wa
Katarina [22]

Answer:

A 93%

Explanation:

P_1=P_2 = Pressure will be equal at inlet and outlet

\rho = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

v_1 = Velocity at inlet = 1.2 m/s

v_2 = Velocity at outlet

r_1 = Radius of inlet = \dfrac{1.5}{2}=0.75\ cm

r_2 = Radius of outlet

From Bernoulli's relation

P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow \dfrac{1}{2}\rho v_1^2+\rho gh_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}v_1^2+gh_1-gh_2)}\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}1.2^2+9.81\times 15)}\\\Rightarrow v_2=17.19709\ m/s

From continuity equation

A_1v_1=A_2v_2\\\Rightarrow \pi r_1^2v_1=\pi r_2^2v_2\\\Rightarrow r_2=\sqrt{\dfrac{r_1^2v_1}{v_2}}\\\Rightarrow r_2=\sqrt{\dfrac{0.0075^2\times 1.2}{17.19709}}\\\Rightarrow r_2=0.00198\ m

The fraction would be

\dfrac{A_1-A_2}{A_1}\times 100=\dfrac{r_1^2-r_2^2}{r_1^2}\times 100\\ =\dfrac{0.0075^2-0.00198^2}{0.0075^2}\times 100\\ =93.0304\ \%

The fraction is 93.0304%

8 0
3 years ago
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