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insens350 [35]
2 years ago
14

IF YOU ANSWER THESE 3 QUESTIONS! I WILL GIVE YOU BRAINEST!!! 17 POINTS!!!

Physics
1 answer:
gogolik [260]2 years ago
3 0

Answer:

1. a

2. a [im iffy on this but 95% positive its this]

3. b [walking is a form of aerobics, so i would say b]

Explanation:

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How does the angle of launch affect the kinetic energy of a rubber band?​
Lady_Fox [76]

Answer:

The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.

3 0
3 years ago
If it takes 50n of force to lift a 450n what is the ma of the machine
otez555 [7]

If the machine is 100% efficient, then its
Mechanical Advantage is (450/50) = 9 .

If the machine is less than 100% efficient,
then the MA is more than 9 .

7 0
3 years ago
What is the term for producing a current by moving a wire through a magnetic field?.
maria [59]
The answer is electromagnetic Induction.



I hope this answer will help you
6 0
2 years ago
What is the momentum of a 3kg ball that is traveling at 5m/s
marta [7]

Answer:

15 Kg m/s

Explanation:

momentum p = mv

p is momentum

m is mass (kg)

v is velocity (m/s)

p = (3) (5m/s)

p= 15 Kg m/s

5 0
2 years ago
Read 2 more answers
An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1
goblinko [34]

Answer:

v"_{1} = v_{1} tanΘ

v^{"} _{2} = \frac{m_{1}v_{1}}{m_{2}cos}Θ

Θ = tan^{-1}(\frac{v^{"} _{1} }{v_{1} } )

Explanation:

Applying the law of conservation of momentum, we have:

Δp_{x = 0}

p_{x} = p"_{x}

m_{1}v_{1} = m_{2}v"_{2} cosΘ (Equation 1)

Δp_{y} = 0

p_{y} = p"_{y}

0 = m_{1} v"_{1} - m_{2} v"_{2} sinΘ (Equation 2)

From Equation 1:

v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}Θ

From Equation 2:

m_{2} v"_{2}sinΘ = m_{1} v_{1}

v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }

Replacing Equation 3 in Equation 4:

v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}

v"_{1}=v_{1}\frac{sinΘ}{cosΘ}

v"_{1}=v_{1}tanΘ (Equation 5)

And we found Θ from the Equation 5:

tanΘ=\frac{v"_{1}}{v_{1}}

Θ=tan^{-1}(\frac{v"_{1}}{v_{1}})

7 0
3 years ago
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