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chubhunter [2.5K]
2 years ago
5

When an object absorbs light energy, it reflects

Physics
1 answer:
mestny [16]2 years ago
8 0
I’m pretty sure it’s C
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Calculate the momentum of of a 1,5000 kg car traveling at 6 m/s
ANEK [815]

Answer: 90000 kgm/s

Explanation:

Given that,

Momentum of car = ?

Mass of car = 1,5000 kg

Velocity of car = 6 m/s

Recall that momentum is the product of mass of the moving object by its velocity

i.e Momentum = mass x velocity

Momentum = 15000kg x 6m/s

= 90000kgm/s

Thus, the momentum of the car is 90000 kgm/s

4 0
3 years ago
Explain why is the temperature of a hot tea higher than the temperature of iced tea?
melamori03 [73]

Answer:

Because the hot tea is hot from a microwave or coffee machine when iced tea is cold from ice in the tea.

Explanation:

5 0
2 years ago
atoms combine in different ways to make up all of the substances you encounter everyday true or false
amm1812
True hope this helps :)
4 0
2 years ago
You drop a 0.375 kg ball from a height of 1.37 m. It hits the ground and bounces up again to a height of 0.67 m. How much energy
Radda [10]

2.57 joule energy lose in the bounce .

<u>Explanation</u>:

when ball is the height of 1.37 m from the ground  it has some gravitational potential energy with respect to hits the ground  

Formula for gravitational potential energy given by  

Potential Energy = mgh

Where ,

m = mass  

g = acceleration due to gravity  

h = height

Potential energy when ball hits the ground

m= 0.375 kg

h = 1.37 m

g = 9.8 m/s²

Potential Energy = 0.375\times9.8\times1.37

Potential Energy = 5.03 joule

Potential energy when ball bounces up again

h= 0.67 m

Potential Energy = 0.375\times0.67\times9.8

Potential Energy = 2.46 joule

Energy loss = 5.03 - 2.46 = 2.57 joule

2.57 joule energy lose in the bounce

6 0
2 years ago
The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give
Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

4 0
2 years ago
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