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3 years ago

If you launch an object in space, it

Physics

1 answer:

MArishka [77]3 years ago

5 0

Answer:

I believe this represents Newton's first law of motion. Any object in motion will continue to move until a force stops it, be it friction or a physical object.

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explain why a high-speed collision between two cars would cause more damage than a low-speed collision between the same two cars

9966 [12]

The force equation can easily prove this. F=ma. This states that the force on an object is equal to mass times acceleration. If the mass stays the same and the velocity of the cars increases than that means there is a larger force. This is because in both cases the cars are stopping in almost an instant and the times of the crashes are theoretically identical. Acceleration is the change in velocity over time. If the velocity is higher with the same amount of time than that means there is a higher acceleration. If you plug a higher acceleration into the force equation then you wind up with a higher force and in turn a more damaging collision.

8 0

3 years ago

The electric company charges $0.08/kilowatt-hour. If a 2 kilowatt appliance is run for 3hours, how much will you have to pay to

asambeis [7]

Answer:

$ 0.48

Explanation:

We can calculate this quantity easily using successive products and taking into account the units.

$\frac{0.08}{kw*h}*2[kw]*3[hr]\\ \\=0.48$

The amount is $ 0.48

3 0

3 years ago

A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carouse

Sidana [21]

Answer:

m = 35.98 Kg ≈ 36 Kg

Explanation:

I₀ = 125 kg·m²

R₁ = 1.50 m

ωi = 0.600 rad/s

R₂ = 0.905 m

ωf = 0.800 rad/s

m = ?

We can apply The law of conservation of angular momentum as follows:

Linitial = Lfinal

⇒ Ii*ωi = If*ωf (I)

where

Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m

If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m

Now, we using the equation (I) we have

(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800

⇒ m = 35.98 Kg ≈ 36 Kg

5 0

3 years ago

Sam receives the kicked football on the 3 yd line and runs straight ahead toward the goal line before cutting to the right at th

Pie

Answer:

Distance: 21 yd, displacement: 15 yd, gain in the play: 12 yd

Explanation:

The distance travelled by Sam is just the sum of the length of each part of Sam's motion, regardless of the direction. Initially, Sam run from the 3 yd line to the 15 yd line, so (15-3)=12 yd. Then, he run also 9 yd to the right. Therefore, the total distance is

d = 12 + 9 = 21 yd

The displacement instead is a vector connecting the starting point with the final point of the motion. Sam run first 12 yd straight ahead and then 9 yd to the right; these two motions are perpendicular to each other, so we can find the displacement simply by using Pythagorean's theorem:

$d=\sqrt{12^2+9^2}=15 yd$

Finally, the yards gained by Sam in the play are simply given by the distance covered along the forward-backward direction only. Since Sam only run from the 3 yd line to the 15 yd line along this direction, then the gain in this play was

d = 15 - 3 = 12 yd

7 0

3 years ago

ASAP HELPPPPPP POP PLEASE

drek231 [11]

Answer:

The answer is in the attachment

Explanation:

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5 0

2 years ago