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Debora [2.8K]
4 years ago
7

Kid A has jumped off the sled! (RIP)

Physics
1 answer:
balandron [24]4 years ago
5 0

Answer:

46.45 m/s

Explanation:

Total momentum before jump = Total momentum after jump

11.7 * ( 36.7 + 45.2 ) = ( 36.7 * (-31.1) ) + ( 45.2 * v )

v*45.2 - 1141.37 = 958.23

v = 2099.6/45.2 = 46.45

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Which forces do not require contact between objects in order to exert a push or pull?
arsen [322]

Answer:

Gravitational force and magnetic force. The other forces on the list require contact (tension requires contact with the rope and friction requires the object to be in contact with the surface)

Explanation:

6 0
3 years ago
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What physical quantity is a measure of the amount of inertia an object has?
allochka39001 [22]
<span>mass and only mass ................</span>
3 0
3 years ago
Why is the fuse used in a circuit called safety fuse?<br>​
Sunny_sXe [5.5K]

Answer:

The maximum current which can flow through a fuse without melting it, is called its rating. ... If current higher than 8 A flows through the fuse, it would melt and circuit gets broken. ... Hence, fuse acts as a safety device

4 0
3 years ago
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A 1250-kg car moves at 20.0 m/s. How much work must be done on the car to increase its speed to 30.0 m/s.
bulgar [2K]

Answer:

312,497.5Joules

Explanation:

Work done = force × distance

W = FS

Get the force

F = ma

F = 1250×9.8

F = 12250N

Get the distances using the equation of motion

v² = u² +2gS

30² =20²+2(9.8)S

900 =400+19.6S

900-400 =19.6S

500 = 19.6S

S = 500/19.6

S = 25.51m

Get the work done

Work done = 12250×25.51

Workdone = 312,497.5Joules

6 0
3 years ago
Find an expression for the kinetic energy of the car at the top of the loop.Express the kinetic energy in terms of m, g, h, and
lyudmila [28]

Answer:

K.E₂ = mg(h - 2R)

Explanation:

The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:

K.E₁ + P.E₁ = K.E₂ + P.E₂

where,

K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)

P.E₁ = Initial Potential Energy = mgh

K.E₂ = Final Kinetic Energy at the top of the loop = ?

P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)

Therefore,

0 + mgh = K.E₂ + mg(2R)

<u>K.E₂ = mg(h - 2R)</u>

3 0
3 years ago
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