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Marrrta [24]
3 years ago
14

Max (15 kg) and Maya (12 kg) are ice-skating on a frozen pond. While standing at the center of the pond, Maya throws a 1.5-kg sn

owball at Max and, as a result, recoils away from Max at 2.5 m/s. With what speed did Maya throw the snowball at Max
Physics
1 answer:
pychu [463]3 years ago
4 0

Answer: The speed at which Maya threw the snowball at Max is 20m/s

Explanation:

Given that

​Mass of snowball=   1.5-kg.

Mass of Maya= 12kg

Initial  speed of Maya=2.5m/s

We use the law of conservation of  momentum to calculate the  speed of the snowball which depends on its own mass,  the mass of Maya who threw the snowball and the initial velocity at which the snowball was thrown by Maya

Ms x Us = Mm x Um

Mass of snowball  x Initial speed of Snow ball =Mass of Maya x Initial  speed of Maya

1.5kg  x Initial speed of Snow ball =12 kg x 2.5m/s

Initial speed of Snow ball=( 12 kg x 2.5m/s) /1.5kg

​ Initial speed of Snow ball  =20m/s

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A circular loop carrying a current of 1.6 A is oriented in a magnetic field of 0.30 T. The loop has an area of 0.14 m 2 and is m
konstantin123 [22]

Answer:

The torque on the loop is 2.4 \times 10^{-2} Nm

Explanation:

Given:

Current I = 1.6 A

Magnetic field B = 0.30 T

Area of loop A = 0.14 m^{2}

Angle between magnetic field and area vector \theta = 21°

Form the formula of  torque in case of magnetic field,

 г = MB \sin \theta

Where M = magnetic moment

  M = IA

 г = IAB \sin 21

 г = 1.6 \times 0.30 \times 0.14 \times 0.3583

 г =2.4 \times 10^{-2} Nm

Therefore, the torque on the loop is 2.4 \times 10^{-2} Nm

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3 years ago
A 5.0 kg rock is dropped from the top of a building. The speed of the rock after it has fallen for 2.2 seconds is?
PSYCHO15rus [73]
5x2.2/3 Because that's how you calculate velocity.

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An automobile of mass 1.46 cross times to the power of blank 10 cubed kg rounds a curve of radius 25.0 m with a velocity of 15.0
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The centripetal force exerted on the automobile while rounding the curve is 1.31\times10^4 N

<u>Explanation:</u>

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Objects moving around a circular track will experience centripetal force towards the center of the circular track.

centripetal\ force=mv^2/r\\=1.46\times10^3\times15^2/25\\=1.46\times 10^3\times 225/15\\=1.31\times 10^4 N

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