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grin007 [14]
3 years ago
11

A hockey puck is set in motion across a frozen pond. If ice friction and air resistance are neglected, the forcerequired to keep

the puck sliding at constant velocity isA) the weight of the puck divided by the mass of the puck.B) the mass of the puck multiplied by 9.8 meters per second per second.~qual to the weight of the puck.'i.£l/ero newtons.E) none of these.
Physics
1 answer:
yulyashka [42]3 years ago
7 0

Answer:

Zero Newtons

Explanation:

Newton's second law of motion states that the net force applied to an object is equal to the product between the object's mass and its acceleration:

F=ma

In this case, we want the hockey puck to slide at constant velocity - constant velocity means zero acceleration:

a = 0

And so this means also that the net force is zero:

F = 0

However, the problem says that ice friction and air resistance are negligible - so there are no forces acting on the hockey puck. This means that the puck will continue its motion at constant velocity if we don't apply any other force on it.

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How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (grav
JulsSmile [24]

Answer:

maximum height on moon is 6 times more than the maximum height on Earth

Explanation:

Let the Astronaut has its maximum speed by which he can jump is "v"

now for the maximum height that it can jump is given as

v_f^2 - v_i^2 = 2 aH

now from above equation we will have

0 - v^2 = 2(-g)H

now we have

H = \frac{v^2}{2g}

now if Astronaut jump on the surface of moon with same speed

then we know that the acceleration of gravity on surface of moon is 1/6 times the gravity on earth

so at surface of moon we have

0 - v^2 = 2(-g/6) H

now we have

H = \frac{6v^2}{2g}

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3 years ago
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katen-ka-za [31]

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3 years ago
How are mountains and plateaus alike?
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Una barra de plata de 335.2 g con una temperatura de 100 ºC se introduce un calorímetro de aluminio de 60 g de masa que contiene
sdas [7]

Respuesta:

0,0560 cal / gºC.

Explicación:

Cantidad de calor; (Q)

Q = mcΔt; Δt = t2 - t1

m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura

c de agua = 1 cal / gºC

c de aluminio = 0,22 cal / gºC

QTotal = Q de agua + Q de aluminio

Q de agua = 450 * 1 * (26 - 23) = 1350 cal

Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal

QTotal = 1350 + 39,6 = 1389,6 cal

Calor perdido = calor ganado

QTotal = calor perdido

- 1389,6 = 335,2 * c * (26 - 100)

-1389,6 = −24804,8 * c

c = 1389,6 / 24804,8

c = 0,056021 cal / gºC.

Capacidad calorífica específica de la plata = 0,0560 cal / gºC.

8 0
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