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ipn [44]
3 years ago
12

A train passes a stationary observer. Which of

Physics
1 answer:
Lyrx [107]3 years ago
4 0

Explanation:

Sorry but thee is none options for me to choose

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Blizzard [7]

Answer:

probably the trip where it took u 5 seconds

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3 years ago
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An object is placed in a fluid and then released. Assume that the object either floats to the surface (settling so that the obje
Irina18 [472]

Answer:

A. Always true

Explanation:

This is because, the buoyancy force is always present whenever and object is placed in a fluid. The magnitude of this  buoyancy force is always equal to the weight of the fluid    displaced by the object according to Archimedes' principle. This principle is true irrespective of whether the object floats or not. When any object is inserted in a fluid, the buoyancy force is always present irrespective of whether it floats or not.  

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3 years ago
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A hockey player applies an average force of 80.0 N to a
denpristay [2]

Answer:

Impulse =  8.0Ns

Explanation:

Given

Force = 80.0N

Mass = 0.25kg

Time = 0.10s

Required

Determine the impulse

The impulse is calculated as follows:

Impulse =  Force * Time

Substitute values for Force and Time

Impulse =  80.0N * 0.10s

Impulse =  8.0Ns

<em>Hence, the impulse experienced is 8.0Ns</em>

5 0
3 years ago
Which sentence is true about Polaris and the Sun?
Andrei [34K]
Its B polaris seems smaller than the Sun
3 0
4 years ago
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A 2 kg rock is at the edge of a cliff 20 meters above a lake The rock becomes loose and falls toward the water below. Calculate
natima [27]

Answer:

The potential energy (P.E) at the top is 392 J

The kinetic energy (K.E) at the top is 0 J

The potential energy (P.E) at the halfway point is 196 J.

The kinetic energy (K.E) at the halfway point is 196 J.

Explanation:

Given;

mass of the rock, m = 2 kg

height of the cliff, h = 20 m

speed of the rock at the halfway point, v = 14 m/s

The potential energy (P.E) and kinetic energy (K.E) when its at the top;

P.E = mgh

P.E = (2)(9.8)(20)

P.E= 392 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the top of the cliff = 0

K.E = ¹/₂(2)(0)²

K.E = 0

The potential energy (P.E) and kinetic energy (K.E) at the halfway point;

P.E = mg(¹/₂h)

P.E = (2)(9.8)(¹/₂ x 20)

P.E = 196 J

K.E = ¹/₂mv²

where;

v is velocity of the rock at the halfway point = 14 m/s

K.E = ¹/₂(2)(14)²

K.E = 196 J.

4 0
3 years ago
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