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3 years ago

A child is riding in a wagon. What reference frame might have been used if an observer said the child was not moving.

Physics

1 answer:

GarryVolchara [31]3 years ago

3 0

Answer:

the wagon should be used as frame of reference if an observer said the child was not moving.

Explanation:

The state of motion of a body depends upon the frame of reference. It is the set of co-ordinates according to which the motion is analyzed. If a child is riding in a wagon, then he will be considered in motion to a person standing outside the wagon. Hence, if we take a frame of reference outside the wagon then the child must be in motion with respect to the observer. On the other hand if the observer is inside the wagon, then the child must be in rest with respect to the observer. Hence, if we take the wagon to be the frame of reference, then the child will be at rest with respect to the observer.

<u>Therefore, the wagon should be used as frame of reference if an observer said the child was not moving.</u>

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The distance from earth to teh sun usb approxximately 93 million miles. a scientist would write that number as

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A scientist would write that number as 1.49 x 10⁸ kilometers .

(Or, if the scientist is in France or the UK, he might write it as 1.49 x 10⁸ kilometres .)

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3 years ago

I need help with these. Please show workings<br>

Sauron [17]

Answer:

Imp = 25 [kg*m/s]

v₂= 20 [m/s]

Explanation:

In order to solve these problems, we must use the principle of conservation of linear momentum or momentum.

$(m_{1}*v_{1})+(F*t)=(m_{1}*v_{2})$

where:

m₁ = mass of the object = 5 [kg]

v₁ = initial velocity = 0 (initially at rest)

F = force = 5 [N]

t = time = 5 [s]

v₂ = velocity after the momentum [m/s]

$(5*0) +(5*5) = (m_{1}*v_{2}) = Imp\\Imp = 25 [kg*m/s]$

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2 years ago

Read 2 more answers

A train whose proper length is 1200 m passes at a high speed through a station whose platform measures 900 m, and the station ma

TiliK225 [7]

Answer:

0.66c

Explanation:

Use length contraction equation:

L = L₀ √(1 − (v²/c²))

where L is the contracted length,

L₀ is the length at 0 velocity,

v is the velocity,

and c is the speed of light.

900 = 1200 √(1 − (v²/c²))

3/4 = √(1 − (v²/c²))

9/16 = 1 − (v²/c²)

v²/c² = 7/16

v = ¼√7 c

v ≈ 0.66 c

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3 years ago

Starting from rest, a particle that is confined to move along a straight line is accelerated at a rate of 5.0 m/s2. Which one of

Elanso [62]

Answer:

c) The slope is not constant and increases with increasing time.

Explanation:

The equation for the position of this particle (starting from rest is)

$s = at^2/2 = 5t^2/2 = 2.5t^2$

We can take derivative of this with respect to time t to get the equation of slope:

$s' = (2.5t^2)' = 2*2.5t = 5t$

As time t increase, the slope would increases with time as well.

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3 years ago

A particle is moving with SHM of period pie . initially it is 10 cm from The center of the motion and moving in the positive dir

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Answer:

y = 10.44cos(2t - 0.291) cm

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v = y' = -2Αsin(2t + φ)

10 = Acos(2(0) + φ) = Acosφ

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φ = -0.291 radians

10 = Acos(-0.291)

A = 10/cos(-0.291) = 10.44

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3 years ago