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2 years ago

Discuss what it means to view a moving objectt from a frame of reference. Supply an example that illustrates your explanation.

1 answer:

5 0

what is frame of reference:
it is set of fixed points which are inter related on the basis of rest and motion..,, or an explanation of what an observer is doing..
examples:
1 ,, if you are in motion and observing something then frame of reference is motion... and if you are at rest then frame of motion is rest...
2 frame of reference may be place like if a passenger is sitting in a car then he is in motion with respect to the ground but he is at rest with respect to the car in which he is sitting ...
3 also if a passenger is in car and he is tossing a coin and you are on the ground ,, then for passenger it is in vertical motion but for you it is in parabolic arc .. hence there are two different views for two persons for the same thing i.e coin... :)

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faust18 [17]

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2 years ago

A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc

Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

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2 years ago

While on a playground, you and your niece take turns sliding down a frictionless slide. Your mass is 75 kg while your little nie

Olenka [21]

Answer:

Explanation:

In absence of friction, total mechanical energy must be conserved.
This means that the change in gravitational potential energy (in magnitude) must be equal to the change in kinetic energy:

$\Delta K = \Delta U \\ \\ \frac{1}{2} * m* v^{2} = m*g*h$

As it can be seen, the mass m is on the both sides of the equation, which means that it can be simplified.
We can solve this equation for v (speed at the bottom) as follows:

$v = \sqrt{2*g*h}$

As it can be seen, the mass has no part in the equation, so, due both are starting from the same height, both people have the same speed at the bottom.
The option c is the right one.

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2 years ago

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Explanation:

Below is an attachment containing the solution.

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