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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.

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

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Two cars move down a hill at a constant velocity. Car c is much larger than car d. The car that has a greater momentum is:

babunello [35]

Answer:

Explanation:

When an object has a greater mass than the drag force pushing it back then it is moving forward therefore making car C faster than car D. If car D was much smaller it wouldn't have enough mass to push past the drag force.

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3 years ago

Read 2 more answers

Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acti

Tems11 [23]

Answer:

The unknown force will be 18.116 lb.

Explanation:

Given that,

Three forces act on a flange as shown in figure.

The net force acting on the flange is a minimum.

$\dfrac{dF_{net}}{df}=0$

We need to calculate the unknown force

Using formula of net force

$\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}$

Put the value into the formula

$\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}$

$\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}$

The magnitude of net force,

$F_{net}=\sqrt{F_{x}^2+F_{y}^2}$

$F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}$

$F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}$

$F_{net}=\sqrt{F^2+4899.78+36.232F}$

On differentiating w.r.to F

$(\dfrac{dF_{net}}{dF})^2=2F+36.232$

$0=2F+36.232$

$F=-\dfrac{36.232}{2}$

$F=-18.116\ lb$

Negative sign shows the direction of force which is downward.

Hence, The unknown force will be 18.116 lb.

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3 years ago

A 3-liter container has a pressure of 4 atmospheres. The container is sent underground, with resulting compression into 2 L. App

monitta

Answer:

6 atm

Explanation:

PV = PV

(4 atm) (3 L) = P (2 L)

P = 6 atm

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3 years ago

Consider that a ray of light is travelling from glass to water. The refractive index of water is 1.30 (i e n . ., 1.30 w = ) and

Bess [88]

Answer:

$\theta_i=49.88^{\circ}$

Explanation:

Total internal reflection can happen when light goes from a medium with higher refractive index (in this case, glass) to a medium with lower refractive index (in this case, water).

Snell's Law tells us that $n_isin\theta_i=n_rsin\theta_r$ , where the i stands for incident (in this case, glass) and the r for refracted (in this case, water). We want to know when $\theta_r=90^{\circ}$ , that is, when $n_isin\theta_i=n_r$ , and this happens when the incident angle is: