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2 years ago

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A snowboarder goes down the hill with a slope of 28° if friction acts on him as he slides down which of the following is the cor

rect free body diagram for this situation

1 answer:

xxMikexx [17]2 years ago

6 0

Answer:

A

Explanation:

All of the frictions are the same, but weight always goes straight down so it can only be A or B. Since they are going down a slope, then the normal force must be sloped. A is the only one out of A and B with a sloped normal force, so it has to be A

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A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu

Montano1993 [528]

Answer:

The angle of separation is $\Delta \theta = 0.93 ^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta _ i = 56^o$

The refractive index of violet light in diamond is $n_v = 2.46$

The refractive index of red light in diamond is $n_r = 2.41$

The wavelength of violet light is $\lambda _v = 400nm = 400*10^{-9}m$

The wavelength of red light is $\lambda _r = 700nm = 700*10^{-9}m$

Snell's Law can be represented mathematically as

$\frac{sin \theta_i}{sin \theta_r} = n$

Where $\theta_r$ is the angle of refraction

=> $sin \theta_r = \frac{sin \theta_i}{n}$

Now considering violet light

$sin \theta_r__{v}} = \frac{sin \theta_i}{n_v}$

substituting values

$sin \theta_r__{v}} = \frac{sin (56)}{2.46}$

$sin \theta_r__{v}} = 0.337$

$\theta_r__{v}} = sin ^{-1} (0.337)$

$\theta_r__{v}} = 19.69^o$

Now considering red light

$sin \theta_r__{R}} = \frac{sin \theta_i}{n_r}$

substituting values

$sin \theta_r__{R}} = \frac{sin (56)}{2.41}$

$sin \theta_r__{R}} = 0.344$

$\theta_r__{R}} = sin ^{-1} (0.344)$

$\theta_r__{R}} = 20.12^o$

The angle of separation between the red light and the violet light is mathematically evaluated as

$\Delta \theta = \theta_r__{R}} - \theta_r__{V}}$

substituting values

$\Delta \theta =20.12 - 19.69$

$\Delta \theta = 0.93 ^o$

6 0

3 years ago

A pendulum that has a period of 2.67000 s and that is located where the acceleration due to gravity is 9.77 m/s2 is moved to a l

zalisa [80]

Answer:

Explanation:

Expression for time period of pendulum is given as follows

$T=2\pi\sqrt{\frac{l}{g} }$

where l is length of pendulum and g is acceleration due to gravity .

Putting the given values for first place

$2.67=2\pi\sqrt{\frac{l}{9.77} }$

Putting the values for second place

$T=2\pi\sqrt{\frac{l}{9.81} }$

Dividing these two equation

$\frac{T}{2.67} =\sqrt{\frac{9.77}{9.81} }$

T = 2.66455 s.

5 0

2 years ago

Technician A says that valve overlap is the number of crankshaft degrees that both valves are closed. Technician B says that val

8_murik_8 [283]

Answer:

B. Technician B only

Explanation:

Technician A says that valve overlap is the number of crankshaft degrees that both valves are closed. Technician B says that valve overlap is the number of degrees that both valves are open. Who is right? A. Technician A only B. Technician B only C. Both technicians A and B D. Neither technician A nor B

Valve overlap is when both the intake and exhaust valves are open. This occurs towards the end of the exhaust stroke, the intake valves are opened just before all the exhaust gases are released, providing more time for the intake air to enter the engine. Valve overlap must be shorter for a petrol/gasoline engine because the engine is drawing in fuel as well as air through the intake. ...it is not so for a sport car

the inlet valve and exhaust valve are located at the head of the engine. fuel enters the engine through the intake valve.

3 0

2 years ago

An object in free fall near the surface of the earth accelerates at a rate of 78979.4 mi/hr What is the rate of acceleration for

Nina [5.8K]

Answer:

32.176 ft/s²

9.807 m/s²

Explanation:

Data provided in the question:

Acceleration of the free falling object = 78979.4 mi/hr²

a) In ft/s

Now,

1 mi = 5280 foot

and,

1 hour = 3600 seconds

thus,

78979.4 mi/hr = $\frac{\textup{78979.4}\times\textup{5280}}{\textup{1}\times\textup{3600}^2}$

or

78979.4 mi/hr = 32.176 ft/s² ...............(1)

b) In m/s²

Now,

1 foot = 0.3048 m

thus,

we have from 1

32.17 ft/s² = $\frac{\textup{32.17}\times\textup{0.3048}}{\textup{1}\times\textup{1}^2}$

or

78979.4 mi/hr = 32.176 ft/s² = 9.807 m/s²

7 0

2 years ago

Graphing: Michelle climbs a tree and drops the toy car once she has reached the top. Please create an Energy-Time graph to show

Vlada [557]

Answer:

Refer to the attachment for the graph. The shape of both functions should resemble part of a parabola. Assumption: air resistance on the car is negligible.

Explanation:

The toy car started with a large amount of (gravitational) potential energy (PE) when it is at the top of the tree. Since it wasn't moving (as it was within Michelle's grip,) its kinetic energy (KE) would be equal to zero.
As the car falls to the ground, its PE converts to KE.
When the car was about to reach the ground, its PE is almost zero, while its KE is at its maximum.

<h3>PE of the car over time</h3>

The size of gravitational PE depends on both the mass and the height of the object. In this case, assume that the mass of the car stayed the same, PE should be proportional to the height of the car.

Assume that air resistance on the car is negligible. The height $h$ of the car at time $t$ could be found with the equation:

$\displaystyle h = -\frac{1}{2}\, g\, t^2 + h_0 \, (\text{Initial height})$ ,

where

$g \approx \rm 9.81\; m \cdot s^{-2}$ near the surface of the earth, and
$h_0$ is the initial height of the car.

On the other hand, $\displaystyle \text{GPE} = m \, g \, h = -\frac{m \cdot g^2}{2}\, t^2 + \underbrace{m \cdot g \cdot h_0}_{\text{Initial GPE}}$ .

In other words, plotting the gravitational PE of the car against time would give a parabola. Since $\displaystyle -\frac{m \cdot g^2}{2} < 0$ (the quadratic coefficient is smaller than zero,) the parabola should open downwards. Besides, since at $t = 0$ the initial GPE is positive, the $y$ -intercept of this parabola should also be positive.

<h3>KE of the car over time</h3>

Assume that the air resistance on the car is negligible. The mechanical energy (ME) of the toy car should conserve (stay the same.) The mechanical energy of an object is the sum of its PE and KE. The PE of the toy car has already been found as a function of time. Therefore, simply subtract the expression of PE from mechanical energy to find an expression for KE.

To find the value of mechanical energy, consider the PE of the toy car before it was dropped. Since initially KE was equal to zero, the mechanical energy of the toy car would be equal to its initial PE. That's $m \cdot g \cdot h_0$ . If there's no air resistance, the value of ME would stay at

Subtract PE from ME to obtain an expression for KE:

$\begin{aligned} \text{KE} &= \text{ME} - \text{PE} \cr &= m \cdot g \cdot h_0 - \left(-\frac{m \cdot g^2}{2}\, t^2 + m \cdot g \cdot h_0\right)\cr &= \frac{m \cdot g^2}{2}\, t^2\end{aligned}$ .

That's also a parabola when plotted against $t$ . Note that since the quadratic coefficient $\displaystyle \frac{m \cdot g^2}{2}$ is positive, the parabola shall open upwards.

7 0

3 years ago