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3 years ago

When white light passes through a prism, a red light is least changed in direction? True or false

Physics

1 answer:

Goryan [66]3 years ago

4 0

True because the picture below proves this....

* from which red color is least deviated and violet most.

* Hopefully this helps:) Mark me the brainliest:) !!

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A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

Nady [450]

The first collision because a greater amount of momentum must be taken and used in order to push the cart back, giving it a greater mass and impulse

6 0

3 years ago

What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (see section 17.6;

antiseptic1488 [7]

<h3><u>Answer;</u></h3>

volume = 6.3 × 10^-2 L

<h3><u>Explanation</u>;</h3>

Volume = mass/density

Mass = 0.0565 Kg,

Density = 900 kg/m³

= 0.0565 kg/ 900 kg /m³

= 6.3 × 10^-5 M³

but; 1000 L = 1 m³

Hence, <u>volume = 6.3 × 10^-2 L</u>

8 0

3 years ago

If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed

beks73 [17]

Answer:

v = 5.9 x 10⁷ m/s

Explanation:

The kinetic energy of the electron in terms of potential difference is given as:

$K.E = eV$ --------------- equation (1)

where,

e = charge on electron = 1.6 x 10⁻¹⁹ C

V = Potential Difference = 9.9 KV = 9900 Volts

The kinetic energy in general is given as:

$K.E = \frac{1}{2}mv^{2}\\$ --------- equation (2)

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of electron = ?

Therefore, comparing equation (1) and equation (2), we get:

$\\\frac{1}{2}mv^{2} = eV\\\\\frac{1}{2}(9.1\ x\ 10^{-31}\ kg)v^{2} = (1.6\ x\ 10^{-19}\ C)(9900\ volts)\\\\v = \sqrt{34.81\ x\ 10^{14}} \\$

8 0

3 years ago

9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o

Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

$m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.$

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

$F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).$

Now the force of friction is $F_s=\mu*N,$ where $N$ is the normal force and from the diagram it is $F_y=mg*cos(\theta).$

Thus $F_s=\mu*N=\mu*mg*cos(\theta).$

Therefore,

$F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).$

Substituting the value for $F_H,m,\mu, and \:\theta$ we get:

$F_{tot}= -38.63N.$

Now acceleration is simply

$a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.$

The negative sign indicates that the acceleration is directed up the incline.

Part B:

$d=\frac{1}{2} at^2$

Which can be rearranged to solve for t:

$t=\sqrt{\frac{2d}{a} }$

Substitute the value of $d=0.50m$ and $a=7.7m/s$ and we get:

$t=0.36s.$

which is our answer.

Notice that in using the formula to calculate time we used the positive value of $a$ , because for this formula absolute value is needed.

5 0

3 years ago

already did the work, i just need someone to see if i did the tangent line right? the line has to touch the point 0.6! thank you

Dmitrij [34]

The tangent looks good.

The curve is a bit crooked, at the 0.9 and 1.

But overall, cool graph.

5 0

3 years ago