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2 years ago

The angle between incident ray and reflected ray is 80 degrees. What is the value of angle of incidence?

Physics

1 answer:

saul85 [17]2 years ago

8 0

Answer:

The angle of incident ray is 40°.

Explanation:

Given that the angle of incident and reflected ray are the same. In this question, we had given that both angles added up will gives you 80° so you have to divide it by 2 :

incident + reflected = 80°

Let incident = reflected = θ

θ + θ = 80°

2θ = 80°

θ = 80° ÷ 2

= 40°

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6th grade science pls help

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Answer:

Explanation:

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2 years ago

Biologists use ball-and-stick models to study complex molecules by representing atoms with balls and bonds with sticks. DNA is a

AleksAgata [21]

One of the major limitations of using the ball and stick model for DNA, is that within a single double stranded segment of DNA, one would have to use many many balls to represent atoms that are present in the sugar phosphate backbone, along with all of the main atoms that compose the nitrogenous bases of DNA, we also cannot construct or show the helical form of DNA, by using balls and sticks as well.

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2 years ago

True or False. The pressure inside a piston is the same in all directions.

Dominik [7]

Answer:

True

Explanation:

Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container. The pressure at any point in the fluid is equal in all directions.

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2 years ago

A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =

natali 33 [55]

With acceleration

$\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j$

and initial velocity

$\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i$

the velocity at time t (b) is given by

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

We can get the position at time t (a) by integrating the velocity:

$\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du$

The particle starts at the origin, so $\mathbf x(0)=\mathbf0$ .

$\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du$

$\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}$

$\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j$

Get the coordinates at t = 8.00 s by evaluating $\mathbf x(t)$ at this time:

$\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j$

$\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j$

so the particle is located at (x, y) = (64.0, 64.0).

Get the speed at t = 8.00 s by evaluating $\mathbf v(t)$ at the same time:

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j$

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

This is the velocity at t = 8.00 s. Get the speed by computing the magnitude of this vector:

$\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}$

5 0

2 years ago

A girl moves quickly to the center of a spinning merry-go-round, traveling along the radius of the merry-go-round. Which of the

DochEvi [55]

Answer:

The angular speed of the system increases.

The moment of inertia of the system decreases.

Explanation:

As we know that the girl is going towards the center of the circle so here the moment of inertia of the girl is given as

$I = mr^2$

here we know that

r = position of the girl from the center of the disc

now we know that the girl is moving towards the center so its distance will continuously decreasing

So the moment of inertia of the girl will decrease

Now we know that that with respect to the center of the disc there is no torque on the disc + girl system

So here we can use angular momentum conservation

So we have

$I\omega = constant$

since moment of inertia is decreasing for the system

so angular speed will increase

3 0

3 years ago