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4 years ago

A vector is 9.55 m long and points in a -48.0 degree direction. find the x-component of the vector.

Physics

1 answer:

jeyben [28]4 years ago

6 0

Answer:

freshwater 9.5 is a long points in the 1.4 degrees direction in the response water is the only one that can be solve the problems of the anyone is

Explanation:

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A 0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary fo

frutty [35]

Answer:

90 N

Explanation:

The force applied to the ball is given by:

$F=\frac{\Delta p}{\Delta t}$

where

$\Delta p$ is the change in momentum of the ball

$\Delta t$ is the time taken

The change on momentum of the ball is:

$\Delta p=m\Delta v=(0.45 kg)(20 m/s)=9 kg m/s$

So, the force applied is

$F=\frac{9 kg m/s}{0.10 s}=90 N$

6 0

3 years ago

Read 2 more answers

A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude

kondaur [170]

Answer:

81.8 m/s

Explanation:

The initial velocity of the plane is:

$v_0=115 m/s$ (toward east)

So, decomposing along the x- and y- directions:

$v_{x0} = 115 m/s\\v_{y0} = 0$

(we took east as positive x-direction and north as positive y-direction)

The acceleration is

$a=2.88 m/s^2$ (northwest, so the angle with the positive x-direction is 135 degrees)

Decomposing it along the two directions:

$a_x = a cos 135^{\circ} = (2.88 m/s^2)(cos 135^{\circ})=-2.04 m/s^2\\a_y = a sin 135^{\circ} = (2.88 m/s^2)(sin 135^{\circ})=2.04 m/s^2$

So the two components of the velocity after a time t = 25.0 s will be

$v_x = v_{x0} + a_x t = 115 m/s + (-2.04 m/s^2)(25.0 s)=64 m/s\\v_y = v_{y0} + a_y t = 0 m/s + (2.04 m/s^2)(25.0 s)=51 m/s$

So, the magnitude of the velocity of the plane will be

$v=\sqrt[v_x^2+v_y^2}=\sqrt{(64 m/s)^2+(51 m/s)^2}=81.8 m/s$

6 0

3 years ago

A solid sphere of radius R carries a fixed, uniformly distributed charge q. Obtain an expression for the magnitude of the electr

NISA [10]

Answer:

The electric field outside the sphere will be $\dfrac{qr}{4\pi\epsilon_{0}R^3}$ .

Explanation:

Given that,

Radius of solid sphere = R

Charge = q

According to figure,

Suppose r is the distance between the point P and center of sphere.

If $\rho$ be the volume charge density,

Then, the charge will be,

$q=\rho\times\dfrac{4}{3}\pi R^3$ .....(I)

Consider a Gaussian surface of radius r.

We need to calculate the electric field outside the sphere

Using formula of electric field

$\oint{\vec{E}\cdot \vec{dA}}=\dfrac{Q}{\epsilon_{0}}$

$E\times4\pi r^2=\dfrac{\rho\dotc \dfrac{4}{3}\pi r^3}{\epsilon_{0}}$

Put the value from equation (I)

$E\times4\pi r^2=\dfrac{qr^3}{\epsilon_{0}R^3}$

$E=\dfrac{qr}{4\pi\epsilon_{0}R^3}$

Hence, The electric field outside the sphere will be $\dfrac{qr}{4\pi\epsilon_{0}R^3}$ .

4 0

3 years ago

witch of the following is the least reliable source of background information for a scientific project.

Tems11 [23]

Primary sources are most often produced around the time of the events you are studying.

hopwe this helps

3 0

3 years ago

Read 2 more answers

In 5 or more sentences, explain how you can make everyday count ?

IceJOKER [234]

You can make everyday count by accomplishing all the tasks you have for that day. Also you can try and find the positive throughout the day. Another thing you could do is eat a healthy meal. You can do a workout and that will make you feel better about your self. The last thing you can do is always have a good mindset about that day.

8 0

3 years ago