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3 years ago

you push the ball with a force of 22.8N which induces a -2.3 frictional force what is the net force while you push the ball

1 answer:

4 0

Force is a vector quantity in which direction matters. This means that the forces will act in opposing directions so the net force will be
22.8 - 2.3 =
20.5 Newtons

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A student is watching waves come in from the ocean. He noticed that the first wave he saw (Wave A) had twice the amplitude of th

Alika [10]

Answer:

Wave A

I hope this helps

8 0

2 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

8 0

3 years ago

reviews the approach taken in problems such as this one. A bird watcher meanders through the woods, walking 0.916 km due east, 0

Verizon [17]

Answer:

Displacement: 2.230 km Average velocity: 1.274 $\frac{km}{h}$

Explanation:

Let's represent displacement by the letter S and the displacement in direction 49.7° as A. Displaement is a vector, so we need to decompose all the bird's displacement into their X-Y compoments. Let's go one by one:

0.916 km due east is an horizontal direction and cane be seen as direction towards the negative side of X-axis.

0.928 km due south is a vertical direction and can be seen as a direction towards the negative side of Y-axis.
3.52 km in a direction of 49.7° has components on X and Y axes. It is necessary to break it down using trigonometry,

First of all. We need to sum all the X components and all the Y componets.

∑ $Sx = Ax -0.916$ ⇒ ∑ $Sx = [tex]3.52cos(49.7) - 0.916$

∑ $Sx = 1.361 km$

∑ $Sy = Ay - 0.918$ ⇒ ∑ $Sy = 3.52sin(49.7) - 0.918$

∑ $Sy = 1.767$

The total displacement is calculated using Pythagoeran therorem:

$S_{total} =\sqrt{Sx^{2}+ Sy^{2} }$ ⇒

$S_{total} = 2.230 km$

With displacement calculated, we can find the average speed as follows:

$V = S/t$ ⇒ $V = \frac{2.230}{1.750}$

$V = 1.274\frac{km}{h}$

7 0

2 years ago

What is meant by the statement,the linear expansivity of copper is 0.000017k

Aliun [14]

Answer:

The change in length per unit length per degree rise in temperature of copper is 0.000017k

Explanation:

Given that :

The linear expansivity of copper is 0.000017k. This simply means that ; for a given copper length, the length of such copper will increase by 0.000017k for every degree rose in temperature of the copper rod.

Therefore, the change in length per unit length per degree rise in temperature (k) is 0.000017

4 0

2 years ago

EXPLAIN HOW ENERGY IS TRANSMITTED THROUGH A MEDIUM

jek_recluse [69]

I say it helped then because TrueType had room

5 0

3 years ago

Read 2 more answers