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Vadim26 [7]
3 years ago
10

A sled is pushed with 30 Newtons and sliding friction is 10 Newtons find the net force on the sled​

Physics
1 answer:
mixas84 [53]3 years ago
5 0

The net force is basically the speed or we could call it the centre of gravity. So it's simple:

to find the centre you minus the 2 numbers -

30 - 10 = 20

so the net force is 20

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A ball accelerates from rest down a ramp at 2.4 m/s^2. Write an equation that could be used to determine the balls finals positi
Alik [6]

Answer:

x=2.4t+4.9t^2

Explanation:

This equation is one of the kinematic equations to solve for distance. The original equation is as follows:

X=Xo+Vt+1/2at^2

We know that the ball starts at rest meaning that its initial velocity and position is zero.

X=0+Vt+1/2at^2

Since it is going down the ramp, you can use the acceleration of gravity constant. (9.81 m/s^2) and simplify that with the 1/2.

X=Vt+4.9t^2

Note: Since the positive direction in this problem is down, you are adding the 4.9t^2, but if a question says that the downward direction is negative, you would subtract those values.

Now, substitute in your velocity value.

X=2.4t+4.9t^2

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2 years ago
Which is an example of a scientist using a conceptual model to describe a volcano?
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3 years ago
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The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
the cross section area of a hole is 725cm^2. Given that the area of a circle is A=3.14r^2 , find the radius of the hole.
asambeis [7]
The\ area\ of\ a\ circle\ =  \pi r^2 \\ 725\ cm^2 =  3.14r^2 \\
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