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muminat
3 years ago
5

Your friends sit in a sled in the snow. If you apply a force of 95 N to them, they have an acceleration of 0.8 m/s2. What is the

ir mass?
Physics
2 answers:
ss7ja [257]3 years ago
5 0

F=ma
m=F/a=95/0.8= 118.75kg
your friend is pretty heavy XD
sasho [114]3 years ago
5 0
Net force= mass*acceleration so if we move this around we can isolate mass and get
Net force/acceleration=mass
Which means
95N/0.8 m/s^2 =118.75 kg
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Velocity of the electron in the beam.

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where:

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B= intensity of the magnetic field in tesla

\theta= angle between the velocity and direction of magnetic field

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F=m.\frac{v^2}{r}..........................................(2)

where:

m= mass of the charged particle

v= velocity of projection of charge into the magnetic field

r= radius of the path traced  by the charge in the magnetic field

From eq. (1) and (2) we can calculate the magnetic field .

Now,

Using Ampere's Law we have:

B = \frac{\mu_0 .I}{2 \pi r}

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\mu_0= The permeability of free space.

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The current through inductance L is given by I=I0e−t/τ
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t= Os
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6 0
4 years ago
A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav
gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2        (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

v^2=v_o^2+2gy

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y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

With this value you can find the spring constant k from the equation (1):

mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

5 0
3 years ago
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