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Anna [14]
3 years ago
15

As the launch force increase the launch velocity will

Physics
1 answer:
FrozenT [24]3 years ago
5 0

Answer:

As the launch force increase the launch velocity will

<em><u>Increase</u></em>

The reason for your answer to number six is because

<em><u>There is a direct relationship between force and acceleration.</u></em>

<em><u /></em>

Explanation:

<em>It is known all over the place that, there is a direct relationship between Force and acceleration of an object leading to an increase in force being directly proportional to the increase in the acceleration of the given object and vice versa.</em>

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A proton is moving in a circular orbit of radius 12 cm in a uniform 0.31-T magnetic field perpendicular to the velocity of the p
PIT_PIT [208]

Answer:

Explanation:

A proton of charge

q=+1.609×10^-19C

Orbit a radius of 12cm

r=0.12m

Magnetic Field of 0.31T

Angle between velocity and field is 90°

a. Because the magnetic force F supplies the centripetal force Fc.

The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by

F = qvB sin θ

And the centripetal force is given as

Fc=mv²/r

Where m is mass of proton

m=1.673×10^-27kg

Then, F=Fc

qvB sin θ=mv²/r

qBSin90=mv/r

rqB=mv

Then, v=rqB/m

v=0.12×1.609×10^-19×0.31/1.673×10^-23

v=3577692.78m/s

v=3.58×10^6m/s

b. Since,

F=qVBSin90

F=1.609×10^-19×3.58×10^6×0.31

F=1.785×10^-13 N.

6 0
3 years ago
An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be
Inessa05 [86]

Explanation:

The given data is as follows.

    Inner wheel Radius = 20 m,

   Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

    Angular velocity of automobile = w = \frac{V}{R}

where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

and,   u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

                      = \frac{u}{(R - d)}

                  u = V \times \frac{(R - d)}{R}

                    = V \times (1 - \frac{d}{R})

If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus,             u = 2\pi rn = V \times (1 - \frac{d}{R})

Now, rotation per minute of inner wheel is calculated as follows.

         n = \frac{V}{2 \pi r \times (1 - \frac{d}{R})}

            = \frac{V}{2 \pi r \times (1 - \frac{0.75}{20})} (since 2d = 1.5m given, d = 0.75m),

             = \frac{V}{r} \times 0.1532

So, rotation per minute of outer wheel; n' =  

                   = \frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}

                   = \frac{V}{r} \times 0.1651

5 0
3 years ago
Which is greater, the gravitational force between earth and the moon, or the force between earth and the sun?
Karo-lina-s [1.5K]
Earth and the Sun
GF = 3.647x10^22 N
8 0
3 years ago
Se aplica una fuerza neta de 25 N a una masa de 10 kg. ¿Cuál es la aceleración dada a la masa?
lys-0071 [83]

Answer:

Se queda

Explanation:

Porque 10 kg es más que 25N

4 0
3 years ago
Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person
sertanlavr [38]

Answer:

 v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

           f ’= f (v + v₀) / (v-v_{s})

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

                v₀ = v_{s} = v ’

we substitute

               f ’= f (v + v’) / (v - v ’)

               f ’/ f (v-v’) = v + v ’

               v (f ’/ f -1) = v’ (1 + f ’/ f)

               v ’= (f’ / f-1) / (1 + f ’/ f) v

               v ’= (f’-f) / (f + f’) v

let's calculate

                v ’= (3400 -3000) / (3000 +3400) 343

                v ’= 400/6400 343

                v ’= 21.44 m / s

3 0
3 years ago
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