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3 years ago

11. एक समान चुम्बकीय क्षेत्र में लम्बवत प्रवेश करने वाले किसी आवेशित कण द्वारा प्राप्त वृत्तीय पथ की

Physics

1 answer:

mojhsa [17]3 years ago

5 0

Answer:

Explanation:

If a positive charge enters a magnetic field at 90 degrees the charge is deflected in a circular path by a force that acts perpendicular to it in line with Flemings right-hand rule

to derive the radius of the path of the charge we apply

F= mv^2/r=Bev

where

m= mass of the electronic charge

e=charge

B=magnetic field

v=average speed

r=radius

rearranging we have

r=mv^2/Bev

r=mv/Be

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A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base

umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388 m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388 m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0

3 years ago

A 1560-kilogram truck moving with a speed of 28.0 m/s runs into the rear end of a 1070-kilogram stationary car. If the collision

Nimfa-mama [501]

Answer:

Δ KE = 249158.6 kJ

Explanation:

given data

Truck mass M = 1560 Kg

Truck initial speed, u = 28 m/s

mass of car m = 1070 Kg

initial speed of car u1 = 0 m/s

solution

first we get here final speed by using conservation of momentum that is express as

Mu = (M+m) V .......................1

put here value we get

1560 × 28 = (1560 + 1070 ) V

solve it we get

final speed V = 16.60 m/s

and

Change in kinetic energy will be here

Δ KE = $\frac{1}{2} Mu^2 - \frac{1}{2}(M+m)V^2$ .................2

put here value and we get

Δ KE = $\frac{1}{2}\times 1560\times 28^2 - \frac{1}{2}\times (1560 + 1070)\times 16.60^2$

solve it we get

Δ KE = 249158.6 kJ

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What would happen if the moon was hit by an asteroid

Semmy [17]

If the moon was hit by an asteroid there would be a crater mark and possible movement.

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Mass is indirectly proportional to acceleration, so, lighter the object greater would be it's acceleration...

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Answer:

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