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3 years ago

An induced charge is when charged particles can be attracted to neutral objects. True or false?

Physics

1 answer:

Sindrei [870]3 years ago

5 0

Answer:

the answer is true Induction charging is a method used to charge an object without actually touching the object to any other charged object.

Explanation:

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What will be net force be if net forces are unbalanced?

malfutka [58]

Answer: If the forces on an object are balanced, the net force is zero. If the forces are unbalanced forces, the effects don't cancel each other. Any time the forces acting on an object are unbalanced, the net force is not zero, and the motion of the object changes.

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3 years ago

Adam and Natalie push on a cart using 150 N of force.The cart does not move.What is static friction on the cart?

Roman55 [17]

When force is applied on an object the object will not slide

So here object will remain in equilibrium after applying the force

So in this case the condition of equilibrium is maintained due to friction force that is applied on the object.

So here for equilibrium be can say

$F_{net} = 0$

also we know that

$F_{net} = F - F_f$

we are given that

$F = 150 N$ ( applied force on the box)

$F_{net} = 150 - F_f = 0$

$F_f = 150 N$

So here this friction force on the cart is known as static friction

$F_s = 150 N$

4 0

4 years ago

Read 2 more answers

The center of the Milky Way galaxy lies in the direction of the _constellation, aboutlight-years away.

Elza [17]

Answer:

Sagittarius

Explanation:

The center of the Milky Way galaxy lies in the direction of the Sagittarius constellation, about 26,000 light-years away.

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4 years ago

Read 2 more answers

The positions of the consellations appear to change throughout the year because

slava [35]

The earth is revolving around the sun and when the seasons change the earth tilts at a different angle and it changes your perspective on the constellations<span />

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3 years ago

Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed o

Mrac [35]

Answer:

a)α= 53.13°

b)The velocity at the highest point = 15 m/s

The acceleration at the highest point = 9.8 $m/s^2$

c)h=15 m

V=18.02 m/s

Explanation:

Speed of water ,u= 25 m/s

So the horizontal component of speed u = u cos α

Given that horizontal distance cover by water in 3 s is 45 m.

So We know that in projectile motion horizontal acceleration is zero.

In horizontal direction

Distance = Velocity x time

45 = u cos α x 3

u cos α = 45

45 = 25 cos α x 3

cos α = 45/75

α= 53.13°

So the velocity at the highest point = u cos α

The velocity at the highest point = 15 m/s

The acceleration at the highest point = 9.8 $m/s^2$

Now the velocity along vertical direction(Vo) = u sin α

Vo= 25 sin 53.13°

Vo =20 m/s

$h=V_o.t-\dfrac{1}{2}gt^2$

$h=20\times 3-\dfrac{1}{2}\times 10\times 3^2$

h=15 m

So at 15 m above the ground water will strike .

The y-component of velocity after 3 sec

Vy= Vo - g t

Vy = 20 - 10 x 3

Vy= -10 m/s

The horizontal component of velocity will remain 15 m/s.

The resultant velocity

$V=\sqrt{10^2+15^2}\ m/s$

V=18.02 m/s

5 0

4 years ago