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3 years ago

What is the effect on current through a circuit of steady resistance when the voltage is doubled?

1 answer:

5 0

If voltage is doubled then let the new voltage be such that . ... Thus, the current flowing throughthe circuit will be doubled if voltage applied across the circuit is doubled.

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Heavy crate sits at rest on the floor of a warehouse. you push on the crate with a force of 400 n, and it doesn't budge. what is

maks197457 [2]

Heavy crate sits at rest on the floor of a warehouse. you push on the crate with a force of 400 N, and it doesn't budge. The magnitude of the friction force on the crate in Newton is 400N

This is due to Friction force, which is defined as the resisting force that acts on a body when it is at rest (Static friction) or when it is in motion (Kinetic friction).

When a force is applied on a stationary body, the force of static friction starts to act on the body which prevents any relative motion between the object and surface. The magnitude of friction increases up to μsN, where μs is the coefficient of static friction. As the crate didn't budge, it means the amount of force applied was less than μsN. Hence the force applied was canceled by an equal and opposite amount of frictional force which was equal to 400N.

Learn more about frictional force here

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8 0

2 years ago

A 50-kg box is being pushed along a horizontal surface. The coefficient of static friction between the box and the ground is 0.6

dsp73

To solve the exercise it is necessary to apply the concepts related to Newton's Second Law, as well as the definition of Weight and Friction Force.

According to the problem there is a movement in the body and it is necessary to make a sum of forces on it, so that

$\sum F = ma$

There are two forces acting on the body, the Force that is pushing and the opposing force that is that of friction, that is

$F - F_f = ma$

To find the required force then,

$F=F_f+ma$

By definition we know that the friction force is equal to the multiplication between the friction coefficient and the weight, that is to say

$F = \mu mg +ma$

$F = 0.35*50*0.8+50*1.2$

$F=(171.5N)+(50Kg)(1.2m/s^2)$

$F=231.5N$

$F\approx 230N$

Therefore the horizontal force applied on the block is B) 230N

6 0

4 years ago

Elect the correct equations that show that when a 3.0-kg book is lifted 2.6 m its increase in gravitational potential energy is

grigory [225]

The increase in gravitational potential energy for an object of mass m is given by
$\Delta U = mg \Delta h$
where $\Delta h$ is the increase in altitude of the object.

In our problem, m=3.0 kg, $\Delta h= 2.6 m$ and $g=10 m/s^2$ (approximated value), so we have
$\Delta U = mg\Delta h=(3.0 kg)(10 m/s^2)(2.6 m)=78 J$

5 0

3 years ago

An attacker at the base of a castle wall 3.95 m high throws a rock straight up with speed 5.00 m/s from a height of 1.60 m above

s344n2d4d5 [400]

Answer:

a) No

b) the rock must have a minimum initial speed of 6.79m/s for it to reach the top of the building.

Explanation:

Given:

Height of the wall = 3.95m

Initial height = 1.60m

Initial speed = 5.00m/s

distance between the initial height and wall top = 3.95 - 1.60 = 2.35m

Using the formula;

v^2 = u^2 + 2as ....1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance travelled

From equation 1

s = (v^2 - u^2)/2a ...2

Since the rock t moving up,

the acceleration = -g = -9.8m/s2

s = maximum height travelled

v = 0 (at maximum height velocity is zero)

Substituting into equation 2

s = (0 - 5^2)/(2×-9.8) = 1.28m

Therefore, the maximum height is 1.28 from his initial height Which is less than the 2.35m of the wall from his initial height. So the rock will not reach the top of the wall

b) Using equation 1:

u^2 = v^2 - 2as

v = 0

a = -9.8m/s

s = 2.35m. (distance between the initial height and wall top)

u^2 = 0 - 2(-9.8 × 2.35)

u^2 = 46.06

u = √46.06

u = 6.79m/s

Therefore, the rock must have a minimum initial speed of 6.79m/s

8 0

3 years ago

Round your decimal answer to the nearest hundredth.

trasher [3.6K]

Son of c is .72

Have a good night

3 0

3 years ago