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shusha [124]
3 years ago
8

The work done to lift a 10kg object to a height of 1m above the ground is

Physics
1 answer:
vlada-n [284]3 years ago
8 0
A = mg∆h = 10•10•1 = 100J
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Unit 5 lesson 7 physical science 12 question
Kryger [21]
I just took it 100% 11/11
1.D
2.A
3.A
4.A
5.B
6.C
7.D
8C
9A
10B
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8 0
3 years ago
How do I solve this step by step? I’m really confused
LekaFEV [45]

Step-#1:

Ignore the wire on the right.

Find the strength and direction of the magnetic field at P,

caused by the wire on the left, 0.04m away, carrying 5.0A

of current upward.

Write it down.


Step #2:

Now, ignore the wire on the left.

Find the strength and direction of the magnetic field at P,

caused by the wire on the right, 0.04m away, carrying 8.0A

of current downward.

Write it down.


Step #3:

Take the two sets of magnitude and direction that you wrote down

and ADD them.


The total magnetic field at P is the SUM of (the field due to the left wire)

PLUS (the field due to the right wire).


So just calculate them separately, then addum up.

4 0
3 years ago
At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There
katen-ka-za [31]

a) Constant

b) Constant

Explanation:

a)

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

\tau_{net} = I \alpha (1)

where

\tau_{net} is the net torque acting on the object in rotation

I is the moment of inertia of the object

\alpha is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

\alpha = \frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time interval

So we can rewrite eq.(1) as

\tau_{net}=I\frac{\Delta \omega}{\Delta t}

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

\tau_{net}=0

From the previous equation, this implies that

\Delta \omega =0

Which means that the angular velocity at that instant does not change anymore.

b)

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

\tau_{net}=0

Therefore, this means that

\Delta \omega =0

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0
3 years ago
A 65-kg person walks from the ground to the roof of a 100 m tall building. How much gravitational potential energy does she have
ivolga24 [154]
I,think potential energy is mgh so 65*100*9,81
7 0
3 years ago
What force is needed to accelerate an object 5 m/s2 if the object has a mass of 10 kg?
Helga [31]
We know, F = m * a
F = 10 * 5
F = 50 N

In short, Your Answer would be 50 Newtons

Hope this helps!
5 0
3 years ago
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