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Andre45 [30]
3 years ago
9

21. If the Sun's rays were at 45° to a vertical pillar, how would

Physics
1 answer:
diamong [38]3 years ago
4 0

Answer:

Let the height of the pole AB = x m. ∴ Length of shadow OB ol the pole AB = x m. Let the angle of elevation be ө, i.e. Hence, the angle of elevation of the Sun's altitude is 45°.

Explanation:

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Where an electric field line crosses an equipotential surface, the angle between the field line and the equipotential is
n200080 [17]

Answer:

90 degree

Explanation:

Electric field line is vertical to the electric field line.

4 0
3 years ago
Electric Charge (guided inquiry)
Sedaia [141]

Answer:

It looks like you have some writing to do ,(-:

Explanation:

8 0
3 years ago
an object has a mass of 50kg, a final height of 20m and an initial height of 8m. what is the amount of work done
Andrei [34K]

amount of work done is 5880 J

Given:

mass of object = 50kg

Final height = 20m

initial height = 8m

To Find:

amount of work done

Solution:

work is done when a force acts upon an object to cause a displacement. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

The work done by gravity is given by the formula,

W = mgh

W = 50 x 9.8 x ( 20-8)

= 5880 J

So the work done is 5880 J

Learn more about Work done here:

brainly.com/question/25239010

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7 0
1 year ago
Technician A says that the battery must be in good condition and at least 75% charged to accurately test an alternator. Technici
aalyn [17]

Answer:Technician A

Explanation:

Technician A statement is correct as  

The battery is required to start the vehicle which, in effect, rotates the alternator at sufficient speed to keep the battery charged. This means if the battery is low it is not possible to start the vehicle and thus we are unable to test the alternator.  

That is the battery is pre-requisite to test the alternator. So the battery must be at least a 75 % charge to test the alternator.

5 0
3 years ago
A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 m/s just before hitting the ground.
Nadusha1986 [10]
<h2>Thus the force of friction is 235 N</h2>

Explanation:

When the bear was at the height of 14 m . Its potential energy = m g h

here m is the mass of bear , g is acceleration due to gravity and h is the height .

Thus P.E =  27 x 10 x 14 = 3780 J

The K.E of the bear just before hitting = \frac{1}{2} m v²

=   \frac{1}{2} x 27 x ( 6.1 )²  = 490 J

The force of friction f = P.E - K.E = 3290 J

Because the work done = Force x Distance

Thus frictional force = \frac{3290}{14} = 235 N

3 0
3 years ago
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