PLEASE HELPPP IF YOU CAN!

An unusually high tide is called a _______ tide.

Monica [59]

Proxigean Spring <span>Tide</span>

8 0

3 years ago

2.5

Anit [1.1K]

Answer:

what is this I don't understand the question

5 0

3 years ago

A 5 kg rock is lifted 3m find amount of work done <br><br> a= 98 J

Levart [38]

Answer:

How long would it take a machine to do 13,000 joules of work if the power rating ... An object gains 15 joules of potential energy as it is lifted vertically 5.0 meters. ... As the time required to do a given quantity of work decreases, the power ... How much work is being done to the system by the person of the sled moves 10 m?

6 0

3 years ago

Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from the front wall

Allisa [31]

Answer:

Flemmings' left hand rule and Downward direction

Explanation:

The person could able to determine the path of force by applying the left-hand rule of Fleming.

Via holding the thumb the forefinger and second finger at right angles for one other

The left-hand rule is applied in the electric motors

On the other hand, the right-hand rule is applied for electric generators.

So different hands are used for the different thing as there is a difference between the cause and effect

Therefore in this case there is a left hand rule used and the direction is downward

7 0

3 years ago

The engine starter and a headlight of a car are connected in parallel to the 12.0-V car battery. In this situation, the headligh

stepladder [879]

Answer:

The total power they will consume in series is approximately 2.257 W

Explanation:

The connection arrangement of the headlight and the engine starter = Parallel to the battery

The voltage of the battery, V = 12.0 V

The power at which the headlight operates in parallel, $P_{headlight}$ = 38 W

The power at which the kick starter operates in parallel, $P_{kick \ starter}$ = 2.40 kW

We have;

P = V²/R

Where;

R = The resistance

V = The voltage = 12 V (The voltage is the same in parallel circuit)

For the headlight, we have;

R₁ = V²/ $P_{headlight}$ = 12²/38 = 72/19

R₁ = 72/19 Ω

For the kick starter, we have;

R₂ = V²/ $P_{kick \ starter}$ = 12²/2.4 = 60

R₂ = 60 Ω

When the headlight and kick starter are rewired to be in series, we have;

Total resistance, R = R₁ + R₂

Therefore;

R = ((72/19) + 60) Ω = (1212/19) Ω

The current flowing, I = V/R

∴ I = 12 V/(1212/19) Ω = (19/101) A

We note that power, P = I²R

In the series connection, we have;

$P_{headlight}$ = I² × R₁

∴ $P_{headlight}$ = ((19/101) A)² × 72/19 Ω = 1368/10201 W ≈ 0.134 W

The power at which the headlight operates in series, $P_{headlight, S}$ ≈ 0.134 W

$P_{kick \ starter}$ = ((19/101) A)² × 60 Ω = 21660/10201 W ≈ 2.123 W

The power at which the kick starter operates in series, $P_{kick \ starter, S}$ ≈ 2.123 W

The total power they will consume, $P_{Total}$ = $P_{headlight, S}$ + $P_{kick \ starter, S}$

Therefore;

$P_{Total}$ ≈ 0.134 W + 2.123 W = 2.257 W

4 0

3 years ago