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melamori03 [73]
2 years ago
14

4. A lamp is connected to a 230 V mains supply. A current of 4 A flows through the

Physics
2 answers:
White raven [17]2 years ago
4 0

Explanation:Example

A 500 W television set is switched on for 4 hours. Calculate the energy transferred.

500 W = 500/1000 = 0.5 kW

energy transferred = power × time

= 0.5 × 4

= 2 kWh

sasho [114]2 years ago
3 0

Answer:

I= Q ÷ T

Explanation:

I: Current measured in amperes

Q: Charge flow measured in coulombs

T: Time measured in seconds

I hope you find this helpful.

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Answer:

Can i have more context

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3 years ago
Light in vacuum travels at a speed of 3.00 x 10^8 m ^-1 s^-1 on average earth is 93,000,000 miles from the sun how many minutes
NemiM [27]
The correct unit for the speed of light is  [ m s⁻¹ ].

Time = (distance) / (speed)

Time = (9.3 x 10^7 miles) x (1609 m/mile) / (3 x 10^8 m/s) = 498.8 seconds .

That would be  <em>8.31 minutes</em>.  
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2 years ago
A constant net force acts on an object. which of the following best describes the object's motion?
Anton [14]
The object<span> is moving with a decreasing acceleration. The </span>object<span> is moving with </span>a constant<span> velocity.</span>
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3 years ago
Which characteristic surface of the moon contains the same rock as the ocean floor
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3 0
3 years ago
To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
2 years ago
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