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2 years ago

The potential difference across the terminals of a battery is 8. 4v when there is a current of 1. 50 a in the ba

Physics

1 answer:

Vikentia [17]2 years ago

7 0

Answer:

The potential difference across the terminals of a battery is 8. 4v when there is a current of 1. 50 a in the ba

Explanation:

The potential difference across the terminals of a battery is `8.4 V` when there is a current of `1.50 A` in the battery from the negative to the positive terminal. When the current is `3.50 Air`, the reverse direction, the potential difference becomes `9.4 V`. <br>(a) What is the internal resistance of the battery? <br>(b) What is the emf of the battery?

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2 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind.

RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

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1 year ago

Rafe is placing his car on a lift to change the oil and oil filter. He needs to drive up the lift 1.5 meters and the lift raises

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Answer:

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Explanation:

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2 years ago

What is kinematics . Give two examples

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Answer:

kinematics explains the terms like acceleration, velocity, and position of the objects while in motion.

Some important parameters in kinematics are displacement, velocity, and time.

Explanation:

your welcome,can you give me the brainliest? plssss

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2 years ago

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An apple drops from the tree and falls freely. The apple is originally at rest a height H0 above the ground. The ground is cover

zzz [600]

Answer:

Explanation:

The distance travelled in the free fall is H - h

Since the apple originally started from rest we can use v^2 = u^2 + 2 x g x s where v is the final velocity, g the accln due to gravity and s the distance travelled and u is the initial velocity = 0

So the velocity just before it enters the grass is sq rt [2 x g x (H - h)]

Once in the grass, it slows down at a constant rate which means that the acceleration (a) during this period is constant.

So once again using the same formula we have v = O and u = sq rt[2 x g x (H-h)]

so since v^2 = u^2 + 2 x a x s then

O^2 = 2 x g x (H-h) + 2 x a x h

{O^2 - 2 x g x (H - h)}/(2 x h) = a

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2 years ago