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3 years ago

Ocean water conducts electrical current because it contains _____.

Physics

2 answers:

xeze [42]3 years ago

8 0

Answer:

electrolytes

Explanation:

icang [17]3 years ago

3 0

<span>Ocean water conducts electrical current because it contains electrolytes. Electrolytes are good conductors of electricity, and are naturally present in appropriate amounts in bodies of water. However, pure water (without electrolytes or metals) is not a conductor of electricity. </span>

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A kicked soccer ball has an initial velocity of 25 meters per second at an angle of 40 degrees, above the horizontal.

Zinaida [17]

(1) The vertical magnitude of the ball's initial velocity is 16.075 m/s,

(2) The maximum height of the ball is 13.18 m

(1) To Calculate the vertical component of the ball's initial velocity, we use the formula below.

Formula:

Vy = Vsin∅........................ Equation 1

Where:

Vy = vertical component of the ball's initial velocity
V = Initial velocity of the ball
∅ = angle to the horizontal.

From the question,

Given:

V = 25 m/s
∅ = 40°

Substitute these values into equation 1

Vy = 25(sin40°)
Vy = 25×0.643
Vy = 16.075 m/s.

(2) To calculate the maximum height reached by the ball, we use the formula below.

Formula:

H = (Vy)²/2g............... Equation 2

Where:

g = Acceleration due to gravity = 9.8 m/s²

Substitute the value above into equation 2

H = (16.075)²/(2×9.8)
H = 13.18 m

Hence, (1) The vertical magnitude of the ball's initial velocity is 16.075 m/s, (b) The maximum height of the ball is 13.18 m.

Learn more about maximum height here: brainly.com/question/13665920

8 0

2 years ago

Read 2 more answers

A cannonball is catapulted toward a castle. The cannonball’s velocity when it leaves the catapult is 51.6 m/s at an angle of 37.

Andrei [34K]

Answer:

$v_{y}=35.21m/s$

Explanation:

From the exercise we know the cannonball's <u>initial velocity</u>, the <u>angle</u> which its released with respect to the horizontal and its <u>initial height</u>

$v_{o}=51.6m/s\\\beta =37.0º\\y_{o}=7m$

If we want to know whats the <u>y-component of velocity</u> we need to use the following formula:

$v_{y}^2=v_{oy}^2+ag(y-y_{o})$

Knowing that $g=-9.8m/s^2$

$v_{y}=\sqrt{((51.6m/s)sin(37))^2-2(9.8m/s^2)(0m-7m)}=35.21m/s$

So, the cannonball's y-component of velocity is $v_{y}=35.21m/s$

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3 years ago

A lamp has a power of 120 w and is left on for 24 hours. A television has a power of 350 W and is left on for 3 hours. Which sta

soldi70 [24.7K]

<h2>Hello!</h2>

The answer is C. The television uses 6,588,000 J less than the lamp.

First, we must know that 1 Wh( Watt consumed in 1 hour) = 3600 J

Wh is the unit that express the energy amount of consumed in 60 minutes/ 1 hour.

So,

From the enunciate information we can calculate the energy consumed by the lamp and the television.

For the lamp, we have that it left on for 24 hours and it has a power of 120

For the television, we have that it left on for 3 hours and it has a power of 350W

Calculating the energy consumed (Wh) we have:

$Wh= Watt*Time= 120 Watt* 24 hours= 2880 Wh$

and converting it to Joule, we have:

$2880 Wh*\frac{3600 J}{1 Wh}= 10,368,000 J$

Doing the same for the television, we have:

$Wh= 350 Watt* 3 hours= 1050 Wh$

and converting it to Joule, we have:

$1050 Wh*\frac{3600 J}{1 Wh}= 3,780,000 J$

Finally, we have

$10,368,000J(Lamp)-3,780,000J(Television)= 6,588,000J$ :

Then having calculated both consume, we have that the television uses 6,588,000 J less than the lamp.

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3 years ago

Read 2 more answers

What will the magnitude of the field be if the 10 nc charge is replaced by a 20 nc charge? Assume the system is big enough to co

MArishka [77]

Answer:

Same magnitude of the 10 nc charge cause the electric field is external.

Explanation:

To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.

As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.

F = qE

If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.

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3 years ago

The strength of a magnetic field around a wire carrying a current of 20 A is 0.004 T. What is the strength of the

Alex777 [14]

Answer:

D. 0.008 T

Explanation:

Eng 2021

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3 years ago