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labwork [276]
3 years ago
9

A shark is swimming at 10 m/s west when it spots some floating bait caught in some seaweed straight ahead. After 10 s, and trave

ling at a velocity of 15 m/s, the shark reaches the bait. What was the shark's average acceleration? A. 0.5 m/s² B. 0.5 m/s² east C. 0.5 m/s² west D. 10 m/s²
Physics
2 answers:
Volgvan3 years ago
8 0
<h2>Answer:</h2>

<u>The average acceleration is </u><u>0.5 m/s² west</u>

<h2>Explanation:</h2>

As we know that

Acceleration =  

Putting the values

Acceleration =

Acceleration = 0.5

Since the shark was moving in west so the Acceleration will be in west too

Andreyy893 years ago
3 0

Answer:

C. 0.5 m/s² west

Explanation:

The shark average acceleration is given by the following equation:

a=\frac{v-u}{t}

where:

u = 10 m/s west is the initial velocity of the shark

v = 15 m/s west is the final velocity

t = 10 s is the time taken

Plugging numbers into the formula, we find

a=\frac{15 m/s-10 m/s}{10 s}=0.5 m/s^2

and the direction is west, because the shark speeds up in the same direction of the initial velocity, so the acceleration is positive and therefore in the same direction.

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Write with all the steps and formulas and drawing if needed.​
lianna [129]

<u>The answer is not contained detail explanation, just a solution and the required values. </u>

All the details are in the pictures, the answers are marked with orange colour.

Note,

in the task no 20.:

m_A - the \ mass \ of \ A; \ m_B-the \ mass \ of \ B \ balls.\\V_A \ and \ V_B-the \ velocities \ of \ the \ A&B \ balls \ before \ collision.\\V'_A \ and \ V'_B-the \ velocities \ of \ the \ A&B \ balls \ after \ collision.

V - the velocity of the pair of the balls after collision.

in the task no 21:

m₁ - the mass of the copper ball; m₂ - the mass of the copper calorimeter; m₃ - the mass of the water; t₀ - the initial temperature of water in the copper calorimeter; θ - the final temperature in the calorimeter after the copper ball is transferred into a copper calorimeter; t₁ - the required initial temperature of the copper ball before it is transferred into the calorimeter.

7 0
3 years ago
Three Small Identical Balls Have Charges -3 Times 10^-12 C, 8 Times 10^12 C And 4 Times 10^-12 C Respectively. They Are Brought
IgorLugansk [536]

Answer:

The charge in each ball will be 3 * 10^-12 C

Explanation:

(Assuming the correct charge of the second ball is 8 * 10^-12)

When the balls are brought in contact, all the charges are split evenly among then.

So first we need to find the total charge combined:

(-3 * 10^-12) + (8 * 10^-12) + (4 * 10^-12) = 9 * 10^-12 C

Then, when the balls are separated, each ball will have one third of the total charge, so in the end they will have the same charge:

(9 * 10^-12) / 3 = 3 * 10^-12 C

So the charge in each ball will be 3 * 10^-12 C

8 0
3 years ago
When Karl Kaveman adds chilled grog to his new granite mug, he removes 10.9 kJ of energy from the mug. If it has a mass of 625 g
mrs_skeptik [129]

Answer:

3°C

Explanation:

We can that heat Q=mc_p dT

Where m is the mass c_p = specific heat capacity

dT = Temperature difference

here we have given m=625 g =.625 kg

specific heat of granite =0.79 J/(g-K) = 0.79 KJ/(kg-k)

T_1 =25°C

T_2 we have to find

we have also given Q=10.9 KJ

10.9=0.625×0.79×(25-T_2)

25-T_2 =22

T_2=3°C

7 0
3 years ago
Read 2 more answers
The unit for measuring electric power is the <br> A. ampere.<br> B. volt.<br> C. ohm.<br> D. watt.
Drupady [299]
The correct answer is D: Watt. This unit was named after James Watt, and is used to express the equivalent of one joule per second in energy. In experiments and on the packaging for electrical products such as light-bulbs, the measurement will usually be written in its abbreviated format: W.
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6 0
3 years ago
An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?
hjlf

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

rating of resistor R= 8 ohms

power P= 100W

Required

The current I

Step two

Yet this power is also given by

P = I^2R

make I subject of the formula we have

I= \sqrt{\frac{P}{R} }

substitute

I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps

8 0
3 years ago
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