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Lelu [443]
2 years ago
15

Please help today exam

Physics
1 answer:
Alex Ar [27]2 years ago
6 0

Answer:

The answer is given in the picture

Explanation:

PLEASE MARK ME BRAINLIEST IF MY ANSWER IS CORRECT PLEASE let

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An arrow movirg 48.3 m/s has 5.22<br> kg•m/s of momentum. What is its<br> mass?
spin [16.1K]

Answer:

0.11 kg

Explanation:

Ft = MV

Ft = momentum 5.22kg m/s

M = mass

V = velocity 48.3m/s

Therefore

5.22 = M x 48.3

Divide both sides by 48.3

5.22/48.3 = M x 48.3/48.3

0.11 = M

M = 0.11kg

6 0
3 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.33 times a second. A tack is stuck in the tire a
Troyanec [42]

Answer: 6.47m/s

Explanation:

The tangential speed can be defined in terms of linear speed. The linear speed is the distance traveled with respect to time taken. The tangential speed is basically, the linear speed across a circular path.

The time taken for 1 revolution is, 1/3.33 = 0.30s

velocity of the wheel = d/t

Since d is not given, we find d by using formula for the circumference of a circle. 2πr. Thus, V = 2πr/t

V = 2π * 0.309 / 0.3

V = 1.94/0.3

V = 6.47m/s

The tangential speed of the tack is 6.47m/s

7 0
3 years ago
Read 2 more answers
Which as more weight a kilogram of bricks in air or a kilogram of feathers in water? Explain your answer.
ale4655 [162]
A kilogram is a unit of weight. So a kilogram of bricks would weigh the same as a kilogram of feathers despite if its in water or air since weight is determined by gravity in relation to mass and not what substance the object is in.
3 0
3 years ago
PLEASE ANSWER QUICK!
svet-max [94.6K]
I think it’s C sorry if it’s wrong :(
5 0
3 years ago
A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas
Nookie1986 [14]

The maximum amount of work performed is

W_{max}=\frac{T_H-T_C}{T_C}Q_C

Explanation:

The efficiency of a real heat engine is given by the equation:

\eta = 1-\frac{T_C}{T_H} (1)

where

T_C is the temperature of the cold reservoir

T_H is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

\eta = \frac{W_{max}}{Q_H}

where

W_{max} is the maximum work done

Q_H is the heat absorbed from the hot reservoir

Q_H can be written as

Q_H=W_{max}+Q_C

where

Q_Cis the heat released to the cold reservoir

So the previous equation can be also written as

\eta=\frac{W_{max}}{W_{max}+Q_C} (2)

By combining eq.(1) and (2) we get

1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}

And re-arranging the equation and solving for W_{max}, we find

W_{max}=\frac{T_H-T_C}{T_C}Q_C

Learn more about work and heat:

brainly.com/question/4759369

brainly.com/question/3063912

brainly.com/question/3564634

#LearnwithBrainly

8 0
3 years ago
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