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gayaneshka [121]
3 years ago
13

Which two factors does the power of a machine depend on?

Physics
1 answer:
soldi70 [24.7K]3 years ago
7 0
B

~ hope this helps ~
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If a wooden ball has a mass of 2.4g and a volume of 6cm3. What is it’s density?
Luba_88 [7]

Answer:

<h3>The answer is 0.4 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 2.4 g

volume = 6 cm³

We have

density =  \frac{2.4}{6}  \\

We have the final answer as

<h3>0.4 g/cm³</h3>

Hope this helps you

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True

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2 years ago
An atom has an atomic number 17 and atomic mass number 35. How many neutrons are present in its nucleus?
Anna [14]
The number of <span>neutrons that are present in its nucleus is 18. The correct option among all the options that are given in the question is the second option or option "B". Mass number minus the atomic number gives the number of neutrons. I hope that this is the answer that has come to your help.</span>
7 0
3 years ago
Read 2 more answers
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

5 0
2 years ago
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