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Stels [109]
3 years ago
9

Hahah I look pregnant don’t I

Physics
1 answer:
n200080 [17]3 years ago
8 0
I see a pillow

But you see...this bobcat
Cute ain’t it?

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A 35 N force makes a 10 degree angle with the positive x-axis. What is the magnitude of the vertical component of the force?
Snowcat [4.5K]

Answer:

6.07 N

Explanation:

Given that,

Force, F = 35 N

It makes 10 degree angle with the positive x-axis.

We need to find the magnitude of the vertical component of the force. It can be given by :

F_y=F\sin\theta\\\\=35\times \sin(10)\\\\=6.07\ N

So, the magnitude of the vertical component of the force is 6.07 N.

5 0
3 years ago
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How does the diameter of the disk of milky way galaxy compare to its thickness?
weqwewe [10]
The diameter is about 100 times as great as the thickness.
7 0
4 years ago
An object is accelerating if there is a change in speed and/or ________.
Alborosie

Acceleration means any change in the speed or direction of motion.
8 0
3 years ago
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If you wanted to increase your muscular endurance, it would be important to participate in activities that __________.
Maurinko [17]
In order to increase muscular endurance, it is important to take part in activities that increase the size of muscle cells and the amount of blood that is delivered to the cells<span>. The increased size of the muscle cells reuires them to contract less to accomplish the same task; moreover, the increased blood flow helps deliver oxygen and nutrients to the blood.</span>
3 0
3 years ago
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a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}

\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
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