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1 year ago

What is the y component of a vector defined as 12.2m at 81.5°?

Physics

1 answer:

sergejj [24]1 year ago

3 0

Answer:

Explanation:

This is a displacement vector since it is defined in terms of distance (meters, to be exact). The way you find the y-component is

$V_y=Vsin\theta$ which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):

$V_y=12.2sin(81.5)$ and get

$V_y=$ 12.1 m

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A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is $14,000 + 10,000x − 26,000x^{2}$ , where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

W = $\int_{0}^{0.54} F dx$

= $\int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx$

= $[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}$

= 7560 + 1458 - 1364.69

= 7653.31 J

or, = 7.65 kJ (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0

2 years ago

How much gravitational potential energy does a 45.2 kg object have when it is 21.9m above the ground?

Blizzard [7]

Answer:

Explanation:

The formula for gravitational potential energy is

Ep = m · g · h Assuming that the acceleration is g = 10m/s²

Ep = 45.4 · 10 · 21.9 = 9,942.6 J

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6 0

1 year ago

The chart shows data for four moving objects. A 4 column table with 4 rows. The first column is labeled Object with entries, W,

KatRina [158]

Answer:

Explanation:

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3 0

1 year ago

Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a

Alchen [17]

The height of the roof is <u>3.57m</u>

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

$s_2=\frac{1}{2} g(4t)^2=8gt^2\\ s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2$

The length of the window s is given by,

$s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2$

The first drop is at the bottom and it takes 5t seconds to reach down.

The height of the roof h is the distance traveled by the first drop and is given by,

$h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m$

the height of the roof is 3.57 m

8 0

2 years ago

Read 2 more answers

When balancing an assembly line, which of the following is not a way to reduce the longest task time below the required workstat

tiny-mole [99]

Answer:

D. Speed up the assembly line transfer mechanism

Explanation:

All of the options are logic and very intelligent ways to reduce the longest task time, but if you just speed up the assembly line transfer you will end up with two possible outcomes, frustrated workers cause they always have to be rushing up to finish their task and thus a lower quality in the product you are producing, or that the workers on that specific part of the assembly line definetely can´t perform the task and endu up not doing their job and you end up with incomplete products.

6 0

2 years ago