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3 years ago

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

Physics

1 answer:

sergejj [24]3 years 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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12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it

vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

$v_{f} = v_{o}+a\cdot (t-t_{o})$ (1)

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$v_{f}$ - Final velocity, measured in meters per second.

$a$ - Acceleration, measured in meters per square second.

$t_{o}$ - Initial time, measured in seconds.

$t$ - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust ( $v_{o} = 0\,\frac{m}{s}$ , $a = 30\,\frac{m}{s^{2}}$ , $t_{o} = 0\,s$ , $0\,s\le t< 30\,s$ )

$v = 30\cdot t$ (2)

Free fall ( $v_{o} = 900\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s}$ , $t_{o} = 30\,s$ , $30\,s \le t \le 120\,s$ )

$v = 900-9.81\cdot (t-30)$ (3)

Now we created the graph speed-time, which can be seen below.

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3 years ago

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3 years ago

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3 years ago

Read 2 more answers

Please help me anyone i will mark brainliest

Mashcka [7]

Answer:

for 4.567 I can't tell if it x $10^{3}$ or x $10^{5}$ so:

answer using x $10^{3}$ = 0.0000644697N

answer using x $10^{5}$ = 0.00644697

Explanation:

use equation F = GMm/ $R^{2}$

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3 years ago

A planet has a gravitational acceleration on its surface of 2.2 times Earth's gravitational acceleration on its surface. The pla

lesantik [10]

Answer:

The mass of the planet is 55 times the mass of earth.

Explanation:

From the inverse-square gravitation law,

F = (GMm/r²)

If the weight of a body (the force with which the earth attracts a body to its centre) is to be calculated,

F = mg

m = mass of the body,

g = acceleration due to gravity

mg = (GMm/r²)

G = Gravitational constant

M = mass of the earth

m = mass of body

r = distance between the body and the centre of the earth = radius of the earth

The acceleration due to gravity is given by

g = (GM/r²)

Making the mass of the earth, the subject of formula

M = (gr²/G) (eqn 1)

So, the planet described,

Let the acceleration due to gravity on the planet be g₁

Mass of the planet be M₁

Radius of the planet be r₁

g₁ = 2.2g

r₁ = 5r

M₁ = ?

Note that the gravitational constant is the same for both planets.

So, we can write a similar expression for the planet's acceleration due to gravity

g₁ = (GM₁/r₁²)

Substituting all the parameters known in terms of their corresponding earth values

2.2g = [GM₁/(5r)²]

2.2g = [GM₁/25r²]

M₁ = (55gr²/G)

Recall the expression for the mass of the earth

M = (gr²/G)

M₁ = 55 M

The mass of the planet, in terms of Earth masses = 55M

The mass of the planet is 55 times the planet of earth.

Hope this Helps!!!

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4 years ago