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3241004551 [841]
3 years ago
8

Can someone help me?!!

Physics
1 answer:
Fantom [35]3 years ago
5 0
<h2>Hello!</h2>

The answer is:

The first option, the force tending to lift Rover is equal to 14.5 N.

<h2>Why?</h2>

To calculate the force that is tending to lift Rover vertically, we need to calculate the vertical component force.

Since we know that the angle between the force and the ground is 29°, we can calculate the vertical component of the force using the following formula:

F_y=Force*Sin(29\°)

We are given that the force is equal to 30.0 N, so, calculating we have:

F_y=Force*Sin(29\°)

F_y=30N*Sin(29\°)=14.5N

Also, we can calculate the horizontal component of the force using the following formula:

F_x=Force*Cos(29\°)

F_x=30N*Cos(29\°)=26.24N

Hence, we have that the correct option is the first option, the force tending to lift Rover is equal to 14.5 N.

Have a nice day!

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We have given velocity of the car is decreases from 32 m /sec to 24 m/sec in 4 second

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And finally car reaches to a velocity of 24 m/sec

Time taken to change in velocity = 4 sec

So final velocity v = 24 m/sec

From first equation of motion v = u+at

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In the simulation above, as the projectile travels upward, how does the vertical velocity change? Question 9 options:
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Vertical velocity decreases.

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In the vertical motion, there is a constant acceleration directed downward: this means that the vertical velocity decreases as the ball goes higher. In fact, it decreases following the equation

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Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m
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<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

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