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schepotkina [342]
2 years ago
11

please solve this for me ,A garden roller is pulled with a force of 200N acting at an angle of 50 degree with the ground level.f

ind the force pulling the roller along the ground...​
Physics
1 answer:
Bingel [31]2 years ago
8 0

Answer:

The force pulling the roller along the ground is 128.55 N

Explanation:

A force of 200 N acting at an angle of 50° with the ground level

This force is pulled a garden roller

We need to find the force pulling the roller along the ground

The force that pulling the roller along the ground is the horizontal

component of the force acting

→ The force acting is 200 N at direction 50° with ground (horizontal)

→ The horizontal component = F cosФ

→ F = 200 N , Ф = 50

→ The horizontal component = 200 cos(50) = 128.55 N

128.55 N is the horizontal component of the force that pulling the

roller along the ground

<em>The force pulling the roller along the ground is 128.55 N</em>

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Answer:

a. the force between them quadruples

Explanation:

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In this problem, the charges on both objects are doubled, so

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Answer:

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Explanation:

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So we can create a linear model in the for y=b+mx, where the intercept b=\rho_{center}=13000 \frac{kg}{m^3} and the slope would be given by m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}

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Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be dV=r^2 sin\theta d\phi d\theta dr.

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Replacing dV we have the following result:

M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)

We can solve the integrals one by one and the final result would be the following

M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})

Simplyfind this last expression we have:

M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]

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M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg

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