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2 years ago

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A construction worker is pushing a 50.0-kg box with a force of 150.0 N to the right. If the box is moving at a constant velocity

answer the following questions.
Draw a free body diagram of the box on the box provided. Include directions for velocity and acceleration if known.

Create an equation of motion for the x-direction.

Create an equation of motion for the y-direction.

Calculate the normal force that the ground exerts on the box.

Calculate the magnitude of the force of friction.
Calculate the magnitude of the coefficient of friction.

1 answer:

9966 [12]2 years ago

7 0

Yes omg yes I literally have the same question and need to find the answer

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Primary, secondary

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Which explains the information needed to calculate speed and velocity?

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Over [174]

What Kepler's constant ? ? ! ?

The only constant in Kepler's laws is in the third one, where it says something to the
effect that (square of a body's period) / (cube of its distance from the central body)
is a constant.

That means it's a constant for multiple little ones orbiting the same central body.
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4 0

3 years ago

I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite

e-lub [12.9K]

Answer:

The range of powers is $- 5 \ D \le P \le - 2.667\ D$

Explanation:

From the question we are told that

The far point of the left eye is $n_f = 20 cm$

The near point of the left eye is $n = 15cm$

The near point with the glasses on is $n_g =25 \ cm$

From these parameter we can see that with the glass on that for near point the

Object distance would be $u = -25 \ cm$

Image distance would be $v = -15 \ cm$

To obtain the focal length we would apply the lens formula which is mathematically represented as

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

substituting values

$\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}$

$f = - \frac{75}{2} cm$

converting to meters

$f = - \frac{75}{2} * \frac{1}{100}$

$f = - \frac{75}{200} \ m$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f}$

Substituting values

$P = - \frac{200}{75} m$

$P = - 2.667 \ D$

From these parameter we can see that with the glass on that for far point the

Object distance would be $u_f = - \infty \ cm$

Image distance would be $v_f = -20 \ cm$

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

$\frac{1}{f_f} = \frac{1}{v_f} - \frac{1}{u_f}$

substituting values

$\frac{1}{f} = \frac{1}{-20} - \frac{1}{- \infty}$

$\frac{1}{f} = \frac{1}{-20} - 0$

$f_f = \frac{20}{1} \ cm$

converting to meters

$f_f = - \frac{20}{1} * \frac{1}{100}$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f_f}$

Substituting values

$P = - \frac{100}{20} m$

$P = - 5 \ D$

This implies that the range of powers of the lens in his glass is

$- 5 \ D \le P \le - 2.667\ D$

3 0

3 years ago

In the Bohr model of the hydrogen atom, the speed of the electron is approximately 2.2 106 m/s.

Murrr4er [49]

The central force acting on the electron as it revolves in a circular orbit is $9.52 \times 10^{-8} \ N$ .

The given parameters;

<em>speed of electron, v = 2.2 x 10⁶ m/s</em>
<em>radius of the circle, r = 4.63 x 10⁻¹¹ m</em>

<em />

The central force acting on the electron as it revolves in a circular orbit is calculated as follows;

$F = \frac{M_e v^2}{r} \\\\$

where;

$M_e$ is mass of electron = 9.11 x 10⁻³¹ kg

$F = \frac{(9.11 \times 10^{-31}) \times(2.2\times 10^6)^2 }{4.63 \times 10^{-11}} \\\\F = 9.52 \times 10^{-8} \ N$

Thus, the central force acting on the electron as it revolves in a circular orbit is $9.52 \times 10^{-8} \ N$ .

Learn more about centripetal force here:brainly.com/question/20905151

8 0

3 years ago