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3 years ago

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A person pushes a 50kg box to the right with 100N of force at a constant velocity of 5m/s. Calculate the coefficient of friction

between the box and the surface.

1 answer:

Anna11 [10]3 years ago

8 0

The weight of the box is 50x9.8 = 490 N. The force of friction is 100N. F= μΝ so coefficient = 100/490 = 0.20

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Consider that a ray of light is travelling from glass to water. The refractive index of water is 1.30 (i e n . ., 1.30 w = ) and

Bess [88]

Answer:

$\theta_i=49.88^{\circ}$

Explanation:

Total internal reflection can happen when light goes from a medium with higher refractive index (in this case, glass) to a medium with lower refractive index (in this case, water).

Snell's Law tells us that $n_isin\theta_i=n_rsin\theta_r$ , where the <em>i</em> stands for incident (in this case, glass) and the <em>r</em> for refracted (in this case, water). We want to know when $\theta_r=90^{\circ}$ , that is, when $n_isin\theta_i=n_r$ , and this happens when the incident angle is:

$\theta_i=arcsin(\frac{n_r}{n_i})$

Which for our values means:

$\theta_i=arcsin(\frac{1.3}{1.7})=arcsin(0.76470588235)=49.88^{\circ}$

6 0

3 years ago

A frame hanging on a wall is held by two cables. The tension in each cable is 30 N, and the cables make an angle of 45° with the

Dimas [21]

Answer: option D) 42.4 N

The weight of the frame is balanced by the vertical component of tension.

W = T sin θ + T sin θ = 2 T sin θ

The tension in each cable is T = 30 N

Angle made by the cables with the horizontal, θ = 45°

⇒ W = 2×30 N × sin 45° = 2 × 30 N × 0.707 = 42.4 N

Hence, the weight of the frame is 42.4 N. Correct option is D.

6 0

3 years ago

The equation for density is mass divided by volume. the mass of a substance is changed so that it is made of fewer particles. If

Finger [1]

Density is directly proportional to mass. So if there's less matter inside object, its density will also reduce.

6 0

3 years ago

Read 2 more answers

A chicken crosses a 7.50 m wide road at a constant speed of 0.367 m/s. How much time does it take to cross (in seconds)?

mars1129 [50]

<h3><u>Answer;</u></h3>

= 20.436 seconds

<h3><u>Explanation;</u></h3>

Speed = Distance × time

Therefore;

Time = Distance/speed

Distance = 7.50 m, speed = 0.367 m/s

Time = 7.50/0.367

<u>= 20.436 seconds </u>

7 0

3 years ago

Read 2 more answers

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

3 years ago