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mote1985 [20]
2 years ago
10

A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W

hat is the coefficient of friction between the box and the ramp?
Physics
1 answer:
maxonik [38]2 years ago
5 0

Answer:

0.466 (3 sig. fig.)

Explanation:

Frictional force acting on the box = 5.00×10^2xsin25

Normal force acting on the box = 5.00×10^2xcos25

coefficient of friction = 0.466 (3 sig. fig.)

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List 3 examples in which friction helps us or makes things easier in our daily life. Explain the effect of friction for each.
Kisachek [45]

<span>Friction can help us play tennis, knock down bowling pins, and cook dinner.</span>

8 0
3 years ago
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A ball starts from rest. It rolls down a ramp and reaches the ground after 4 seconds. Its final velocity when it reaches the gro
Hunter-Best [27]
If it starts at rest the initial velocity is 0.
For an acceleration, a, and time, t, the velocity is v=at. Since at t=4, v=7, then a=7/4=1.75m/s^2
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3 years ago
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g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Veseljchak [2.6K]

Answer:

The speed of q₂ is 4\sqrt{10}\ m/s

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

E_{i}=E_{f}

\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2

\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})

Put the value into the formula

\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})

0.00075(400-v_{f}^2)=0.18&#10;

400-v_{f}^2=\dfrac{0.18}{0.00075}

-v_{f}^2=240-400

v_{f}^2=160

v_{f}=4\sqrt{10}\ m/s

Hence, The speed of q₂ is 4\sqrt{10}\ m/s

7 0
2 years ago
A Michelson interferometer uses red light with a wavelength of 656.45 nm from a hydrogen discharge lamp. Part A How many bright-
kati45 [8]

Answer:

51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7cm

Explanation:

The number of maxima appearing when the mirror M moves through distance \Delta L is given as follows,

\Delta m = \frac{\Delta L}{\frac{\lambda}{2}}

Here,

\Delta L= is the distance moved by the mirror M

\lambda is the wavelenght of the light used.

\Delta L= 0.017m

\lambda = 656.45*10^{-9}m

\Delta m = \frac{0.017}{\frac{656.45*10^{-9}}{2}}

\Delta m = 51793.72

Therefore, 51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7

3 0
3 years ago
A uniform stationary ladder of length L and mass M leans against a smooth vertical wall, while its bottom legs rest on a rough h
ikadub [295]

Answer:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

Explanation:A uniform ladder of mass and length leans at an angle against a frictionless wall .If the coefficient of static friction between the ladder and the ground is , determine a formula for the minimum angle at which the ladder will not slip.

3 0
3 years ago
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