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stira [4]
3 years ago
5

A cart is moving to the right with a constant speed of 20 m s . A box of mass 80 kg moves with the cart without slipping. The co

efficient of static friction between the box and cart is 0.30, and the coefficient of kinetic friction between the box and cart is 0.15. What are the magnitude and direction of the frictional force acting on the box as it moves with the cart?
Physics
1 answer:
nignag [31]3 years ago
4 0

Answer:

Explanation:

Given

mass of box m=80 kg

coefficient of kinetic friction \mu _k=0.15

coefficient of Static friction \mu _s=0.30

cart is moving with constant velocity therefore Net Force is zero

Since there is no net acceleration therefore friction force will be zero

mathematically

f_r=ma

where f_r=frictional\ Force

a=acceleration

a=0

f_r=0

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Vedmedyk [2.9K]

Answer: 8 years

Explanation:

According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”:</em>

<em />

T^{2}\propto a^{3} (1)

In other words: this law states a relation between the orbital period T of a body (moon, planet, satellite, comet, asteroid) orbiting a greater body in space (the Sun, for example) with the size a of its orbit.  

However, if T is measured in years (Earth years), and a is measured in astronomical units (equivalent to the distance between the Sun and the Earth: 1AU=1.5(10)^{8}km), equation (1) becomes:

T^{2}=a^{3} (2)

This means that now both sides of the equation are equal.

Knowing a=4 AU and isolating T from (2):

T=\sqrt{a^{3}} (3)

T=\sqrt{(4 AU)^{3}} (4)

Finally:

T= 8 yearsThis is the period of the asteroid

3 0
3 years ago
Suggest at least two reagents other than sodium borohydride that could be used to carry out the reduction of 9-fluorenone to 9-h
vampirchik [111]
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8 0
3 years ago
Approximately 80% of the energy used by the body must be dissipated thermally. The mechanisms available to eliminate this energy
Semenov [28]

Answer:

the correct answer is c) 23 g

Explanation:

The heat lost by the runner has two parts: the heat absorbed by sweat in evaporation and the heat given off by the body

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The latent heat is

      Q_absorbed = m L

The heat given by the body

      Q_lost = M c_{e} ΔT

       

where m is the mass of sweat and M is the mass of the body

       m L = M c_{e} ΔT

        m = M c_{e} ΔT / L

let's replace

        m = 90  3.500  1.8 / 2.42 10⁶

 

        m = 0.2343 kg

reduced to grams

        m = 0.2342 kg (1000g / 1kg)

        m = 23.42 g

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8 0
3 years ago
A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball
Akimi4 [234]

Answer:

Approximately 122.625\; {\rm m} (assuming that g = 9.81\; {\rm m\cdot s^{-2}}, the ball was launched from ground level, and that the drag on the ball is negligible.)

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If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly (-v_{0}). The velocity of the ball would be changed from v to (-v_{0})\! (such that \Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})) within t = 10\; {\rm s}.

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Since \Delta v = (-2\, v_{0}):

-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}.

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The ball reaches maximum height when its velocity is v_{1} = 0\; {\rm m\cdot s^{-1}}. Apply the SUVAT equation x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a) to find the displacement x between the original position (ground level, where v_{0} = 49.05\; {\rm m\cdot s^{-1}}) and the max-height position of the ball (where v_{1} = 0\; {\rm m\cdot s^{-1}}.)

\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}.

7 0
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Correct answer gets marked brainly
daser333 [38]

Answer:

CHEMICAL WHETEARING

Explanation:

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5 0
3 years ago
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