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weqwewe [10]
3 years ago
5

A box of unknown mass is sliding with an initial speed vi = 4.70 m/s across a horizontal frictionless warehouse floor when it en

counters a rough section of flooring d = 4.50 m long. the coefficient of kinetic friction between the rough section of flooring and the box is 0.100. using energy considerations, determine the final speed of the box after sliding across the rough section of flooring.
Physics
1 answer:
malfutka [58]3 years ago
8 0

Here Change in Kinetic Energy = Work Done by Friction

Therefore, substituting the given values to the equation, we get

0.5 * m * (vFinal^2 - vInitial^2) = µ m g * d

Therefore

0.5*( 5.90^2 - Vfinal^2 ) = 0.100*9.8*2.10

Therefore

vfinal = 5.54 m/sec

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A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.3 m/s rel
Sonja [21]

consider east-west direction along X-axis  and north-south direction along Y-axis

V_{ra} = velocity of migrating robin relative to air = 12 j m/s

(where "j" is unit vector in Y-direction)

V_{ag} = velocity of air relative to ground = 6.3 i m/s

(where "i" is unit vector in X-direction)

V_{rg} = velocity of migrating robin relative to ground = ?

using the equation

V_{rg} = V_{ra} + V_{ag}

V_{rg} = 12 j + 6.3 i

V_{rg} = 6.3 i + 12 j

magnitude : sqrt((6.3)² + (12)²) = 13.6 m/s

direction : tan⁻¹(12/6.3) = 62.3 deg north of east

4 0
3 years ago
A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. 6
Rina8888 [55]

Answer:

a) 0.147 N

b) 9.408 N

c) 9.261 N

Explanation:

The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force \frac{mv^{2} }{r}. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.

7 0
2 years ago
A Net Force of 9.0 N acts through a distance of 3.0 m in a time of 3.0 s. The work done is?
JulijaS [17]
Work done is the distance a force acts over.

So, the work done here is 9.0N * 3.0m = 27 J
5 0
3 years ago
Read 2 more answers
If the distance between a neutral atom and a point charge is quadrupled, by what factor does the force on the atom by the point
IgorLugansk [536]
New force or Old Force 
8 0
3 years ago
39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
3 years ago
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