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Setler [38]
3 years ago
8

A 2.44 kg block is pushed 1.55 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle o

f 69.6 ◦ with the horizontal. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of kinetic friction between the block and wall is 0.691, find the work done by F.
Physics
1 answer:
svlad2 [7]3 years ago
6 0

Answer:

The work done is L= 49.83 Joules

Explanation:

m= 2.44kg

d= 1.55m

α= 69.6°

g= 9.8 m/s²

μ= 0.691

F= ?

Fy= F*sin(α)

Fx= F*cos(α)

Fr= μ * Fx

Fr= μ * F*cos(α)

W= m*g

W= 23.91 N

Fy - W - Fr = 0

F*sin(α) - W -  μ * F*cos(α) = 0

F* ( sin(α) - μ *cos(α) ) = W

F= W /  ( sin(α) - μ *cos(α) )

F= 34.3 N

L= Fy * d

L= 49.83 J

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LuckyWell [14K]

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

Thus, the expression for final pressure in the two containers will be:

PV=P_1V_1+P_2V_2

P=\frac{P_1V_1+P_2V_2}{V}

where,

P_1 = pressure of N₂ gas = 4.45 atm

P_2 = pressure of Ar gas = 2.75 atm

V_1 = volume of N₂ gas = 3.00 L

V_2 = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}

P=2.62atm

Thus, the final pressure in the two containers is, 2.62 atm

8 0
2 years ago
How does an increase in temperature generally affect the rate of a reaction?
Zigmanuir [339]

Most reactions are exothemic. If the forward reaction of an equilibrium reaction is exothemic then the reverse reaction must be endothermic.

If a system in equilibrium is heated, it will move in exothermic direction to give out heat energy.

7 0
3 years ago
A boy standing on a bridge above a river throws stone A vertically upward with an initial velocity =15m/s.2 seconds later he dro
Anastaziya [24]

I'm not sure if this is correct but it's what I'll do

This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.

Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )

Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12

Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2

4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t

Time for Stone B is 4s
Time for Stone A is 6s

7 0
3 years ago
(a) What is the entropy change of a 14.6 g ice cube that melts completely in a bucket of water whose temperature is just above t
Alex73 [517]

Answer:

a) 17.81 J/K

b) 33.325 J/K

Explanation:

The expression to use here is the following:

ΔS = Q/T

Where:

Q: heat released or absorbed

T: Temperature in K

Now, in order to do this, we need to gather the data. We know that the temperature in part a) is above the freezing temperature of water, which is 0 ° C or 273 K. and the mass of the ice cube is 14.6 g.

a) Using the water heat of fusion (Cause it's melting), we can calculated the heat released using the following expression:

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Solving for Q first we have:

Q = (14.6 / 1000) * 333,000

Q = 4,861.8 J

Now, the entropy change is:

ΔS = 4,861.8 / 273

ΔS = 17.81 J/K

b) In this part, we follow the same procedure than in part a) but using the water heat of boiling (Lv = 2,256,000 J/kg), the temperature of boiling which is 100 °C (or 373 K) and the mass of 5.51 g (0.00551 kg)

Calculating the heat:

Q = 0.00551 * 2,256,000 = 12,430.56 J

Now the entropy change:

ΔS = 12,430.56 / 373

ΔS = 33.325 J/K

8 0
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noname [10]

Answer:

b. The electric field points away from the source charge, if the source charge is positive

d. The electric field points toward the source charge, if the source charge is negative.

Explanation:

A positive source charge would create an electric field that would exert a repulsive effect upon a positive test charge. Thus, the electric field vector would always be directed away from positively charged objects. On the other hand, a positive test charge would be attracted to a negative source charge. Therefore, electric field vectors are always directed towards negatively charged object.

Also electric field strength depends only on test charge

The correct options include b and d

The electric field points away from the source charge, if the source charge is positive.

Also, the electric field points toward the source charge, if the source charge is negative.

3 0
2 years ago
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