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3 years ago

Which of the following statements accurately describes the sign of the work done on the box by the force of the push?

Physics

1 answer:

USPshnik [31]3 years ago

4 0

Answer: positive

Explanation: according to newton's second law of motion, the applied force (pull) will produce a motion for the box defined by the formulae

f = ma, where m is the mass of the box and a is the acceleration.

But due to friction, a force of the same magnitude but opposite in direction to the push is applied, hence we have that

f - fr = ma

Where fr is the frictional force.

As we can see from the formulae, f (push) is responsible for a positive work while fr (friction) will bring about a negative work done.

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Electrical power companies sell electrical energy

mixer [17]

Heat used by electric heater :

Q = m • c • ∆T

Q = (75 kg)(4200 J/kg°C)(43°C - 15°C)

Q = 8.82 × 10⁶ J

Cost of electrical energy :

Cost = (8.82 × 10⁶ J)/(3.6 × 10⁶ J) • ($ 0.15)

Cost = $ 0.3675

4 0

2 years ago

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

5 0

3 years ago

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

Sladkaya [172]

Answer:

0.767

Explanation:

The work done on the truck by the frictional drag force is given by

$W=-Fd$

where

F is the magnitude of the frictional force

d = 38.0 m is the maximum displacement allowed for the truck

The negative sign is due to the fact that the force of friction is opposite to the motion of the truck

The force of friction can also be written as:

$F=\mu mg$

where

$\mu$ is the coefficient of kinetic friction between the truck and the lane

m is the mass of the truck

g is the acceleration of gravity

So we can rewrite the work done as

$W=-\mu mg d$ (1)

According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (2)

where

v = 0 is the final velocity of the truck

u = 23.9 m/s is the initial velocity of the truck

By combining (1) and (2) we get

$-\frac{1}{2}mu^2 = -\mu mg d$

And solving for $\mu$ , we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:

$\mu = \frac{u^2}{2gd}=\frac{23.9^2}{2(9.8)(38.0)}=0.767$

7 0

3 years ago

Sketch the resultant field pattern around the following current carrying conductors and

Novay_Z [31]

Cccccccvvvvvcjjjjjjjjjjkllk ki e

8 0

2 years ago

The gauge pressure in the tires of your car is 210 kPa (30.5 psi) when the temperature is 25°C (77 °F). Several days later it is

DaniilM [7]

Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:

P/T = const.

P = pressure, T = temperature, the quotient of P/T must stay constant.

Initial P and T values:

P = 210kPa + 101.325kPa

P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)

T = 25°C = 298.15K

Final P and T values:

P = ?, T = 0°C = 273.15K

Set the initial and final P/T values equal to each other and solve for the final P:

311.325/298.15 = P/273.15

P = 285.220kPa

Subtract 101.325kPa to find the final gauge pressure:

285.220kPa - 101.325kPa = 183.895271kPa

The final gauge pressure is 184kPa or 26.7psi.

8 0

3 years ago