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3 years ago

14

A wagon is pulled at a speed of 0.40metets/seconds by a horse exerting an 1,800 Newton's horizontal force. what is the power of

this horse?

1 answer:

Tatiana [17]3 years ago

6 0

Given:

speed of 0.40meters/seconds

1,800 Newton's horizontal force

Required:

Power of the horse

Solution:

P = F(D/T) where P is power in watts, F is the force, D is the distance and T is time

P = (1,800N) (0.40 meters/seconds)

P = 720 Watts

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The plane of a5cm*8cm rectangular loop of wire is parallel to a 0.19T magnetic field the loop carries a current of 6.2 A. What t

Sedaia [141]

Answer:

Torque; τ = 4.712 × 10^(-3) J

Magnetic moment; M = 0.0248 J/T

Explanation:

Torque is gotten from the formula;

τ = BIA

Where;

B is magnetic field

I is current

A is area

We are given;

B = 0.19T

I = 6.2A

Rectangle dimensions = 5cm by 8cm = 0.05m by 0.08m

Thus;

Area; A = 0.05m × 0.08m = 0.004 m²

Thus;

τ = 0.19 × 6.2 × 0.004

τ = 4.712 × 10^(-3) J

Formula for the magnetic moment is given by;

M = IA

M = 6.2 × 0.004

M = 0.0248 J/T

5 0

2 years ago

A car initially traveling at 27.2 m/s undergoes a constant negative acceleration of magnitude 1.90 m/s2 after its brakes are app

zubka84 [21]

Answer:

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

Explanation:

We can use the following equation:

$\omega_{f}^{2}=\omega_{i}^{2}-2\alpha \Delta \theta$ (1)

The angular acceleration is:

$a_{tan}=\alpha R$

$\alpha=\frac{1.9}{0.325}$

$\alpha=5.85\: rad/s^{2}$

and the initial angular velocity is:

$\omega_{i}=\frac{v}{R}$

$\omega_{i}=\frac{27.2}{0.325}$

$\omega_{i}=83.69\: rad/s$

Now, using equation (1) we can find the revolutions of the tire.

$0=83.69^{2}-2*25.85 \Delta \theta$

$\Delta \theta=135.47\: rad$

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

I hope it helps you!

6 0

2 years ago

The form of energy that can move from place to place across the universe is . On Earth, the main source of this energy is .

serg [7]

The form of energy that can move from place to place across the universe is light energy. On earth, the main source of this energy is Sun. Most of the light energy comes from the sun because it is the primary source of all the energies. The food, fossil fuels, movement of winds, etc all exists due to Sun. Without sun, there won't be any light energy on the earth. In all the processes which occur on earth has a direct or indirect involvement of light energy which comes from sun.

5 0

2 years ago

Read 2 more answers

two long parrallel wires that are 0.30 m apart carry currents of 5.0 A and 8.0 A in the opposite direction.Find the magnitude of

Vera_Pavlovna [14]

Answer: The magnitude of force per length that each wire exert on the other wire is 2.67×10^-5 N/m.

The force is repulsive.

Explanation: Please see the attachments below

3 0

2 years ago

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

erastova [34]

Complete Question

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).

J

(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)

MW

Answer:

The heat transferred is $Q = 5.866 * 10^9 J$

The power is $P = 5866\ MW$

Explanation:

From the question we are told that

Mass of the water per second is $m = 1917 \ kg$

The initial temperature of the water is $T_i = 35^oC$

The boiling point of water is $T_b = 100^oC$

The final temperature $T_f = 450^oC$

The latent heat of vapourization of water is $c__{L}} = 2256*10^3 J/kg$

The specific heat of water $c_w = 4184 J/kg^oC$

The specific heat of stem is $C_s =1520 \ J/kg ^oC$

Generally the heat needed each second is mathematically represented as

$Q = m[c_w (T_i - T_b) + m* c__{L}} + m* c__{S}} (T_f - T_b)]$

Then substituting the value

$Q = m[c_w [T_i - T_b] + c__{L}} + C__{S}} [T_f - T_b]]$

$Q = 1917 [(4184) [100 - 35] + [2256 * 10^3] +[1520] [450 - 100]]$

$Q = 1917 * [3.05996 * 10^6]$

$Q = 5.866 * 10^9 J$

The power required is mathematically represented as

$P = \frac{Q}{t}$

From the question $t = 1\ s$

So

$P = \frac{5.866 *10^9}{1}$

$P = 5866*10^6 \ W$

$P = 5866\ MW$

6 0

2 years ago