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PtichkaEL [24]
3 years ago
10

A wagon is pulled at a speed of 0.40 meters/sec by a horse exerting an 1,800-newton horizontal force. what is the power of this

horse?
Physics
1 answer:
Alex787 [66]3 years ago
5 0
The answer is: 720W
The definition of Power is:P =L/t =→F⋅→s/t =→F⋅→s/t = →F⋅→v
The ⋅ in the last formulation designate a scalar product, well-defined by:
→v . →w = |v| |w| cos(ˆvw), the angle ˆvw is 0 because →F and →v are parallel, so the cosinus is 1.
So: P=1800N⋅ 0,40m/s=720W.
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3 years ago
The __________-second rule applies to any speed in ideal weather and road conditions.
SOVA2 [1]
The two-second rule.

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8 0
3 years ago
A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. T
sergij07 [2.7K]

Answer:

0.98kW

Explanation:

The conservation of energy is given by the following equation,

\Delta U = Q-W

\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)

Where

\dot{m} = Mass flow

h_1 =Specific Enthalpy (IN)

h_2 = Specific Enthalpy (OUT)

g = Gravity

z_{1,2} = Heigth state (In, OUT)

V_{1,2} =Velocity (In, Out)

Our values are given by,

T_i = 10\°C

P_1 = 100kPa

\dot{m} = 5kg/s

z_2 = 20m

For this problem we know that as pressure, temperature as velocity remains constant, then

h_1 = h_2

V_1 = V_2

Then we have that our equation now is,

\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}

\dot{W} = \frac{(5)(9.81)(0-20)}{1000}

\dot{W} = -0.98kW

8 0
3 years ago
Please help on this one?
diamong [38]

Answer:

ITS C

Explanation:

7 0
2 years ago
An applied force of 50 N is used to accelerate an object, that weighs 73 N, to the right across a frictional surface. The object
Hunter-Best [27]

Answer:

5.38 m/s^2

Explanation:

NET force causing the object to accelerate  =  50 -10 = 40 N

Mass of the object =  73 N / 9.81 m/s^2 = 7.44 kg

F = ma

40 = 7.44 * a         a = 5.38 m/s^2

6 0
1 year ago
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