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zmey [24]
2 years ago
10

As a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used

to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 5.15 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W.
Physics
1 answer:
velikii [3]2 years ago
8 0

Answer:

128 N

Explanation:

Power can be defined as

P = F\cdot v

<em>F</em> is the force and <em>v</em> is the velocity.

F = \dfrac{P}{v}

According to the question, <em>P</em> = 5.15 hp = 5.15 × 746 W.

<em>v</em> = 30 m/s

F = \dfrac{5.15\times 746\text{ W}}{30\text{ m/s}} = 128 \text{ N}

Since most of the power is used to compensate for the energy transformations due to friction forces, the total frictional is 128 N

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Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra
Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

s^2 = (x_A - x_B)^2+(y_A - y_B)^2  (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

s^2 = x_B^2+y_A^2 (2)

Taking the differential with respect to time:

\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}  (3)

where \displaystyle{\frac{dx_B}{dt}}=v_B and \displaystyle{\frac{dx_A}{dt}}=v_A are the respective given velocities of the boats. To find s and x_B we make use of the given position for A, y_A=30, the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h

with this time, we know can now calculate the distance at which B is:

\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi

and applying Pythagoras:

\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}

Now substituting all the values in (3) and solving for  \displaystyle{\frac{ds}{dt} } we get:

\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}

4 0
3 years ago
what velocity must a 1340kg car have in order to havw the same momentum as a 2680 kg truck traveling at a velocity of 15m/s to t
kykrilka [37]
Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.

Hope i helped
Have a good day :)

 
6 0
3 years ago
Suppose you apply a force of 75 N to a 25-kg object. What will the acceleration of the object be?
Serjik [45]

Answer:

I belive it would be ture

Explanation:

It's been a while since I learned this but I think that is right.

6 0
2 years ago
Read 2 more answers
A ball is thrown upward with a speed of 40 m/s. Approximately how much time does it take the ball to travel from the release loc
zvonat [6]

I'm going to assume that this gripping drama takes place on planet Earth, where the acceleration of gravity is 9.8 m/s².  The solutions would be completely different if the same scenario were to play out in other places.

A ball is thrown upward with a speed of 40 m/s.  Gravity decreases its upward speed (increases its downward speed) by 9.8 m/s every second.

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Just like so many other aspects of life, the downward fall is an exact "mirror image" of the upward trip.  After another 4.08 seconds, the ball has returned to the height of the hand which flung it.  In total, the ball is in the air for <em>8.16 seconds</em> up and down.

4 0
3 years ago
Why do you see objects when you shine a flashlight in a dark room?​
Afina-wow [57]

Answer:

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Explanation:

3 0
2 years ago
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