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vekshin1
3 years ago
7

Question 4 A car of mass 820 kg has a maximum power of 30 kW and moves against a constant resistance of motion to 910 N. Calcula

te the maximum speed of the car in the following situation; b)Up an incline (hill) of 8.75° to the horizontal?​
Physics
1 answer:
wolverine [178]3 years ago
7 0

Answer:

P = W / t = m g s / t = m g v       where work by auto = m g s

30 kw = 30000 watts = 30000 J / s

Work wasted = F v       as shown above relating work and power

Work done against incline = m g s sin 8.75  and power against incline

= m g v sin 8.75 = 1222 v Joules / sec

power in moving auto = power available - power lost to friction

power in moving auto = 30000 - resistance = 30000 - 910 v

1222 v = 30000 - 910 v

v = 30000 / 2132 = 14 m/s

Note: constant resistance to motion must mean P = W / t = F s / t = F v

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Answer with Explanation:

We are given that

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\theta=75^{\circ}

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Using the formula

F=0.6\times 10^{-4}\times 15\times 19sin75

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15 degree above the horizontal  in the northward direction.

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3 years ago
A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic
Ilia_Sergeevich [38]

Answer:

\mu_k=0.27

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\sum F_x:-F_k=ma\\\sum F_y:N=mg

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F_k=\mu_k N

Replacing and solving for the coefficient of kinetic friction:

-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}

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4 0
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SVETLANKA909090 [29]
This is a great problem if you like getting tied up in knots
and making smoke come out of your brain.

I found that it makes the problem a lot easier if I give the objects some
numbers. I'm going to say that the mass of Object 5 is 20 clods.

Let the mass of Mass of Object 5 be 20 clods .

Then . . .

-- The mass of Object 2 is double the mass of Object 5 = 40 clods.

-- The mass of Object 4 is half of the mass of Object 5 = 10 clods.
and
-- the mass of Object 3 is half of the mass of Object 4 = 5 clods.

So now, here are the masses:

Object #1 . . . . . unknown
Object #2 . . . . . 40 clods
Object #3 . . . . . 5 clods
Object #4 . . . . . 10 clods
Object #5 . . . . . 20 clods .

Now let's check out the statements, and see how they stack up:

Choice-A:
Object 3 and Object 5 exert the same gravitational force on Object 1.
Can't be.
Objects #3 and #5 have different masses, so they can't both
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Choice-B.
Object 2 and Object 4 exert the same gravitational force on Object 1.
Can't be.
Objects #2 and #4 have different masses, so they can't both
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Choice-C.
The gravitational force between Object 1 and Object 2 is greater than
the gravitational force between Object 1 and Object 4.
Yes ! Yay !
Object-2 has more mass than Object-4 has, so it must exert more force on
ANYTHING than Object-4 does, (as long as the distances are the same).

Choice-D.
The gravitational force between Object 1 and Object 3 is greater than the gravitational force between Object 1 and Object 5.
Can't be.
Object-3 has less mass than Object-5 has, so it must exert less force on
ANYTHING than Object-4 does, (as long as the distances are the same).

Conclusion:
If the DISTANCE is the same for all the tests, then Choice-C is
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