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Alex
3 years ago
10

A 60 N force is applied over a distance of 15 m. How much work was done?

Physics
1 answer:
GuDViN [60]3 years ago
3 0

Answer:

900J

Explanation:

w =f×s

60×15

=900J

thus the k.e of the body is 900j

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A man attaches a divider to an outdoor faucet so that water flows through a single pipe of radius 9.25 mm into four pipes, each
irinina [24]

Answer:

1.24 m/s

Explanation:

Metric unit conversion:

9.25 mm = 0.00925 m

5 mm = 0.005 m

The volume rate that flow through the single pipe is

\dot{V} = vA = 1.45 * \pi * 0.00925^2 = 0.00039 m^3/s

This volume rate should be constant and divided into the 4 narrower pipes, each of them would have a volume rate of

\dot{V_n} = \dot{V} / 4 = 0.00039 / 4 = 9.74\times10^{-5} m^3/s

So the flow speed of each of the narrower pipe is:

v_n = \frac{\dot{V_n}}{A_n} = \frac{\dot{V_n}}{\pi r_n^2}

v_n = \frac{9.74\times10^{-5}}{\pi 0.005^2} = 1.24 m/s

8 0
3 years ago
Trick question (really easy) just for fun! If u get this right u get brainliest!
Maurinko [17]

Answer:

piper duh

Explanation:

6 0
3 years ago
Read 2 more answers
A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo
CaHeK987 [17]

Answer:\omega =93.51 rad/s

Explanation:

Given

mass of Flywheel m_1=1440 kg

mass of bus m_b=10200 kg

radius of Flywheel r=0.63 m

final speed of bus v=21 m/s

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let \omegabe the angular velocity of Flywheel

0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}

I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2

0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}

\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}

\omega =21\times 4.45=93.51 rad/s

8 0
3 years ago
Explain a situation in which you can accelerate even though your speed doesn’t change.
Serga [27]
"Acceleration" does NOT mean speeding up.  It also doesn't mean
slowing down.  Acceleration means ANY change in the speed
OR DIRECTION of motion.

The only kind of motion that's NOT accelerated is motion at a steady
speed AND in a straight line.

Even when your speed is steady, you're accelerating if your direction
is changing.

A few examples:
(no speeds are changing):

-- driving on a curved road, or turning a corner
-- going around a curve on a skateboard, a bike, or a Segway
-- running on a quarter-mile track
-- an Indy car cruising a practice lap around the track
-- water spinning, getting ready to go down the drain
-- any point on the blade of a fan
-- the little ball going around the inside of a Roulette wheel
-- the Moon in its orbit around the Earth
-- the Earth in its orbit around the sun
4 0
3 years ago
PLEASE HELP ASAP
77julia77 [94]

Answer:

<em>The first choice (32m/s) is the closest to the answer</em>

Explanation:

The magnitude of a vector is the distance between the initial and the end point of the vector.

Being Vx and Vy the horizontal and vertical components of the vector V respectively, the magnitude of V is calculated as:

\mid \vec{V} \mid =\sqrt{V_x^2+V_y^2}

The components of the velocity of the physics student's projectile launcher are Vx=28 m/s and Vy=15 m/s.

Calculate the magnitude of the velocity:

\mid \vec{V} \mid =\sqrt{28^2+15^2}

\mid \vec{V} \mid =\sqrt{784+225}

\mid \vec{V} \mid =\sqrt{1009}

\mid \vec{V} \mid =31.8 \ m/s

The first choice (32m/s) is the closest to the answer

8 0
3 years ago
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