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dimulka [17.4K]

2 years ago

9

a fly wheel of mass 12kg and radius 0.5m with wrapped around it.on the end of the rope is 2kg load.intially it is at rest.calcul

ate the torque of the flywheel.

1 answer:

leva [86]2 years ago

6 0

Answer:

19.6 N of torque. The 2kg load is being affected by acceleration due to gravity which is 9.8 m/s^s

Explanation:

2×9.8=19.6

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Ksivusya [100]

26-8/3 = 6 seconds
to do it

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3 years ago

A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the

Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

v = u + at

0 = 19.09 - 9.81 t

t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

S = ut + 0.5 at²

H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

Maximum height of the ball = 18.57 m

6 0

3 years ago

Two samples of dirt are collected from a suspect's tread in his shoe and a crime scene. She notes very similar characteristics.

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4 years ago

Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 ohm resistor, a 10 ohm resistor, a 15 ohm r

MakcuM [25]

Answer:

Current = 0.33 A

Explanation:

For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

$\dashrightarrow \: \: \sf V = 2 \times 5 = 10 V$

Three resistor of 5 $\Omega$ , 10 $\Omega$ , 15 $\Omega$ are connected in Series, so the net resistance:

$\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}$

$\dashrightarrow \: \: \sf R = 5 + 10 + 15$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \: \Omega}}}}$

According to ohm's law:

$\dashrightarrow \sf\: \: V = IR$

$\dashrightarrow \sf \: \: I = \dfrac{V}{R}$

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 ${\pmb{\sf{\Omega}}}$ we get:

$\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}$

$\therefore$ The electric current passing through the above circuit when the key is closed will be <u>0.33 A</u>

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2 years ago

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3 years ago