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Alinara [238K]
1 year ago
11

Suppose a child swings around her straight arm about her shoulder joint at an angular velocity of 13 rad/s. If her arm is 40. 0

cm long, what is the speed of her hand?.
Physics
1 answer:
Misha Larkins [42]1 year ago
8 0

A child swings her 40 cm (0.4 meter) long, straight arm around her shoulder joint at a 13 rad/s angular velocity. So, her hand is moving at 5.2 m/s every second.

<h3>What does angular velocity mean?</h3>

angular velocity is the speed where an object rotates or revolves around an axis or alters the angle between two bodies. The angle between a line solely on a single body and a line on the other in the illustration serves as a representation of this displacement.

<h3>In what way does angular velocity rise?</h3>

The angular velocity increases with increasing rotational angle over a certain period of time. Degrees per second (o/s), radians per second (rad/s), or revolutions per minute (rpm), if appropriate, are the units for angular velocity. An element of a vector is angular velocity.

To know  more about angular velocity visit:

brainly.com/question/1980605

#SPJ4

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2. A jack exerts a vertical force of 4.5 X 103
skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

\\  \\

Given :-

\star  \sf  \small force = 4.5 \times  {10}^{3}  \: newton

\star  \sf  \small distance = 0.25 \: meter

\\  \\

To find:-

\sf \star \: work = \: ?

\\  \\

Solution:-

we know :-

\bf \dag \boxed{ \rm work = force \times distance}

\\  \\

So:-

\dashrightarrow \sf work = force \times distance

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \frac{0 \cancel.5}{10}  \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \frac{5}{10}  \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \cancel \frac{5}{10}  \\

\\  \\

\dashrightarrow \sf work =  \dfrac{4\cancel.5}{10}  \times 1 {0}^{3} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10}  \times 1 {0}^{3} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10 {}^{0} }  \times 1 {0}^{3 - 1} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10 {}^{0} }  \times 1 {0}^{2} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{1}  \times 1 {0}^{2} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45 \times 10 \times  \cancel{10}}{ \cancel2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45 \times 10 \times 5}{ 1} \\

\\  \\

\dashrightarrow \sf work =225 \times 10

\\  \\

\dashrightarrow \bf work =\red{2250\: joule}

5 0
2 years ago
1) If you release a rubber band that had 10 units of elastic energy, 12 units of movement energy cannot be produced. Why not?
Mila [183]

Answer:

press dat crown for meh

Explanation:

4 0
3 years ago
Read 2 more answers
34.6 cL= (blank) hL convert
shepuryov [24]

<u>Answer</u>

0.00346 hL

<u>Explanation</u>

cL means Centilitre while hL means Hectolitre.

10,000 cL = 1 hL

∴ 34.6 cL = 34.6/10,000  hL

                = <em>0.00346 hL</em>

3 0
3 years ago
When you apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed ().When you apply a f
elena-s [515]

Answer:

Explanation:

When we apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed. So net force on the block is zero .

It implies that a force ( frictional ) acts on it which is equal to 76 N in opposite direction ( friction )

When we apply  a greater force on it it starts moving with acceleration .

This time kinetic friction acts on it due to rough ground equal to 76 N .This is limiting friction ( maximum friction )

Net force on the body in later case

= 89 - 76

= 13 N

Force by ground on the block in horizontal direction = 76 N ( FRICTIONAL FORCE )

=

3 0
3 years ago
May you help me answer this​
Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is 2.2 m/s^2

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

  1. 1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
  2. 2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
  3. 3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, T^2 \propto r^3, where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

M g sin \theta - T = Ma (1)

where

Mg sin \theta is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

g=9.8 m/s^2 (acceleration of gravity)

\theta=35^{\circ}

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

T-mg sin \theta = ma (2)

where

m = 3.5 kg (mass of the block)

\theta=35^{\circ}

From (2) we get

T=mg sin \theta + ma

Substituting into (1),

M g sin \theta - mg sin \theta - ma = Ma

Solving for a,

a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2

2b)

The tension in the string can be calculated using the equation

T=mg sin \theta + ma

where

m = 3.5 kg (mass of lighter block)

g=9.8 m/s^2

\theta=35^{\circ}

a=2.2 m/s^2 (acceleration found in part 2)

Substituting,

T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

U=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

K=\frac{1}{2}(0.5)(3)^2=2.25 J

Learn more about acceleration and forces:

brainly.com/question/11411375

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And about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

7 0
3 years ago
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