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3 years ago

Work can _____ energy between objects and can cause a change in the form of energy.

Physics

1 answer:

zhuklara [117]3 years ago

6 0

Answer:

Work can transfer energy between objects and can cause a change in the form of energy.

Explanation:

The best explanation can be done with an example.Consider an apple of mass m falling from a tree. Initially, the apple is at rest on the tree, at a certain height h above the ground, so it has a form of energy called gravitational potential energy, equal to: where g is the acceleration due to gravity.As the apple falls, the force of gravity acts on it, and so this force does work on the apple. As a result, the potential energy of the apple is converted into another form of energy, kinetic energy, which is the energy due to the motion of the apple: where v is the speed of the apple.Consider another example: a man pushing a box along the floor with a force F. The man is doing work on the box, and this work is equal towhere d is the displacement of the box. As a result of this work, some of the chemical energy contained in the muscles of the man is transferred into the box and converted to kinetic energy of the box, which is put in motion and it gains a certain speed v. In particular, if there are no frictional forces involved, the work done by the man is equal to the kinetic energy gained by the box.

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As a bat flies toward a wall at a speed of 6.0 m/s, the bat emits an ultrasonic sound wave with frequency 30.0 kHz. What frequen

Leto [7]

Answer:

$f_b = 29.98 Hz$

Explanation:

speed of bat = 6 m/s

sound wave frequency emitted by bat = 30.0 kHz

as we know,

speed of sound (c)= 343 m/s

$f_w = f(\dfrac{c+v_r}{c+v_s})$

$f_w = 30(\dfrac{343+0}{343+6})$

$f_w = 29.48 Hz$

now frequency received by bat is equal to

$f_b = f(\dfrac{c+v_r}{c+v_s})$

$f_b = 29.48(\dfrac{343+6}{343+0})$

$f_b = 29.98 Hz$

hence the frequency hear by bat will be 29.98 Hz

7 0

3 years ago

A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when

choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

$F_t - mg = ma$

here since speed is constant or it is at rest

so we will have

$a = 0$

$F_t = mg$

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

$F_t - mg = ma$

$F_t = mg + ma$

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

$F_t - mg = -ma$

$F_t = mg - ma$

so tension force is less than the weight of crate if it is accelerating downwards

4 0

3 years ago

A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are 42◦ from t

Semmy [17]

Answer:

A) P.E = 138.44 J

B) The velocity of swing at bottom, v = 3.33 m/s

C) The work done, W = -138.44 J

Explanation:

Given,

The mass of the child, m = 25 Kg

The length of the swing rope, L = 2.2 m

The angle of the swing to the vertical position, ∅ = 42°

A) The potential energy at the initial position ∅ = 42° is given by the relation

P.E = mgh joule

Considering h = 0 for the vertical position

The h at ∅ = 42° is h = L (1 - cos∅)

P.E = mgL (1 - cos∅)

Substituting the given values in the above equation

P.E = 25 x 9.8 x 2.2 (1 - cos42°)

= 138.44 J

The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J

B) The velocity of the swing at the bottom.

At bottom of the swing the P.E is completely transformed into the K.E

∴ K.E = P.E

1/2 mv² = 138.44

1/2 x 25 x v² 138.44

v² = 11.0752

v = 3.33 m/s

The velocity of the swing at the bottom is, v = 3.33 m/s

C) The work done by the tension in the rope from initial position to the bottom

Tension on string, T = Force acting on the swing, F

$W=L\int\limits^0_\phi{F} \, d \phi$

$=L\int\limits^0_\phi{mg.sin \phi} \, d \phi$

= $-Lmg[cos\phi]_{42}^{0}$

= - 2.2 x 25 x 9.8 [cos0 - cos 42°]

= - 138.44 J

The negative sign in the in energy is that the work done is towards the gravitational force of attraction.

The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J

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3 years ago

if there are infinite universes is it possible that there is a universe where 1+1=3 or 4 or something

zhenek [66]

Answer:

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Explanation:

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3 years ago

Give a specific example of a machine,and describe how its mechanical efficiency might be calculated

natta225 [31]

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3 years ago