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2 years ago

14

Which of these correctly describes whether a girl holding a ball in the same position is doing work on the ball?

1 answer:

aksik [14]2 years ago

4 0

Answer:

The girl is doing work on the ball because the energy in her muscles changed, even though the ball is not displaced.

Explanation:

The complete question is...

Which of these correctly describes whether a girl holding a ball in the same position is doing work on the ball?

-The girl is doing work on the ball because the energy of the ball changed, even though it is not displaced.

-The girl is doing work on the ball because the energy in her muscles changed, even though the ball is not displaced.

-The girl is doing no work on the ball because the ball is not displaced.

-The girl is doing no work on the ball because she is exerting a net force on the ball.

Holding up a ball costs energy, which is used to counter the work that would have otherwise be done on the ball by gravity. Although no physical distance is moved, we should consider the fact that by holding the ball, the girls hand exerts physical force to hold the ball in place. Also, there is a potential gravitational work on the ball due to gravity, but the force exerted by the girls hand does an equivalent of this gravito-potential work in order to counter it and hold the ball in place. All these activities eventually lead to a change in energy in her hand muscle to show that energy is expended.

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How does the table show that the balloon went downwards ?

o-na [289]

Answer:

I am going to guess it shows that the balloon is going downwards because the speed of rise is in the negatives for the last 2.

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2 years ago

The velocity of the water in the pipe at right is given by V1 = 0.5t m/s and V2 = 1.0t m/s, where t is in seconds. Determine the

Nonamiya [84]

Answer:

A) At point 1, local acceleration = 0.5 m/s²

At point 2, local acceleration = 1.0 m/s²

B) Average Eulerian convective acceleration over the two points in the cross section shown = 0.5 m/s²

This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.

Explanation:

Local acceleration at those points is the instantaneous acceleration at those points and it is given as

a = dv/dt

At point 1, v₁ = 0.5 t

a₁ =dv₁/dt = 0.5 m/s²

At point 2, v₂ = 1.0 t

a₂ = dv₂/dt = 1.0 m/s²

b) Average Eulerian convective acceleration over the two points in the cross section shown = (change of velocity between the two points)/time

Change of velocity between the two points = v₂ - v₁ = 1.0t - 0.5t = 0.5 t

Time = t

Average acceleration = 0.5t/t = 0.5 m/s²

This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.

8 0

3 years ago

What is the Net Force?

marshall27 [118]

It is 800 N FN = 600N + 200 N = 800 N Answer to your question: The net force is all Newton's second law. It is the force that acts on a body or a particle. for example: It is the force we make when we push a car or something heavy that is in a straight line. .

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3 years ago

You run a race with your friend. At first you each have the same kinetic energy, but then you find that she is beating you. When

Blizzard [7]

Answer:

52.49 Kg

Explanation:

Let m1 and v1 denote your mass and velocity respectively

Let m2 and v2 denote your friends mass and velocity respectively

Kinetic energy is given by

$KE= 0.5mv^{2}$

Since your kinetic energies are the same hence

$0.5m1(v1)^{2}=0.5m2(v2)^{2}$

$m1(v1)^{2}=m2(v2)^{2}$ and making m2 the subject then

$m2=\frac { m1(v1)^{2}}{(v2)^{2}}$

Since v2 is v1+0.28v1=1.28v1

Substituting m1 for 86 Kg

$m2=\frac { 86 Kg(v1)^{2}}{(1.28v1)^{2}}= 52.49023\approx 52.49 Kg$

3 0

3 years ago

After a check up, a person now has a far point of 100 cm, but with good near point vision. He needs to wear a new pair of correc

lakkis [162]

Answer:

so his far point according to this pair of glass is 200 cm

Explanation:

power of old pair of corrective glasses is given as

$P = -0.5 dioptre$

now we have

$f = \frac{1}{P}$

$f = -2 m$

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now we know that for normal vision the maximum distance of vision is for infinite distance

so by lens formula we have

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$d_i = 200 cm$

so his far point according to this pair of glass is 200 cm

7 0

3 years ago