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umka21 [38]
3 years ago
15

Does an athlete do work on a trophy as she lifts it overhead

Physics
1 answer:
bogdanovich [222]3 years ago
3 0
No she/he is doing work for her muscles
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At a distance r1 from a point charge, the magnitude of the electric field created by the charge is 395 N/C. At a distance r2 fro
Ksenya-84 [330]

Answer:

r₂/r₁ = 1.82

Explanation:

The electric field due to a point charge, has the following expression:

E =\frac{k*q}{r^{2}}

For a distance r₁, the magnitude of the electric field is 395 N/C, so we can solve for r₁², as follows:

r₁² = \frac{k*q}{395 N/C} } (1)

For a distance r2, the magnitude of the electric field is 119 N/C, so we can solve for r₂², as follows:

r₂² = \frac{k*q}{119 N/C} }

We can find the quotient r₂/r₁, from (1) and (2):

r₂/r₁ =\sqrt{395/119} = 1.82

7 0
4 years ago
What happens to potential energy as the car goes up the hill?
True [87]
The potential energy will increase
8 0
3 years ago
Read 2 more answers
square root A 1400 kg car is coasting on a horizontal road with a speed of 18 m/s . After passing over an unpaved, sandy stretch
Lina20 [59]

Answer:

The net force on the car is 2560 N.

Explanation:

According to work energy theorem, the amount of work done is equal to the change of kinetic energy by an object. If 'W' be the work done on an object to change its kinetic energy from an initial value 'K_{i}' to the final value 'K_{f}', then mathematically,

W = K_{f} - K_{i} = \dfrac{1}{2}~m~(v_{f}^{2} - v_{i}^{2})........................................(I)

where 'm' is the mass of the object and 'v_{i}' and 'v_{f}' be the initial and final velocity of the object respectively. If 'F_{net}' be the net force applied on the car, as per given problem, and 's' is the displacement occurs then we can write,

W = F_{net}~.~s.......................................................(II)

Given, m = 1400~Kg,~v_{i} = 18~m~s^{-1}~v_{f} = 14~m~s^{-1}~and~s = 35~m.

Equating equations (I) and (II),

&& - F_{net} \times 35~m = \dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})~m^{2}~s^{-2}\\&or,& F_{net} = \dfrac{\dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})}{35}~N\\&or,& F_{net} = 2560~N

6 0
3 years ago
Suppose a skydiver (mass = 75 kg) is falling toward the Earth. When the skydiver is 100 m above the Earth he is moving at 60 m/s
Andrej [43]

Answer:

  Potential energy = 73.575 kJ

  Kinetic energy = 135kJ

  Total mechanical energy = 208.575 kJ

Explanation:

   The potential energy of a body is given by the expression, PE = mgh, where m is the mass of the body, g is the acceleration due to gravity value and h is the height of the body.

  The kinetic energy of a body is given by KE=\frac{1}{2} mv^2, where v is the velocity and m is the mass of body.

  Total mechanical energy = Kinetic energy + Potential energy

  KE=\frac{1}{2} mv^2=\frac{1}{2} *75*60^2= 135000J = 135kJ

  PE = mgh = 75*9.81*100 = 73575 J = 73.575 kJ

  Total mechanical energy = Kinetic energy + Potential energy = 135+73.575

                                             = 208.575 kJ    

4 0
3 years ago
Mature salmon swim upstream, returning to spawn at their birthplace. During the arduous trip they leap vertically upward over wa
patriot [66]

Answer:

The minimum speed is 7.5 m/s.

Explanation:

Given that,

Height = 2.87 m

We need to calculate the minimum speed

Using equation of motion

v^2=u^2-2gh

u=\sqrt{2gh}

Where, u = minimum velocity

g = acceleration due to gravity

h = height

Put the value into the formula

u=\sqrt{2\times9.8\times2.87}

u=7.5\ m/s

Hence, The minimum speed is 7.5 m/s.

5 0
4 years ago
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