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Jobisdone [24]
3 years ago
12

A weightlifter is bench-pressing 710 N. He raises the weight 0.65m above his chest. How much work does he do with this lift?

Physics
2 answers:
stepan [7]3 years ago
6 0

The work done is 461.5 J.

<u>Explanation: </u>

According to Kinetics, the work done is calculated as the product of force which has done work, and the displacement it has imposed. Hence,

                     \text {Work done}=F \times s

In the problem, the given data, F = 710 N and s = 0.65 m

By applying the given values to the work done equation, we get as follows,

                     \text { Work done } = 710 \times 0.65 = 461.5 \mathrm{J}

The unit is same as the energy, work done defines when 1 N force acts in a 1 m distance along the force’s direction.

Assoli18 [71]3 years ago
4 0

The work done is 6958 J

Explanation:

The work done by the man to raise the weight is the equal to the increase in gravitational potential energy of the weight, so we can write:

W=mg \Delta h

where

(mg) is the weight, with m being the mass and g the acceleration of gravity

\Delta h is the change in height of the weight

In this problem, we have

mg = 710 N (weight)

\Delta h = 0.65 m

Therefore, the work done is

W=(710)(9.8)=6,958 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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Car A hits car B (initially at rest and of equal mass) frombehind while going 35 m/s. Immediately after the collision, car Bmove
kolezko [41]

Answer:

The fraction of kinetic energy lost in the collision in term of the initial energy is 0.49.

Explanation:

As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.

By definition, the kinetic energy is:

k = 0.5*mV^2

Expressing the initial and final kinetic energy for cars A and B:

ki=0.5*maVa_{i}^2+0.5*mbVb_{i}^2

kf=0.5*maVa_{f}^2+0.5*mbVb_{f}^2

Since the masses are equals:

m=ma=mb

For the known velocities, the kinetics energies result:

ki=0.5*mVa_{i}^2

ki=0.5*m(35 m/s)^2=612.5m^2/s^2*m

kf=0.5*mbVb_{f}^2

kf=0.5*m(25 m/s)^2=312.5m^2/s^2*m

The lost energy in the collision is the difference between the initial and final kinectic energies:

kl=ki-kf

kl = 612.5m^2/s^2*m-312.5 m^2/s^2*m=300 m^2/s^2*m

Finally the relation between the lost and the initial kinetic energy:

kl/ki = 300 m^2/s^2 * m / 612.5 m^2/s^2 * m

kl/ki = 24/49=0.49

7 0
3 years ago
You are trying to climb a castle wall so, from the ground, you throw a hook with a rope attached to it at 24.1 m/s at an angle o
Serhud [2]

Answer:

The value is  h  =  13.2 \  m

Explanation:

From the question we are told that

    The speed of the rope with hook is u =  24 .1 \  m/s

     The angle is  \theta = 65.0^o

      The speed at which it hits top of the wall is  v  =  16.3 m/s

Generally from kinematic equation we have that

      v_y^2  =  u_y ^2 + *  2 (-g)* h

Here h is the height of the wall so

      [16.3 sin (65)]^2 =  [24.1 sin (65)] ^2+   2 (-9.8)* h

=>    h  =  13.2 \  m

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A ball with a mass of 5000 g is floating on the surface of a pool of water.
Vitek1552 [10]

Answer:

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Explanation:

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8 0
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A homing pigeon starts from rest and accelerates uniformly at +4.00 m/s squared for 10.0 seconds. What is its velocity after the
zavuch27 [327]
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We can't say anything about its velocity, because we have
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Explanation :

Dispersion forces are also known as London dispersion forces. It is the weakest force. Also, it is the part of the Van der Waals forces.

(1) This force is exhibited by all atoms and molecules.

(2) These forces are the result of the fluctuations in the electron distribution within molecules or atoms. Due to these fluctuations, the electric field is created. The magnitude of this force is explained in terms of Hamaker constant 'A'.

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(4) Dispersion force magnitude depends on the amount of surface area available for interactions. If the area increases, the size of the atom also increase. As a result, stronger dispersion forces.

So, the false statement is "Dispersion forces always have a greater magnitude in molecules with a greater molar mass".  

4 0
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