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marishachu [46]
2 years ago
8

A weightlifter lifts a 1250-N barbell 2 m in 3 s.How much power was used to lift the barbell?

Physics
1 answer:
Vlad [161]2 years ago
5 0

The power is 833.3 W

Explanation:

First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:

W=(mg)h

where

mg = 1250 N is the weight of the barbell

h = 2 m is the change in height

Substituting,

W=(1250)(2)=2500 J

Now we can calculate the power, which is equal to the work done per unit time:

P=\frac{W}{t}

where

W = 2500 J is the work done

t = 3 s is the time taken

Substituting,

P=\frac{2500}{3}=833.3 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

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The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

                  V_x'=0.8c\\V=0.6c\\m=4*10^5kg

  • Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

                  KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}

  • So, to V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
  • We have an expression from Lorents transformation for relativistic law of addition of velocities as,

                      V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V

  • Substituting values, we get,

          V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c

  • Thus, the KE will be,

              KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J

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A solid, uniform disk of radius 0.250 m and mass 45.2 kg rolls down a ramp of length 5.40 m that makes an angle of 17.0° with th
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Explanation:

Given that,

Radius of the disk, r = 0.25 m

Mass, m = 45.2 kg

Length of the ramp, l = 5.4 m

Angle made by the ramp with horizontal, \theta=17^{\circ}

Solution,

As the disk starts from rest from the top of the ramp, the potential energy is equal to the sum of translational kinetic energy and the rotational kinetic energy or by using the law of conservation of energy as :

(a) mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

h is the height of the ramp

sin\theta=\dfrac{h}{5.4}

h=sin(17)\times 5.4=1.57\ m

v is the speed of the disk's center

I is the moment of inertia of the disk,

I=\dfrac{1}{2}mr^2

\omega=\dfrac{v}{r}

mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}\dfrac{1}{2}mr^2\times (\dfrac{v}{r})^2

gh=\dfrac{1}{2}v^2+\dfrac{1}{4}v^2

gh=\dfrac{3}{4}v^2

9.8\times 1.57=\dfrac{3}{4}v^2

v = 4.52 m/s

(b) At the bottom of the ramp, the angular speed of the disk is given by :

\omega=\dfrac{v}{r}

\omega=\dfrac{4.52}{0.25}

\omega=18.08\ rad/s

Hence, this is the required solution.

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