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3 years ago

Ben and Dan both weigh 600 N. They are doing pull-ups

Physics

1 answer:

ivann1987 [24]3 years ago

6 0

Explanation:

Given parameters:

Weight of Ben = 600N

Weight of Dan = 600N

Distance of pull-up = 0.5m

Unknown:

Work done by each pull - up = ?

Solution:

Work done is defined as the product of the force applied to move a body through a certain distance.

Work done = Force x distance

Work done by Ben And Dan will be the same:

Work done = 600 x 0.3 = 180J

Work done by Ben = 180J

Work done by Dan = 180J

learn more:

Work done brainly.com/question/9100769

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Course hero N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches

aleksandrvk [35]

Answer:

Therefore it is save to carry a 62kg adult

Explanation:

From the question we are told that:

Mass $m=380kg$

Height of supporting Rock $X=85cm$

Length of Board $L_r=4.5m$

Mass of board $M_b=22kg$

Mass of adult $M_a=62$

Generally the moment of balance about wedge part about is mathematically given by

$N -Q + R = Mg + mg$

$0.85*N - Mg*2.25 - mg*(2.25 + x) = 0$

$0.85*N = + Mg*2.25 + mg*(2.25 + x)$

where

$N+R=4547$

therefore

$N = 570.70588 + 1608.3529 + 714.823 x$

if N=0 at fallen person

$x=3.04m$

Therefore it is save to carry a 62kg adult

8 0

2 years ago

Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water

tangare [24]

Answer:

$T_b=107.3784\ ^{\circ}C$

Explanation:

Given:

thickness of the base of the kettle, $dx=0.52\ cm=5.2\times 10^{-3}\ m$

radius of the base of the kettle, $r=0.12\ m$

temperature of the top surface of the kettle base, $T_t=100^{\circ}C$

rate of heat transfer through the kettle to boil water, $\dot Q=0.409\ kg.min^{-1}$

We have the latent heat vaporization of water, $L=2260\times 10^3\ J.kg^{-1}$

and thermal conductivity of aluminium, $k=240\ W.m^{-1}.K^{-1}$

So, the heat rate:

$\dot Q=\frac{0.409\times 2260000}{60}$

$\dot Q=15405.67\ W$

From the Fourier's law of conduction we have:

$\dot Q=k.A.\frac{dT}{dx}$

$\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}$

where:

$A=$ area of the surface through which conduction occurs

$T_b=$ temperature of the bottom surface

$15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}$

$T_b=107.3784\ ^{\circ}C$ is the temperature of the bottom of the base surface of the kettle.

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3 years ago

1. Spring force is proportional to stretch distance, F=kd. Do you expect doubling the stretch distance double the force?

Solnce55 [7]

Answer:

Yes

Explanation:

The spring force is given as:

F = kd

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K is the spring constant

d is the magnitude of the stretch

Since k is a constant, therefore, doubling the stretch distance will double the force.

Both stretch distance and force applied can be said to be directly proportional to one another.

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lisov135 [29]

Answer:

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Hope this helped!

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3 years ago