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storchak [24]
4 years ago
12

A horizontal pull A pulls two wagons over a horizontal frictionless floor, the first wagon is 500N, the second is 2000 N. The te

nsion in the light horizontal rope connecting the wagons is.
help explain your answer,
Physics
1 answer:
MrMuchimi4 years ago
5 0
<span>The force will be zero if the wagons are moving at a constant speed (i.e. not accelerating), as there is no frictional force to overcome. If the wagons are accelerating, the force will be proportional to the acceleration, and 20% of the force applied by A.</span>
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A sled with a mass of 25.0 kg rests on a horizontal sheet of essentially frictionless ice. It is attached by a 5.00-m rope to a
Tpy6a [65]

Answer:

The value is  F  =  34.3 \  N

Explanation:

From the question we are told that

    The mass of the shed is  m  =  25 \  kg

    The length of the rope is  l = 5.0 \  m

    The angular speed of the shed is w = 5 rev/minute = \frac{2* \pi * 5}{60 }= 0.524 \  rad/sec

Generally the force exerted on the shed is mathematically represented as

     F  =  m  *  w^2 *  l

=>  F  =  25  *  0.524^2 *  5

=>  F  =  34.3 \  N    

7 0
3 years ago
How much work is done when a girl applies a force of 80 N to move an object 4 m, and then moves the object 1 m back?
eduard

Answer:

w = \underline{{ \boxed{400 \:  J }}}

Explanation:

work \: done =  \boxed{force \times distance} \\ w = 80 \times (4 + 1) = 80 \times 5 \\ w = \underline{{ \boxed{400 \:  J }}}

7 0
3 years ago
8
aev [14]
OD because Boyle’s law specifically states
4 0
3 years ago
From the data table below, which sample shows the largest decrease in average kinetic energy?
Phantasy [73]
D. sample 4 would be your answer
3 0
2 years ago
Uma ema pesa aproximadamente 360 N e consegue desenvolver uma velocidade de 60 km/h, o que lhe confere uma quantidade de movimen
anastassius [24]

Vê quantidade de movimento precisa ta em kg.m/s certo?

 

a questão disse que a gravidade é de 10 int posso dizer q minha massa é de 36kg pois p(360n)= massa . gravidade(10).

 

ja a velocidade esta km/h int vou dividir por 60 por 3,6 ou 600 por 36, aumentando de 10 vezes pra poder ajudar no calculo.

 

quantidade de movimento é  Q= m(36) . velocidade( 600/36) 

 

Q= 36.600/36 se corta 36 de cima com o 36 de baixo vai ficar 600kg . m/s



4 0
3 years ago
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