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MaRussiya [10]
3 years ago
7

One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p

ound per second D. the proper replacement unit for one joule per second E. the same as one horsepower per second
Physics
1 answer:
Naddik [55]3 years ago
8 0

Answer:

D. the proper replacement unit for one joule per second

Explanation:

When energy is divided by the time the energy was used we get power

P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{mv^2}{t}\\\Rightarrow P=\dfrac{kg(\dfrac{m^2}{s^2})}{s}\\\Rightarrow P=kg(\dfrac{m^2}{s^2})}\times \dfrac{1}{s}\\\Rightarrow P=\dfrac{kgm^2}{s^3}

kg\dfrac{m^2}{s^2}=Joule

P=\dfrac{kgm^2}{s^3}

So, the answer is D. the proper replacement unit for one joule per second

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Answer:

Explanation:

Firstly, we have to define momentum.

Momentum is define as the product of mass and velocity.

That is P = mass×velocity

Also considering the third law of motion which states that: For every action, there is equal and opposite reaction.

Moreso, considering the 2nd law of motion which states that the rate of change in the momentum of a body is equal to the applied force and takes place in the direction of the applied force.

Now, applying P = mass×velocity

They both have same mass and velocity definitely, they will both experience same momentum.

Also from the question, the both share same velocity hence, the will both hit the wall with same velocity meaning the will both feel the same impact from the wall as well. Hence the third law of motion proves this right.

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Un pez llamado PARGO ROJO vive a grandes profundidades. Si se pesca, al salir a la superficie puede tomar el aspecto de la foto
zalisa [80]

Answer:

Hay diversas leyes que podemos usar acá.

Acá sabemos que la vejiga aumenta su tamaño al reducir la presión, esto tiene sentido, pues al haber menos presión, hay menos fuerza que comprime la vejiga, lo que le permite aumentar su volumen.

Acá tenemos una relación inversa de la forma: V = K/P

Una relación inversa donde la presión esta en el denominador y K es un termino que no depende ni del volumen ni de la presión.

Entonces, a medida que aumenta P, el denominador aumenta, por lo que el valor del volumen decrece.

Un ejemplo de una ecuación similar es la del gas ideal, por ejemplo, para un gas ideal dentro de un globo de volumen V para una dada presión P:

V = nRT/P

donde n es el numero de moles, R es la constante termodinámica y T es la temperatura, acá podemos ver que esta ecuación tiene la misma forma fundamental que la escrita arriba.

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Which of the following molecules is correctly paired with its macromolecule class? (2 points)
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Which moon has a thick atmosphere made mostly of nitrogen?
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A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio
Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

F - External force exerted on the sled, measured in newtons.

f - Friction force, measured in newtons.

N - Normal force from the ground on the mass, measured in newtons.

W - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

W = m\cdot g (1)

Where:

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

If we know that m = 50\,kg and g = 9.807\,\frac{m}{s^{2}}, the weight of the sled is:

W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

W = 490.35\,N

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

F_{min,s} = \mu_{s}\cdot W (2)

Where:

\mu_{s} - Static coefficient of friction, dimensionless.

W - Weight of the sled, measured in newtons.

If we know that \mu_{s} = 0.3 and W = 490.35\,N, then the force needed to start the sled moving is:

F_{min,s} = 0.3\cdot (490.35\,N)

F_{min,s} = 147.105\,N

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

F_{min,k} = \mu_{k}\cdot W (3)

Where \mu_{k} is the kinetic coefficient of friction, dimensionless.

If we know that \mu_{k} = 0.1 and W = 490.35\,N, then the force needed to keep the sled moving at a constant velocity is:

F_{min,k} = 0.1\cdot (490.35\,N)

F_{min,k} = 49.035\,N

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

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3 years ago
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