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3 years ago

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SI unit of specific heat is

1 answer:

mote1985 [20]3 years ago

5 0

Answer:

The S.I unit of heat is Joule .

Hope it helps you :--)

Explanation:

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La resultante de dos fuerzas perpendiculares aplicadas a un mismo cuerpo es 11.18 N y el módulo de una de ellas es de 10 N. ¿Cuá

Dmitrij [34]

Answer:

English?

Explanation:

3 0

3 years ago

Humans first traveled to the Moon in a spacecraft in 1969. What happened as the spacecraft traveled farther from Earth?

horsena [70]

Answer:

The answer is C)The force of gravity from Earth acting on the spacecraft decreased because the distance from Earth increased.

Explanation:

Gravity, a force, is dependent on the mass of the object exerting the gravity and the distance of an outside object from that object. The larger the object, the more gravity it will exert on an outside object. This force decreases as you move away from the object, but it will always still exist and never be equal to 0.

4 0

3 years ago

Read 2 more answers

A rock is lifted by a machine to a height of 10m. If it has a mass of 22 kilograms

zmey [24]

Answer:

2156J

Explanation:

Given parameters:

Height of lift = 10m

Mass = 22kg

Unknown:

Work done by the machine = ?

Solution:

Work done is the force applied to move a body through a certain distance.

So;

Work done = Force x distance

Here;

Work done = mass x acceleration due to gravity x height

Work done = 22 x 9.8 x 10 = 2156J

5 0

3 years ago

Which one of the following is not equivalent to 2.50 miles?

lisabon 2012 [21]

Answer:

Choice C is not equivalent to 2.50 miles.

Explanation:

The given data is now converted into feet, inches, kilometers, yards and centimeters:

mi - ft

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)$

$x = 13200\,ft$

$x = 1.320\times 10^{4}\,ft$ (Choice A)

mi - in

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)\cdot \left(12\,\frac{in}{ft} \right)$

$x = 158400\,in$

$x = 1.584\times 10^{5}$ (Choice B)

mi - km

$x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)$

$x = 4.025\,km$ (Different from Choice C)

mi - yd

$x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi}\right) \cdot \left(\frac{1}{3}\,\frac{yd}{ft} \right)$

$x = 4400\,yd$

$x = 4.40\times 10^{3}\,yd$ (Choice D)

mi - cm

$x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)\cdot \left(100000\,\frac{cm}{km}\right)$

$x = 402500\,cm$

$x = 4.025\times 10^{5}\,cm$ (Choice E)

Choice C is not equivalent to 2.50 miles.

6 0

4 years ago

A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,

muminat

Answer:

The speed of the plank relative to the ice is:

$v_{p}=-0.33\: m/s$

Explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

$m_{g}v_{g}+m_{p}v_{p}=0$ (1)

Where:

m(g) is the mass of the girl
m(p) is the mass of the plank
v(g) is the speed of the girl
v(p) is the speed of the plank

Now, as we have relative velocities, we have:

$v_{g/b}=v_{g}-v_{p}=1.55 \: m/s$ (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

$45v_{g}+168v_{p}=0$

$v_{g}-v_{p}=1.55$

$v_{p}=-0.33\: m/s$

I hope it helps you!

8 0

3 years ago