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2 years ago

14

A 10kg oil is heated by a source of heat that supplies 800J of heat every minute. Calculate The heat added to oil after 30 minut

es.

1 answer:

goblinko [34]2 years ago

4 0

Answer:

4,200 joules per kilogram per degree Celsius

Explanation:

The specific heat capacity of a material is the energy required to raise one kilogram (kg) of the material by one degree Celsius (°C). The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.

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Artist 52 [7]

Answer:

I don't do physics , I'm sorry can't help you

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3 years ago

Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

<h3>a.</h3>

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

so

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

<h3>b.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

<h3>c.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

5 0

2 years ago

A cement truck of mass 14,000 kg moving 10m/s slams into a wall and comes to a halt in .2s. What is the force of impact on the t

skelet666 [1.2K]

And it’s x10 is 100000

6 0

3 years ago

In comparison with other ocean basins, major sedimentary features such as continental rises and abyssal plains are relatively ra

Maksim231197 [3]

Answer:

Why are continental rises and abyssal plains relatively rare in the Pacific? This is because the extensive system of trenches along the active margins of the Pacific, trap much of the sediments flowing off the continents, preventing them from building the broad, flat abyssal plains typical of the Atlantic ocean basins.

3 0

2 years ago

A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.

alexdok [17]

Here when an object is placed on the level floor then in that case there are two forces on the object

1). Weight of object downwards (mg)

2). Normal force due to floor which will counterbalance the weight (N)

so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

$F_g = F_n + F_a$

given that

$F_g = mg$

here

m = 30 kg and

g = acceleration due to gravity = 10 m/s^2

$F_n = 150 N$

now from above equation

$30*10 = 150 + F_a$

$F_a = 300 - 150 = 150 N$

So force applied by the person must be 150 N

7 0

3 years ago