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2 years ago

What is specific heat? List the formula

1 answer:

5 0

An example of a high specific heat is water’s specific heat, which requires 4.184 joules of heat to increase the temperature of 1 gram of water 1 degree Celsius. Scientifically, water’s specific heat is written as: 1 calorie/gm °C = 4.186 J/gm °C.

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An earth satellite travels in a circular orbit at 20,000 mph if the radius of the orbit is 4,300 mi what angular velocity is gen

Bumek [7]

Answer:

0.00129rad/s

Explanation:

The angular velocity is expressed as;

v = wr

w is the angular velocity

r is the radius

Given

v = 20,000 mph

r = 4300mi

Get w;

w = v/r

w = 20000* 0.44704/4300*1609.34

w = 8940.8/6,920,162

w = 0.00129rad/s

Hence the angular velocity generated is 0.00129rad/s

5 0

3 years ago

In a collision that is not perfectly elastic, what happens to the mechanical energy of the system?

joja [24]

C.

Thanks me later, that's my answer.

5 0

3 years ago

You are asked to explain what happens to individual atoms when a chemical change takes place in lab you are doing in class. What

ss7ja [257]

B. They are rearranged.

The First Law of Thermodynamics states that matter and energy can not be created or destroyed.

3 0

3 years ago

Read 2 more answers

If a ball is thrown into the air with a velocity of 36 ft/s, its height (in feet) after t seconds is given by y = 36t − 16t2. fi

Juliette [100K]

Height (y) = 36t - 16t^2, where t = time in seconds (s).
Our height (y) after 1s = 36(1) - 16(1)^2
y = 36 - 16 = 20 ft
So it reached a height of 20 ft during that 1 second, which means that at that 1 second it had a velocity of 20ft/s, since v = d(distance)/t = 20ft/1s

6 0

3 years ago

Ricardo, of mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30 kg canoe. When the canoe is at r

Deffense [45]

Answer:

m=57.65 kg

Explanation:

Given Data

Ricardo mass m₁=80 kg

Canoe mass m₂=30 kg

Canoe Length L= 3 m

Canoe moves x=40 cm

When Canoe was at rest the net total torque is zero.

Let the center of mass is at x distance from the canoe center and it will be towards the Ricardo cause. So the toque around the center of mass is given as

$m_{1}(L/2-x)=m_{2}x+m_{2}(L/2-x)$

We have to find m₂.To find the value of m₂ first we need figure out the value of.As they changed their positions the center of mass moved to other side by distance 2x.

2x=40

x=40/2

x=20 cm

Substitute in the above equation we get

$m_{x}=\frac{m_{1}(L/2-x)-m_{2}x }{L/2+x}\\m_{x}=\frac{80(\frac{3}{2}-0.2 )-30*0.2}{3/2+0.2}\\m_{x}=57.65 kg$

4 0

3 years ago