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poizon [28]
3 years ago
14

Which of the following statements describes a compound machine?

Physics
2 answers:
ASHA 777 [7]3 years ago
6 0
From the word compound, the compound machine is already a combination of two or more types of simple machine. Thus, the answer is letter C. Because of its complexity, it is able to perform several other functions than a simple one. 
larisa86 [58]3 years ago
6 0

It is C. It is made up of more than one simple machine

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Date<br>Page<br>What are the advantages of alcohol<br>thermometric liquid?<br>​
Leya [2.2K]

Answer:

low freezing point. high vapour pressure.

<em>HOPE</em><em> </em><em>IT</em><em> </em><em>WILL</em><em> </em><em>HELP</em><em> </em><em>U</em><em>! </em><em>!</em><em>!</em><em>!</em><em>!</em><em>!</em>

3 0
3 years ago
How are the different firework shapes produced
AleksAgata [21]

To create the shapes, stars are arranged on a piece of cardboard in the desired configuration. If the stars are placed in a smiley face pattern on the cardboard, for example, they will explode into a smiley face in the sky. In fact, you may see several smiley faces in the sky at one time.

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8 0
3 years ago
A fixed mass of an ideal gas is heated from 50°C to 80°C (a) at constant volume and (b) at constant pressure. For which case do
soldi70 [24.7K]

Answer:

Specific heat at constant pressure is =  1.005 kJ/kg.K

Specific heat at constant volume is =  0.718 kJ/kg.K

Explanation:

given data

temperature T1 =  50°C

temperature T2 = 80°C

solution

we know energy require to heat the air is express as

for constant pressure and volume

Q  = m ×  c × ΔT     ........................1

here m is mass of the gas and c is specific heat of the gas and Δ T is change in temperature of the gas

here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.

and here at constant pressure Specific heat  is greater than the specific heat at constant volume,

so the amount of heat required to raise the temperature of one unit mass by one degree at constant pressure is

Specific heat at constant pressure is =  1.005 kJ/kg.K

and

Specific heat at constant volume is =  0.718 kJ/kg.K

3 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
State the law of conservation of linear momentum using Newton's third law of deduce this ​
MArishka [77]

Answer:

Derivation of Conservation of Momentum

Applying Newton's third law, these two impulsive forces are equal and opposite i.e. is equal to the change in momentum of the first object. is equal to the change in momentum of the second object. This relation suggests that momentum is conserved during the collision.

Explanation:

Hope it helps!!!

7 0
2 years ago
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