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AveGali [126]
3 years ago
15

Figure 2.31 shows how the velocity of four cars changed as they

Physics
2 answers:
masya89 [10]3 years ago
8 0
I think it’s c but not sure about a
elena55 [62]3 years ago
7 0
It’s c I think so aro no
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3. When driving a car, the chemical energy in gasoline is converted to heat (thermal energy) in the engine. Most of this thermal
tekilochka [14]

Explanation:

As we know that the maximum of thermal energy will be converted into mechanical energy.( The thermal energy will be given to us by engine.)

But the total thermal energy will not be converted into the mechanical energy,  there will be some lost of energy. The lost of mechanical energy will be in multiple forms. 2 of them are

1.Heat : The mechanical energy heats up the engine and then it will lost some energy.

2.Noise : Some energy also lost in the sound or the noise comes out from the engine. It will consume some energy.

Also, the engine never be 100% efficient some amount of energy lost in environment also.

6 0
3 years ago
The immersed volume of body in a liquid depends on density of the liquid. TRUE or FALSE?​
miss Akunina [59]

Answer:

True - Archimedes Principle states that the buoyant force on an object is equal to the weight of the liquid displaced.

8 0
2 years ago
Gravity can be defined as a force of attraction between objects. The effects of gravity can be witnessed in many systems. Which
murzikaleks [220]
C, planets orbit around the sun because of gravity
5 0
3 years ago
Where is the chart option found in the Excel application?
vodomira [7]

Answer:

its under the insert table

Explanation:

5 0
4 years ago
Calculate the wavelength and frequency at which the intensity of the radiation is a maximum for a blackbody at 298 K. You will n
Juliette [100K]

Answer:

Wavelength, \lambda=9.72\times 10^{-6}\ m

Frequency, f=3.08\times 10^{13}\ Hz

Explanation:

We need to find the intensity of the radiation is a maximum for a black body at 298 K. It can be calculated using Wein's displacement law. It is given by :

\lambda T=2.898\times 10^{-3}\ m-K

\lambda=\dfrac{2.898\times 10^{-3}}{T}

Here, T = 298 K

\lambda=\dfrac{2.898\times 10^{-3}}{298}

\lambda=9.72\times 10^{-6}\ m

If f is the frequency of black body radiation. It is given by :

f=\dfrac{c}{\lambda}

f=\dfrac{3\times 10^8}{9.72\times 10^{-6}}

f=3.08\times 10^{13}\ Hz

Hence, this is the required solution.                                              

6 0
3 years ago
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