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3 years ago

A 4000kg truck has a head-on inelastic collision with a 2500kg truck.

Physics

1 answer:

iogann1982 [59]3 years ago

7 0

Answer:it could be B

Explanation:

im not sure

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What is the energy equivalent of an object with a mass of 1.83 kg?

xxMikexx [17]

To determine the energy equivalent of an object, we use the famous equation of Einstein which is E=mc^2 where m is the mass of the object and c is the speed of light (3x10^8 m/s). We calculate as follows:

E = mc^2
E = 1.83 kg (3x10^8 m/s)^2
E = 1.647x10^17 J

6 0

3 years ago

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Canola oil is less dense than water, so it floats on water, but its index of refraction is 1.47, higher than that of water. When

kupik [55]

Answer:

therefore critical angle c= 69.79°

Explanation:

Canola oil is less dense than water, so it floats over water.

Given $n_{canola}= 1.47$

which is higher than that of water

refractive index of water $n_{water}=1.33$

to calculate critical angle of light going from the oil into water

we know that

$sinc= \frac{n_{water}}{n_{canola}}$

now putting values we get

$sinc= \frac{1.33}{1.47}$

c= $sin^{-1}(\frac{1.33}{1.47} )$

c=69.79°

therefore critical angle c= 69.79°

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3 years ago

You are walking up an icy slope. suddenly your feet slip, and you start to slide backward. will you slide at a constant speed, o

Aleks04 [339]

You will accelerate down the slope.

6 0

3 years ago

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Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Bl

shepuryov [24]

Answer:

b) TA = TB = TC

Explanation:

When put in contact each other, and isolated, both blocks will exchange heat till they reach to thermal equilibrium.
During this process, the body at a higher temperature, will loss heat, tat it will be gained by the other body.
The equilibrium condition will be reached when the following equation be met:

$\Delta Q = c_{st}* m_{A} * (T_{fin} - T_{0A} ) = c_{st}* m_{B} * (T_{0B} - T_{fin} )$

Replacing by the values of T₀A = 300º C, and T₀B = 400ºC, and simplifying common terms as mA = mB, we can solve for Tfin, as follows:

$(400 \ºC - T_{fin}) = (T_{fin} - 300 \ºC) \\ \\ 2* T_{fin} = 700\ºC\\ \\ T_{fin} = 350\ºC$

So, when both blocks reach to equilibrium, they will be at a common final temperature, 350ºC.
When put in contact with block C, at the same temperature, at that instant, the three blocks will have the same common temperature of 350 ºC.
So, option b) is the right one.

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3 years ago

Durning which type of process does pressure remain consistent

Simora [160]

Answer:

$\fbox {D. Isobaric}$

Explanation:

The process during which pressure remains constant is called an isobaric process.

6 0

2 years ago

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