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drek231 [11]
3 years ago
8

А A van accelerates from Amst to zomst in 8s. How far does ittravel inthis time?​

Physics
1 answer:
8_murik_8 [283]3 years ago
6 0

Answer:

1663m

Explanation:

mark brainliest

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How many significant figures does the following number have: 0.002040?
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Answer: 4

Explanation: because 0s aren’t significant and after the decimal point, there was to be a value greater than 0 than the rest are sig figs.

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What is the acceleration of a 4,000 kg car pushed with a<br> force of 12,000 N?
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Consider a hydrogen atom in the n = 1 state. The atom is placed in a uniform B field of magnitude 2.5 T. Calculate the energy di
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Answer:

E=29\times 10^{-5}eV

Explanation:

For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,

E^{+}=E+\mu B,

E^{-}=E-\mu B,

E^{0}=E

Here, E is the energy in the absence of electric field.

And

E^{+} and E^{-} are the highest and the lowest energies.

The difference of these energies

\Delta E=2\mu B

\mu=9.3\times 10^{-24}J/T is known as Bohr's magneton.

B=2.5 T,

Therefore,

\Delta E=2(9.3\times 10^{-24}J/T)\times 2.5 T\\\Delta E=46.5\times 10^{-24}J

Now,

Delta E=46.5\times 10^{-24}J(\frac{1eV}{1.6\times 10^{-9}J } )\\Delta E=29.05\times 10^{-5}eV\\Delta E\simeq29\times 10^{-5}eV

Therefore, the energy difference between highest and lowest energy levels in presence of magnetic field is E=29\times 10^{-5}eV

6 0
3 years ago
A car maintains a constant speed v as it traverses the hill and valley as shown below. Both the hill and valley have a radius of
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Answer:

As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.

(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.

(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.

(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.

(iv) At point A, the driver will feel the lightest.

(v)The car can go that much fast without losing contact with the road at A can be determined as follow:

Fn=0 - lose contact with road

Fg= mv²/r

mg=mv²/r

v=sqrt (gr)

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