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pav-90 [236]
3 years ago
5

How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and

Physics
1 answer:
aev [14]3 years ago
4 0

Answer:

h= 46.66 m

Explanation:

Given that

Initial speed of the car ,u = 110 km/h

We know that

1 km/h= 0.277 m/s

u= 30.55 m/s

lets height gain by car is h.

The final speed of the car will be zero at height h.

v²=u²- 2 g h

v= 0 m/s

0²=30.55²- 2 x 10 x h           ( g = 10 m/s²)

h= 46.66 m

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Answer:

Pressure

Explanation:

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There is no loss of energy in the collisions involving the gas molecules and that is why their movement can be described as elastic. The descriptions of the behavior of gas molecules in the Kinetic Molecular Theory, give rise to Charles law, Boyle's Law, Avogadro's Laws, Dalton's Law, and Amonton's Law.

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3 years ago
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2 years ago
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Question 10 of 10
Margaret [11]

Answer: D. 5.0(10)^{-38} J

Explanation:

The energy E of an electromagmetic wave is given by:

E=\frac{hc}{\lambda}

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h=6.626(10)^{-34}J.s is the Planck constant

c=3(10)^{8} m/s is the speed of light in vacuum

Then hc=1.99 (10)^{-25}Jm

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E=\frac{1.99 (10)^{-25}Jm}{4.0(10)^{12} m}

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7 0
3 years ago
A professor sits on a rotating stool that is spinning at 10.0 rpm while she holds a heavy weight in each of her hands. Her outst
USPshnik [31]

Answer:

0.3950m

Explanation:

Use conservation of angular momentum:

Let L be the angular momentum(a vector).

We know that:p=mv, L=l\omega, \ \ \ l=\sum (m_ir_i^2)} and that the two masses have the same radius:

L_o=2mr_o^2\omega_o=L_f=2mr_f^2\omega_f\\\\r_o=0.795m,\omega_o=10\times2\pi \ rad/m , \omega_f=40.5\times 2\pi \ rad/m\\\\\therefore r_f=r_o\sqrt{(w_o/w_f)}\\\\r_f=0.795\sqrt{(10.0/40.5)}\\\\r_f=0.3950m

Hence, the weights are 0.3950m away .

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3 years ago
Dropping a stone from a height is work or not?​
kvasek [131]

Answer:

Dropping a stone from a height is work  

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