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erastovalidia [21]
3 years ago
9

A small car of mass m and a large car of mass 4m drive along a highway at constant speeds VS and VL. They approach a curve of ra

dius R. The small and large cars have accelerations as and aL respectively, as they travel around the curve. The magnitude of as is twice of that of aL. How does the speed of the small car VS compare to the speed of the large car VL as they move around the curve
Physics
1 answer:
Arte-miy333 [17]3 years ago
4 0

Answer:

v_S=\sqrt{2}v_L

Explanation:

The acceleration experimented while taking a curve is the centripetal acceleration a=\frac{v^2}{r}. Since a_S=2a_L, we have that: \frac{v_S^2}{r_S}=\frac{2v_L^2}{r_L}

They take the same curve, so we have: r_S=r_L=R

Which means: v_S^2=2v_L^2

And finally we obtain: v_S=\sqrt{2}v_L

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At this radius, what is the magnitude of the net force that maintains circular motion exerted on the pilot by the seat belts, th
Ainat [17]

Answer:

Fc=5253 N

Explanation:

Answer:

Fc=5253 N

Explanation:

sequel to the question given, this question would have taken precedence:

"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."

so we derive centripetal acceleration first

ac (centripetal acceleration) = v^2/r

make r the subject of the equation

r= v^2/ac

 ac is 6.23*g which is 9.81

v is 101m/s

substituing the parameters into the equation, to get the radius

(101^2)/(6.23*9.81) = 167m

Now for part

( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.

he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.

Fc (Centripetal Force) = m*v^2/r  

So (86kg* 101^2)/(167) =

Fc=5253 N

4 0
3 years ago
Calculate the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s.
Aleonysh [2.5K]

De broglie wavelength, \lambda = \frac{h}{mv}, where h is the Planck's constant,  m is the mass and v is the velocity.

h = 6.63*10^{-34}

Mass of hydrogen atom,  m = 1.67*10^{-27}kg

v = 440 m/s

Substituting

   Wavelength \lambda = \frac{h}{mv} = \frac{6.63*10^{-34}}{1.67*10^{-27}*440} = 0.902 *10^{-9}m = 902 *10^{-12}m

1 pm = 10^{-12}m\\ \\ So , \lambda =902 pm

So  the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm

7 0
3 years ago
Complete the sentences below using the words below, you may use each word more than once
satela [25.4K]
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3 years ago
What is density?
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<span>a. the amount of matter in a given volume </span>
8 0
3 years ago
Read 2 more answers
Suppose a blood vessel's radius is decreased to 87% of its original value by plaque deposits and the body compensates by increas
dexar [7]

Answer:

The pressure difference will increase by the factor of 1.75

Explanation:

For constant flow rate, coefficient of viscosity, length of the vessel and the pressure difference is inversely proportional to the fourth power of the radius of the blood vessel

Apply the principle of Poiseuille’s law.

Q = (P2 - P1)/R

Pls check the attached file for step by step solution of the question. It is submitted in this way as typing the equation may not be explanatory.

3 0
3 years ago
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