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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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Monochromatic light of variable wavelength is incident normally on a thin sheet of plastic film in air. The reflected light is a
BartSMP [9]

Answer:

thickness t = 528.433 nm

Explanation:

given data

wavelength λ1 = 477.1 nm

wavelength λ2 = 668.0 nm

n = 1.58

solution

we know for constructive interference condition will be

2 × t × μ = (m1+0.5) × λ1     ....................1

2 × t × μ = (m2+0.5) × λ2     ....................2

so we can say from equation 1 and 2

(m1+0.5) × λ1 = (m2+0.5) × λ2

so

\frac{\lambda 2}{\lambda 1} = \frac{m1+0.5}{m2+0.5}     ..............3

put here value and we get  

\frac{668.0}{477.1} = \frac{m1+0.5}{m2+0.5}  

\frac{m1+0.5}{m2+0.5}   = 1.4

\frac{m1+0.5}{m2+0.5}  = \frac{7}{5}   ...................4

so we here from equation 4

m1+0.5  = 7

m1 = 3    .................5

m2+0.5 = 4

m2 = 2    .................6

so now put value in equation  1

2 × t × μ = (m1+0.5) × λ1

2 × t × 1.58 = (3+0.5) ×  477.1

solve it we get

thickness t = 528.433 nm

4 0
2 years ago
What is a wave period?
Kamila [148]

A wave period is the time it takes to complete one cycle


I hope that's help:0

8 0
2 years ago
An 12 N force is applied to a 1 kg object. What is the magnitude of the objects acceleration?
Sauron [17]

Answer:

a=12 m/s²

Explanation:

Newton's second law of motion states that the acceleration of a body is directly proportional to the force applied and takes place in the direction of force.

This can be summarized as: F=ma, where m is the mass of the object on which force F acts. a is the acceleration due to the force applied.

12N= 1kg×a

a=12N/1kg

a=12m/s²

6 0
3 years ago
Since friction is a force,<br> what unit is friction<br> measured In???
dimaraw [331]
It would be measured in newtons :)
4 0
2 years ago
Read 2 more answers
When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) 1.21\times 10^{15}Hz

Explanation:

We have given Wavelength of the light  \lambda = 240 nm

According to plank's rule ,energy of light

E = h\nu = \frac{hc}{}\lambda

E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

E = KE_{max}+\Phi _{0}, here \Phi _0 is work function.

\Phi _{0}=E - KE_{max}= 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}

\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }

\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm

Part (C) In this part we have to find the cutoff frequency

\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz

5 0
3 years ago
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