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wlad13 [49]
4 years ago
12

A policeman starts giving chase 60 seconds after a stolen car zooms by at 108 km/hr. At what minimum speed should he drive if he

has to catch up with the driver before he manages to get onto an expressway 60 km away?
Physics
1 answer:
Iteru [2.4K]4 years ago
6 0

Answer:

30.93 m/s

Explanation:

Given that, the speed of stolen car is,

v_{s} =108km/hr\\v_{s} =108\times \frac{5}{18}m/s\\ v_{s} =30m/s

As policeman start chasing the stolen car after 60 seconds.

Now suppose the speed of policeman car is, v_{p}

The policeman catches the stolen car at a distance of,

S=60km\\S=60000m

Now the distance covered by the policeman in time t is v_{p}\times t

And the distane cover by the thief in stolen car in time(t+60s) is v_{s}\times (t+60sec).

And these distances are equal and they are equal to 60000 m.

Therefore,

v_{p}\times t=v_{s}\times (t+60sec)=60000m

Therfore,

v_{s}\times (t+60sec)=60000m\\30m/s\times (t+60sec)=60000m\\(t+60s)=2000s\\t=1940s

Now use this value to solve for minimum speed of policeman's car.

v_{p}\times 1940=60000\\v_{p}=30.93 m/s

Therefore minimum speed of policeman's car is 30.93 m/s.

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Answer:

492.183 nm

Explanation:

x = Distance from the central maximum to the first minimum = 1.35 mm

l = Distance of screen = 2.13 m

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We have the relation

tan\theta=\dfrac{x}{l}\\\Rightarrow \theta=tan^{-1}\dfrac{1.35\times 10^{-3}}{2.13}\\\Rightarrow \theta=0.04^{\circ}

dsin\theta=m\lambda\\\Rightarrow \lambda=\dfrac{dsin\theta}{m}\\\Rightarrow \lambda=\dfrac{0.705\times 10^{-3}\times sin0.04}{1}\\\Rightarrow \lambda=4.92183\times 10^{-7}=492.183\ nm

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Two stationary positive point charges, charge 1 of magnitude 3.05 nC and charge 2 of magnitude 1.85 nC, are separated by a dista
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To solve this problem we will apply the concepts related to voltage as a dependent expression of the distance of the bodies, the Coulomb constant and the load of the bodies. In turn, we will apply the concepts related to energy conservation for which we can find the speed of this

V = \frac{kq}{r}

Here,

k = Coulomb's constant

q = Charge

r = Distance to the center point between the charge

From each object the potential will be

V_1 = \frac{kq_1}{r_1}+\frac{kq_2}{r_2}

Replacing the values we have that

V_1 =  \frac{(9*10^9)(3.05*10^{-9})}{0.41/2}+\frac{(9*10^9)(1.85*10^{-9})}{0.41/2}

V_1 = 215.12V

Now the potential two is when there is a difference at the distance of 0.1 from the second charge and the first charge is 0.1 from the other charge, then,

V_1 =  \frac{(9*10^9)(3.05*10^{-9})}{0.1}+\frac{(9*10^9)(1.85*10^{-9})}{0.41-0.1}

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Applying the energy conservation equations we will have that the kinetic energy is equal to the electric energy, that is to say

\frac{1}{2} mv^2 = q(V_2-V_1)

Here

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Rearranging to find the velocity

v = \sqrt{ \frac{2q(V_2-V_1)}{m}}

Replacing,

v = \sqrt{ \frac{-2(1.6*10^{-19})(328.2-215.12)}{9.11*10^{-3}}}

v = 6.3*10^6m/s

Therefore the speed final velocity of the electron when it is 10.0 cm from charge 1 is 6.3*10^6m/s

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