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3 years ago

2) Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from

Physics

1 answer:

OLEGan [10]3 years ago

4 0

Answer:

$Relative\ Velocity = 105m/s$

Explanation:

Given

$V_A = 45m/s$

$V_B = 60m/s$

Required

Determine the speed of B w.r.t A

The question implies that, we determine the relative velocity of B w.r.t A

Because both trains are moving towards one another, the required velocity is a $sum\ of\ velocities\ of$ both trains:

This is shown below:

$Relative\ Velocity = V_A + V_B$

$Relative\ Velocity = 45m/s + 60m/s$

$Relative\ Velocity = 105m/s$

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In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the met

mixer [17]

Answer:

The kinetic energy of the ejected electrons increases.

Explanation:

As we know that electrons are only ejected from a metal surface if the frequency of the incident light increases the work function of the metal. If the frequency of the incident light is less than the work function of the metal no matter how intense the beam the electrons will not be ejected from the surface.

Using conservation of energy principle we have

$E_{incident}=h\nu +\frac{1}{2}mv^{2}$

If we increase the intensity of incident light the term on the LHS of the above equation increases this increase appears in the kinetic energy term in RHS of the equation since $h\times \nu$ remains constant.

5 0

4 years ago

Read 2 more answers

An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th

neonofarm [45]

Answer:

Electric force, $F=2.24\times 10^{-14}\ N$

Explanation:

It is given that,

Charge on an electron is $-1.6\times 10^{-19}\ C$

Electric field, $E=1.4\times 10^5\ N/m$

We need to find the magnitude of the electric force on this electron due to this field. The electric force is given by :

$F=qE\\\\F=1.6\times 10^{-19}\times 1.4\times 10^5\\\\F=2.24\times 10^{-14}\ N$

So, the electric force is $2.24\times 10^{-14}\ N$ .

6 0

3 years ago

A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are

igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

$\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2$

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

$I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2$

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

$T = I\alpha = 0.303*0.6454 = 0.196 Nm$

So the force acting on the other end to generate this torque mush be:

$F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N$

4 0

3 years ago

A. Group of cross country runners decided to go on an hour and a half run. During the first hour, they ran a total of 13 kilomet

pshichka [43]

Answer:

The average speed for the entire run is 12 km/h.

Explanation:

The average speed is given by the following equation:

$\overline{v} = \frac{d_{T}}{t_{T}}$

Where:

$d_{T}$ : is the total distance

$t_{T}$ : is the total time

If during the first hour, they ran a total of 13 kilometers and then, they ran 5.0 kilometers during the next half an hour we have:

$d_{T} = 13 km + 5 km = 18 km$

$t_{T} = 1 h + \frac{1}{2} h = 1.5 h$

Hence, the average speed is:

$\overline{v} = \frac{d_{T}}{t_{T}} = \frac{18 km}{1.5 h} = 12 km/h$

Therefore, the average speed for the entire run is 12 km/h.

I hope it helps you!

3 0

3 years ago

Find the ratio of the lengths of the two mathematical pendulums, if the ratio of periods is 1.5

juin [17]

Answer:

The ratio of lengths of the two mathematical pendulums is 9:4.

Explanation:

It is given that,

The ratio of periods of two pendulums is 1.5

Let the lengths be L₁ and L₂.

The time period of a simple pendulum is given by :

$T=2\pi \sqrt{\dfrac{l}{g}}$

$T^2=4\pi^2\dfrac{l}{g}\\\\l=\dfrac{T^2g}{4\pi^2}$

Where

l is length of the pendulum

$l\propto T^2$

$\dfrac{l_1}{l_2}=(\dfrac{T_1}{T_2})^2$ ....(1)

ATQ,

$\dfrac{T_1}{T_2}=1.5$

Put in equation (1)

$\dfrac{l_1}{l_2}=(1.5)^2\\\\=\dfrac{9}{4}$

So, the ratio of lengths of the two mathematical pendulums is 9:4.

3 0

3 years ago