What are some good biotechnology topics?

A railroad car of mass M moving at a speed v1 collides and couples with two coupled railroad cars, each of the same mass M and m

marshall27 [118]

Answer:

vf = v₁/3 + 2v₂/3

Explanation:

Using the law of conservation of linear momentum,

momentum before impact = momentum after impact

So, Mv₁ + 2Mv₂ = 3Mv (since the railroad cars combine) where v₁ = initial velocity of first railroad car, v₂ = initial velocity of the other two coupled railroad cars, and vf = final velocity of the three railroad cars after impact.

Mv₁ + 2Mv₂ = 3Mvf

dividing through by 3M, we have

v₁/3 + 2v₂/3 = vf

vf = v₁/3 + 2v₂/3

6 0

3 years ago

The battery of a flashlight develops 3 V, and the current through the bulb is 200 mA.What power is absorbed by the bulb?

Verizon [17]

Answer : The power absorbed by the bulb is, 0.600 W

Explanation :

As we know that,

Power = Voltage × Current

Given:

Voltage = 3 V

Current = 200 mA = 0.200 A

Conversion used : (1 mA = 0.001 A)

Now put all the given values in the above formula, we get:

Power = Voltage × Current

Power = 3V × 0.200 A

Power = 0.600 W

Thus, the power absorbed by the bulb is, 0.600 W

3 0

4 years ago

A man is flying in a hot-air balloon in a straight line at a constant rate of 5 feet per second, while keeping it at a constant

icang [17]

Answer:

$x = 220.85 ft$

Explanation:

Let at any moment of time the friend's car is at some horizontal distance "x" from the position of balloon.

Now if the altitude of the balloon is fixed and it is at height "h"

so here we will have

$tan \theta = \frac{h}{x}$

now we know that

initially the angle of the friend's car is 35 degree

so the horizontal distance will be

$x_1 = h cot35$

similarly if the angle after passing the car position is 36 degree

then we have

$x_2 = h cot36$

now the speed of the balloon is constant

so we have

$v = \frac{x_1 + x_2}{\Delta t}$

$5 ft/s = \frac{h cot35 + h cot36}{90 s}$

$5 ft/s = \frac{2.8h}{90}$

$h = 160.45 ft$

so the final position of friend when the angle is 36 degree

$x = \frac{h}{tan36}$

$x = \frac{160.45}{tan36}$

$x = 220.85 ft$

4 0

3 years ago

Four identical masses of 2.5 kg each are located at the corners of a square with 1.0-m sides. What is the net force on any one o

Ghella [55]

Answer:

F=8.0*10^{-10}N

Explanation:

See the attached file for the masses distributions

The force between two masses at distance r is expressed as

$F=\frac{Gm_{1}m_{2} }{r^{2} }\\ G=Gravitional constant \\$

since the masses are of the same value, the above formula can be reduce to

$F=\frac{Gm^{2}}{r^{2} }\\$

using vector notation,Let use consider the force on the lower left corner of the mass due to the upper left side of the mass is

$F_{12} =\frac{Gm^{2}}{r^{2} }j\\$

The force on the lower left corner of the mass due to the lower right side of the mass is

$F_{14} =\frac{Gm^{2}}{r^{2} }i\\$

The force on the lower left corner of the mass due to the upper right side of the mass is

$F_{13} =\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\$

The net force can be express as

$F=\frac{Gm^{2}}{r^{2} }j +\frac{Gm^{2}}{r^{2} }i +\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\\\F=Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}cos\alpha }]i + Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}sin\alpha }]j\\\alpha=45^{0}, G=6.67*10^{-11}Nmkg^{-2}$

if we insert values we arrive at

$F=6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}cos45 }]i + 6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}sin45}]j\\F=5.643*10^{-10}i+5.643*10^{-10}j$