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Mazyrski [523]
3 years ago
15

A 1,600 kg train car rolling freely on level track at 16 m/s bumps into a 1.0 × 103 kg train car moving at 10.0 m/s in the same

direction, and the two latch onto each other and continue together. What is their final speed?
A.) 16 m/s
B.) 6.0 m/s
C.) 10 m/s
D.) 14 m/s
Physics
1 answer:
emmainna [20.7K]3 years ago
6 0
This item is solved through the concept of the conservation of momentum which states that the momentum before and after collision should be equal. 
                                   momentum = mass x velocity
            (1,600 kg)(16 m/s) + (1.0x10^3 kg)(10 m/s) = (1600 + 1000 kg)(x)
The value of x is 13.69 m/s. Thus, their final speed is approximately letter D. 14 m/s. 
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5 0
1 year ago
What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 22 km/h and the coef
Aleks [24]
First, let's put 22 km/h in m/s:

22 \frac{km}{h} \times  \frac{1000m}{1km}  \times  \frac{1h}{3600s}=6.11 \frac{m}{s}

Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by

F_{rad}=m \frac{v^2}{R}

The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:

F_{fric}= mg \mu_{s}

Notice we don't use the kinetic coefficient even though the bike is moving.  This is because when the tires meet the road they are momentarily stationary with the road surface.  Otherwise the bike is skidding.

Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:

m\frac{v^2}{R} = mg \mu_{s} \\ \\ \frac{v^2}{R} = g \mu_{s} \\ \\R= \frac{v^2}{g \mu_{s}} \\ \\ R= \frac{6.11}{9.8 \times 0.37}=1.685m

3 0
3 years ago
An electron moves with a speed of 8.0×106m/s along the -z-axis. It enters a region where there is a uniform magnetic field B = (
Crazy boy [7]

Answer:

Acceleration, a=9.36\times 10^{18}\ m/s^2

Explanation:

It is given that,

Speed of electron, v=8\times 10^6\ m/s

Charge on an electron, q=1.6\times 10^{-19}\ C

Mass of electron, m=9.1\times 10^{-31}\ kg

Magnetic field, B=5.5i-3.7j

Magnitude, |B|=\sqrt{5.5^2+(-3.77)^2}=6.66\ T

Magnetic force is given by :

F=qvB

Also, F = ma

a=\dfrac{qvB}{m}

a=\dfrac{1.6\times 10^{-19}\times 8\times 10^6\times 6.66}{9.1\times 10^{-31}}

a=9.36\times 10^{18}\ m/s^2

So, the acceleration of the electron is 9.36\times 10^{18}\ m/s^2. Hence, this is the required solution.

5 0
3 years ago
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