A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling

Physics

1 answer:

Setler [38]2 years ago

6 0

m1=1500kg
m_2=3000kg
v_1=5m/s
v_2=7m/s

Using law of conservation of momentum

$\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3$

$\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3$

$\\ \sf\Rrightarrow 7500-21000=4500v_3$

$\\ \sf\Rrightarrow -13500=4500v_3$

$\\ \sf\Rrightarrow v_3=-3m/s$

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In this circuit (see picture), which resistor will draw the least power?

Basile [38]

A few different ways to do this:

Way #1:
The current in the series loop is (12 V) / (total resistance) .
(Turns out to be 2 Amperes, but the question isn't asking for that.)

In a series loop, the current is the same at every point, so it's
the same current through each resistor.

The power dissipated by a resistor is (current)² · (resistance),
and the current is the same everywhere in the circuit, so the
smallest resistance will dissipate the least power. That's R1 .

And by the way, it's not "drawing" the most power. It's dissipating it.

Way #2:
Another expression for the power dissipated by a resistance is

(voltage across the resistance)² / (resistance) .

In a series loop, the voltage across each resistor is

[ (individual resistance) / (total resistance ] x battery voltage.

So the power dissipated by each resistor is

(individual resistance)² x [(battery voltage) / (total resistance)²]

This expression is smallest for the smallest individual resistance.
(The other two quantities are the same for each individual resistor.)
So again, the least power is dissipated by the smallest individual resistance.
That's R1 .

Way #3: (Einstein's way)
If we sat back and relaxed for a minute, stared at the ceiling, let our minds
wander, puffed gently on our pipe, and just daydreamed about this question
for a minute or two, we might have easily guessed at the answer.

===> When you wire up a battery and a light bulb in series, the part
that dissipates power, and gets so hot that it radiates heat and light, is
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3 0

2 years ago

A student on an amusement park ride moves in a circular path with a radius of 3.5 meters once every 8.9 seconds. What is the ave

Ulleksa [173]

Distance traveled by him = circumference of that circular path = 2πr = 2π(3.5)
= 7π = 7×3.14 = 21.98 m
time = 8.9 s [ Given ]

Now, Average speed = distance / time
s = 21.98 / 8.9
s = 2.46 m/s

Hope this helps!

7 0

2 years ago

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

C) $emf=5.4090\times 10^{-4}V$

Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

$\Phi=B.a\,cos \theta$ ..................................(1)

$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.