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n200080 [17]
4 years ago
11

Problem I Marcella (see warmup problem, above) gets her car moving steadily at 4m/s but suddenly someone stops ahead to assist h

er and parks their car 14 meters from the front of her car. Marcella grabs the car bumper and pulls very hard, with 200 N of force. The work she does transfers energy out, it reduces the K of the car, as it gradually approaches the other car. a) What is the initial kinetic energy before she tries to stop the car? b) What is the final kinetic energy, when her car hits the other car? What is the speed? c) Suppose the other person also slowed her car, pushing it from the front. How much force would be needed to stop her car 1 meter from the other car? [1 m allows the person not to be crushed!]
Physics
1 answer:
Nady [450]4 years ago
3 0

Answer:

Explanation:

a) KE = (1/2) * m * (v^{2}) = F * d = 14m * 200N = 2800 m/N or 2.8 * 10^{3} m/N

b) 0J and 0m/s (if Marcella stopped after going 14 meters)

c)  Known from part (a) that KE = 2800 J = F1 * d1,

    2800J = F1 * (14m - 1m)  => F1 = 2800J/13m = 215.384 N

   

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polet [3.4K]

Answer: 71.16\ Hz

Explanation:

Given

Capacitance C=100\ \mu F

Resistance R=500\ \Omega

Inductance L=50\ mH

In LCR circuit, current is maximum at resonance frequency i.e.

X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}

Insert the values

\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}

Also, frequency is given by

\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}

\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz

8 0
3 years ago
A car drives to the east in a time of 6 hours. Then, immediately (not realistic, but just assume this is the case for this probl
garri49 [273]

Answer:

Average speed of car in the first trip is 10 km/hr    

Explanation:

It is given that first the car drives 6 hours to the east

Then travels 12 km to west in 3 hours

Average speed for the entire trip = 8 km/hr

Total time = 3+6 = 9 hour

So distance traveled in 9 hour = 9×8 = 72 km

As the car travel 12 km in west so distance traveled in east = 72-12 = 60 km

Time by which car traveled in east = 6 hour

So speed =\frac{distance}{time}=\frac{60}{6}=10km/hr

So average speed of car in the first trip is 10 km/hr

7 0
3 years ago
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The spring to launch a pinball in a pinball machine is compressed 25 cm and has a spring constant of 140 N/m.
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Calculate Neptune's mass given the acceleration due to gravity at the north pole is 11.529 m/s2 and the radius of Neptune at the
dolphi86 [110]

Answer:

The mass of Neptune is 1.023\times 10^{26} kilograms.

Explanation:

From Newton's Law of Gravitation, the gravitational acceleration of Neptune is determined by the following formula:

g = \frac{G\cdot M}{R^{2}} (1)

Where:

G - Gravitational constant, measured in cubic meters per kilogram-square second.

M - Mass of the planet, measured in kilograms.

R - Radius of the planet, measured in meters.

g - Gravitational acceleration, measured in meters per square second.

If we know that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, g = 11.529\,\frac{m}{s^{2}} and R = 24.340\times 10^{6}\,m, then the mass of Neptune is:

M = \frac{g\cdot R^{2}}{G}

M = \frac{\left(11.529\,\frac{m}{s^{2}}\right)\cdot (24.340\times 10^{6}\,m)^{2} }{6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} }

M = 1.023\times 10^{26}\,kg

The mass of Neptune is 1.023\times 10^{26} kilograms.

5 0
3 years ago
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