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3 years ago

what will the stopping distance be a a 3000-kg car if -3000N of force are applied when the car is traveling 10 m/s

Physics

1 answer:

Inga [223]3 years ago

5 0

Answer:

50 m

Explanation:

Acceleration= force/mass

3000/3000=1m/s^-2

Applying equation of motion:

V^2=U^2+2as; V is final velocity, u is initial velocity, a is acceleration and s is the distance covered.

0=10^2 -2*1s;

Solve for s

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Electricity?

Explanation:

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3 years ago

A 57 kg person in a rollercoaster moving through the bottom of a curved track of radius 42.7 m feels a normal force of 995 N. Ho

natita [175]

Answer:

Use Fc centripetal force as positive and W the weight as negative

N = m v^2 / R + m g

v^2 = (N - m g) R / m

v^2 = (995 - 57 * 9.8) 42.7 / 57 = 327 m^2/s^2

v = 18.1 m/s

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2 years ago

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3 years ago

Read 2 more answers

A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when

Marianna [84]

Answer:

A) $P=13.92\ J.s^{-1}$

B) $v=3730.9912\ m.s^{-1}$

C) $v=74.44\ mm.s^{-1}$

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

Distance form the sound source, $s=50\ m$

sound intensity level at the given location, $\beta=114\ dB$

diameter of the eardrum membrane in humans, $d=8.4 \times 10^{-3}\ m$

We have the minimum detectable intensity to the human ears, $I_0=10^{-12}\ W.m^{-2}$

(A)

Now the intensity of the sound at the given location is related mathematically as:

$\beta=10\ log(\frac{I}{I_0} )$ ..........................................(1)

$114=10\ log\ (\frac{I}{10^{-12}} )$

$11.4=log\ I+12\ log\ 10$

$I=0.2512\ W.m^{-2}$

As we know :

$I=\frac{P}{A}$

$0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }$

$P=13.92\ J.s^{-1}$ is the energy transferred to the eardrums per second.

(B)

mass of mosquito, $m=2\times 10^{-6}\ kg$

Now the velocity of mosquito for the same kinetic energy:

$KE=\frac{1}{2} m.v^2$

$13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2$

$v=3730.9912\ m.s^{-1}$

(C)

Given:

Sound intensity, $\beta = 20\ dB$

Using eq. (1)

$20=10\ log\ (\frac{I}{10^{-12}} )$

$2=log\ I+12\ log\ 10$

$I=10^{-10}\ W.m^{-2}$

Now, power:

$P=I.A$

$P=10^{-10}\times \pi\times \frac{8.4^2}{4}$

$P=5.54\times 10^{-9}\ J.s^{-1}$

Hence:

$KE=\frac{1}{2} m.v^2$

$5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2$

$v=0.07444\ m.s^{-1}$

$v=74.44\ mm.s^{-1}$

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0

3 years ago

A signal source that is most conveniently represented by its Th´evenin equivalent has vs = 10 mV and Rs = 1 k. If the source fee

gogolik [260]

Answer:

RL=100K → Vo=9.90 mV

RL=10K → Vo=9.09 mV

RL=1K → Vo=5 mV

RL=100 → Vo=909.09 μV

In order to obtain 80% of the power source we have to put a resistor of 4 KOhm.

Explanation:

Here we have a power source in serie with a resistor of 1K and RL, in order to obtain the Vo voltage we have to apply the voltage divider rule, that states:

$Vo=Vin*\frac{RL}{RL+R1} \\R1=1kOhm$

Substituing the resistor values of RL we obtained the following results:

RL=100K → Vo=9.90 mV

RL=10K → Vo=9.09 mV

RL=1K → Vo=5 mV

RL=100 → Vo=909.09 μV

In order to find the lowest value that gives us 80% of the source voltage we have to use the voltage divider rule again and make the Vo equal to 0.8 Vin:

$0.8*Vin=Vin*\frac{RL}{RL+R1}$

The result of the last equation is 4000, so in order to obtain 80% of the power source we have to put a resistor of 4 KOhm.

8 0

3 years ago