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1 year ago

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Centripetal force Fc acts on a car going around a curve. If the speed of the car were twice as great, the magnitude of the centr

ipetal force necessary to keep the car moving in the same path would be

1 answer:

kondaur [170]1 year ago

6 0

Answer:

We need 4 times more force to keep the car in circular motion if the velocity gets double.

Explanation:

Lets take the mass of the car = m

The radius of the arc = r

$F=\frac{m\times v^2}{r}$

Given that speed of the car gets double ,v' = 2 v

Then the force on the car = F'

$F'=\frac{m\times v'^2}{r}$ ( radius of the arc is constant)

$F'=\frac{m\times (2v)^2}{r}$

$F'=4\times \frac{m\times v^2}{r}$

We know that $F=\frac{m\times v^2}{r}$

Therefore F' = 4 F

So we can say that we need 4 times more force to keep the car in circular motion if the velocity gets double.

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I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =

Helen [10]

Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)

7 0

1 year ago

Michael Phelps needs to swim at an average speed of 2.00 m/s in order to set a new world record in the 200 m freestyle. If he sw

Natasha_Volkova [10]

Answer:

Explanation:

Given

average speed of Phelps $v_{avg}=2\ m/s$

total distance $d=200\ m$

he swims first 100 m at an average speed if $1.8 m/s$

so time taken is $t_1=\frac{100}{1.8}=55.55\ s$

suppose $t_2$ is the time taken to swim remaining half

average velocity is $v_{avg}=\frac{displacement}{total\ time}$

$v_{avg}=\frac{100+100}{55.55+t_2}$

$t_2+55.55=\frac{200}{2}=100$

$t_2=44.45\ s$

so velocity in the second half is

$v_2=\frac{100}{45.45}$

$v_2=2.19\approx 2.2\ m/s$

3 0

1 year ago

The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em> $1.83\times 10^{7}\,kg$ <em>. The wheels of the locomotive push back on the tracks with a constant net force of </em> $7.50\times 10^{5}\,N$ <em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force ( $F$ ), measured in newtons, then acceleration ( $a$ ), measured in meters per square second, must be constant and can be found by the following expression:

$a = \frac{F}{m}$ (1)

Where $m$ is the mass of the freight train, measured in kilograms.

If we know that $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , then the acceleration experimented by the train is:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now, the time taken to accelerate the freight train from rest ( $t$ ), measured in seconds, is determined by the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - Final speed of the train, measured in meters per second.

$v_{o}$ - Initial speed of the train, measured in meters per second.

If we know that $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ and $v = 22.222\,\frac{m}{s}$ , the time taken by the freight train is:

$t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }$

$t = 542.265\,s$

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0

1 year ago

Two bar magnets are labeled A and B. The ends of each magnet are numbered 1 or 2, but the poles are not labeled. When A1 is brou

vichka [17]

It is definitely letter D. <span>A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

A1 and B1 is either both north poles or both south poles. Repulsion of both magnets says it all--like poles always repel while opposite poles always attract. Thus, the best conclusion to this would be choice D.</span>

3 0

2 years ago

Read 2 more answers

A proton is confined in an infinite square well of width 10 fm. (The nuclear potential that binds protons and neutrons in the nu

kvasek [131]

Answer:

First Question

$E = 1.065*10^{-12} \ J$

Second Question

The wavelength is for an X-ray

Explanation:

From the question we are told that

The width of the wall is $w = 10\ fm = 10*10^{-15 }\ m$

The first excited state is $n_1 = 2$

The ground state is $n_0 = 1$

Gnerally the energy (in MeV) of the photon emitted when the proton undergoes a transition is mathematically represented as

$E = \frac{h^2 }{ 8 * m * l^2 [ n_1^2 - n_0 ^2 ] }$

Here h is the Planck's constant with value $h = 6.62607015 * 10^{-34} J \cdot s$

m is the mass of proton with value $m = 1.67 * 10^{-27} \ kg$

So

$E = \frac{( 6.626*10^{-34})^2 }{ 8 * (1.67 *10^{-27}) * (10 *10^{-15})^2 [ 2^2 - 1 ^2 ] }$

=> $E = 1.065*10^{-12} \ J$

Generally the energy of the photon emitted is also mathematically represented as

$E = \frac{h * c }{ \lambda }$

=> $\lambda = \frac{h * c }{E }$

=> $\lambda = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 1.065 *10^{-15 } }$

=> $\lambda = 1.87*10^{-10} \ m$

Generally the range of wavelength of X-ray is $10^{-8} \to 1)^{-12}$

So this wavelength is for an X-ray.

8 0

1 year ago