Ask question

Ask question

All categories

4 years ago

12

A motorist is driving at 20m/s when she sees that a traffic light 200m ahead has just turned red. She knows that this light stay

s red for 15s, and she wants to reach the light just as it turns green again. It takes her 1.0s to step on the brakes and begin slowing. What is her speed as she reaches the light at the instant it turns green?

1 answer:

yuradex [85]4 years ago

4 0

Answer:

5.71428571422 m/s

Explanation:

u = Initial velocity = 20 m/s

v = Final velocity

s = Displacement

a = Acceleration

Time taken = 15-1 = 14 s

Distance traveled in 1 second = $20\times 1=20\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 200-20=20\times 14+\frac{1}{2}\times a\times 14^2\\\Rightarrow a=\frac{2(180-20\times14)}{14^2}\\\Rightarrow a=-1.02040816327\ m/s^2$

$v=u+at\\\Rightarrow v=20-1.02040816327\times 14\\\Rightarrow v=5.71428571422\ m/s$

The speed as she reaches the light at the instant it turns green is 5.71428571422 m/s

You might be interested in

A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,

egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0

4 years ago

A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin

Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

$\epsilon = \frac{NBA}{t}$

Where,

$\epsilon =$ electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

$R = R_c + R_d$

And that the current can be expressed as function of charge and time, then

$I = \frac{q}{t}$

Equation Faraday's law and Ohm's law we have,

$V = \epsilon$

$IR = \frac{NBA}{t}$

$(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}$

Re-arrange for Magnetic Field B, we have

$B = \frac{q(R_c+R_d)}{NA}$

Our values are given as,

$R_c = 58.7\Omega$

$R_d = 45.5\Omega$

$N = 120$

$q = 3.53*10^{-5}C$

$A = 3.21cm^2 = 3.21*10^{-4}m^2$

Replacing,

$B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}$

$B = 0.095T$

Therefore the magnetic field in the System is 0.095T

3 0

3 years ago

What is the ideal banking angle for a gentle turn of 1.20-km radius on a highway with a 105 km/h speed limit (about 65 mi/h), as

Mnenie [13.5K]

Answer:

4.14°

Explanation:

given:

r = 1.2 km

v = 105 km/h

1) <em>convert your given </em>

a) r = 1.2 km to m = 1200m

b) v = 105 km/h to m/s = 29.2 m/s

2) <em>plug into your ideal banking angle equation</em>

$tan^-1$ ( $\frac{v^2}{rg}$ ) = $\frac{29.2^2}{(1200)(9.8)}$ = 4.14°

8 0

3 years ago

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. You have a stopped pipe of adjustable length clo

faltersainse [42]

Answer:

Length of pipe $= 0.057$ meter

Explanation:

Speed of a transverse wave on a string

$v = \sqrt{\frac{F}{\mu} }$

where F is the tension in string and $\mu$ is the mass per unit length

Thus,

$\mu = \frac{m}{L}$

Substituting the given values we get -

$\mu = \frac{7.25 * 10^{-3}}{0.62}\\mu = 0.0117 \frac{Kg}{m}$

Speed of a transverse wave on a string

$v = \sqrt{\frac{4510}{0.0117} } \\v = 620.86 \frac{m}{s}$

For third harmonic wave , frequency is equal to

$f = \frac{nv}{2L}$

Substituting the given values, we get -

$f = \frac{3 * 620.86}{2 * 0.62} \\f = 1502.08$

Length of pipe

$L = \frac{nv}{4 f}$

Substituting the given values we get

$n = 1$ for first harmonic wave

$L = \frac{344* 1}{4*1502.08} \\L = 0.057$

Length of pipe $= 0.057$ meter

5 0

3 years ago

During a solar eclipse, the Moon is positioned directly between Earth and the Sun. Find the magnitude of the net gravitational f

dezoksy [38]

Answer:

$F= 2.3733 x10^{20} N$

Explanation:

Let's define the variables to proceed with the operations,

So,

The masses

$M_ {sun} = 1.99 * 10^{30} Kg$

$M_ {Earth} = 5.98 * 10 ^ {24} Kg$

$M_ {Moon} = 7.36 * 10 ^{22} Kg$

Average distances

$\bar {x} _ {Sun \rightarrow Earth} = 1.5 * 10 ^ {11} m$

$\bar {x} _ {Earth \rightarrow Moon} = 3.84 * 10 ^ 8m$

Gravitational constant

$G = 6.67 * 10^ {-11} \frac {Nm ^ 2} {kg ^ 2}$

The formula of the Gravitational Force between the Moon and the Earth would be,

$F = \frac {GM_ {Earth} M_ {Moon}} {(\bar {x} _ {Earth \rightarrow Moon}) ^ 2}$

$F= \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(5.98*10^{24}Kg)(7.36*10^{22}Kg)}{(3.84*10^8m)^2}$

$F = 1.9908 * 10 ^{20} N$

This force is in the direction of the earth.

We perform the same process but now between the Sun and the Moon, like this,

$F_2 = \frac {GM_ {Sun} M_ {Moon}} {(\bar {x} _ {Sun \rightarrow Earth} - \bar {x} _ {Earth \rightarrow Moon}) ^ 2}$

$F_2 = \frac{(6.67*10^{-11} \frac{Nm^2}{kg^2})(1.99*10^{30}Kg)(7.36*10^{22}Kg)}{(1.5*10^{11}m-3.84*10^8m)^2}$

$F_2 = 4.3641*10^-{ 20} N$

This force is in the direction of the Sun

The net force must be

$F_ {net} = F_2-F$

$F_ {net} = 2.3733*10^{20} N$

This in the direction of the Sun.

8 0

3 years ago