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3 years ago

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A policeman investigating an accident measures the skid marks left by a car on the horizontal road. He determines that the dista

nce was 23.74 n. The coefficient of kinetic friction between the tires and the road is μk = 0.29 How fast was the car going when the driver applied the brakes

1 answer:

sleet_krkn [62]3 years ago

3 0

<h2>The velocity of car was m m/s </h2>

Explanation:

When the car skids , its centripetal force is provided by frictional force .

Here centripetal force F = $\frac{mv^2}{r}$

here v is the velocity of car and r is the radius of the curve .

The frictional force f = μ R

here μ is the coefficient of kinetic friction and R is normal reaction

Thus f = m g

On horizontal road R = mg

Thus F = f at the time of skid .

Thus v = $\sqrt{\mu rg}$

= $\sqrt{0.29x23.74x10}$ = 9 m/s

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Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

a)

The intensity of electromagnetic radiation is given by

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

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

c)

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$

Now we can find the radiation intensity with the equation

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

Where the area is

$A=4\pi r'^2 = 4\pi(2.4\cdot 10^{11})^2=7.24\cdot 10^{23} m^2$

And substituting,

$I=\frac{3.910\cdot 10^{26}}{7.24\cdot 10^{23}}=540 W/m^2$

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3 years ago

Metamorphic rock is not formed from

Mashcka [7]

Metamorphic rock is not formed from A. erosion

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3 years ago

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Yuki888 [10]

Answer:

0.2 m

Explanation:

PHASE 1

First, we calculate the distance the tongue moved in the first 20 ms (0.02 secs). We use one of Newton's equations of linear motion:

$s = ut + \frac{1}{2}at^2$

where u = initial velocity = 0 m/s

a = acceleration = $250 m/s^2$

t = time = 0.02 s

Therefore:

$s = 0 + \frac{1}{2} * 250 * (0.02)^2\\\\\\s = 0.05 m$

PHASE 2

Then, for the next 30 ms (0.03 secs), we use the formula:

$distance = speed * time$

This speed is the same as the final velocity of the tongue after the first 20 ms.

This can be obtained by using the formula:

$v = u + at\\\\=> v = 0 + (250 * 0.02)\\\\\\v = 5 m/s$

Therefore:

distance = 5 * 0.03 = 0.15 m

Therefore, the total distance moved by the tongue in the 50 ms interval is:

0.05 + 0.15 = 0.2 m

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3 years ago

Determine the distance between a newly discovered planet and its single moon if the orbital period of the moon is 1.2 Earth days

Vinil7 [7]

Answer:

The distance is $r = 55430496 \ m$

Explanation:

From the question we are told that

The period of the moon $T = 1.2 days = 1.2 * 24 * 3600 = 103680 \ s$

The mass of the planet is $m_p = 9.38*10^{24} kg$

Generally the period of the moon is mathematically represented as

$T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }$

Here G is the gravitational constant with value

$G = 6.67 *10^{-11} \ N \cdot m^2/kg^2$

=> $T = 2 * \pi * \sqrt{ \frac{r^3 }{ G * m_p } }$

=> $103680 = 2 * 3.142 * \sqrt{ \frac{r^3 }{ 6.67*10^{-11} * 9.38*10^{24} } }$

=> $272218492.31 = \frac{r^3}{ 6.67 *10^{-11} * 9.38*10^{24}}$

=> $r = \sqrt[3]{ 1.7031241*10^{23}}$ j

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3 years ago

Figure 21.62 shows a box on whose surfaces the electric field is measured to be horizontal and to the right. On the left face (3

natima [27]

(a) The electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

(b) The total flux is 1.24 Vm.

(c) The total amount of charge that is inside the box is 1.1 x 10⁻¹¹ C.

<h3>Area of the left face</h3>

The area of the left face is calculated as follows;

A1 = 0.03 m x 0.02 m = 0.0006 m²

<h3>Electric flux on the left face</h3>

Ф1 = EA1

Ф1 = (400 V/m)( 0.0006 m²) = 0.24 Vm

Let the dimension of the right face = 5 cm by 2 cm

<h3>Area of the right face</h3>

A2 = 0.05 m x 0.02 m = 0.001 m²

<h3>Electric flux on the right face</h3>

Ф2 = EA2

Ф2 = (1000 V/m)( 0.001 m²) = 1 Vm

<h3>Total flux</h3>

Ф = Ф1 + Ф2

Ф = 0.24 Vm + 1 Vm = 1.24 Vm

<h3>Total charge inside the box</h3>

Ф = Q/ε

Q = εФ

Q = (8.85 x 10⁻¹²)(1.24)

Q = 1.1 x 10⁻¹¹ C

Thus, the electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

The total flux is 1.24 Vm and the total amount of charge that is inside the box is 1.1 x 10⁻¹¹ C.

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