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3 years ago

11

A car moving at 60 mph slams on its brakes to stop before hitting a deer. Another identical car traveling at 60 mph slows to a s

top at a stop sign. Who experiences the greatest impulse during the stop

1 answer:

ankoles [38]3 years ago

8 0

Happy Holidays!

Recall that:

Impulse = Change in Momentum = mass × change in velocity

Since both cars are identical and have the same initial velocity of 60 mph, them breaking to a stop means that they both experience the same change in velocity.

Thus, both of the cars' impulses are equal.

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What do you think will happen to Charlie now that he is smart? Explain.

postnew [5]

.... I don’t know but, he will be able to make smarter choices, he will be able to think before he does something, honestly don’t know

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3 years ago

Define mass .Also write the table showing the relation of 'kg' with it's sub- multiples and multiples.

lys-0071 [83]

Answer:

the quantity of matter contained in a body is called mass .si unit of mass is kilogram.weight is effected by gravity but mass is not effected by gravity.

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2 years ago

Read 2 more answers

A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no apprecia

masha68 [24]

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

$V(t) = \epsilon. e^{-t.\frac{L}{R} }$

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

$0.8*19 = 19. e^{-4.\frac{22}{R} }$

$0.8 = e^{-\frac{88}{R} }$

$ln(0.8) = ln(e^{-\frac{88}{R} })$

$ln(0.8) = -\frac{88}{R}$

$R = -\frac{88}{ln(0.8)}$

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.

7 0

3 years ago

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,

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

Learn more about electromagnetic radiation:

brainly.com/question/9184100

brainly.com/question/12450147

#LearnwithBrainly

4 0

3 years ago

Mt. Asama, Japan, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the c

Nata [24]

Answer:

a) 69.3 m/s

b) 18.84 s

Explanation:

Let the initial velocity = u

The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively

uᵧ = u sin 40° = 0.6428 u

uₓ = u cos 40° = 0.766 u

We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m

The range of a projectile motion is given as

R = uₓt

where t = total time of flight

1000 = 0.766 ut

ut = 1305.5

The vertical distance travelled by the projectile rocks,

y = uᵧ t - (1/2)gt²

y = - 900 m (900 m below the crater's level)

-900 = 0.6428 ut - 4.9t²

Recall, ut = 1305.5

-900 = 0.6428(1305.5) - 4.9 t²

4.9t² = 839.1754 + 900

4.9t² = 1739.1754

t = 18.84 s

Recall again, ut = 1305.5

u = 1305.5/18.84 = 69.3 m/s

7 0

3 years ago