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2 years ago

A sports car skids off of a wet mountain road at 48.5 m/s and lands in the river, 110

Physics

1 answer:

Oxana [17]2 years ago

3 0

Answer:

about 4.74 seconds

Explanation:

The time to fall distance d from height h is given by ...

t = √(2d/g)

t = √(2·110 m/(9.8 m/s^2)) ≈ 4.74 s

It will take the car about 4.74 seconds to fall 110 meters to the river.

We assume the car's speed is horizontal, so does not add or subtract anything to/from the time to fall from the height.

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A motorist traveling at 12 m/s encounters a deer in the road 39 m ahead. If the maximum acceleration the vehicle’s brakes are ca

Vera_Pavlovna [14]

Answer:

Explanation:

Given

Motorcyclist speed=12 m/s

maximum acceleration $=-6 m/s^2$

distance=39 m

Let x be the distance traveled by motorist in his reaction time

therefore remaining 39-x will be traveled with $-6m/s^2$ acceleration

$v^2-u^2=2as$

s=39-x

v=0

u=12 m/s

$0-12^2=2\left ( -6\right )\left ( 39-x\right )$

x=27 m

Therefore he traveled 27 m in his reaction time

$27=12\times t$

t=2.25 s

(b)If his reaction time is 2.56 sec

then distance traveled in his reaction time

$x_0=12\times 2.56=30.72 m$

Remaining distance 39-30.72=8.28 m

therefore its velocity when it reaches the deer

$v^2-u^2=2as$

$v^2=12^2+2\times \left ( -6\right )\times 8.28=44.64$

v=6.681 m/s

8 0

3 years ago

What do scientists do to easily share measurement data they can understand?

malfutka [58]

These days, scientists all over the world use a standard system of measurements. It's the SI or metric system.

What about scientists in the United States, Liberia, and Burma ?
These three countries are the only ones in the world that haven't
adopted the metric system. What do THEY do ?

Easy. When scientists in those countries are off work, they use the
inches, yards, feet, quarts, miles and gallons that everybody around
them is using. But when they go to work, they use the same metric
system that everyone else in the world is using.

3 0

3 years ago

Read 2 more answers

Why is it safe to watch an eclipse of the Moon but not an eclipse of the Sun

gavmur [86]

A solar eclipse occurs when the moon crosses in front of the Sun, blocking some or all of its rays. A lunar eclipse happens when the moon is directly behind the earth, blocking the moon from receiving light. The only light comes from the light on earth's reflected shadow.

You can look at a lunar eclipse because there is very little light or none at all. You can't look at a solar eclipse because you are looking directly at the sun unless it is complete. Before totality, only some of the Sun is blocked, causing your pupils dilate to let in more light. Since they do this, more of the Sun's rays can be let in to the eye, which effectively allows your eyes to burn.

Some doctors and eye care specialists say that after someone complains of blindness after looking at a solar eclipse unaided, they can see what the Sun and moon looked like at the time that they looked at it, as it is burned onto their retinas.

8 0

3 years ago

The velocity of a 48.0 g shell leaving a 2.95 kg rifle is 391. m/s. What is the recoil velocity of the rifle?

Monica [59]

Hi there!

$\large\boxed{ -6.36m/s}$

Use the equation:

$v_{2} = -\frac{m_{1}}{m_{2}} v_{1}$

Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:

v2 = ?

m1 = 0.048 kg (converted)

m2 = 2.95

v1 = 391

$v_{2} = -\frac{0.048}{2.95} *391$

$v_{2} = -6.36m/s$

8 0

3 years ago

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago