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2 years ago

A car is moving at a constant speed of 100 miles per hour. how long does it take for the car to travel 200 miles.

Physics

1 answer:

Pepsi [2]2 years ago

4 0

Answer:

d. 2 hours

Explanation:

because if it travels 100 miles per hour in 1 hour it would travel 200 miles in 2 hours and so fourth.

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A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen

NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

$F_1 + F_2 = W$

$F_1 + F_2 = (1530\times 9.81)$

$F_1 + F_2 = 15009.3 N$

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

$F_1(1.15) = F_2(2.70 - 1.15)$

$F_1 = 1.35 F_2$

now from above equation

$F_2 + 1.35F_2 = 15009.3$

now we have

$F_2 = 6392.85 N$

now the other force is given as

$F_1 = 8616.45 N$

4 0

3 years ago

Question 5 help please

IrinaVladis [17]

Answer:

It is another machine that helps the main machine. Hope that helps!

5 0

3 years ago

The Astronomer Carl Sagan is famous. Why?

Zolol [24]

Mainly because he was Johnny Carson's advisor and consultant
on space, astronomy, and science in general, and he appeared
on The Tonight Show Starring Johnny Carson many times.

6 0

3 years ago

Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×10^7 light-years from Earth. If the lifetime of a human

kupik [55]

Answer:

0.9999986*c

Explanation:

The ship would travel 2.54*10^7 light years, which means that at a speed close to the speed of light the trip would take 2.54*10^7 years from the point of view of an observer on Earth. However from the point of view of a passenger of that ship it will take only 70 years if the speed is close enough to the speed of light.

$\Delta t = \Delta t' * \sqrt{1 - (\frac{v}{c})^2}$

Where

Δt is the travel time as seen by a passenger

Δt' is the travel time as seen by someone on Earth

v is the speed of the ship

c is the speed of light in vacuum

We can replace the fraction v/c with x

$\Delta t = \Delta t' * \sqrt{1 - x^2}$

$\sqrt{1 - x^2} = \frac{\Delta t}{\Delta t'}$

$1 - x^2 = (\frac{\Delta t}{\Delta t'})^2$

$x^2 = 1 - (\frac{\Delta t}{\Delta t'})^2$

$x = \sqrt{1 - (\frac{\Delta t}{\Delta t'})^2}$

$x = \sqrt{1 - (\frac{70}{2.54*10^7})^2} = 0.9999986$

It would need to travel at 0.9999986*c

5 0

3 years ago

You have a device that takes temperature measurements and runs off of solar power. How often it is programmed to take a measurem

Tcecarenko [31]

Answer:

In the Equator:

As far as the temperature is concerned equator is more or less the same throughout the year, however, there are some fluctuations also, I will set this device in March and September because, in this month, the Sun is exactly over the equator and I would be able to get the more results in this month.

In Antarctic:

As far as the climatic conditions of Antarctic are concerned, it is all the same while it fluctuates in December because the Sun is very much close to Antarctic that's why I will choose this month.

Explanation:

8 0

2 years ago