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4 years ago

A crate falls from an airplane flying horizontally at an altitude of 2,000 m.

Physics

1 answer:

Hatshy [7]4 years ago

7 0

Answer:

20.2 seconds

Explanation:

The airplane (and therefore the crate) initially has no vertical velocity, so v₀ = 0 m/s.

The crate is in free fall, so a = -9.8 m/s².

The crate falls downward, so Δx = -2000 m.

Find: t, the time it takes for the crate to land.

Δx = v₀ t + ½ at²

-2000 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 20.2 s

It takes 20.2 seconds for the crate to land.

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3 years ago

Briefly explain how einstein's special theory of relativity explains the perpendicular (right hand screw like) behavior of movin

ddd [48]

Einstein's special theory of relativity explains that the electric and magnetic fields are both can formulate together in mathematically.

It is given Einstein's special theory of relativity.

It is find the Einstein's special theory of relativity explains the perpendicular behavior of moving charges without recourse to invoking the concept of a magnetic field.

<h2>What is Einstein's special theory of relativity?</h2>

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3 0

2 years ago

Please help me with this (with explanation)

Sergeeva-Olga [200]

Suppose the cyclist travels for a total time of t hours.

For 20 min = 1/3 hr, the cyclist does not move.

Over the remaining (t - 1/3) hr, the cyclist is moving at a constant speed of 22.0 km/hr, so that the cyclist would travel a distance of

x = (22.0 km/hr) • ((t - 1/3) hr) ≈ (22.0 km/hr) t - 7.33 km

If the cyclist's average speed over the total time t was 17.5 km/hr, then by the definition of average speed,

17.5 km/hr = x / t

Replace x with the distance expression from earlier:

17.5 km/hr = ((22.0 km/hr) t - 7.33 km) / t

Solve for t :

17.5 km/hr = 22.0 km/hr - (7.33 km) / t

(7.33 km) / t = 4.5 km/hr

t = (7.33 km) / (4.5 km/hr)

t ≈ 1.62963 hr

Then the distance the cyclist traveled must have been

x ≈ (22.0 km/hr) (1.62963 hr) - 7.33 km ≈ 28.5 km

and so the answer is A.

Alternatively, as soon as you arrive at

17.5 km/hr = x / t

you can instead solve for t in terms of x, then plug that into the distance equation.

t = x / (17.5 km/hr)

then

x ≈ (22.0 km/hr) (x / (17.5 km/hr)) - 7.33 km

x ≈ 1.25714 x - 7.33 km

0.25714x ≈ 7.33 km

x = (7.33 km) / 0.25714 ≈ 28.5 km

6 0

3 years ago

A distracted driver is driving towards a turn where the edge of the road leads into a 75.0 m cliff. The velocity of the vehicle

Vlada [557]

As long as the car is on the road, it moves with a constant speed of 80km/h.

As soon as the car starts to fall down the cliff, it follows a parabolic motion. It means that it still moves with constant speed along the x axis, but it also starts to move along the y axis, with constant acceleration (i.e. the acceleration due to gravity).

The good thing about parabolic motions is that the two motions along the x and y axes are completely separable.

So, first of all, we need to know how long it takes for an object to fall for 75m. The equation of a constantly accelerated motion is

$s=s_0+v_0t+\dfrac{1}{2}at^2$

Where $s_0$ is the initial position, $v_0$ is the initial speed, and $a$ is the constant rate of acceleration. In our case, we start from an initial height of 75m, an initial (vertical!) speed of zero, and our acceleration is $-g$ . So, our equation becomes

$s=75-\dfrac{g}{2}t^2$

And we want to solve for the time when $s=0$ (i.e. we want to know how long will it take for the object to reach the ground). We have

$0=75-\dfrac{g}{2}t^2 \iff 75=\dfrac{g}{2}t^2 \iff \dfrac{2\cdot75}{g}=t^2 \iff t=\sqrt{\dfrac{150}{g}}$

(I'm discarding the negative solution because it wouldn't make sense)

Now that we've used the vertical motion to find out the falling time, we can go back to the horizontal motion. We know that the car moves for a certain amount of time at a certain speed. So, we simply have to plug our values in the $s=vt$ equation, to get

$s=80\sqrt\dfrac{150}{g}}$

This is how far from the base of the cliff the vehicle lands.

3 0

3 years ago