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3 years ago

How would one be able to travel in time?

1 answer:

7 0

The most important thing that one would need to have is a clear, vivid, active, convincing imagination. In reality, time travel is not possible.

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PLZ ANSWER THE QUESTION

adelina 88 [10]

Answer:

Explanation:

7 0

3 years ago

A roller coaster has a mass of 275 kg. It sits at the top of a hill with height 85 m. If it drops from this hill, how fast is it

Minchanka [31]

kinematic equation

v squared = u squared + 2 a x s

v= sq root (0 + 2 10 x 65)

i thimk

4 0

3 years ago

Read 2 more answers

A major difference between AC and DC electricity is that A) the current in AC electricity varies in magnitude and direction. B)

galina1969 [7]

Answer:

A) the current in AC electricity varies in magnitude and direction.

C) the voltage in AC electricity varies in magnitude and direction.

Explanation:

In DC current and voltage the direction of current will not change with time and it always remains the same.

So here in DC voltage and DC current the magnitude may change with time but the direction will always remain same

While in AC voltage and AC current the direction of AC will change with time

periodically.

So here magnitude and direction both will change in AC current and AC voltage.

so the correct answer is

A) the current in AC electricity varies in magnitude and direction.

C) the voltage in AC electricity varies in magnitude and direction.

7 0

3 years ago

Read 2 more answers

A train is travelling towards the station on a straight track. It is a certain distance from the station when the engineer appli

AleksandrR [38]

Answer:

500 m

Explanation:

t = Time taken

u = Initial velocity = 50 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -2.5 m/s²

Equation of motion

$v=u+at\\\Rightarrow 0=50-2.5\times t\\\Rightarrow \frac{-50}{-2.5}=t\\\Rightarrow t=20\ s$

Time taken by the train to stop is 20 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=50\times 20+\frac{1}{2}\times -2.5\times 20^2\\\Rightarrow s=500\ m$

∴ The engineer applied the brakes 500 m from the station

4 0

3 years ago

An object is released from rest and falls in free fall motion. The speed v of the object after it has fallen a distance y is giv

konstantin123 [22]

Answer:

8.91 %

Explanation:

Since v² = 2gy

By the relative error formula,

2Δv/v = Δg/g + Δy/y multiplying by 100%, we have

2Δv/v × 100% = Δg/g × 100 % + Δy/y × 100%

2(Δv/v × 100%) = Δg/g × 100 % + Δy/y × 100%

Δg/g × 100 % = 2(Δv/v × 100%) - Δy/y × 100%

Since Δv/v × 100% = 3.69 % and Δy/y × 100% = 5 %

Since we have a difference for the percentage error in g, we square the percentage errors and add them together. So,

[Δg/g × 100 %]² = [2(Δv/v × 100%)]² + [Δy/y × 100%]²

[Δg/g × 100 %]² = [2(3.69)]² + [5%]²

[Δg/g × 100 %]² = [4)(3.69 %)² + [5%]²

[Δg/g × 100 %]² = 54.4644 %² + 25%²

[Δg/g × 100 %]² = 79.4644 %²

taking square-root of both sides, we have

[Δg/g × 100 %] = 8.91 %

So, the percent uncertainty in the calculated value of g is 8.91 %

6 0

3 years ago