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3 years ago

If you know the distance traveled and the time traveled, you can determine an object’s

1 answer:

3 0

If I tell you that I traveled 30 miles in the last 2 hours, the only thing you can calculate is my average speed during those two hours. You can't tell anything about my acceleration or the direction in which I traveled.

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What will happen to the current if the voltage is reduced to one half?

stira [4]

We use v=IR and assuming the resistance doesn’t change we can also say that the voltage and current (I) are directly proportional which means the voltage also decreases by 1/2

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2 years ago

What accurately describes what happens when water vapor condenses into dew in terms of energy

Andreyy89

The awnser is condensation

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3 years ago

A wave is produced from a vibrating string on a violin. Two observers are standing 20 m and 40 m away from the musician. The obs

max2010maxim [7]

Well the one that is closer can see and hear more

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3 years ago

Achilles and the tortoise are having a race. The tortoise can run 1 mile (or whatever the Hellenic equivalent of this would be)

fenix001 [56]

Answer:

Surely Achilles will catch the Tortoise, in 400 seconds

Explanation:

The problem itself reduces the interval of time many times, almost reaching zero. However, if we assume the interval constant, then it is clear that in two hours Achilles already has surpassed the Tortoise (20 miles while the Tortoise only 3).

To calculate the time, we use kinematic expression for constant speed:

$x_{final}=x_{initial}+t_{tor}v_{tor}=1+t_{tor}\\x_{final}=x_{initial}+t_{ach}v_{ach}=10t_{ach}$

The moment that Achilles catch the tortoise is found by setting the same final position for both (and same time as well, since both start at the same time):

$1+t=10t\\t=1/9 hour=0.11 hours$

7 0

3 years ago

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field

Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength at the midpoint between the two rings is given as;

$E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0$

Therefore, the electric field strength at the mid-point between the two rings is zero.

7 0

3 years ago