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Galina-37 [17]
3 years ago
13

A conversion factor is a ratio with a value of

Physics
1 answer:
Nikitich [7]3 years ago
7 0
The answer would be
B
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How do you graph motion in physics? How do you graph motion in physics? I've seen problems state that an object is in free fall
Dahasolnce [82]

This is Kinematics and the equations in your book.

A speed time graph would plot the speed of something against the teime it was at a speed.

If it were changing it speed constantly, that would be a straight line if acclerating. Total distrance would be the area under the graph.

5 0
3 years ago
What is the resistivity of a wire of 1.3 mm diameter, 2.3 m length, and 61 ms resistance. Number Units The number of significant
kakasveta [241]

Answer:

\rho = 3.68 *10^{-8} Ω. m

Explanation:

The resistance is given as

R = \frac{\rho L}{A}

Where A IS Cross sectional area of wire= \pi r^{2}

therefore resistivity \rho can be wrtten as

\rho = \frac{\pi Rd^{2}}{4L}

Putting all value to get resistivity value\rho =\frac{\pi* 61*10^{-3}* (1.33*10^{-3})^{2}}{4*2.3}

\rho = 3.68 *10^{-8} Ω. m

3 0
3 years ago
If the battery of your phone can provide 2 mA of current to your phone and holds a charge of 130 C, how long will it take a full
Naya [18.7K]

Answer: 65000 seconds

Explanation:

Given that,

Current (I) = 2 mA

(Since 1 mA = 1 x 10^-3A

2 mA = 2 x 10^-3A)

Charge (Q) = 130 C

Time taken for a fully charged phone to die (T) = ?

Recall that the charge is the product of current and time taken.

i.e Q = I x T

130C = 2 x 10^-3A x T

T = 130C / (2 x 10^-3A)

T = 65000 seconds (time will be in seconds because seconds is the unit of time)

Thus, it will take a fully charged phone 65000 seconds to die

5 0
3 years ago
Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side
pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m  + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

8 0
3 years ago
500km is equal to how many millimeters
Sindrei [870]

Answer:

500000000

if you can give me brainliest that would be great

7 0
3 years ago
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