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3 years ago

If an object accelerates at 40 m/s^2 in four minutes ("careful this is in minutes),

Physics

1 answer:

yaroslaw [1]3 years ago

3 0

Change in speed = (acceleration) x (time)

4 minutes = 240 seconds

Change in speed = (40 m/s²) x (240 seconds)

Change in speed = 9,600 m/s

What you're actually describing here is a car pulling 4 G's for 4 minutes, and ending up going 21,475 miles per hour.

The driver would definitely NOT get a speeding ticket, because nobody could catch him.

Also, his car would heat up and shoot flames from atmospheric friction.

(He could avoid this with some fancy steering, leave the atmosphere, and end up in low-Earth-orbit.)

Actually, I hope there's nobody in the car. His experience wouldn't be pretty.

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All of the following would be useful in trying to obtain procedural information to replicate an experiment previously published

Lera25 [3.4K]

Answer:

I am pretty sure that the answer would be a peer reviewed article.

Explanation:

I saw this because, an encyclopedia, published scientific journal, and a lab journal used in the original experiment, are all reliable sources of information.

7 0

3 years ago

Read 2 more answers

A coil 4.20 cm radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×1

erma4kov [3.2K]

Explanation:

Given that,

Radius of the coil, r = 4.2 cm

Number of turns in the coil, N = 500

The magnetic field as a function of time is given by :

$B=1.2\times 10^{-2}t+2.6\times 10^{-5}t^4$

Resistance of the coil, R = 640 ohms

We need to find the magnitude of induced emf in the coil as a function of time. It is given by :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(NBA)}{dt}\\\\\epsilon=N\pi r^2\dfrac{-dB}{dt}\\\\\epsilon=N\pi r^2\times \dfrac{-d(1.2\times 10^{-2}t+2.6\times 10^{-5}t^4)}{dt}\\\\\epsilon=N\pi r^2\times (1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\\\\\epsilon=500\pi \times (4.2\times 10^{-2})^2\times (1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\\\\\epsilon=2.77(1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\ V$

Hence, this is the required solution.

4 0

3 years ago

Read 2 more answers

You are walking from your math class to your science class. You are carrying books

kolezko [41]

Answer:

1800J

Explanation:

Given parameters:

Weight of the book = 20N

Total distance covered = 45m + 15m + 30m = 90m

Unknown:

Total work performed on the books = ?

Solution:

To solve this problem we must understand that work done is the force applied to move a body through a certain distance.

So;

Work done = Force x distance

Work done = 20 x 90 = 1800J

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3 years ago

Lymph ducts are more similar to arteries than veins. A. True B. False

Strike441 [17]

Answer:

False

Explanation:

5 0

4 years ago

If f3=0 and f1=12n, what does the magnitude of f⃗ 2 have to be for there to be rotational equilibrium?

QveST [7]

I found out F2 and its correct just need help in solving the other parts 
F2 = 4.5 N

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4 years ago