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4 years ago

WHAT IS THE NET FORCE REQUIRED TO GIVE AN AUTOMBILE OF MASS 1600KG AN ACCELERATION OF 4.5M/S2?

Physics

1 answer:

Fed [463]4 years ago

4 0

Answer:

The required net force has a magnitude of 7200 N

Explanation:

Use Newton's 2nd Law to obtain the answer:

$F_{net}= m\,*\,a\\F_{net}=1600 \,*\,4.5 \,= 7200\,\,N$

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Light travels fastest when moving through

vitfil [10]

A vaccum unlike sound,light can travel through any matter including a great vacuum of nothing (space)

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3 years ago

The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of +Q is located at one of the cor

WINSTONCH [101]

Answer:

option d) 12.0 V

Explanation:

Data provided in the question:

Potential difference due to a single charge (+Q), E = 3.0 V

Now,

The Electric potential for the system of charges is given as:

$E=\frac{1}{4\pi \epsilon_o}[\Sigma\frac{Q}{r}]$

given for single charge, E = 3.0 V = $\frac{1}{4\pi \epsilon_o}[\frac{Q}{r}]$ .......(1)

so thus, for 4 charges, we have

$E=\frac{1}{4\pi \epsilon_o}[4\frac{Q}{r}]$ ...............(2)

from 1 and 2 we have

E = 4 × 3.0 V = 12 V

Hence, option d) 12 V is the correct answer.

8 0

3 years ago

Speed has what properties?

BlackZzzverrR [31]

Answer:

I think it's (C) magnitude and direction

Explanation:

scalar is only magnitude and no direction so that answer makes no since, and i would think speed can't go anywhere without direction so i think (C)

hope it's right

brainliest please

7 0

3 years ago

Read 2 more answers

A flexible loop has a radius of 10.5 cm and it is in a magnetic field of B = 0.117 T. The loop is grasped at points A and B and

MAXImum [283]

Explanation:

Given that,

Radius = 10.5 cm

Magnetic field = 0.117 T

Time = 0.243 s

After stretched, area is zero

(I). We need to calculate the magnetic flux through the loop before stretched

Using formula of magnetic flux

$\phi=B\times A$

$\phi=B\times \pi r^2$

Where, B = magnetic field

r = radius

Put the value into the formula

$\phi=0.117\times3.14\times(10.5\times10^{-2})^2$

$\phi=4.05\times10^{-3}\ Tm^2$

(II). We need to calculate the magnetic flux through the loop after stretched

$\phi=B\times A$

Here, A = 0

$\phi=0$

So, The magnetic flux through the loop after stretched is zero.

(III). We need to calculate the magnitude of the average induced electromotive force

Using formula of the induced electromotive force

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{\phi_{after}-\phi_{before}}{t}$

$\epsilon=-\dfrac{0-4.05\times10^{-3}}{0.243}$

$\epsilon =16.67\times10^{-3}\ V$

Hence, This is the required solution.

3 0

3 years ago

How do I solve this Physics problem?

schepotkina [342]

Like this:

1). Ignore the 22.7 m/s horizontal speed. It doesn't make a bit of difference, and the answer doesn't depend on it. (A bullet fired horizontally from a rifle and another bullet dropped from the end of the rifle barrel at the same time both hit the ground at the same time.)

2). Calculate how long it takes an object dropped from rest to fall 42.3 m. Use the "dropped from rest" formula:

Distance = (1/2) (acceleration) (time)²

3). In the formula, "acceleration" is the acceleration of gravity. If this story happened on Earth, then use the acceleration of gravity on Earth. That's 9.81 m/s² .

I wish I could make it more complicated for you, but I don't know how.

3 0

3 years ago