A car of mass 200 kg, moving with a forward acceleration of 3 m/s-ıs acted upon by

A ball is dropped from the ceiling and free falls to the floor. Which one of the

Alex73 [517]

Answer:

A

Explanation:

Gravity

5 0

3 years ago

What’s the power if a student does 2240j of work for 2.8 seconds

Alex787 [66]

Answer:

800 Watts

Explanation:

Power = Work/time

Working in SI units, Power = Watts, Work = Joules, Time = seconds.

Power = 2240J/2.8s = 800 Watts.

7 0

2 years ago

Calculate the force of a particle with a net charge of 170 coulombs traveling at a speed of 135 meters/second perpendicular to t

amm1812

Answer:

F=1.14N j

Explanation:

The magnitude of the magnetic force over a charge in a constant magnetic field is given by the formula:

$|\vec{F}|=|q\vec{v} \ X\ \vec{B}|=qvsin\theta$ (|)

In this case v and B vectors are perpendicular between them. Furthermore the direction of the magnetic force is:

-i X k = +j

Finally, by replacing in (1) we obtain:

$\vec{F}=(170C)(135\frac{m}{s})(5.0*10^{-5}T)=1.14N\ \hat{j}$

hope this helps!

6 0

3 years ago

Read 2 more answers

We see a bolt of lightning and 4 s later we hear the thunderclap. If the speed of

Alex73 [517]

Answer:

i guess 0.8 miles away

Explanation:

mathematically:-

speed is given in question

Time is given

and we have to find distance

simply by using speed formula (s) = d/t

we get answer

6 0

2 years ago

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at x = -2.7 m or at x = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

k is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the x-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be Q. It is positive.

Let the distance from charge X be x m. Then the distance from charge Y will be (0.60 - x) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at x = -2.7 m or at x = 0.27 m

3 0

3 years ago