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2 years ago

7

How much force does it take to accelerate a 2000 kg car at 1m/s^2

1 answer:

Dmitry [639]2 years ago

4 0

Answer:

2000 N

Explanation:

F=ma

m=2000 kg

a=1m/s^2

F=(2000 kg)(1m/s^2)

F=2000 N

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Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

I really need help.

Zielflug [23.3K]

2/5 = .4
.4*100= 40%
Alex spends more time

4 0

3 years ago

Whats the answer???????????

Leviafan [203]

Answer: Less than 4 ohms

Explanation:

We have three resistors with the following resistance:

$R_{1}=4\Omega$

$R_{2}=6\Omega$

$R_{3}=8\Omega$

Now, when the resistors are connected in parallel, the total resistance $R$ is calculated as follows:

$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

Isolating $R$ :

$R=\frac{R_{1}R_{2}R_{3}}{R_{3}(R_{1}+R_{2})+R_{1}R_{2}}$

Rewriting with th known values:

$R=\frac{(4\Omega)(6\Omega)(8\Omega)}{8\Omega(4\Omega+6\Omega)+(4\Omega)(6\Omega)}$

Finally:

$R=1.84 \Omega$

Hence, the correct option is less than 4 ohms.

4 0

3 years ago

Is the bike rider in the picture above demonstrating kinetic energy or potential energy? you need to explain your answer.

soldi70 [24.7K]

Answer:

Kinetic energy

Explanation:

Kinetic energy is a function of velocity. Since the rider is moving at a certain speed, he's demonstrating kinetic energy. It can't be potential energy because potential energy encompass mgh

4 0

3 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 <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

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

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

<em>k</em> 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 <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) 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 <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

3 0

3 years ago