PLEASE HELP ME!!!!!!!!!!!!!!!!

10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when

SVEN [57.7K]

Answer:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

Explanation:

For this case we can use the second law of Newton given by:

$\sum F = ma$

The friction force on this case is defined as :

$F_f = \mu_k N = \mu_k mg$

Where N represent the normal force, $\mu_k$ the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

$F_f = ma$

Replacing the friction force we got:

$\mu_k mg = ma$

We can cancel the mass and we have:

$a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}$

And now we can use the following kinematic formula in order to find the distance travelled:

$v^2_f = v^2_i - 2ad$

Assuming the final velocity is 0 we can find the distance like this:

$d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m$

5 0

4 years ago

Calculate the average speed (in km/hr) of Charlie whonruns to the store 4 km away in 30 minutes

max2010maxim [7]

I don't know the answer to that

4 0

3 years ago

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

7 0

4 years ago

A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads sh

Dmitry_Shevchenko [17]

Answer:

Explanation:

Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = $\frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}$

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

3 0

3 years ago

Two billion people jump up in the air at the same time with an average velocity of 7.0 m/sec. If the mass of an average person i

faust18 [17]

Well first of all, you must realize that it depends on how the jumpers are distributed on the earth's surface. If,say, one billion of them are in the eastern hemisphere and the other billion are in the western one, then the sum of all of their momenta could easily be zero, and have no effect at all on the planet. I'm pretty sure what you must have in mind is to consider the Earth to be a block, with a flat upper surface, and all the people jump in the same direction.

average mass per person = 60 kg.
jump velocity = 7 m/s straight up and away from the block, all in the same direction
one person's worth of momentum = (m) (v) = 420 kg.m/s
sum of two billion of them = 8.4 x 10¹¹ kg-m/s all in the same direction

Earth's "recoil" momentum = 8.4 x 10¹¹ in the opposite direction = (m) (v)

Divide each side by 'm' :   v = (momentum) / (mass) =

The Earth's "recoil" velocity is (8.4 x 10¹¹) / (5.98 x 10²⁴) =

   1.405 x 10⁻¹³ m/s =

   <em> 0.00000000014 millimeter per second

</em>I have no intuitive feeling for this kind of thing, so can't judge whether
the answer is reasonable. But my math and physics felt OK on the
way to the solution, so that's my answer and I'm sticking to it.

4 0

4 years ago