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3 years ago

How far from the castle wall does the launched rock hit the ground?

2 answers:

7 0

Khan Academy has a really good video on this: https://m.youtube.com/watch?v=a5QnXi_lZII&ebc=ANyPxKqwT2wePxBeuoMvGuAgopefPsAAET3fCVrN7oTr7j7jOobG5sJS_Db_czLaVyZJioXM2Pt-yDUZ_tzi1nX6bCAZWTpZUQ

I'll get you started really quick (but it's way past my bedtime, so... ;)).

30 sin(28) = 14.1
30 cos(28) = 26.5

This means, when it was launched, it was launched 14.1 m/s upwards and 26.5 m/s outwards/horizontally. If you recal horizontally-launched projectile motion, there is negligible resistance (we just have to find the force of gravity against the vertical component). Calculate the time it takes by plugging in the vertical displacement into one of your mechanics equations, and then using a range/distance formula for the rest.

(The video is more helpful than I'm being at 2:08 am. Have fun and good luck!)

lina2011 [118]3 years ago

4 0

King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 14 m above the moat. The rock is launched at a speed of 27 m/s and an angle of 32degrees above the horizontal.

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The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1

ahrayia [7]

Answer:

The temperature is $T = 168.44 \ K$

Explanation:

From the question ewe are told that

The rate of heat transferred is $P = 13.1 \ W$

The surface area is $A = 1.55 \ m^2$

The emissivity of its surface is $e = 0.287$

Generally, the rate of heat transfer is mathematically represented as

$H = A e \sigma T^{4}$

=> $T = \sqrt[4]{\frac{P}{e* \sigma } }$

where $\sigma$ is the Boltzmann constant with value $\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.$

substituting value

$T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }$

$T = 168.44 \ K$

7 0

4 years ago

What is capacitance?

just olya [345]

the amount of charge stored per volt

3 0

3 years ago

A mass on a horizontal surface is connected to the spring and pulled to the right along the surface stretching the spring by 25

solniwko [45]

Answer:

320 N/m

Explanation:

From Hooke's law, we deduce that

F=kx where F is applied force, k is spring constant and x is extension or compression of spring

Making k the subject of formula then

$k=\frac {F}{x}$

Conversion

1m equals to 100cm

Xm equals 25 cm

25/100=0.25 m

Substituting 80 N for F and 0.25m for x then

$k=\frac {80}{0.25}=320N/m$

Therefore, the spring constant is equal to 320 N/m

3 0

4 years ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago

A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an

olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = $\pi r^2$

V = Potential applied = 2 mV

k = Dielectric constant = 40

$\epsilon_0$ = Electric constant = $8.854\times 10^{-12}\ \text{F/m}$

Capacitance is given by

$C=\dfrac{k\epsilon_0A}{d}$

Charge is given by

$Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}$

Number of electron is given by

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}$

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

6 0

3 years ago