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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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Which group of elements are shiny, opaque, and have a high melting point?

vlabodo [156]

The answer is A. Metals

3 0

3 years ago

May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

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

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

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

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

$K=\frac{1}{2}(0.5)(3)^2=2.25 J$

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

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And about kinetic energy:

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#LearnwithBrainly

7 0

4 years ago

Review. From a large distance away, a particle of mass 2.00 g and charge 15.0σC is fired at 21.0 i^ m/s straight toward a second

MissTica

(a)

Determine the system's initial configuration at ri = infinite particle separation and the system's final configuration at the point of closest approach.

Since the two-particle system is not being affected by any outside forces, we may treat it as an isolated system for momentum and use the momentum conservation law.

m1v1 + m1v2 = (m1+m2)v

The second particle's starting velocity is zero, so:

m1v1 = (m1+m2)v

After substituting the values we get,

v = 6i m/s

(b)

Since the two particle system is also energy-isolated, we may use the energy-conservation principle.

dK + dU = 0

Ki +Ui = Kf + Uf

Substituting the values,

1/2m1v1^2i + 1/2 m2v2^2i + 0 = 1/2m1v1^2f + 1/2m2v2^2f +ke q1q2/rf

The second particle's initial speed is 0 (v2 = 0). Additionally, both the first and second particle's final velocity have the same value, v. Put these values in place of the preceding expression:

1/2m1v1^2i = 1/2m1v1^2 + 1/2m2v2^2 +ke q1q2/rf

After solving we get,

rf = 2ke q1q2 / m1v1^2 - (m1+m2)v^2

Substituting the values we get,

rf = 3.64m

(c)

v1f = (m1-m2 / m1 + m2) v1i

v1f = -9i m/s

(d)

v2f = (2m1/ m1 +m2) v1i

After substituting the values,

v2f = 12i m/ s

Question :

Review. From a large distance away, a particle of mass 2.00 g and charge 15.0 \muμC is fired at 21.0 m/s straight toward a second particle, originally stationary but free to move, with mass 5.00 g and charge 8.50 \muμC. Both particles are constrained to move only along the x axis. (a) At the instant of closest approach, both particles will be moving at the same velocity. Find this velocity. (b) Find the distance of closest approach. After the interaction, the particles will move far apart again. At this time, find the velocity of (c) the 2.00-g particle and (d) the 5.00-g particle. \hat{i}

To learn more about momentum conservation law click on the link below:

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5 0

2 years ago

Which of the following is not a benefit of increased energy efficiency?

Rzqust [24]

From what I can see it's D, I did this by simply examining the other answers and seeing that they are beneficial, so, from that information, this one must not be.

7 0

3 years ago

Which of the following is most closely related to the specific heat of a substance?A. its temperatureB. its melting pointC. its

swat32

Answer:

option (C)

Explanation:

The amount of heat required to raise the temperature of substance of unit mass by unit degree is called specific heat of that substance.

Its SI unit is Joule / Kg °C.

Every material has a constant value of specific heat.

So, option (c) is correct.

4 0

3 years ago