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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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Then it is called

DiKsa [7]

Answer:

It is called force of friction

Explanation:

The force of friction is a force that acts between two objects whose surfaces are in contact with each other.

Consider the typical case of an object sliding along a certain surface. There are two types of frictions:

- Static friction: this is the force of friction that acts when the object is not in motion yet. If you push the object forward with a force F, the object will not move immediately, but it will "oppose" to this motion with a force of static friction exactly equal to the push applied:

$F_f = F$

However, this force of static friction has a maximum value, which is given by

$F_{max} = \mu_s N$

where

$\mu_s$ is the coefficient of static friction

N is the normal reaction exerted by the surface on the object

So, when $F$ becomes greater than $F_{max}$ , the static friction is no longer able to balance the push applied, and the object will start sliding forward.

- Kinetic friction: this is the force of friction that acts when the object is already in motion. Its magnitude is given by

$F_f = \mu_k N$

where

$\mu_k$ is the coefficient of kinetic friction, and its value is generally smaller than $\mu_s$ . The direction of this force is also opposite to the direction of motion of the object.

8 0

2 years ago

Josh has a toy car of mass 3 kg tied to a string of length 2 m. He ties the string to a pole and has the toy car drive in a circ

gulaghasi [49]

Part a)

Centripetal acceleration is defined as

$a_c = \frac{v^2}{R}$

now here we know that

$v = 3m/s$

$R = 2 m$

m = 3 kg

now from above formula we have

$a_c = \frac{3^2}{2}$

$a_c = 4.5 m/s^2$

Part b)

Maximum possible tension in the string is given as

$T = 50 N$

now by force equation we have

$F = ma$

$50 = 3 a$

$a = \frac{50}{3} m/s^2$

now again by above formula

$\frac{v^2}{R} = \frac{50}{3}$

$v = \sqrt{\frac{50 \times 2}{3}}$

$v = 5.77 m/s$

5 0

3 years ago

Read 2 more answers

If your weight is force created by gravity on your body, and therefore is the Earth pulling you down, what is the reaction force

IRISSAK [1]

Answer:

The reaction force is the force exerted by your body on the earth pulling upwards.

Explanation:

This can be explained using Newton's third law of motion which states that if object A exerts a force on object B, object B exerts a force of equal magnitude and opposite direction on object A. One characteristics of these force are that they act o 2 different bodies. In this example, the earth exerts a force on you. The Newton's reaction force would be you exerting a force of equal magnitude on the earth.

3 0

2 years ago

In the design of a supermarket, there are to be several ramps connecting different parts of the store. Customers will have to pu

tatuchka [14]

Answer:

3.90 degrees

Explanation:

Let g= 9.81 m/s2. The gravity of the 30kg grocery cart is

W = mg = 30*9.81 = 294.3 N

This gravity is split into 2 components on the ramp, 1 parallel and the other perpendicular to the ramp.

We can calculate the parallel one since it's the one that affects the force required to push up

F = WsinΘ

Since customer would not complain if the force is no more than 20N

F = 20

$294.3sin\theta = 20$

$sin\theta = 20/294.3 = 0.068$

$\theta = sin^{-1}0.068 = 0.068 rad = 0.068*180/\pi \approx 3.90^0$

So the ramp cannot be larger than 3.9 degrees

6 0

3 years ago

N 1800kg car has an<br> of 3.8m/s? What is it<br> on the car?<br> acceleration<br> force acting

nevsk [136]

Answer:

6840 N

Explanation:

The force acting on the car can be found by using Newton's second law:

F = ma

where

F is the net force on the car

m is the mass of the car

a is its acceleration

For the car in this problem,

m = 1800 kg

$a=3.8 m/s^2$

Substituting,

$F=(1800)(3.8)=6840 N$

7 0

3 years ago