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arsen [322]
3 years ago
8

In art, the material used by an artist to create a work of art is called the ___________.

Physics
2 answers:
kotykmax [81]3 years ago
6 0
Answer - D = medium
Katen [24]3 years ago
3 0
D. Medium that is your answer
You might be interested in
Erica (38 kg ) and danny (43 kg ) are bouncing on a trampoline. just as erica reaches the high point of her bounce, danny is mov
AfilCa [17]
<span>2.5 m/s going upward.
   In the situation described, Erica and Danny undergo a non-elastic collision which will conserve their combined momentum. Since Erica is stationary, her momentum is 0. And since Danny is moving upward at 4.7 m/s his momentum is 43 kg * 4.7 m/s = 202.1 kg*m/s. Assuming that both Erica and Danny will be moving as a joined system, their combined mass is 38 kg + 43 kg = 81 kg. Since the momentum will be the same, their velocity will be 202.1 kg*m/s / 81 kg = 2.495061728 m/s. Since we only have 2 significant figures in the provided data, rounding the result to 2 significant figures gives a velocity of 2.5 m/s going upward.</span>
4 0
3 years ago
Please help! Will give brainliest. 10 points. Show work!
Natasha_Volkova [10]

Answer:

421.83 m.

Explanation:

The following data were obtained from the question:

Height (h) = 396.9 m

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

First, we shall determine the time taken for the ball to get to the ground.

This can be calculated by doing the following:

t = √(2h/g)

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 396.9 m

Time (t) =.?

t = √(2h/g)

t = √(2 x 396.9 / 9.8)

t = √81

t = 9 secs.

Therefore, it took 9 secs fir the ball to get to the ground.

Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:

Time (t) = 9 secs.

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

s = ut

s = 46.87 x 9

s = 421.83 m

Therefore, the horizontal distance travelled by the ball is 421.83 m

5 0
2 years ago
A diver jumps off a cliff 50m high and needs to clear the rock that extend outward 5.0m from the base of the cliff. The diver ju
igor_vitrenko [27]

Answer:

He should run at least at 1.5 m/s

The diver will enter the water at an angle of 87° below the horizontal.

Explanation:

Hi there!

The position and velocity of the diver are given by the following vectors:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector at time t

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the  upward direction as positive)

v = velocity vector at time t

Please, see the attached figure for a description of the problem. Notice that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0.

We know that, to clear the rocks, the position vector r final (see figure) should be:

r final = ( > 5.0 m, -50 m)

So let´s find first at which time the y-component of the vector r final is - 50 m:

y = y0 + v0y · t + 1/2 · g · t²

-50 m = 2.1 m/s · t - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 2.1 m/s · t + 50 m

Solving the quadratic equation

t = 3.4 s

Now, we can calculate the initial horizontal velocity using the equation of the x-component of the position vector knowing that at t =3.4 the horizontal component should be greater than 5.0 m:

x = x0 + v0x · t      (x0 = 0)

5.0 m < v0x · 3.4 s

v0x > 5.0 m / 3.4 s

v0x > 1.5 m/s

The initial horizontal velocity should be greater than 1.5 m/s

To find the angle at which the diver enters the water, we have to find the magnitude of the final velocity (vector vf in the figure). We already know the magnitude of the x-component of the vector vf, since the horizontal velocity is constant. So:

vfx > 1.5 m/s

Now, let´s calculate vfy:

vfy = v0y + g · t

vfy = 2.1 m/s - 9.8 m/s² · 3.4 s

vfy = -31 m/s

Let´s calculate the minimum magnitude that the final velocity will have if the diver safely clears the rocks. Let´s consider the smallest value allowed for vfx: 1.5 m/s. Then:

|v| = \sqrt{(1.5 m/s)^{2} + (31m/s)^{2}} = 31 m/s

Then the final velocity of the diver will be greater or equal than 31 m/s.

To find the angle, we have to use trigonometry. Notice in the figure that the vectors vf, vfx and vy form a right triangle in which vf is the hypotenuse, vfx is the adjacent side and vfy is the opposite side to the angle. Then:

cos θ = adjacent / hypotenuse = vfx / vf = 1.5 m/s / 31 m/s

θ = 87°

The diver will enter the water at an angle of 87° below the horizontal.

8 0
3 years ago
Aldis is swinging a ball tied to the end of a string over his head. Suddenly, the string breaks and the ball flies away. Arrow b
prisoha [69]

Answer:

Straight line in the direction of the tangential velocity the ball had at the moment the string broke

Explanation:

After the string breaks, the ball now disconnected from the centripetal force that was exerted via the string, continues its travel in a straight line in the direction of the tangential velocity it had at the moment the string broke.

3 0
3 years ago
Read 2 more answers
HELP PLEASE <br> Find the net force necessary for a 15 kg object to accelerate at 20 m/s/s.
tensa zangetsu [6.8K]

Answer:

<h3>The answer is 300 N</h3>

Explanation:

The force acting on an object given the mass and acceleration we use the formula

<h3>force = mass × acceleration</h3>

We have

force = 15 × 20

We have the final answer as

<h3>300 N</h3>

Hope this helps you

3 0
3 years ago
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