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nignag [31]
2 years ago
15

Which of the following is an example of a physical change? burning paper digesting food o ripping paper​

Physics
1 answer:
tekilochka [14]2 years ago
4 0
Ripping paper, because you are changing the physical appearance. The other answer choices are an example of chemical change.
You might be interested in
brainly an andean condor with a wingspan and a mass soars along a horizontal path. model its wings as a rectangle with a width.
Anna35 [415]

The difference between the pressure at the top surfaces of the condor's wings and the pressure at the bottom surfaces is 101,204 Pa.

<h3>Difference in pressure between the top and bottom of the wingspan</h3>

The difference in pressure between the top and bottom of the wingspan is calculated as follows;

ΔP = P(top) - P(bottom)

<h3>Area of the wingspan</h3>

A = bh

A = 2.7 m x 0.27 m

A = 0.729 m²

<h3>Weight of the Andean condor</h3>

W = mg

W = 9 x 9.8

W = 88.2 N

<h3>Pressure at the top surface of condor's wings</h3>

The pressure at the top surface of condor's wings is due to atmospheric pressure

P(top) = 14.7 Psi = 101,325 Pa

<h3>Pressure at the bottom surface of condor's wings</h3>

The pressure at the bottom surface is due to weight of andean condor.

P = W/A

P(bottom) = 88.2/0.729

P(bottom) = 120.99 Pa

The difference between the pressure at the top surfaces of the condor's wings and the pressure at the bottom surfaces is calculated as;

ΔP = P(top) - P(bottom)

ΔP = 101,325 Pa - 120.99 Pa

ΔP = 101,204 Pa

The complete question is below;

An Andean condor with a wingspan of 270 cm and a mass of 9.00 kg soars along a horizontal path. Model its wings as a rectangle with a width of 27.0 cm.

Learn more about pressure here: brainly.com/question/25736513

5 0
2 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What
Crank

Answer:

\mu = \frac{r (2\pi f)^{2}}{g}

Explanation:

N = normal force acting on the coin

Normal force in the upward direction balances the weight of the coin, hence

N = mg

f = frequency of rotation

Angular velocity of turntable is hence given as

w = 2\pi f

r = distance from the axis of rotation

\mu = minimum coefficient of static friction

static frictional force is given as

f = \mu N\\f = \mu mg

The  static frictional force provides the necessary centripetal force , hence

Centripetal force = Static frictional force

m r w^{2} = \mu mg\\r w^{2} = \mu g\\\\\mu = \frac{r w^{2}}{g} \\\mu = \frac{r (2\pi f)^{2}}{g}

3 0
3 years ago
20 pts !!!!
garri49 [273]
Am not really sure but what i see is D
4 0
2 years ago
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it i
lianna [129]

Answer:

<em>A) the moment of inertia of the system decreases and the angular speed increases. </em>

Explanation:

The complete question is

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the  merry-go-round. As he does this, It is true to say that

A) the moment of inertia of the system decreases and the angular speed increases.

B) the moment of inertia of the system decreases and the angular speed decreases.

C) the moment of inertia of the system decreases and the angular speed remains the same.

D) the moment of inertia of the system increases and the angular speed increases.

E) the moment of inertia of the system increases and the angular speed decreases

In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as

I_{1} w_{1} = I_{2} w_{2}    ....1

where I_{1} and I_{2} are the initial and final moment of inertia respectively.

and w_{1} and w_{2} are the initial and final angular speed respectively.

Also, we know that the moment of inertia of a rotating body is given as

I = mr^{2}    ....2

where m is the mass of the rotating body,

and r is the radius of the rotating body from its center.

We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.

From equation 1, we see that in order for the angular momentum to be conserved, the decrease from I_{1} to I_{2} will cause the angular speed of the system to increase from w_{1} to w_{2} .

From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.

7 0
3 years ago
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