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Andreyy89
3 years ago
6

Can you hear this Picture???If you know it, you get it

Physics
1 answer:
Tcecarenko [31]3 years ago
4 0

Answer:

Ah yes.. Beetlejuice

Explanation:

I hear it so well ngl

You might be interested in
a tire with inner volume of 0.0250m^3 is filled with air at a gauge pressure of 36.0 psi. If the tire valve is opened to the atm
enyata [817]

Answer: Escaped volume = 0.0612m^3

Explanation:

According to Boyle's law

P1V1 = P2V2

P1 = initial pressure in the tire = 36.0psi + 14.696psi = 50.696psi (guage + atmospheric pressure)

P2 = atmospheric pressure= 14.696psi

V1 = volume of tire =0.025m^3

V2 = escaped volume + V1 ( since air still remain in the tire)

V2 = P1V1/P2

V2 = 50.696×0.025/14.696

V2 = 0.0862m^3

Escaped volume = 0.0862 - 0.025 = 0.0612m^3

5 0
3 years ago
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
3 years ago
A tennis ball is served horizontally from 2.4m above the ground at net is 12m away and point 0.9 high will be ball clear the net
Schach [20]

Explanation:

Let us first calculate  long does it take to go 12m at 30m/s( assumed speed)

12/30 = 0.4 seconds

horizontal distance the ball drop in that time

H= (0)(0.4)+1/2(-9.8)(0.4)2

H= -0.78m

negative sign shows that the height of the ball at the net from the top.

Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m

As 1.62m>0.9m so the ball will clear the net.

H_1= V0y t’ + ½ g t’^2

-2.4= (0)t’ + ½ (-9.8) t’^2

t’= 0.69s

X’=V0x t’

X’=(30)(0.96)

X’= 20.7m

3 0
3 years ago
Maggie completed a 10000-m race at an average speed of 160
Gala2k [10]

Answer: 200m/min

Explanation:

Divide 10000m by 160m/min, you will get the answer 62.5. You then subtract 12.5 from 62.5 to understand what you will need your answer for the other person’s speed will be. 10000m divided by 50min is 200m/min.

3 0
3 years ago
Mitch throws a 100-g lump of clay at a 500-g target, which is at rest on a horizontal surface. After impact, the target, includi
max2010maxim [7]

Answer:

27.22 m/s

Explanation:

Let the speed of clay before impact is u.

the speed of clay and target is v after impact.

use conservation of momentum

momentum before impact  momentum after impact

mass of clay x u = (mass of clay + mass of target) x v

100 x u = (100 + 500) x v

u = 6 v .....(1)

distance, s = 2.1 m

μ = 0.5

final velocity is zero. use third equation of motion

v'² = v² + 2as

0 = v² - 2 x μ x g x s

v² = 2 x 0.5 x 9.8 x 2.1 = 20.58

v = 4.54 m/s

so by equation (1)

u = 6 x 4.54 = 27.22 m/s

thus, the speed of clay before impact is 27.22 m/s.

3 0
3 years ago
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