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Novosadov [1.4K]

4 years ago

11

A small ball of mass m is dropped immediately behind a large one of mass M from a height h mach larger then the size of the ball

s. What is the relationship between m and M if the large ball stops at the floor? Under this condition, how high does the small ball rise? Assume the balls are perfectly elastic and use an independent collision model in which the large ball collides elastically with the floor and returns to strike the small ball in a second collision that is elastic and independent from the first.

1 answer:

Mumz [18]4 years ago

3 0

Answer:

a) (M / m -1), b) $h_{f}$ = h (M / m -1)²

Explanation:

For this exercise we will begin by looking for the speed of the balls when they reach the floor, for this we use the concepts of energy conservation

Initial. Highest point

Em₀ = U = m g h

Final. Floor

$Em_{f} }$ = K = ½ m v²

Em₀ = $Em_{f} }$

m gh = ½ m v²

v = √ 2g h

The speed is the same for the two balls since it does not depend on the mass.

First shock

The heaviest ball (M) with the floor, as the floor does not move the ball bounces with the same speed as it arrives, but in the opposite direction

Second shock

Between the large ball with velocity upwards of value v and the small ball of mass (m) with velocity v downwards. Let's use the moment

Initial. Before the crash

p₀ = Mv - m v

After the crash

$p_{f}$ = 0 + m $v_{f}$

p₀ = $p_{f}$

Mv - m v = m $v_{f}$

$v_{f}$ = (M-m) / m v

$v_{f}$ = (M / m -1) v

Let's look with energy to where e raises the small ball with speed vf

Em₀ = $Em_{f}$

½ m vf² = m g $h_{f}$

$h_{f}$ = ½ $v_{f}$ ² / g

$h_{f}$ = ½ v² / g (M / m-1)²

We substitute in value of v

$h_{f}$ = ½ 2gh / g (M / m -1)²

$h_{f}$ = h (M / m -1)²

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Note that the simulation allows you to also display the force of the smaller moon

Lelu [443]

the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet

Explanation:

In this problem we are analzying the gravitational force acting between a planet and its moon.

The magnitude of the gravitational attraction between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, we are considering a planet and its moon. According to Newton's third law of motion,

"When an object A exerts a force (action force) on an object B, then object B exerts an equal and opposite force (reaction force) on object A"

If we apply this law to this situation, this means that the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

Solve this physics for me <br>please with steps<br>

Mars2501 [29]

Answer:

The answers are located in each of the explanations showed below

Explanation:

a)

(i) Surface Tension: The tensile force that causes this tension acts parallel to the surface and is due to the forces of attraction between the molecules of the liquid. The magnitude of this force per unit of length is called surface tension.

σ = F/l [N/m]

where:

F = force [N]

l = length [m]

σ = Surface Tension [N/m]

(ii) Frequency is the number of repetitions per unit of time of any periodic event.

f = 1/T [1/s] or [s^-1] or [Hz]

where:

T = period [s] or [seconds]

f = frecuency [Hz] or [hertz]

(iii) Each of the units will be shown for each variable

v = velocity [m/s]

a = accelertion [m/s^2]

s = displacement [m]

$[\frac{m}{s} ]^{2} =[\frac{m}{s} ]^{2} + 2* [\frac{m}{s^{2} } ]*[m]\\$

$[\frac{m^2}{s^2} ] =[\frac{m^2}{s^2} ] + [\frac{m^{2} }{s^{2} } ]$

$[\frac{m^2}{s^2} ]$

b) To find the velocity we must derivate the function X with respect to t because this derivate will give us the equation for the velocity, it means:

$v=\frac{dx}{dt} \\v = 0.75*2*t+5*t$

$(i) X = 0.75*t^{2} +5*t+1\\X = 0.75*(4)^{2} +5*(4)+1\\X = 33 [m]$

ii) replacing in the derivated equation.

$v=1.5*(4)+5\\v=11[m/s]$

iii) the average velocity is defined by the expresion v = x/t

$v = \frac{x-x_{0} }{t-t_{0} } \\$

$x_{0}=0.75(2)^{2}+5(2)+1 \\ x_{0}=14[m]\\x=0.75(7)^{2}+5(7)+1\\x=72.75[m]\\t = 7 [s]t0= 2[s]Now replacing:[tex]v_{prom} = \frac{72.75-14}{7-2} \\v_{prom} = 11.75 [m/s]$

2

a) Pascal's principle or Pascal's law, where the pressure exerted on an incompressible fluid and in balance within a container of indeformable walls is transmitted with equal intensity in all directions and at all points of the fluid.

Therefore:

P1 = pressure at point 1.

P2 = pressure at point 2.

P1 = F1/A1

P2= F2/A2

$\frac{F_{1} }{A_{1} }=\frac{F_{2}}{A_{2} } \\F_{1}=A_{1}*(\frac{F_{2}}{A_{2} })$

b) One of the applications of the surface tension is the <u>capillarity</u> this is a property of liquids that depends on their surface tension (which, in turn, depends on the cohesion or intermolecular force of the liquid), which gives them the ability to climb or descend through a capillary tube.

Other examples of surface tension:

The mosquitoes that can sit on the water.

A clip on the water.

Some leaves that remain floating on the surface.

Some soaps and detergents on the water.

5 0

3 years ago

In 1999, Robbie Knievel was the first to jump the Grand Canyon on a motorcycle. At a narrow part of the canyon (65 m wide) and t

vfiekz [6]

Answer:

His launching angle was 14.72°

Explanation:

Please, see the figure for a graphic representation of the problem.

In a parabolic movement, the velocity and displacement vectors are two-component vectors because the object moves along the horizontal and vertical axis.

The horizontal component of the velocity is constant, while the vertical component has a negative acceleration due to gravity. Then, the velocity can be written as follows:

v = (vx, vy)

where vx is the component of v in the horizontal and vy is the component of v in the vertical.

In terms of the launch angle, each component of the initial velocity can be written using the trigonometric rules of a right triangle (see attached figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In our case, the side opposite the angle is the module of v0y and the side adjacent to the angle is the module of vx. The hypotenuse is the module of the initial velocity (v0). Then:

sin angle = v0y / v0 then: v0y = v0 * sin angle

In the same way for vx:

vx = v0 * cos angle

Using the equation for velocity in the x-axis we can find the equation for the horizontal position:

dx / dt = v0 * cos angle

dx = (v0 * cos angle) dt (integrating from initial position, x0, to position at time t and from t = 0 and t = t)

x - x0 = v0 t cos angle

x = x0 + v0 t cos angle

For the displacement in the y-axis, the velocity is not constant because the acceleration of the gravity:

dvy / dt = g ( separating variables and integrating from v0y and vy and from t = 0 and t)

vy -v0y = g t

vy = v0y + g t

vy = v0 * sin angle + g t

The position will be:

dy/dt = v0 * sin angle + g t

dy = v0 sin angle dt + g t dt (integrating from y = y0 and y and from t = 0 and t)

y = y0 + v0 t sin angle + 1/2 g t²

The displacement vector at a time "t" will be:

r = (x0 + v0 t cos angle, y0 + v0 t sin angle + 1/2 g t²)

If the launching and landing positions are at the same height, then the displacement vector, when the object lands, will be (see figure)

r = (x0 + v0 t cos angle, 0)

The module of this vector will be the the total displacement (65 m)

module of r = $\sqrt{(x0 + v0* t* cos angle)^{2} }$

65 m = x0 + v0 t cos angle ( x0 = 0)

65 m / v0 cos angle = t

Then, using the equation for the position in the y-axis:

y = y0 + v0 t sin angle + 1/2 g t²

0 = y0 + v0 t sin angle + 1/2 g t²

replacing t = 65 m / v0 cos angle and y0 = 0

0 = 65m (v0 sin angle / v0 cos angle) + 1/2 g (65m / v0 cos angle)²

cancelating v0:

0 = 65m (sin angle / cos angle) + 1/2 g * (65m)² / (v0² cos² angle)

-65m (sin angle / cos angle) = 1/2 g * (65m)² / (v0² cos² angle)

using g = -9.8 m/s²

-(sin angle / cos angle) * (cos² angle) = -318.5 m²/ s² / v0²

sin angle * cos angle = 318.5 m²/ s² / (36 m/s)²

(using trigonometric identity: sin x cos x = sin (2x) / 2

sin (2* angle) /2 = 0.25

sin (2* angle) = 0.49

2 * angle = 29.44

<u>angle = 14.72°</u>

3 0

3 years ago

How much voltage is in the primary coil if there are 3200 windings in the

Lesechka [4]

Answer:

Voltage in primary coil is 3.91 V

Explanation:

For transformer we know that the working principle is given as

$\frac{V_1}{V_2} = \frac{N_1}{N_2}$

here we know that

$V_1 [tex] = voltage in primary coil[tex]V_2 = 25 V$

$N_1 = 500$

$N_2 = 3200$

Now we have

$\frac{V_1}{25} = \frac{500}{3200}$

$V_1 = 3.91 V$

8 0

3 years ago

Imagine a Rip-van-Winkle type who lives in the mountains. Just before going to sleep he yells "WAKE UP" and the sound echoes off

stiv31 [10]

Answer:

4939200 m

Explanation:

v = Velocity of sound in air = 343 m/s (general value)

For the echo to reach Rip van Winkle it must cover the distance to the mountain twice. So, time taken for the sound traveling to the mountain once

$t=\dfrac{8}{2}=4\ hr$

Distance is given by

$s=vt\\\Rightarrow s=343\times 4\times 60\times 60\\\Rightarrow s=4939200\ m=4939.2\ km$

The distance to the mountain is 4939200 m

6 0

3 years ago