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erastova [34]
2 years ago
5

Question #6a) You were told to assume that the ball bounced to a height of 1.2 m and that the distance between the bounces measu

red along the floor was 0.5 m. Assume that the height of the bounce were less than before, but the distance between bounces were still the same, and complete the following predictions.The horizontal velocity would be (greater/same/less) before.The time between bounces would be (greater/same/less) before.The magnitude of the acceleration of the ball would be (greater/same/less) before.The maximum vertical velocity (right after the first bounce) would be (greater/same/less) before.b) Which of the following statements are evidence that the motion in the x and y coordinates can be treated separately? Select each accordingly.Evidence/NOT evidence : From the graphs of x(t) and y(t), we can see that acceleration happens only in the y coordinate.Evidence/NOT evidence : The time of flight of a bounced ball depends only on the initial y velocity after the first bounce.Evidence/NOT evidence : The acceleration of free-fall is constant.Evidence/NOT evidence : It takes the same amount of time for a dropped ball to reach the floor as it does for a ball that is launched horizontally from the same height.Evidence/NOT evidence : The component equations of x(t) and vx(t) contain no functions of y or vy, and the component equations of y(t) and vy(t) contain no functions of x or vx.Evidence/NOT evidence : A line fit to vy(t) can be used to find the acceleration in the y direction.
Physics
1 answer:
Inessa [10]2 years ago
4 0

Answer:

Explanation:

In the problem they give the case of a ball that bounces on the vertical axis the height is getting smaller, on the x axis it is the same distance all the time.

Questions about the aspect of the x-axis

- Since the distance traveled between the rebounds is the same, the speed must be the same or constant throughout the entire journey.

- Since the distance and speed are equal, the time between rebounds is the same

- There can be no acceleration because the bounce gap should change, the acceleration is zero all the time

Questions about the Y axis

- The vertical speed of the first boat would be the greatest of all, so it has the highest height

b) Evidence or not Evidence

1  evidence, because the graphs are different, one is a straight line and he gives a parable

2 no evidence, nothing involved on the x axis

3  no evidence, the acceleration on the y axis is independent of the acceleration on the x axis

4  no evidence, the time that is cast is the only change that is the same for both movements

5  evidence, the fact that no mixture of components is found allows the variable to be separated into different equations

6  no evidence, says nothing about the x-axis

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In the standing waves experiment, the string has a mass of 31.2 g and a length of 0.7 m. The string is connected to a mechanical
mestny [16]

Answer:

linear density of the string = 4.46 × 10⁻⁴ kg/m

Explanation:

given,

mass of the string = 31.2 g

length of string = 0.7 m

linear density of the string = \dfrac{mass\ of\ string}{length}

linear density of the string = \dfrac{31.2\times 10^{-3}\ kg}{0.7\ m}

linear density of the string = 44.57 × 10⁻³ kg/m

linear density of the string = 4.46 × 10⁻⁴ kg/m

7 0
3 years ago
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mezya [45]

Answer:

wave frequency

Explanation:

i took the test, trust me

7 0
2 years ago
For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

4 0
3 years ago
Read 2 more answers
Two rigid tanks of equal size and shape are filled with different gases. The tank on the left contains oxygen, and the tank on t
fredd [130]

Answer:

The number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.

Explanation:

Given:

Molar mass of oxygen, M_O=32

Molar mass of hydrogen, M_H=2

We know ideal gas law as:

PV=nRT

where:

P = pressure of the gas

V = volume of the gas

n= no. of moles of the gas molecules

R = universal gs constant

T = temperature of the gas

∵n=\frac{m}{M}

where:

m = mass of gas in grams

M = molecular mass of the gas

∴Eq. (1) can be written as:

PV=\frac{m}{M}.RT

P=\frac{m}{V}.\frac{RT}{M}

        as: \frac{m}{V}=\rho\ (\rm density)

So,

P=\rho.\frac{RT}{M}

Now, according to given we have T,P,R same for both the gases.

P_O=P_H

\rho_O.\frac{RT}{M_O}=\rho_H.\frac{RT}{M_H}

\Rightarrow \frac{\rho_O}{32}=\frac{\rho_H}{2}

\rho_O=16\rho_H

∴The molecules of oxygen are more densely packed than the molecules of hydrogen in the same volume at the same temperature and pressure. So, <em>the number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.</em>

5 0
3 years ago
An object is 50 cm from a converging lens with a focal length of 40 cm . A real image is formed on the other side of the lens, 2
Leno4ka [110]

Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

M=-\frac{q}{p}

where

M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

M=-\frac{200 cm}{50 cm}=-4.0

7 0
3 years ago
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