PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (30pts)

A Ping-Pong ball with mass 2.5 g is attached by a thread to the bottom of a beaker. When the beaker is filled with water so that

dimaraw [331]

Answer:

0.022m or 2.2cm

Expxlanation:

Step 1:

Data obtained from the question. This includes:

Mass (m) = 2.5g = 2.5/1000 = 2.5x10^-3Kg

Tension (T) = 0.029 N

Density (ρ) = 1000 kg/m3

Acceleration due to gravity (g) = 9.81 m/s2

Diameter (d) =?

Step 2:

Finding an expression to calculate the diameter of the ball. This is illustrated below:

Tension = weight displaced - weight of the ball

Weight displaced = Mass of water x acceleration due to gravity

Mass of water = Density x volume

Mass of water = ρxV

Weight displaced = ρxVxg = ρVg

Weight of the ball = Mass of the ball x acceleration due to gravity

Weight of the ball = mg

Therefore,

Tension = weight displaced - weight of the ball

T = ρVg - mg

Make V the subject of the formula

T = ρVg - mg

T + mg = ρVg

Divide both side by ρg

V = ( T + mg) /ρg. (1)

Recall that the ball is spherical in shape and the Volume of a sphere is given by

V = 4/3πr^3

Radius (r) = diameter (d) /2

V = 4/3π(d/2)^3

V = 4/3πd^3/8

V = πd^3 /6

Substituting the value of V into equation 1, we have

V = ( T + mg) /ρg

πd^3 /6 = ( T + mg) /ρg.

Making d the subject of the formula, we have:

πd^3 /6 = (T + mg) /ρg.

d^3 = 6(T + mg) /πρg.

Taking the cube root of both sides

d = [6(T + mg) /πρg]^1/3

Step 3:

Determination of the diameter of the ball. This is illustrated below:

T = 0.029 N

m = 2.5x10^-3Kg

g = 9.81 m/s2

ρ = 1000 kg/m3

d =?

d = [6(T + mg) /πρg]^1/3

d = [6(0.029 + 2.5x10^-3x9.81)/ πx1000x9.81]^1/3

d = 0.022m

Therefore, the diameter of the ball is 0.022m or 2.2cm

6 0

3 years ago

Consider the circuit shown in the figure below. (Let R1 = 3.00 Ω, R2 = 8.00 Ω, and = 10.0 V.). . (a) Find the voltage across R1.

Goryan [66]

You take the inverse of the total resistances of each branch and add them up. So if you have 5ohm, 7 ohm, and 10ohm, you would add 1/5 + 1/7 + 1/10 = 31/70 Then flip it back by either using the x−1 (inverse) key on your calculator or simply dividing 70 by 31 to get a total of 2.26ohms

8 0

3 years ago

A grating has 2000 slits/cm/cm. How many full spectral orders can be seen (400 to 700 nmnm) when it is illuminated by white ligh

Anton [14]

With the use of the formula SinФ = nλ / d, there are 16 spectral orders which can be seen when it is illuminated by white light.

Given that a grating has 2000 slits/cm. That is,

d = 0.01 / 2000

d = 5 x $10^{-6}$ m

The wavelength λ = (700 - 400) nm

λ = 300 x $10^{-9}$ m

To calculate how many full spectral orders that can be seen (400 to 700 nm) when it is illuminated by white light, we will use the below formula

SinФ = nλ / d

Φ = $Sin^{-1}$ (nλ / d)

When n = 1

Φ = $Sin^{-1}$ (300 x $10^{-9}$ / 5 x $10^{-6}$ )

Φ = 3.4 degrees

when n = 2

Φ = $Sin^{-1}$ (2 x 300 x $10^{-9}$ / 5 x $10^{-6}$ )

Ф = 6.9 degrees

When n = 3

Ф = $Sin^{-1}$ (3 x 300 x $10^{-9}$ / 5 x $10^{-6}$ )

When n = 16

Ф = $Sin^{-1}$ (16 x 300 x $10^{-9}$ / 5 x $10^{-6}$ )

Ф = $Sin^{-1}$ (0.96)

Ф = 73.7 degrees

when n = 17

Ф = $Sin^{-1}$ (17 x 300 x $10^{-9}$ / 5 x $10^{-6}$ )

Ф = $Sin^{-1}$ (1.05)

Ф = Error ( that is, it does not exist)

Therefore, there are 16 spectral orders which can be seen when it is illuminated by white light.

Learn more about double slit here: brainly.com/question/4449144

6 0

2 years ago

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,

eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

$x(t) = A cos (\omega t + \phi)$ (1)

where

A is the amplitude

$\omega$ is the angular frequency

$\phi$ is the phase

t is the time

The displacement of the piston in the problem is given by

$x(t) = (5.00 cm) cos (5t+\frac{\pi}{5})$ (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

$x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm$

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

$v(t) = x'(t) = -\omega A sin (\omega t + \phi)$ (3)

Differentiating eq.(2), we find

$v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})$

And substituting t=0, we find the velocity at time t=0:

$v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s$

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

$a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)$

Differentiating eq.(3), we find

$a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})$

And substituting t=0, we find the acceleration at time t=0:

$a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2$

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

$\omega=\frac{2\pi}{T}$

Using $\omega = 5.0 rad/s$ and solving for T, we find

$T=\frac{2\pi}{5 rad/s}=1.26 s$

4 0

3 years ago

a bicycle accelerates at 1 m/s² from an initial velocity of 4 m/s² for 10s. Find the distance moved by it during this interval o

BaLLatris [955]

Answer:

90 m

Explanation:

Acceleration, $a=\frac {v-u}{t}$ where v and u are final and initial velocities respectively, t is the time taken

Substituting $1 m/s^{2}$ for a, 4 m/s for u and 10 s for t then

1*10=v-4

v=14 m/s

From kinematic equations

$v^{2}=u^{2}+2as$

Making s the subject then

$s=\frac {v^{2}-u^{2}}{2a}=\frac {14^{2}-4^{2}}{2\times 1}=90 m$

6 0

3 years ago