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3 years ago

Relate a real life phenomenon with each branch of physics

Physics

1 answer:

anastassius [24]3 years ago

4 0

Answer:

Branches of physics with real life examples

In measuring and understanding nuclear fission (a real life phenomenon), all branches of theoretical and experimental physics have to be employed. Physics branches needed in it are, radiation detection and measurement, nuclear physics, statistical physics, thermodynamics, and almost all others.

Explanation:

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A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0

MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

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

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

$\omega=\frac{2\pi}{1 s}=6.28 rad/s$

Now we can find the linear speed of the ladybug, which is given by

$v=\omega r$

where:

$\omega=6.28 rad/s$ is the angular speed

r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation

Substituting, we find

$v=(6.28)(0.65)=4.1 m/s$

Learn more about angular speed:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

7 0

4 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

8 0

3 years ago

An emf of 28.0 mV is induced in a 501 turn coil when the current is changing at a rate of 12.0 A/s. What is the magnetic flux th

dmitriy555 [2]

Answer:

Φ = 5.589×10⁻⁵ Wb

Explanation:

The inductance of a coil is given as

L = e/(di/dt) ..................... Equation 1

Where L = inductance of the coil, e = induced e.m.f, di/dt = rate of change of current in the coil.

Also,

The inductance of each turn of the coil when a magnetic field is step up in the coil is

L = NΦ/i ................. Equation 2

Where N = number of turns, Φ = magnetic field, i = current.

equating equation 1 and equation 2

e/(di/dt) = NΦ/i

making Φ the subject of the equation,

Φ = (e×i)/N.(di/dt) .................. Equation 3

Given: e = 28.0 mV = 0.028 V, N = 501 turns, di/dt = 12.0 A/s, i = 4.00 a

Substitute into equation 3,

Φ = (0.028×4)/(12×501)

Φ = 0.112/2004

Φ = 5.589×10⁻⁵ Weber

Φ = 5.589×10⁻⁵ Wb

6 0

3 years ago

find the image of (4 "-1)" obtained by translating 4 units up followed by a reflection over the y-axis

Harrizon [31]

Answer:(-4,3)

Explanation: They didn’t show the whole graph so it looks confusing but it’s not.

4 0

2 years ago

A player holds two baseballs a height h above the ground. He throws one ball vertically upward at speed v0 and the other vertica

Degger [83]

Answer:

a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g

Explanation:

For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.

The velocity of each ball is

ball 1. thrown upwards vo is positive

v² = v₀² - 2 g (y-y₀)

in this case the height y is zero and the height i = h

v = √(v₀² + 2g h)

ball 2 thrown down, in this case vo is negative

v = √(v₀² + 2g h)

The times to get to the ground

ball 1

v = v₀ - g t₁

t₁ = $\frac{v_{o} - v }{ g}$

ball 2

v = -v₀ - g t₂

t₂ = - \frac{v_{o} + v }{ g}

From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is

Δt = t₂ -t₁

Δt = $\frac{1}{g} \ [(v_{o} - v) - ( - v_{o} - v) ]$

Δt = 2 v₀ / g

6 0

3 years ago