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4 years ago

6

Who studies this branch of physics? What types of careers are there in this branch of physics? What particular subjects should s

omeone study if they want a job in this branch of physics?

1 answer:

kolezko [41]4 years ago

8 0

The answers to all three questions depend on WHICH branch

of Physics you're talking about . . . a piece of information you've

neglected to specify.

You might be interested in

Câu 1. Trường hợp nào dưới đây không phải là vật sáng?

Marianna [84]

Answer:

A

Explanation:

A. The pencil is on the table in broad daylight

5 0

3 years ago

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

Starting from rest, a 6.79 kg block slides 2.82 m down a rough 20.7 ◦ incline. The coefficient of kinetic friction between the b

Veronika [31]

Answer:

23.52092 J

Explanation:

m = Mass of block = 6.79 kg

s = Sliding distance = 2.82 m

$\theta$ = Angle of slide = 20.7°

$\mu$ = Coefficient of kinetic friction = 0.425

g = Acceleration due to gravity = 9.8 m/s²

Work done by the force of gravity is given by

$W=mgsin\theta\\\Rightarrow W=6.79\times 9.8\times sin20.7\\\Rightarrow W=23.52092\ J$

The work done by the force of gravity is 23.52092 J

8 0

3 years ago

A bucket filled with water has a mass of 54 kg and is hanging from a rope that is wound around a 0.050 m radius stationary cylin

Lubov Fominskaja [6]

Answer:

The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

Explanation:

Given that,

Mass of bucket = 54 kg

Radius = 0.050 m

We need to calculate the magnitude of the torque the bucket produces around the center of the cylinder

Using formula of torque

$\tau=F\times r$

$\tau=mg\times r$

Where, m = mass

g = acceleration due to gravity

r = radius

Put the value into the formula

$\tau=54\times9.8\times0.050$

$\tau=26.46\ N-m$

Hence, The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

3 0

3 years ago

If the speed of sound in air is 343 m/s and the wavelength of this note is 2.62 m, what is the frequency of this C3 note? (Round

snow_lady [41]

Answer:

<em>The frequency of of the note = 131 Hz.</em>

Explanation:

<em>Frequency:</em><em> Frequency can be defined as the number of complete oscillation completed by a wave in one seconds. The S.I unit of frequency is Hertz ( Hz)</em>

v = λf ............................ Equation 1

Making f the subject of the equation,

f = v/λ .......................... Equation 2

Where v = Speed, λ = wavelength, f = frequency

<em>Given: v = 343 m/s, λ = 2.62 m.</em>

<em>Substituting these values into equation 2</em>

<em>f = 343/2.62</em>

<em>f = 131 Hz</em>

<em>Thus the frequency of of the note = 131 Hz.</em>

8 0

3 years ago