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Viefleur [7K]
3 years ago
5

Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds 5.0

0 m/s, four have speeds of 7.00 m/s, three have speeds of 9.00 m/s, and two have speeds of 12.0 m/s. find (a) the average speed, (b) the rms speed, and (c) the most probable speed of these particles.
Physics
1 answer:
Tems11 [23]3 years ago
5 0
A. Average speed is weighted mean (1 × 2 + 2 × 3 + 3 × 5 + 4 × 7 + 3 × 9 + 2 × 12.5)/15 = (2 + 6 + 15 + 28 + 27 + 25)/15 = 103/15 = 6.867 b. RMS is square root of 1/15 times sum of squares of speeds Sum of squares is 4 + 9 + 9 + 25 + 25 + 25 + 49 + 49 + 49 + 49 + 81 + 81 + 81 +156.25 + 156.25 = 848.5 
c. RMS speed = √(848.5/15) = 7.521 
Most likely the speed is the peak in the speed distribution, which is 7.
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A bird takes off from the tree and after 5 s is flying at a speed of 10 m/s. Calculate
Volgvan

Answer:

10miles/second

Explanation:

Change in velocity of the bird is expressed as the difference between the final velocity and initial velocity of the body.

Change in velocity = Final velocity - initial velocity

Since the bird takes off from the tree, the initial velocity of the bird = 0miles/sec

Final velocity = 10miles/secs

Change in velocity = 10-0

Change in velocity = 10miles/second

6 0
2 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
Why can work be thought of as the transfer of energy? Please help it's physics
deff fn [24]
That is because work requires energy. According to the law of conservation of energy, it cannot be created or destroyed. When doing work, energy change forms and gets transferred to the object until it is released.
for example, when you lift up an object and place it on a higher elevation, you transferred energy to it and gave it potential energy. The potential energy is transformed into kinetic energy when the object falls down, and if it hits a surface, the energy will scatter, vibrating the areas around it and producing sound.
Also, work= force X distance. The energy does not go away, but rather get changed into some other form of energy
7 0
3 years ago
Read 2 more answers
A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =
castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

emf=NBA\omega sin(\omega t)

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V

(b) the maximum rate of change of the magnetic flux is given by:

\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}

(c) emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V

(d) The torque is given by:

\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm

3 0
3 years ago
A body of mass 2 kg is moving in the positive X-Direction with a speed of 4 m/s collides head on with an another body of mass 3
Inga [223]
m_1=2 \\ m_2=3 \\ v_1=4 \\ v_2=1 \\ v\text{ =speed after collision (to be determined)}.

The momentul of the system preserves:

m_1v_1-m_2v_2=(m_1+m_2)v \ \ \ \ \ \Rightarrow \ \ \ \ \ v=\dfrac{m_1v_1-m_2v_2}{m_1+m_2}.

Ok, we found the speed after the collision.
Now, because the impact is plastic, it produces heat, sound energy and who knows what other forms of energy. We denote all this wasted energy with E.

Now, we write the energy conservation law:

\dfrac{m_1v_1^2}{2}+\dfrac{m_2v^2_2}{2}=\dfrac{(m_1+m_2)v^2}{2}+E

From the above equation, you find E,  and then conclude that the sound energy can certainly not be greater than this.
8 0
3 years ago
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