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o-na [289]
2 years ago
5

Find the frequency of a wave produced by a generator that emits 100 pulses in 2.0 s.

Physics
1 answer:
kobusy [5.1K]2 years ago
7 0

The frequency of the wave is 50 Hz

Explanation:

The frequency of the wave is defined as the number of cycles per second of the wave:

f=\frac{N}{t}

where

N is the number of cycles completed in a time t.

Frequency is measured in Hertz (Hz).

In this problems, the wave has

N = 100 pulses

in

t = 2.0 s

Therefore, its frequency is

f=\frac{100}{2.0}=50 Hz

Learn more about waves and frequency here:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

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A mass of 30.0 grams hangs at rest from the lower end of a long vertical spring. You add different amounts of additional mass ΔM
ra1l [238]

Answer:

K = 373.13 N/m

Explanation:

The force of the spring is equals to:

Fe - m*g = 0     =>    Fe = m*g

Using Hook's law:

K*X = m*g    Solving for K:

K = m/X * g

In this equation, m/X is the inverse of the given slope. So, using this value we can calculate the spring's constant:

K = 10 / 0.0268 = 373.13N/m

3 0
3 years ago
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Control rods in a nuclear reactor are used to:
kiruha [24]
B.absorb neutrons to prevent chain reactions which become uncontrollable
7 0
2 years ago
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A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what hei
tekilochka [14]
Assume no air resistance, and g = 9.8 m/s².

Let
x =  angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity

The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x)  s

With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
 55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0

Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°

The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s

If t = 5.8096 s,
  u*t = 9.467*5.8096 = 55 m (Correct)
or
 u*t = 28.469*15.8096 = 165.4 m (Incorrect)

Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s

The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m

Answer: h = 110.4 m

7 0
3 years ago
A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3
Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,  

\text { Electric power }(p)=\frac{V^{2}}{R}

\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}

\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}

For calculating number of turns

\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}

Given that,

80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }

\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}

\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .

We need to find the number of turns in the secondary winding \left(N_{S}\right) to run the bulb at 120W \left(P_{2}\right)

Firstly find the secondary voltage in the transformer use, \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}

\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}

V_{2}^{2}=\frac{120^{2} \times 120}{80}

V_{2}^{2}=\frac{1728000}{80}

V_{2}^{2}=21600

V_{2}=\sqrt{21600}

V_{2}=146.9 \mathrm{V}=V_{S}

Now, finding the number of turns in secondary coil. Use, \frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}

\frac{30}{N_{S}}=\frac{65}{146.9}

N_{S}=\frac{30 \times 146.9}{65}

N_{S}=\frac{4407}{65}N_{S}=67.8

The number of turns in the secondary winding are 67.8 turns.

6 0
3 years ago
Here's a question from ~ [ AIEEE 2002 ]
lyudmila [28]
  • r=150m
  • coefficient of friction=\mu=0.6

As car is avoid skidding

\\ \sf\hookrightarrow \dfrac{mv^2}{r}=\mu mg

  • Cancel m

\\ \sf\hookrightarrow \dfrac{v^2}{r}=\mu g

\\ \sf\hookrightarrow v^2=\mu rg

\\ \sf\hookrightarrow v^2=0.6(10)(150)

\\ \sf\hookrightarrow v^2=60(150)

\\ \sf\hookrightarrow v^2=900

\\ \sf\hookrightarrow v=30ms^{-1}

Done

7 0
2 years ago
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