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3 years ago

calculate the period of a wave whose frequency is 5 Hz and whose wavelength is 1 cm give your answer in decimal form

Physics

2 answers:

Pavlova-9 [17]3 years ago

8 0

Period T = 1/f, where f = frequency = 5 Hz

T = 1/5 = 0.20 s

Period = 0.20 seconds.

trapecia [35]3 years ago

4 0

The period is the reciprocal of frequency.

Period = 1 / (frequency) = 1 / 5 per sec = 0.2 second.

The wavelength doesn't matter.

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A bicyclist covers the first leg of a journey that is d1meters in t1seconds at a speed of v1m/s and the second leg of d2meters i

Murljashka [212]

According to motion in straight line t1≠t2

A biker travels d1 meters in t1 seconds at v1 m/s for the first leg and d2 meters in t2 seconds at v2 m/s for the second leg. It's possible that t1t2 if his average speed is equal to the average of v1 and v2.
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2 years ago

A sled is pulled up to the top of a hill. At thetop of the hill the sled is released from rest and allowed to coastdown the hill

Stolb23 [73]

Answer:

a.) the speed at the bottom is greater for the steeperhill

Explanation:

since the energy at the bottom of the steeper hilis greater

$mgh =\frac{1}{mv^2}$

As we can see from above that v is higher when h ishigher.

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3 years ago

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

GaryK [48]

We will determine the wavelength through the relationship given by the distance between slits, this relationship is given under the function

$y = \frac{m\lambda}{d}$

Here,

m = Number of order bright fringe

$\lambda$ = Wavelength

d = Distance between slits

Both distance are the same, then

$y_1 = y_2$

$\frac{m_1\lambda_1 r}{d} = \frac{m_2\lambda_2 r}{d}$

$\frac{m_1\lambda_1}{m_2\lambda_2} =1$

Rearranging to find the second wavelength

$m_1 \lambda_1 = m_2 \lambda _2$

$\lambda_2 = \frac{m_1\lambda_1}{m_2}$

$\lambda_2 = \frac{7(629)}{8}$

$\lambda_2 = 550.3nm$

Therefore the wavelength of the light coming from the second monochromatic light source is 550.3nm

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2 years ago

n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat

musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

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3 years ago

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

In conclusion, the direction of the induced current will be Counterclockwise.