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3 years ago

The distance between two successive maximaof

Physics

1 answer:

denpristay [2]3 years ago

4 0

Answer:

v = 1.224 m/s

Explanation:

given,

distance between the two successive maxima = 1.70 m

number of crest = 8

time = 11 s

frequency is equal to number of cycle per secod

$f = \dfrac{8}{11}$

$f = 0.72\ Hz$

velocity of wave

v = f x λ

v = 0.72 x 1.70

v = 1.224 m/s

Hence, the wave speed is equal to v = 1.224 m/s

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A car of mass 800kg travels a distance of 40m at constant speed in a duration of 2.0s. The car exerts a forward force of 15kN.

Alex17521 [72]

W = F × s

W = 15kN × 40 m

W = 15.000 N × 40 m

W = 600.000 J

P = W/t

P = 600.000 J/2 s

P = 300.000 Watt

P = 300kWatt

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6 0

2 years ago

I need some help with this!

Sonja [21]

Explanation:

ummm I believe it's frequency

5 0

3 years ago

Unpolarized light passes through two polarizers whose transmission axes are at an angle # with respect to each other. What shoul

Yuki888 [10]

Answer:

$63.4^{\circ}$

Explanation:

When unpolarized light passes through the first polarizer, the intensity of the light is reduced by a factor 1/2, so

$I_1 = \frac{1}{2}I_0$ (1)

where I_0 is the intensity of the initial unpolarized light, while I_1 is the intensity of the polarized light coming out from the first filter. Light that comes out from the first polarizer is also polarized, in the same direction as the axis of the first polarizer.

When the (now polarized) light hits the second polarizer, whose axis of polarization is rotated by an angle $\theta$ with respect to the first one, the intensity of the light coming out is

$I_2 = I_1 cos^2 \theta$ (2)

If we combine (1) and (2) together,

$I_2 = \frac{1}{2}I_0 cos^2 \theta$ (3)

We want the final intensity to be 1/10 the initial intensity, so

$I_2 = \frac{1}{10}I_0$

So we can rewrite (3) as

$\frac{1}{10}I_0 = \frac{1}{2}I_0 cos^2 \theta$

From which we find

$cos^2 \theta = \frac{1}{5}$

$cos \theta = \frac{1}{\sqrt{5}}$

$\theta=cos^{-1}(\frac{1}{\sqrt{5}})=63.4^{\circ}$

6 0

3 years ago

How do odd-shaped ceilings, decorative panels, draperies, and glass windows affect echo and noise?

vesna_86 [32]

I think they decrease echo and reduce noise, they do this by either absorbing vibrations or by scattering the sound so that echoes arrive at different times rather than reverberating as a standing wave. An echo is a reflection of a sound that arrives at the listener with a delay after the direct sound. The delay is usually proportional to the distance of the reflecting surface from the source and the listener.

6 0

2 years ago

Two point charges of equal magnitude are 8.0 cm apart. At the midpoint of the line connecting them, their combined electric fiel

bagirrra123 [75]

Answer:

r = 8/2 = 4cm = 0.04m

k = 9×10^9

Enet = 51 N/C

Enet = E1 + E2

since E1 = E2

E1 = Enet/2 = 51/2

E/2 = kq/r²

q = Er²/2k

q = (51 × 0.04²)/(2×9×10^9)

q = 4.5×10^-12 C

q1 = q2 = 4.5 pC

Explanation:

The electric field is a region around a

charge in which it exerts electrostatic force

on another charges. While the strength of

electric field at any point in space is called

electric field intensity. It is a vector

quantity. Its unit is NC¯¹.

According to coulomb’s law ,if a unit

positive charge q (call it a test charge) is

brought near a charge q (call a field

charge) placed in space,the charge q will

experience a force. The value of this force

depends upon the distance between the

two charges. If the charge q is moved

away from q ,this force would decrease till

at a certain distance the force would be

practically reduced to zero. The charge q

is then out of the influence of charge q.

The region of space surrounding the charge

q in which it exerts a force on the charge

q is known as E.F of the charge

q. Mathematically it is expressed as:

E =F/q

The direction of the vector E is the same

as the direction of F,because q is a

positive scalar. Dimensionally,the E.F is

force per unit charge,and its SI unit is the

newton/coulomb (N/C).

7 0

2 years ago