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Alexxandr [17]
3 years ago
9

The number of waves produced in a given time is the ____ of the wave

Physics
1 answer:
vodka [1.7K]3 years ago
8 0
Hi friend,

The number of waves produced in a given time is the FREQUENCY of the wave.

Hope it helps....
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An aeroplane accelerates from rest to 65 m/s for take-off. It travels for
Virty [35]

Answer:

1.76m/s²

Explanation:

Given parameters:

Initial velocity  = 0m/s

Final velocity  = 65m/s

Distance traveled  = 1200m

Unknown:

Acceleration  = ?

Solution:

This is linear velocity and we apply the appropriate motion equation to solve this problem.

      V²  = U² + 2as

S is the distance

u is the initial velocity

V is the final velocity

a is the acceleration

Now, insert the parameters and solve;

             65²  =  0² + 2 x a x 1200

             4225 = 2400a

                  a  = 1.76m/s²

5 0
3 years ago
A 100 kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is 10.0 m/s2. A 20 kg ma
DIA [1.3K]

Answer

given,

mass = 100 kg

acceleration = 10 m/s²

A mass 20 kg slides over 100 kg block

acceleration = 3 m/s²

horizontal friction exerted by the 100 kg block on 20 kg

using newton's second law

F - f = 0

F = f

f = ma

f = 20 × 3

f = 60 N

now net force acting on the 100 kg block

F_net = m a

F_net = 100 x 10

F_net = 1000 N

after 20 kg block falls the acceleration of the bock

F = 1000 +60

F = 1060 N

acceleartion on the block

a = \dfrac{F}{m}

a = \dfrac{1060}{100}

a = 10.60 m/s²

3 0
3 years ago
How do particles move when a transverse wave passes through a medium?
bagirrra123 [75]
<span> In </span>transverse waves<span>, </span>particles<span> of the</span>medium<span> vibrate </span>to<span> and from in a direction perpendicular </span>to<span> the direction of energy transport. </span>
4 0
3 years ago
A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re
lbvjy [14]

Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

\frac{dy}{dt} = \frac{80\pi}{3}

7 0
2 years ago
Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at
dedylja [7]

Answer:

50.91 \mu C

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for q_{1} and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted byq_{1} on q. Then we can know the magnitude of the force exerted by q_{2} about q, finally this will allow us to know the magnitude of q_{2}

q_{1} exerts a force on q in +y direction, and q_{2} exerts a force on q in -y direction.

F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\

The net force on q is:

F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}

Rewriting for q_{2}:

q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C

8 0
3 years ago
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