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3 years ago

What is the speed of a wave with a frequency of 100 hz and a wave length of .5 m?

Physics

2 answers:

____ [38]3 years ago

7 0

Answer:

Explanation:

There are 2 ways to help with this. Explain the details, which are fairly simple in this topic, or give the formula. My hope is that an explanation will last longer than memorizing the formula. I give you both.

If a wave has frequency, f, of 3 Hz, its period, T, is

. The wavelength,

, is 5 meters. That means that in the time of one period, the wave travels 5 m.

In general,

distance

time

In applying this general definition of speed

↑

to a wave, we have

speed of the wave

wavelength

period

Note: we generally use v for speed of a wave. Using the variable names, then that last formula is written

Since

, we can also say that

⋅

So, using that last formula

⋅

Note: the unit Hz is equivalent to what it was called 100 years ago,

cycles

second

(

also cps

)

. Cycles is not a true unit, so the Hz contributed only the "per second" to the result

ElenaW [278]3 years ago

4 0

Answer: the speed is =30ms^-1 The speed of a wave is given by "speed" (ms^-1)= "frequency(Hz)" xx "wavelength (m)" The frequency is f=100Hz The wavelength is lambda=0.3m The speed is v=lambdaf=0.3*100=30ms^-1

Explanation:

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If you are asked to modify the robots or drones that are currently used , what kind of modifications you would do and suggest an

Eddi Din [679]

Explanation:

In recent times of pandemic, robots can be use as replacement of labor in the industries. Mundane tasks can be programmed in their system so that they can used readily.

Drones can used delivery for essential goods and services, so that human interference can be least and the spread of virus can be curbed.

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4 0

3 years ago

An electron with a speed of 1.2 × 107 m/s moves horizontally into a region where a constant vertical force of 5.2 × 10-16 N acts

Aliun [14]

Answer: 0.642mm

Explanation: F= force = 5.2×10^-16 N,

v = velocity of electron = 1.2×10^7 m/s,

m = mass of electron = 9.11×10^-31 kg.

We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.

Recall that f = ma.

Where a = acceleration

This acceleration of vertical because it occurred when the object deflected.

5.2×10^-16 = 9.11×10^-31 (ay)

ay = 5.2×10^-16 / 9.11×10^-31

ay = 5.71×10^14 m/s²

For the horizontal motion, x = vt

Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,

By substituting the parameters, we have that

0.019 = 1.27×10^7 × t

t = 0.019 / 1.27 × 10^7

t = 1.5×10^-9 s

The vertical distance (y) is gotten by using the formulae below

y = ut + at²/2

but u = 0

y = at²/2

y = 5.71×10^14 × (1.5×10^-9)²/2

y = 0.00128475/2

y = 0.000642m = 0.642mm

7 0

3 years ago

Read 2 more answers

A solid cylinder has a diameter of 17.4 mm and a length of 50.3mm. It's mass is 49g . What is its density of the cylinder in met

Tamiku [17]

Answer:

4 tonne/m³

Explanation:

ρ = m / V

ρ = 49 g / (π (17.4 mm / 2)² (50.3 mm))

ρ = 0.0041 g/mm³

Converting to tonnes/m³:

ρ = 0.0041 g/mm³ (1 kg / 1000 g) (1 tonne / 1000 kg) (1000 mm / m)³

ρ = 4.1 tonne/m³

Rounding to one significant figure, the density is 4 tonne/m³.

6 0

3 years ago

Keisha is making a diagram of a simple machine in the body.

Amiraneli [1.4K]

Answer:

Sorry for being late. It is...

A.) X: Load, Y: Fulcrum, Z: Lever

3 0

3 years ago

Read 2 more answers

A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric

Liula [17]

The electric field produced by a single-point charge is given by

$E(r)=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.

1) The first charge is $q=-3.00 nC=-3.00 \cdot 10^{-9} C$ , and it is located at x=0, so its distance from the point x=0.200 m is

$r=0.200 m-0=0.2 m$

Therefore, the electric field is

$E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C$

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.

2) The second charge is $q=-5.50 nC=-5.5 \cdot 10^{-9}C$ and it is located at x=0.800 m, so its distance from the point is

$r=0.800 m-0.200 m=0.6 m$

Therefore, the electric field is

$E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N$

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.

3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

$E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C$

and the sign tells us that the field is directed toward negative x-direction.

7 0

3 years ago