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3 years ago

A wave has a wavelength of 0.02m and period of 8.0*10^-2 seconds. calculate it's speed. (showing working)

Physics

1 answer:

erik [133]3 years ago

4 0

Answer:

$v = 0.25m/s$

Explanation:

Given

$\lambda = 0.02m$ --- Wavelength

$T = 8.0 * 10^{-2}s$ --- Period

Required

Determine the velocity of the wave

This is calculated using:

$v = \frac{\lambda}{T}$

$v = \frac{0.02m}{8.0*10^{-2}s}$

$v = \frac{0.02m}{8.0*0.01s}$

$v = \frac{0.02m}{0.08s}$

$v = 0.25m/s$

<em>Hence, the velocity is 0.25m/s</em>

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Explain what happens with the light when you see black and white opaque objects

maksim [4K]

Answer:

When white light strikes a colored opaque object, some colors of light are absorbed and some are reflected. ... Some of the colors are absorbed and some are reflected. When white light shines on a white object, all colors are reflected. When white light strikes a black object, all colors are absorbed.

Explanation:

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3 years ago

Read 2 more answers

Suppose it takes a constant force a time of 6.0 seconds to slow a 2500 kg truck

erastovalidia [21]

Answer:

$3.3\cdot 10^3\:\mathrm{N}$

Explanation:

Impulse on an object is given by $\mathrm{[impulse]}=F\Delta t$ .

However, it's also given as change in momentum (impulse-momentum theorem).

Therefore, we can set the change in momentum equal to the former formula for impulse:

$\Delta p=F\Delta t$ .

Momentum is given by $p=mv$ . Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:

$p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}$ .

Now we plug in our values and solve:

$\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}$ (two significant figures).

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3 years ago

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

3 years ago

When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees

Naya [18.7K]

Answer:

a) F = 1.26 10⁵ N, b) F = 2.255 10³ N, c) F_ {soil} = 3078 N

Explanation:

For this exercise we will use the relationship between momentum and moment

I = Δp

F t = p_f -p₀

a) with stiff legs, final speed is zero, initial velocity is down

Ft = 0-p₀

F = m v / t

let's calculate

F = 84.0 6.82 / 4.56 10⁻³

F = 1.26 10⁵ N

b) bending the legs

let's calculate

F = 84.0 6.82 / 0.254

F = 2.255 10³ N

c) It is requested to calculate the force of the ground on the man

∑ F = F_soil -W

F_soil = F + W

F_ {soil} = 2.255 103 + 84 9.8

F_ {soil} = 3078 N

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3 years ago

use the general formulas for gravitational force and centripetal force to derive the relationship between speed and orbital radi

MrRissso [65]

Answer:

$v = \sqrt{\frac{GM}{r}}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{Mm}{r^{2}}$ (1)

Where G is the gravitational constant, M and m are the masses of the two objects and r is the distance between them.

The centripetal force can be found by means of Newton's second law:

$F = ma$ (2)

Since it is a circular motion, the acceleration can be defined as:

$a = \frac{v^{2}}{r}$ (3)

Where v is the velocity and r is the orbital radius.

Replacing equation (3) in equation (2) it is gotten:

$F = m\frac{v^{2}}{r}$ (4)

Hence,

$m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}$

Then, v can be isolated:

$mv^{2} = G\frac{Mmr}{r^{2}}$

$mv^{2} = G\frac{Mm}{r}$

$v^{2} = G\frac{Mm}{mr}$

$v^{2} = \frac{GM}{r}$

$v = \sqrt{\frac{GM}{r}}$

So the relationship between speed and orbital radius is given by the expression $v = \sqrt{\frac{GM}{r}}$

8 0

3 years ago

A wave has a wavelength of 0.02m and period of 8.0*10^-2 seconds. calculate it's speed. (showing working)​

A wave has a wavelength of 0.02m and period of 8.0*10^-2 seconds. calculate it's speed. (showing working)