What is the wavelength of a 25 Hz wave traveling at 35m/s?

3

djverab [1.8K]

Answer:

it converges to the focal point

Explanation:

6 0

3 years ago

In a young's double-slit experiment, the seventh dark fringe is located 0.028 m to the side of the central bright fringe on a fl

goblinko [34]

Use: dsin(θ) = mλ where d is slit separation, m is fringe order (7), and
θ = tan⁻¹(0.028/1.1) = 1.458deg
Now λ = dsin(θ) /m = (1.6e-4)(sin(1.458))/7 = 5.96e-7 or λ = 596 nm

3 0

3 years ago

The average weight of a particular box of crackers is 26.0 ounces with a standard deviation of 0.5 ounce. The weights of the

Olenka [21]

Answer:

(a) 99.865%

(b) 0.135%

Explanation:

Given that the weight of the boxes are normally distributed.

The average weight of the particular box,

$\mu=26.0$ ounce

The standard deviation of weight,

$\sigma=0.5$ ounce.

Let $z$ be the standard normal variable,

$z=\frac{x-\mu}{\sigma}$

And, the probability of the boxes having weight x ounces is

$P(z)=\frac{1}{2\pi}e^{-\frac{z^2}{2}}$

For $x=24.5$ ,

$z=\frac{x-\mu}{\sigma}=\frac{24.5-26}{0.5}=-3$

(a) For the boxes having weight more than 24.5 ounces: $z>-3$

So, the probability of boxes for $z>-3$ is

$P(z>-3)=\int_{-3}^{\infty}\left(\frac{1}{2\pi}e^{-\frac{z^2}{2}}\right)dx$

=0.99865

So, the percent of the boxes weigh more than 24.5 ounces is 99.865%.

(b) For the boxes having weight less than 24.5 ounces: z<-3

So, the probability of boxes for z<-3 is

$P(z-3}=1-0.99865$

=0.00135

So, the percent of the boxes weigh more than 24.5 ounces is 0.135%.

8 0

4 years ago

What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 m/s²?

Dominik [7]

Force is defined as Mass multiplied by Acceleration, or F = MA.
We have our mass, 15 kg.
We also have our acceleration, 8 m/s^2.
Let's plug in our numbers and solve.

F = 15(8)
Multiply 8 by 15.
8 • 15 = 120.

Your Force is 120.
Remember, the unit of measure for Force is Newtons (N).

Your final answer is:

120N.

I hope this helps!

7 0

3 years ago

Two light bulbs each having a resistance of 3ohms are connected in series to a battery marked 12 volts. how much current will fl

Ludmilka [50]

Givens

Total Resistance (the bulbs are connected in series)

R = r1 + r2
r1 = 3 ohms
r2 = 3 ohms
R = 3 + 3 = 6 ohms
E = 12 volts
I = ??

Formula

E = I * R

Solution

12 = I * 6 Divide both sides by 6

12/6 = I*6/6

2 = I

The current = 2 amperes. Answer

6 0

4 years ago