What is a statement that’s summarizes a pattern found in nature

Calculate the broadcast wavelength of the radio station 99.90 fm. seconds

taurus [48]

The radio waves are electromagnetic wave, so it travels with velocity of light i.e $3\times 10^{8} \ m/s$ .

We can use the relation between frequency, wavelength and speed as

$\lambda =\frac{c}{f}$

Here c is speed of light, $\lambda$ is wavelength and f is frequency and its value is given 99.90 FM, it is actually in megahertz (i.e 99.90 MHz).

Therefore,

$\lambda =\frac{3 \times 10^{8}\ m/s }{99.90 \times 10^{6} \ Hz} = 3.003 \ m$ .

Thus, the broadcast wavelength of the given radio station is 3.003 m.

8 0

3 years ago

Which example describes a nonrenewable resource?

In-s [12.5K]

The refineries that use the oil to put in their cars as gasoline and then after a while the oil will disappear and go away and that's what a nonrenewable resource would be

6 0

3 years ago

Read 2 more answers

Can you explain that gravity pulls us to the Earth & can you calculate weight from masses on both on Earth and other planets

schepotkina [342]

I don't actually understand what your question is, but I'll dance around the subject
for a while, and hope that you get something out of it.

-- The effect of gravity is: There's a pair of forces, in both directions, between
every two masses.

-- The strength of the force depends on the product of the masses, so it doesn't matter whether there's a big one and a small one, or whether they're nearly equal.
It's the product that counts. Bigger product ==> stronger force, in direct proportion.

-- The strength of the forces also depends on the distance between the objects' centers. More distance => weaker force. Actually, (more distance)² ==> weaker force.

-- The forces are equal in both directions. Your weight on Earth is exactly equal to
the Earth's weight on you. You can prove that. Turn your bathroom scale face down
and stand on it. Now it's measuring the force that attracts the Earth toward you.
If you put a little mirror down under the numbers, you'll see that it's the same as
the force that attracts you toward the Earth when the scale is right-side-up.

-- When you (or a ball) are up on the roof and step off, the force of gravity that pulls
you (or the ball) toward the Earth causes you (or the ball) to accelerate (fall) toward the Earth.
Also, the force that attracts the Earth toward you (or the ball) causes the Earth to accelerate (fall) toward you (or the ball).
The forces are equal. But since the Earth has more mass than you have, you accelerate toward the Earth faster than the Earth accelerates toward you.

-- This works exactly the same for every pair of masses in the universe. Gravity
is everywhere. You can't turn it off, and you can't shield anything from it.

-- Sometimes you'll hear about some mysterious way to "defy gravity". It's not possible to 'defy' gravity, but since we know that it's there, we can work with it.
If we want to move something in the opposite direction from where gravity is pulling it, all we need to do is provide a force in that direction that's stronger than the force of gravity.
I know that sounds complicated, so here are a few examples of how we do it:
-- use arm-muscle force to pick a book UP off the table
-- use leg-muscle force to move your whole body UP the stairs
-- use buoyant force to LIFT a helium balloon or a hot-air balloon
-- use the force of air resistance to LIFT an airplane.

-- The weight of 1 kilogram of mass on or near the Earth is 9.8 newtons. (That's
about 2.205 pounds). The same kilogram of mass has different weights on other planets. Wherever it is, we only know one of the masses ... the kilogram. In order
to figure out what it weighs there, we need to know the mass of the planet, and
the distance between the kilogram and the center of the planet.

I hope I told you something that you were actually looking for.

7 0

3 years ago

What measurements will be made to determine the magnitude of the test-mass centripetal acceleration?

DedPeter [7]

Answer:period, spring constant, radius of circular part, velocity of the test mass, mass of the test-mass, mass of the hanging mass

Explanation:

3 0

3 years ago

At the top of a cliff 100 m high, Raoul throws a rock upward with velocity 15 m/s. How much later should he drop a second rock f

Talja [164]

Answer:

d. 1.78s

Explanation:

The total time in the air for the second rock can be found with the next equation:

$h=v_{o}t+\frac{1}{2} gt^2$

where $h$ is the height, in this case 100m

$v_{o}$ the inicitial velocity wich is 0 since it came from rest

g is gravity and t is time

So we have:

$100=\frac{1}{2}(9.8m/s^2)t^2$

$t= \sqrt{\frac{200m}{9.8m/s^2} } =4.52s$

For the fist rock we need to find the time it takes to go up and go back down to the height it was launched:

that time is

$t_{1}=2v_{o}/g =2(15m/s)/9.8m/s^2=3.06s$

and the time the fist rock is going down from that point, we can find in a similar way we did for the fist rock, $t_{2}$ is:

$h=v_{o}t+\frac{1}{2} gt^2$

$100=(15m/s)t_{2}+\frac{1}{2}(9.8m/s^2)t_{2}^2\\0=-100 + (15m/s)t_{2}+\frac{1}{2}(9.8m/s^2)t_{2}^2$

$0=-100 + (15m/s)t_{2}+(4.9m/s^2)t_{2}^2$

solving as a quadratic equation for time we get:

$t_{2}=3.24s$

So, the total time for the first rock is:

$t_{1}+t_{2}= 3.06s + 3.24s =6.3s$

This means that the second rock must be dropped 6.3s - 4.52 s = 1.78 seconds later, wich is the difference in the times that it takes for each rock to get to the bottom if the cliff.

6 0

3 years ago