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2 years ago

What is the period of a wave if the wavelength is 100 m and the speed is 200 m/s?

Physics

2 answers:

Stells [14]2 years ago

8 0

This is simple question and easy to understand.
so first you use a formula to calculate frequency of the wave.
v _ stands for velocity a.k.a speed
f_ stands for frequency which you have to find.
(pronounced lemda)_ stands for wavelength.
Then you gonna place all your values according to you have in the formula as shown then solve for f.
Finally, you use the time formula
where
T_ is period a.k.a time which equal to the reciprocal of frequency or equal to 1/f.
Simply input your values according to the formula and you your answer as 0.5s

dsp732 years ago

5 0

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A carousel that is 5.00 m in radius has a pair of 600-Hz sirens mounted on posts at opposite ends of a diameter. The carousel ro

Gelneren [198K]

Answer:

59cm

Explanation:

angular velocity = 0.8 rad/s

linear velocity = angular velocity * radius

=0.8rad/s * 5m

= 4 m/s

wavelength = (V + U)/F

where,

V is the velocity of the wave

U is the velocity of the source

F is the frequency of the source.

wavelength = (350 m/s + 4 m/s ) / 600 Hz

Wavelength = 0.59m or 59 cm

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2 years ago

A 1000.–kilogram car traveling 20.0 meters per second east experiences an impulse of 2000. newton • seconds west. What is the fi

emmainna [20.7K]

Impulse is equal to change in momentum. So if impulse is 2000 then to solve for new velocity we just set it equal to equation for momentum.

First find original momentum by p=mv
p=1000*20=20000

So then taking that value minus the impulse since it was in opposite direction of original momentum it will slow it down some. To find new velocity we just take

20000-2000=18000=mv

v=18000/1000 =18m/s

Hope this helps!! Any questions please ask!!
Thank you!

5 0

2 years ago

If you used 1000 J of energy to throw a ball, would it travel faster if you threw the ball (ignoring air resistance)

wolverine [178]

To solve this problem it is necessary to apply the kinematic equations of Energy for which the rotation of a circular body is described as

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

Where,

m = Mass of the Vall

v = Velocity

I = Moment of inertia abouts its centre of mass

$\omega =$ Angular speed

Basically the two sums of energies is the consideration of translational and rotational kinetic energy.

a. so that it was also rotating?

The ball is rotating means that it has some angular speed:

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$1000J = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

When there is a little angular energy (and not linear energy to travel faster), translational energy will be greater than the 1000J applied.

$1000J > \frac{1}{2}mv^2$

The ball will not go faster.

c. so that it wasn't rotating?

For the case where the angular velocity does not rotate it is zero therefore

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$1000J = \frac{1}{2}mv^2+\frac{1}{2}I(0)^2$

$1000J = \frac{1}{2}mv^2$

All energy is transoformed into translational energy so it is possible to go faster. This option is CORRECT.

b. It makes no difference.

Although the order presented is different, I left this last option because as we can see with the previous two parts if there is an affectation regarding angular movement, therefore it is not correct.

6 0

1 year ago

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas i

asambeis [7]

This question is incomplete, the complete question is;

The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"

Answer:

the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

Explanation:

Given that;

P₁ = 1.00 atm

P₂ = ?

V₁ = 1 L

V₂ = 1.60 L

the temperature of the gas is kept constant

we know that;

P₁V₁ = P₂V₂

so we substitute

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

5 0

2 years ago

an input force of 50 Newtons is applied through a distance of 10 meters to machine with mechanical advantage of 3. If the work o

gladu [14]

The output of the machine is

(output work) = (output force) x (distance)

450 N-m = (output force) x (3 meters)

Divide each side
by 3 meters: Output force = (450 N-m) / (3 m)

= 150 newtons .

With all the information given about the output work, we don't need
to know anything about the input work, or even the fact that we're
dealing with a machine.

It's comforting, though, to look back and notice that the output work
(450 N-m) is not more than the input work (500 N-m). So everything
is nice and hunky-dory.
___________________________________

Well, my goodness !
I didn't even need to go through all of that.

Given:

-- The input force to the machine is 50 newtons.

-- The mechanical advantage of the machine is 3 .

That right there tells us that

-- The output force of the machine is 150 newtons.

We don't need any of the other given information.

5 0

2 years ago