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3 years ago

A wave x meters long has a speed of y meters per second. The frequency of the wave is

1 answer:

6 0

The correct answer is (b.) y/x hertz. That is because the formula to get the frequency is f = v / w. The following values (v=y meters / second; wavelength = x meters) must be substituted to the equation, which leaves you y/x hertz.

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A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and

alexira [117]

Answer:

A) $\omega_f=17.503\ rad.s^{-1}$

B) $t=55.6822\ s$

C) $\theta=1312\ rad$

Explanation:

Given:

mass of flywheel, $m=40\ kg$

diameter of flywheel, $d=0.72\ m$

rotational speed of flywheel, $N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}$

duration for which the power is off, $t_0=35\ s$

no. of revolutions made during the power is off, $\theta=180\times 2\pi=360\pi\ rad$

<u>Using equation of motion:</u>

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2$

$\alpha=-0.8463\ rad.s^{-2}$

Negative sign denotes deceleration.

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$\omega_f=15\pi-0.8463\times 35$

$\omega_f=17.503\ rad.s^{-1}$ is the angular velocity of the flywheel when the power comes back.

Here:

$\omega_f=0\ rad.s^{-1}$

Now using the equation:

$\omega_f=\omega_i+\alpha.t$

$0=15\pi-0.8463\times t$

$t=55.6822\ s$ is the time after which the flywheel stops.

Using the equation of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2$

$\theta=1312\ rad$ revolutions are made before stopping.

3 0

3 years ago

20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu

Anna71 [15]

Answer:

B. $V_{f}= 10\,cubic\,inches$

Explanation:

Assuming we are dealing with a perfect gas, we should use the perfect gas equation:

$PV=nRT$

With T the temperature, V the volume, P the pressure, R the perfect gas constant and n the number of mol, we are going to use the subscripts i for the initial state when the gas has 20 cubic inches of volume and absolute pressure of 5 psi, and final state when the gas reaches 10 psi, so we have two equations:

$P_{i}V_{i}=n_{i}RT_{i}$ (1)

$P_{f}V_{f}=n_{f}RT_{f}$ (2)

Assuming the temperature and the number of moles remain constant (number of moles remain constant if we don't have a leak of gas) we should equate equations (1) and (2) because $T_{i}=T_{f}$ , $n_{i}=n_{f}$ and R is an universal constant:

$P_{i}V_{i}= P_{f}V_{f}$ , solving for $V_{f}$

$V_{f} =\frac{P_{i}V_{i}}{P_{f}} =\frac{(5)(20)}{10}$

$V_{f}= 10 cubic\,inches$

6 0

3 years ago

Amanda spent 2/5 of her time after school doing homework and ¼ of her remaining time riding her bike. If she rode her bike for 4

Doss [256]

Answer:120 min

Explanation:

Given

Amanda spent $\frac{2}{5}$ of her time after school doing Home work

And $\frac{1}{4}$ of her remaining time riding her bike

It is given that she rode her bike for 45 minutes in a week

Let t be the time after school

therefore Amanda spend $\frac{2t}{5}$ in home work and $\frac{3t}{5}$ time is left

From remaining $\frac{3t}{5}$ time she spends $\frac{1}{4}$ time riding her bike

therefore $\frac{3t}{5}\times \frac{1}{4}=45$

thus $t=300 min$

therefore time spent on home work is $\frac{2}{5}\times 300=120 min$

6 0

3 years ago

Orlat

Nana76 [90]

The maximum force that the athlete exerts on the bag is equal to 1,500 N and in the opposite direction as the force that the bag exerts on the athlete.

<h3>Newton's third law of motion</h3>

Newton's third law of motion states that action and reaction are equal and opposite.

Fa = -Fb

The force exerted by the athlete on the bag is equal to the force the bag exerted on the athlete but in opposite direction.

Thus, the maximum force that the athlete exerts on the bag is equal to 1,500 newtons and in the opposite direction as the force that the bag exerts on the athlete.

Learn more about force here: brainly.com/question/12970081

#SPJ1

8 0

3 years ago

A sample of gold has a mass of 30.94 grams and density of 19.32g/cm^3. What volume of space will this sample of gold occupy?

sdas [7]

Data:
mass, m = 30.94 g
density, d = 19.32 g/cm^3

Formula: d = m / v => v = m / d = 30.94 g / 19.32 g/cm^3 = 1.60 cm^3

Then, the answer is the option C.

6 0

3 years ago

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