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If a wave had a wave speed of 1000 m/s and a frequency of 500 Hz, what is its wavelength?

Reil [10]

Answer:

<h2>2 meters</h2>

Explanation:

<h2>Wavelength = Speed/Frequency </h2><h2>1000 m/s ÷ 500 hz </h2><h2>2 m</h2><h2>hz = s</h2><h2>Hopes this helps. Mark as brainlest plz!</h2>

7 0

3 years ago

A baseball is released at rest from the top of the Washington Monument. It hits the ground after falling for 6 s. What was the h

alukav5142 [94]

Answer:

Total height (s) = 176.4 m

Explanation:

Given:

Initial velocity (u) = 0 m/s

Time taken (t) = 6 sec

Acceleration due to gravity = 9.8 m/s²

Find:

Total height (s)

Computation:

s = ut + [1/2]gt²

s = (0)(6) + [1/2][9.8][6²]

s = 176.4 m

Total height (s) = 176.4 m

6 0

3 years ago

A solid 200-g block of lead and a solid 200-g block of copper are completely submerged in an aquarium filled with water. Each bl

finlep [7]

Answer:

B. The buoyant force on the copper block is greater than the buoyant force on the lead block.

Explanation:

Given;

mass of lead block, m₁ = 200 g = 0.2 kg

mass of copper block, m₂ = 200 g = 0.2 kg

density of water, ρ = 1 g/cm³

density of lead block, ρ₁ = 11.34 g/cm³

density of copper block, ρ₂ = 8.96 g/cm³

The buoyant force on each block is calculated as;

$F_B = mg(\frac{density \ of \ fluid}{density \ of \ object} )$

The buoyant force of lead block;

$F_{lead} = 0.2*9.8(\frac{1}{11.34} )\\\\F_{lead} = 0.173 \ N$

The buoyant force of copper block

$F_{copper} = 0.2*9.8(\frac{1}{8.96})\\\\F_{copper} = 0.219 \ N$

Therefore, the buoyant force on the copper block is greater than the buoyant force on the lead block

3 0

3 years ago

Which location would the selected items BEST fit on the nuclear energy diagram?

Sladkaya [172]

Your answer is c) Fusion

7 0

3 years ago

Read 2 more answers

A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the accelerat

Tanzania [10]

If a Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute then the acceleration of a passenger at his lowest point during the ride is 4.11 $\text{m/s}^{2}$ .

Calculation:

Step-1:

It is given that the radius of the Ferris wheel is r=15 m, and the angular speed of the wheel is $\omega$ =5rev/min.

It is required to find the angular acceleration of a passenger at his lowest point during the ride.

The formula required to calculate the angular acceleration is, $a=\omega ^2 r$ .

Step-2:

Now substituting the given values into the equation to get the value of the angular acceleration.

$\begin{aligned}a &=\omega^{2} r \\&=(5 \mathrm{rev} / \mathrm{min})^{2}(15 \mathrm{~m}) \\&=\left(5 \mathrm{rev} / \mathrm{min} \times \frac{\left(\frac{2 \pi}{60}\right) \mathrm{rad} / \mathrm{sec}}{1 \mathrm{rev} / \mathrm{min}}\right)^{2}(15 \mathrm{~m}) \\&=(0.2741 \times 15) \mathrm{m} / \mathrm{s}^{2} \\&=4.11 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}$

The acceleration is towards upwards that means towards the center of the wheel.

Learn more about the angular acceleration:

brainly.com/question/1592013

#SPJ4

3 0

1 year ago