1) With your pneumatic system, how do you think your results would have been affected if there had been a leak in the tubing?

Anyones good in gr5 science pls help me

oee [108]

Answer:

1. A

sound can't travel through vacuum only

6 0

3 years ago

An engineer who is helping to design a tall building is studying the frequencies that seismic waves travel through the ground th

skad [1K]

The frequencies of seismic waves will check if the presence of the waves have made buildings with weak foundations to be prone to destruction.

<h3>What is Frequency?</h3>

This is defined as the number of occurrences of a repeating event per unit of time and the unit is hertz.

Knowing the frequency of the seismic wave is a good safety precaution which has to be taken to prevent loss of lives and properties.

Read more about Frequency here brainly.com/question/254161

#SPJ1

8 0

2 years ago

You are standing at rest at a bus stop. A bus moving at a constant speed of 5.00 mm/???????? passes you. When the rear of the bu

neonofarm [45]

Answer:

You run 74.1409 mm and you are running at 11.9311 mm/s

Explanation:

If the bus is moving at a constant speed of 5.00mm/s and you start to run when the bus pass you by 12 mm, the equation that describe the position of the bus is:

Xb = 12.0 mm + (5.00 mm/s)*t

Where t is the time in seconds.

If you start to run toward it with a constant acceleration of 0.960 mm/s2, the equation that describe your position is:

$X_y=\frac{1}{2} (0.960\frac{mm}{s^{2}})*t^{2}$

So, the time t when you catch up the rear of the bus is the time when Xb is equal to Xy. This is:

$X_b=X_y\\12+5t=\frac{1}{2} 0.960t^{2} \\0.48t^{2}-5t-12=0$

Then, solving the quadratic equation, we obtain that t is equal to 12.4282 s

So, if we replace this value of t in the equation of Xy, we obtain how far you have run before you catch up with the rear of the bus. This is:

$X_y=\frac{1}{2} (0.960\frac{mm}{s^{2}})*12.4282^{2}$

Xy = 74.1409 mm

Then, the equation of your velocity a time t can be write as:

$Vy=0.960\frac{mm}{s^{2} }*t$

So, the velocity when you catch up the rear of the bus is:

$Vy=0.960\frac{mm}{s^{2} }*12.4282s$

Vy = 11.9311 mm/s

3 0

3 years ago

Read 2 more answers

A 20 newton weight is attached to a spring, causing it to stretch, as shown in the diagram below. What is the spring constant of

Whitepunk [10]

A 20.-N force stretches the spring by 1.00 m - 0.50 m = 0.50 m, so by Hooke's law the spring constant is k such that

20. N = k (0.50 m)

⇒ k = (20. N) / (0.50 m) = 40 N/m

4 0

3 years ago

A sled with no initial velocity accelerates at rate of 5.0 m/s^2 down a hill . How long does it take the sled to go 45 m to the

Leya [2.2K]

The time taken by the sled to reach 45 m to the bottom is 4.24 s.

The sled's initial velocity <em>u</em> is zero. It has an acceleration <em>a</em> of a value 5 m/s² down the hill and it travels a distance <em>s</em> equal to 45 m down the hill.

Use the equation of motion,

$s=ut+\frac{1}{2} at^2$

Substitute 0 m/s for u and rewrite the equation for t.

$t=\sqrt{\frac{2s}{a}} \\ =\sqrt{\frac{2(45 m)}{5.0m/s^2} } \\ =4.24s$

The sled takes a time of 4.24 s to reach the bottom of the hill

6 0

4 years ago