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Goryan [66]
2 years ago
15

What is the central idea for stacking rocks vertically to test the effects of forces

Physics
1 answer:
otez555 [7]2 years ago
8 0

Answer:

You can see how forces can stack on one another by comparing the normal force of the bottom most rock to the normal force of the upper most rock.

Explanation:

Say you have 3 rocks all with mass m=3kg

These 3 rocks will all have the gravitational force of -29.4N (multiply 3kg by -9.8m/s^2)

If none of the rocks are stacked, then they all have the same normal force as well (29.4N)

However, stack them, and see how their normal forces change.

m1 will continue to have a normal force of 29.4N

m2 will now have a normal force of 58.8

m3 will now have a normal force of 88.2

Though they all retain the same gravitational force, their normal force changes depending on the forces acting down on them.

Mass 1 have no forces acting downwards other than gravity.

Mass 2 has the force of mass 1 and gravity acting down on it.

Mass 3 has the force of mass 1, mass 2, and gravity acting down on it.

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A loaded barge has a mass of 1 500 000 kg and is traveling at 3 m/s. If a tugboat applies an opposing force of 12 000 N for 10 s
yan [13]

Answer:

Explanation:

Initial momentum is 1.5e6(3) = 4.5e6 kg•m/s

An impulse results in a change of momentum

The tug applied impulse is 12000(10) = 120000 N•s or 0.12e6 kg•m/s

The remaining momentum is 4.5e6 - 0.12e6 =  4.38e6 kg•m/s

The barge velocity is now 4.38e6 / 1.5e6 = 2.92 m/s

The tug applies 0.012e6 N•s of impulse each second.

The initial barge momentum will be zero in

t = 4.5e6 / 0.012e6 = 375 s or 6 minutes and 15 seconds

To stop the barge in one minute(60 s), the tug would have to apply

4.5e6 / 60 = 75000 N•s /s or 75 000 N

5 0
2 years ago
A transverse wave on a string is described by the wave functiony(x, t) = 0.350 sin (1.25x + 99.6t)where x and y are in meters an
ella [17]

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

<h3>How to solve for the time interval</h3>

We have y = 0.175

y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.5

99.62 = pi/6

t1 = 5.257 x 10⁻³

99.6t = pi/6 + 2pi

= 0.0683

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

b. we have k = 1.25, w = 99.6t

v = w/k

99.6/1.25 = 79.68

s = vt

= 79.68 * 0.0683

= 5.02

Read more on waves here

brainly.com/question/25699025

#SPJ4

complete question

A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?

7 0
1 year ago
Unlike the idealized voltmeter, a real voltmeter has a resistance that is not infinitely large. part a a voltmeter with resistan
Margaret [11]
The EMF of the battery includes the force to to drive across its internal resistance. the total resistance:  
R = internal resistance r + resistance connected rv 
R = r + rv  
Now find the current:  
V 1= IR 
I = R / V1  
find the voltage at the battery terminal (which is net of internal resistance) using  
V 2= IR  
So the voltage at the terminal is:  
V = V2 - V1  
This is the potential difference vmeter measured by the voltmeter.
6 0
3 years ago
Read 2 more answers
Find the correct statement
boyakko [2]

Answer:The disturbance created by a source of sound in the medium travels through the medium and not the particles of the medium

Explanation:i hope this is right

4 0
3 years ago
One runner in a relay race has a 1.50 s lead and is running at a constant speed of 3.25 m/s. The runner has 30.0 m to run before
yarga [219]
The second runner must run 3.3m/s. If the leading runner is 1.5 seconds ahead and there are 30m left, the second runner would need to run slightly faster than the lead in order to finish at the same time. To calculate this I did 30/1.5 which gave me 0.05. I added this onto the speed of the lead runner to get 3.3m/s :)
6 0
3 years ago
Read 2 more answers
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