a moving company is called into a building and asked to relocate 520 bricks to a location 15 meters above the ground if each brick is 4 kilograms what amount of work will be needed to accomplish the task
Answers
Wow ! I'm sweating just thinking about it.
The amount of work required is the additional potential energy
that the bricks will have when they're all up at the higher location.
Increase of potential energy = (mass) x (gravity) x (increase in height)
The total mass is (520 bricks) x (4 kg/brick) = 2,080 kilograms
Potential energy = (2,080 kg) x (9.8 m/s²) x (15 m)
Potential energy = 305,760 Joules.
Grab your hat, put on your sunscreen, and get busy.
Answer:
Work done to lift one brick = 4 Kg x 9.8 m/sec 2 x 15 m = 588 Joules. Then work done to lift 520 bricks = 520 x 588 = 305.760 j
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Answer:
A,E,B
Explanation: touch and taste are just senses.
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Answers
This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
Height at any time 'T' after the drop =
(initial height) +
(initial velocity) x (T) +
(1/2) x (acceleration) x (T²) .
For the balloon problem ...
-- We have both directions involved here, so we have to define them:
Upward = the positive direction
Initial height = +150 m
Initial velocity = + 3 m/s
Downward = the negative direction
Acceleration (of gravity) = -9.8 m/s²
Height when the bag hits the ground = 0 .
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
0 = (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)
-4.9 T² + 3T + 150 = 0
Use the quadratic equation:
T = (-1/9.8) [ -3 plus or minus √(9 + 2940) ]
= (-1/9.8) [ -3 plus or minus 54.305 ]
= (-1/9.8) [ 51.305 or -57.305 ]
T = -5.235 seconds or 5.847 seconds .
(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)
Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.
H(t) = (H₀) + (v₀ T) + (1/2 a T²)
H'(t) = v₀ + a T
The extremes of 'H' (height) correspond to points where h'(t) = 0 .
Set v₀ + a T = 0
+3 - 9.8 T = 0
Add 9.8 to each side: 3 = 9.8 T
Divide each side by 9.8 : T = 0.306 second
That's the time after the drop when the bag reaches its max altitude.
Oh gosh ! I could have found that without differentiating.
- The bag is released while moving UP at 3 m/s .
- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
(3 / 9.8) = 0.306 second .
At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) = 1.5 m/s .
Distance climbed during that time = (average speed) x (time)
= (1.5 m/s) x (0.306 sec)
= 0.459 meter (hardly any at all)
But it was already up there at 150 m when it was released.
It climbs an additional 0.459 meter, topping out at 150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely 5.847 seconds after its release.
We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.
I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work. The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.
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Answer:
The maximum speed of the heart wall during this motion is 0.032 m/s.
Explanation:
Given that,
Amplitude of the simple harmonic motion, A = 1.7 mm = 0.0017 m
Frequency of the fetus's ventricular wall, f = 3 Hz
We need to find the maximum speed of the heart wall during this motion. The maximum speed of the object that is executing SHM is given by :
So, the maximum speed of the heart wall during this motion is 0.032 m/s. Hence, this is the required solution.
Answers
Answer:
I disagree.
Explanation:
Yes, traits may be similar, but it all depends on the dominant and recessive alleles that are passed on. No one person can look alike. Even with twins, a widow's peak or close lobes can be different.
I hope this was the brainliest answer! Thank you for letting me help you.
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The kinetic energy of a body with a mass of 30 kg and a speed of 10 m/s is 1500 kg m²/s².
In physics, the kinetic energy of аn object is the form of energy thаt it possesses due to its motion. In the case above, we are given the formula for kinetic energy:
KE = 0.5 * m * v²
where m is mass and v is speed. And we are given m = 30 kg and v = 10 m/s.
To calculate the kinetic energy, we simply plug in the values of m and v into the formula:
KE = 0.5 * 30 kg * (10 m/s)²
Simplifying the equation gives:
KE = 0.5 * 30 kg * 100 m²/s²
KE = 15 kg * 100 m²/s²
KE = 1500 kg m²/s²
Therefore, the kinetic energy of the body is 1500 kg m²/s².
For more information about kinetic energy refers to the link: brainly.com/question/26472013
#SPJ11
Answer:
The kinetic energy of a body mass of 30 kg traveling in space at a speed of 10 m/s can be calculated with the formula KE = 0.5 * m * v^2, where m is the mass of the body and v is the speed. In this case, the kinetic energy would be equal to 0.5 * 30 * 10^2 = 1500 Joules.