An object with mass 60 kg moved in outer space. When it was at location < 13, -19, -3 > its speed was 3.5 m/s. A single constant force < 220, 320, -120 > N acted on the object while the object moved from location < 13, -19, -3 > m to location < 18, -11, -8 > m. Then a different single constant force < 150, 230, 220 > N acted on the object while the object moved from location < 18, -11, -8 > m to location < 22, -17, -3 > m. What is the speed of the object at this final location?final speed = m/s
Answers
The speed of the object at its' final location is; 38 m/s
What is work energy theorem?
For the first force, we are given;
Force; F₁ = 220i + 320j - 120k
Initial Position; r₁ = 13i - 19j - 3k
Final Position; r₂ = 18i - 11j - 8k
Thus; Displacement; Δr = r₂ - r₁
Δr = 18i - 11j - 8k - (13i - 19j - 3k)
Δr = 5i + 8j - 5k
From work energy theorem, we know that;
F₁ * Δr = ¹/₂m(v₂² - v₁²)
We are given v₁ = 2.5 m/s and m = 60 kg. Thus;
(220i + 320j - 120k) × (5i + 8j - 5k) = ¹/₂ * 60(v₂² - 3.5²)
4260/30 = v₂² - 3.5²
1420 = v₂² - 12.25
Solving gives v₂ = 37.85 m/s
For the second force, we are given;
Force; F₂ = 150i + 230j - 220k
Initial Position; r₁ = 18i - 11j - 8k
Final Position; r₂ = 22i - 17j - 3k
Thus; Displacement; Δr = r₂ - r₁
Δr = 22i - 17j - 3k - (18i - 11j - 8k)
Δr = 4i - 6j + 5k
From work energy theorem, we know that;
F₂ * Δr = ¹/₂m(v₂² - v₁²)
Now, v₁ = 37.85 m/s and m = 60 kg. Thus;
(150i + 230j + 220k) × (4i - 6j + 5k) = ¹/₂ * 60(v₂² - 37.85²)
320/30 = v₂² - 37.85²
10.67 = v₂² - 1,432.6225
Solving gives v₂ = 38 m/s
Read more about Work Energy theorem at; brainly.com/question/14468674
(220, 320, -120).(18-13,-11+19,-8+3) +(150, 230, 220).(22-18,-17+11,-3+8)= 1/2 *60*(V^2- 3.5^2)
220*5+320*8+ -120*-5 + 150*4 + 230* 6 +220* -5= ..
simplify his
Related Questions
the three small spheres shown in (figure 1) carry charges q1 = 3.50 nc , q2 = -7.50 nc , and q3 = 2.05 nc .
The mass is 2.0 kg. If the net force is 20.0 N, what is the acceleration?
Two dogs pull on a toy. One pulls with a force of 10N, the other with a force of 15N. In what relative directions can the dogs act to give the toy the smallest acceleration?
What would likely happen if you were to touch the flask in which an endothermic reaction were occuring? A. The flask would probably feel cooler than before the reaction started. B. The flask would probably feel warmer than before the reaction started. C. The flask would feel the same as before the reaction started. D. none of the above
B. 1.2 m/s
C. 21.3 m/s
D. 12.8 m/s
Answers
Final answer:
The impact speed of the backpack is approximately 12.8 m/s.
Explanation:
To calculate the impact speed of the backpack, we can use the principle of conservation of energy. The potential energy of the backpack at the top of the window is converted into kinetic energy at the bottom. Assuming no air resistance, we can equate the potential energy to the kinetic energy:
mgh = 1/2mv^2
Here, m is the mass of the backpack, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the window (8.3 m). By rearranging the equation and solving for v, we can find the impact speed of the backpack.
Using the given values:
v = sqrt(2gh) = sqrt(2 * 9.8 * 8.3) ≈ 12.8 m/s
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Answers
THREE REASONS WHY IT WOULD BE DIFFICULT TO HAVE NO ELECTRICITY
1) it would be hard to get around
2) would not be able to do homework because no power to light up computers.ipads etc
3) without light we can't read and get an education
THREE REASONS WHY ELECTRICITY IS GREAT
1) we have devices that get lighted up by electricity like computers,ipods,etc
2) we could get around pretty easy with light
3) we could get an education cause we can see and use devices like computers to help us
HOPE THIS HELPED !!
Answers
Well, there you have a very important principle wrapped up in that question.
There's actually no such thing as a real, actual amount of potential energy.
There's only potential relative to some place. It's the work you have to do
to lift the object from that reference place to wherever it is now. It's also
the kinetic energy the object would have if it fell down to the reference place
from where it is now.
Here's the formula for potential energy: PE = (mass) x (gravity) x (height) .
So naturally, when you use that formula, you need to decide "height above what ?"
If you're reading a book while you're flying in a passenger jet, the book's PE is
(M x G x 0 meters) relative to your lap, (M x G x 1 meter) relative to the floor of the
plane, (M x G x 10,000 meters) relative to the ground, and maybe (M x G x 25,000 meters)
relative to the bottom of the ocean.
Let's say that gravity is 9.8 m/s² .
Then a 4kg block sitting on the floor has (39.2 x 0 meters) PE relative to the floor
it's sitting on, also (39.2 x 3 meters) relative to the floor that's one floor downstairs,
also (39.2 x 30 meters) relative to 10 floors downstairs, and if it's on the top floor of
the Amoco/Aon Center in Chicago, maybe (39.2 x 345 meters) relative to the floor
in the coffee shop that's off the lobby on the ground floor.
P.E.= mass×gravity×height, therefore 4kg×10×0= 0 J
K.E. is also equal to 0 because the block is not moving, is in the state of Inertia
Answers
Scientists consider them the smallest form of life. Yes, they are alive
Answers
4.58 m/s
Speed = displacement / time
2. Calculate the pressure in atmospheres using the van der Waals equation. For N2 , a=1.35 (L2⋅atm)/mol2 , and b=0.0387 L/mol
Answers
Answer:
1) 16.88 atm
2) 34.47 atm
Explanation:
Data:
Volume=0.700L
Temperature = 300K
Number of moles=0.480 mol
Ideal gas constant=0.082057 L*atm/K·mol
1) The ideal gas law is:
(1)
with P the pressure, T the temperature, n the number of moles, V the volume and R the ideal gas constant , so solvig (1) for P:
2) The vander Walls equation is:
solving for P
The pressure in atmospheres is 0.974 atm using the ideal gas law and 0.962 atm using the van der Waals equation for N2.
1. To calculate the pressure in atmosphere using the ideal gas law, we can use the equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
Rearranging the equation, we have P = (nRT)/V. Plugging in the given values, we get P = (0.480 mol * 0.0821 L·atm/mol·K * 300 K) / 0.700 L = 0.974 atm.
2. To calculate the pressure in atmosphere using the van der Waals equation, we can use the equation (P + an^2/V^2)(V - nb) = nRT, where a and b are constants specific to the gas being used. Rearranging the equation, we have P = (nRT/(V - nb)) - an^2/V^2.
Plugging in the given values and the constants for N2, we get P = (0.480 mol * 0.0821 L·atm/mol·K * 300 K/(0.700 L - 0.0387 L/mol * 0.480 mol))^2 - 1.35 (L^2·atm)/mol^2 * (0.480 mol)^2/(0.700 L)^2 = 0.962 atm.
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