Answer:
Answer:196 Joules
Explanation:
Hello
Note: I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem
the work is the product of a force applied to a body and the displacement of the body in the direction of this force
assuming that the force goes in the same direction of the displacement, that is upwards
W=F*D (work, force,displacement)
the force necessary to move the object will be
![F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules](https://tex.z-dn.net/?f=F%3Dmg%28mass%20%2Agravity%29%5C%5CF%3D4kgm%2A9.8%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%5C%5C%20F%3D39.2%20Newtons%5C%5Creplace%5C%5C%5C%5CW%3D39.2%20N%2A5m%5C%5CW%3D196%5C%20Joules)
Answer:196 Joules
I hope it helps
The simplest way to do this is to set up equivalent fractions, like this-
![\frac{1}{2.2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2.2%7D%20)
=
![\frac{40}{x}](https://tex.z-dn.net/?f=%20%5Cfrac%7B40%7D%7Bx%7D%20)
Solve for x by using cross multiplication.
40*2.2= 88
1*x=88
x=88
Therefore, the boy weighs 88lbs.
During the falling down, two forces are present: gravitational force and air resistance.
So, the net force = mg - f = 12*9.8 - 27 = 90.6 N, the positive sign indicates the direction is pointing towards the center of the earth, which is the same as gravitational force
Answer: Igneous it forms because of magma but magma is under the earths surface so its Igneous
Explanation:
Answer: Tension = 47.8N, Δx = 11.5×
m.
Tension = 95.6N, Δx = 15.4×
m
Explanation: A speed of wave on a string under a tension force can be calculated as:
![|v| = \sqrt{\frac{F_{T}}{\mu} }](https://tex.z-dn.net/?f=%7Cv%7C%20%3D%20%5Csqrt%7B%5Cfrac%7BF_%7BT%7D%7D%7B%5Cmu%7D%20%7D)
is tension force (N)
μ is linear density (kg/m)
Determining velocity:
![|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }](https://tex.z-dn.net/?f=%7Cv%7C%20%3D%20%5Csqrt%7B%5Cfrac%7B47.8%7D%7B5.47.10%5E%7B-3%7D%7D%20%7D)
![|v| = \sqrt{0.00874 }](https://tex.z-dn.net/?f=%7Cv%7C%20%3D%20%5Csqrt%7B0.00874%20%7D)
0.0935 m/s
The displacement a pulse traveled in 1.23ms:
![\Delta x = |v|.t](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%20%7Cv%7C.t)
![\Delta x = 9.35.10^{-2}*1.23.10^{-3}](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%209.35.10%5E%7B-2%7D%2A1.23.10%5E%7B-3%7D)
Δx = 11.5×
With tension of 47.8N, a pulse will travel Δx = 11.5×
m.
Doubling Tension:
![|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }](https://tex.z-dn.net/?f=%7Cv%7C%20%3D%20%5Csqrt%7B%5Cfrac%7B2%2A47.8%7D%7B5.47.10%5E%7B-3%7D%7D%20%7D)
![|v| = \sqrt{2.0.00874 }](https://tex.z-dn.net/?f=%7Cv%7C%20%3D%20%5Csqrt%7B2.0.00874%20%7D)
![|v| = \sqrt{0.01568}](https://tex.z-dn.net/?f=%7Cv%7C%20%3D%20%5Csqrt%7B0.01568%7D)
|v| = 0.1252 m/s
Displacement for same time:
![\Delta x = |v|.t](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%20%7Cv%7C.t)
![\Delta x = 12.52.10^{-2}*1.23.10^{-3}](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%2012.52.10%5E%7B-2%7D%2A1.23.10%5E%7B-3%7D)
15.4×![10^{-5}](https://tex.z-dn.net/?f=10%5E%7B-5%7D)
With doubled tension, it travels
15.4×
m