Answer:
a) m = 59.63 [kg]
b) Wm = 95.41 [N]
Explanation:
El peso de un cuerpo se define como el producto de la masa por la aceleración gravitacional. DE esta manera tenemos:
W = m*g
Donde:
m = masa [kg]
g = gravedad = 9.81 [m/s^2]
m = W / g
m = 585 / 9.81
m = 59.63 [kg]
Es importante aclarar que la masa se conserva independientemente de la ubicación del cuerpo en el espacio.
Por ende su masa sera la misma en la luna.
El peso en la luna se calcula como Wm y es igual a:
Wm = 59.63 * 1.6 = 95.41 [N]
Answer:
The object in a uniform motion covers same distances in an equal time period. Objects in a non-uniform motion cover dissimilar distances in an equal time period.
Explanation:
The speed of the object traveling in uniform motion is constant, the actual speed and the average speed of the moving body is same.
Answer:
11060M Joules, where M is the mass of the diver in kg
Explanation:
Mass of the skydiver missing, we're assuming it's M.
It's total energy is the sum of the contribution of his kinetic energy (K)- since he's moving at 50 m/s, and it's potential energy (U), since he's subject to earth gravity.
Energy is the sum of the two, so ![E = K+U= \frac 12 M v^2 + Mgh = M (\frac 12 \cdot 50^2 + 9.81\cdot 1000) = M ( 1250 + 9810) = 11060\cdot M](https://tex.z-dn.net/?f=E%20%3D%20K%2BU%3D%20%5Cfrac%2012%20M%20v%5E2%20%2B%20Mgh%20%3D%20M%20%28%5Cfrac%2012%20%5Ccdot%2050%5E2%20%2B%209.81%5Ccdot%201000%29%20%3D%20M%20%28%201250%20%2B%209810%29%20%3D%2011060%5Ccdot%20M)
Answer:
Rate of change of magnetic field is
Explanation:
We have given diameter of the circular loop is 13 cm = 0.13 m
So radius of the circular loop ![r=\frac{0.13}{2}=0.065m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B0.13%7D%7B2%7D%3D0.065m)
Length of the circular loop ![L=2\pi r=2\times 3.14\times 0.065=0.4082m](https://tex.z-dn.net/?f=L%3D2%5Cpi%20r%3D2%5Ctimes%203.14%5Ctimes%200.065%3D0.4082m)
Wire is made up of diameter of 2.6 mm
So radius ![r=\frac{2.6}{2}=1.3mm=0.0013m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B2.6%7D%7B2%7D%3D1.3mm%3D0.0013m)
Cross sectional area of wire ![A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%3D3.14%5Ctimes0.0013%5E2%3D5.30%5Ctimes%2010%5E%7B-6%7Dm%5E2)
Resistivity of wire ![\rho =2.18\times 10^{-8}m](https://tex.z-dn.net/?f=%5Crho%20%3D2.18%5Ctimes%2010%5E%7B-8%7Dm)
Resistance of wire ![R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B%5Crho%20L%7D%7BA%7D%3D%5Cfrac%7B2.18%5Ctimes%2010%5E%7B-8%7D%5Ctimes%200.4082%7D%7B5.30%5Ctimes%2010%5E%7B-6%7D%7D%3D1.67%5Ctimes%2010%5E%7B-3%7Dohm)
Current is given i = 11 A
So emf ![e=11\times 1.67\times 10^{-3}=0.0183volt](https://tex.z-dn.net/?f=e%3D11%5Ctimes%201.67%5Ctimes%2010%5E%7B-3%7D%3D0.0183volt)
Emf induced in the coil is ![e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}](https://tex.z-dn.net/?f=e%3D-%5Cfrac%7Bd%5CPhi%20%7D%7Bdt%7D%3D-A%5Cfrac%7BdB%7D%7Bdt%7D)
![0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}](https://tex.z-dn.net/?f=0.0183%3D5.30%5Ctimes%2010%5E%7B-6%7D%5Ctimes%20%5Cfrac%7BdB%7D%7Bdt%7D)
![\frac{dB}{dt}=3.466\times 10^3=T/sec](https://tex.z-dn.net/?f=%5Cfrac%7BdB%7D%7Bdt%7D%3D3.466%5Ctimes%2010%5E3%3DT%2Fsec)
M= ?
g=9.8 m/s (2)
h=20 m
Eg=362,600 J
Eg/mg
362,600 J/9.8 m/s (2) x 20 m
=1,850 m