Answer:
Imp = -4 [kg*m/s]
Explanation:
To determine the impulse we must know the initial speed and final velocity of the ball, that is before and after hitting the wall. As these data are given in the input information, we can very easily find the value of the impulse given during the hit on the wall.
![m*v_{1}+Imp=mv_{2} \\Imp= m*v_{2}-m*v_{1}\\Imp = (2*4)-(2*6)\\Imp = -4[kg*\frac{m}{s} ] \\](https://tex.z-dn.net/?f=m%2Av_%7B1%7D%2BImp%3Dmv_%7B2%7D%20%20%5C%5CImp%3D%20m%2Av_%7B2%7D-m%2Av_%7B1%7D%5C%5CImp%20%3D%20%282%2A4%29-%282%2A6%29%5C%5CImp%20%3D%20-4%5Bkg%2A%5Cfrac%7Bm%7D%7Bs%7D%20%5D%20%5C%5C)
The negative sign means that the impulse is given in the opposite direction to the one that the ball was thrown.
Answer;
D. The car would begin to move in the direction it was headed in a straight line.
Explanation;
-Centripetal force is any net force causing uniform circular motion. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration.
-The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.
-Therefore,If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular
path were reduced then the car would begin to move in the direction it was headed in a straight line.
Answer:
Explanation:
Technician A is incorrect as in series circuit the current is the same for all the resistors and if one bulb fails in the path then the circuit is incomplete and therefore no flow of current.
Technician B is correct as in parallel circuit there are multiple paths for current through which it can pass and malfunctioning of one resistor does not affect the other.
Answer:
0.0360531138247 V/m
Explanation:
= Resistivity of gold =
(General value)
I = Current = 940 mA
d = Diameter = 0.9 mm
A = Area = ![\dfrac{\pi}{4}d^2](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpi%7D%7B4%7Dd%5E2)
E = Electric field
Resistivity is given by
![\rho=\dfrac{EA}{I}\\\Rightarrow E=\dfrac{\rho I}{A}\\\Rightarrow E=\dfrac{2.44\times 10^{-8}\times 940\times 10^{-3}}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}\\\Rightarrow E=0.0360531138247\ V/m](https://tex.z-dn.net/?f=%5Crho%3D%5Cdfrac%7BEA%7D%7BI%7D%5C%5C%5CRightarrow%20E%3D%5Cdfrac%7B%5Crho%20I%7D%7BA%7D%5C%5C%5CRightarrow%20E%3D%5Cdfrac%7B2.44%5Ctimes%2010%5E%7B-8%7D%5Ctimes%20940%5Ctimes%2010%5E%7B-3%7D%7D%7B%5Cdfrac%7B%5Cpi%7D%7B4%7D%280.9%5Ctimes%2010%5E%7B-3%7D%29%5E2%7D%5C%5C%5CRightarrow%20E%3D0.0360531138247%5C%20V%2Fm)
The electric field in the wire is 0.0360531138247 V/m
F=G(m1*m2)\D^2
so
G=6.67 × 10^(-11)<span> m</span>3<span> kg</span>-1<span> s</span><span>-2
</span>
= G(2*2)\4
=G