Answer:
Iv. The direction of motion of each object is the same.
Explanation:
Momentum of an object is defined as:
![p= mv](https://tex.z-dn.net/?f=p%3D%20mv)
where
m is the mass of the object
v is its velocity
Since v is a vector, it follows that p is also a vector, and it has the same direction of the velocity. This means that if two objects have same momentum, then the direction of their momentum vector is also the same: therefore, the direction of motion of each object must be the same.
The other options are wrong, because they just state that only one of the two quantities involved in the equation (mass or velocity) is the same, but the two objects can actually have the same momentum even if they have different masses and velocities (in fact, the only thing that matters is that the product between mass and velocity is the same).
Answer:
1800W
Explanation:
Power=Current×Potential difference
=15A×120V
=1800W
Answer:
100 N
Explanation:
Given that,
Two forces whose resultant is 100newton are perpendicular to each other.
If one of them makes an angle of 60newton with the resultant.
![F_1=100\times \sin60=86.60\ N](https://tex.z-dn.net/?f=F_1%3D100%5Ctimes%20%5Csin60%3D86.60%5C%20N)
and
![F_2=100\times \cos60=50\ N](https://tex.z-dn.net/?f=F_2%3D100%5Ctimes%20%5Ccos60%3D50%5C%20N)
The magnitude of force,
![F=\sqrt{F_1^2+F_2^2} \\\\F=\sqrt{86.6^2+50^2} \\\\F=99.99\ N](https://tex.z-dn.net/?f=F%3D%5Csqrt%7BF_1%5E2%2BF_2%5E2%7D%20%5C%5C%5C%5CF%3D%5Csqrt%7B86.6%5E2%2B50%5E2%7D%20%5C%5C%5C%5CF%3D99.99%5C%20N)
or
F = 100 N
So, the magnitude of force is 100 N.
Answer:
a) Magnetic field strength, B = 2.397 * 10⁻⁷ T
b) Total energy density, U = 4.58 * 10⁻⁸ J/m³
c) Power flow per unit area, S = 13.71 W/m²
Explanation:
a) Electric field strength, E = 71.9 V/m
The relationship between the Electric field strength and the magnetic field strength in vacuum is:
E = Bc where c = 3.0 * 10⁸ m/s
71.9 = B * 3.0 * 10⁸
B = 71.9 / (3.0 * 10⁸)
B = 23.97 * 10⁻⁸
B = 2.397 * 10⁻⁷ T
b) Total Energy Density:
![U = \frac{1}{2} \epsilon_0E^2 + \frac{1}{2} \frac{B^2}{\mu_0} \\U = \frac{1}{2}* 8.85 * 10^{-12}*71.9^2 + \frac{1}{2} \frac{(2.397*10^{-7})^2}{4\pi*10^{-7}}\\U = 2.29 * 10^{-8} + 2.29 * 10^{-8}\\U = 4.58 * 10^{-8} J/m^3](https://tex.z-dn.net/?f=U%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cepsilon_0E%5E2%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7BB%5E2%7D%7B%5Cmu_0%7D%20%5C%5CU%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A%208.85%20%2A%2010%5E%7B-12%7D%2A71.9%5E2%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%282.397%2A10%5E%7B-7%7D%29%5E2%7D%7B4%5Cpi%2A10%5E%7B-7%7D%7D%5C%5CU%20%3D%202.29%20%2A%2010%5E%7B-8%7D%20%2B%202.29%20%2A%2010%5E%7B-8%7D%5C%5CU%20%3D%204.58%20%2A%2010%5E%7B-8%7D%20J%2Fm%5E3)
c)Power flow per unit area
![S = \frac{1}{\mu_0} EB\\S = \frac{1}{4\pi * 10^{-7} } * 71.9 * 2.397 * 10^{-7}\\S = 13.71 W/m^2](https://tex.z-dn.net/?f=S%20%3D%20%5Cfrac%7B1%7D%7B%5Cmu_0%7D%20EB%5C%5CS%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%2A%2010%5E%7B-7%7D%20%7D%20%2A%2071.9%20%2A%202.397%20%2A%2010%5E%7B-7%7D%5C%5CS%20%3D%2013.71%20W%2Fm%5E2)
The resistance of a conductor is given by
![R= \frac{\rho L}{A}](https://tex.z-dn.net/?f=R%3D%20%5Cfrac%7B%5Crho%20L%7D%7BA%7D%20)
where L is the length of the wire,
![\rho](https://tex.z-dn.net/?f=%5Crho)
the resistivity of the material and A the cross-sectional area.
We can see that if all the other quantities do not change, if the new length of the conductor is 4 times the original length:
![L'=4 L](https://tex.z-dn.net/?f=L%27%3D4%20L)
, then the new resistance is also 4 times the original value: