Answer:
c. Momentum is the product of mass and velocity
e. Momentum is a vector quantity
g. Momentum has unit of kgm/s
Explanation:
Linear momentum P
P = m .v
m =mass
v=Velocity
If mass take in kg and velocity is in m/s then momentum p will be in kg.m/s.
1. momentum is the product of velocity and mass.
2.Momentum is a vector quantity.
3.Momentum have kg.m/s unit.
So the following option are correct.
c. Momentum is the product of mass and velocity
e. Momentum is a vector quantity
g. Momentum has unit of kgm/s.
Note-
1.Joule is the unit of energy.
2.One-half the product of mass and the square of the object's speed is known as kinetic energy.
Answer:
<h3>How did scientists “discover” Pangea and other supercontinents of the past?</h3>
Nowadays, they can study the geologic record and use radioactive dating, seismic surveys, and other technologies to construct maps of how the world looked at various points in Earth's history.
Answer:
parallel circuit
Explanation:
In a parallel circuit, the potential difference across each of the resistors that make up the circuit is the same. This leads to a higher current flowing through each resistor and subsequently the total current flowing through all the resistors is higher.
To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.
PART A)
The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:
. This is the total charge on the inner surface of the conducting shell.
PART B)
The positive charge (of the same value) on the external surface of the conducting shell is:

The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,



The power developed is 500 W ( to the nearest Watt)
Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.


Therefore,

Substitute the given values of force , displacement and time


Thus the Power can be rounded off to the nearest value of 500 W