Answer: Direction
Explanation: A vector is a geometrical representation of physical quantity. It has a particular direction with a specific magnitude. In the geometry of space whether it is two dimensional or three dimensional the vector quantity has a specific direction. Such as a stone is thrown with a velocity in a particular direction.
The path of the stone in three-dimension shows its direction and speed is its magnitude.
Hence, the velocity of stone has two property magnitude mentioned as speed and particular direction. On writing the mathematical expressions for vectors, it is denoted by arrow mark on its top as shown below.
Answer:
Required heat Q = 11,978 KJ
Explanation:
Given:
Mass = 5.3 kg
Latent heat of vaporization of water = 2,260 KJ / KG
Find:
Required heat Q
Computation:
Required heat Q = Mass x Latent heat of vaporization of water
Required heat Q = 5.3 x 2260
Required heat Q = 11,978 KJ
Required heat Q = 12,000 KJ (Approx.)
Answer:
0.35 T
Explanation:
Side, a = 0.132 m, e = 27.1 mV = 0.0271 V, dA / dt = 0.0785 m^2 / s
Use the Faraday's law of electromagnetic induction
e = rate of change of magnetic flux
Let b be the strength of magnetic field.
e = dФ / dt
e = d ( B A) / dt
e = B x dA / dt
0.0271 = B x 0.0785
B = 0.35 T
Answer:
10R/11 and 5R/2
Explanation:
The radius of the conducting shell = R,
Electrostatic potential inside the shell (r<R) = kq/R
Electrostatic potential outside the shell (r>R) = kq/r
If x is the point of zero potential
Electrostatic potential for inner shell,
Electrostatic potential for outer shell,
Electrostatic potential for the thin walled shell,
The values of X=r that satisfy the above equation are 10R/11 and 5R/2
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed
, and an accelerated motion on the y-axis, with initial speed
and acceleration
:
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).
To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
Therefore:
which has two solutions:
is the time of the beginning of the motion,
is the time at which the projectile hits the ground.
Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point: