Answer:
1. Luminosity
2.Apparent brightness
Explanation:
There are two factors on which brightness of star appear to be in the sky
The two factors are
1. Luminosity
2.Apparent brightness
1.Luminosity :It is defined as the total energy emitted by the object in a given time.Luminosity vary with the distance of observer from the star.Luminosity is a intrinsic property which depends on the fundamental chemical composition and structure of the material.Luminosity is depends on the size of star.Lager the star luminosity will be more.
2.Apparent brightness: It is defined as how bright a star appears from an observer on the earth and the amount of starlight reaching the earth.if the distance is large then the brightness decreases.When the distance of star from us small then the brightness of star increases.Distance is inversely proportional to brightness of the star.
Answer;
the potential difference
The magnitude of the electric current is directly proportional to the potential difference of the electric field
Explanation;
An electric current results from the collective movement of free charges under the effect of an electric field. An electric field exists and can be observed in the space around a single charge or a number of charges.
Electric fields cause charges to move. It stands to reason that an electric field applied to some material will cause currents to flow in that material. In other words, the current density is directly proportional to the electric field. The constant of proportionality σ is called the material’s conductivity.
Given
Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied
force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient
of kinetic friction µk = 0.3. If a force P
is applied to the body, no relative motion will take place until the applied
force is equal to the force of friction Ff, which is acting opposite to the
direction of motion. Magnitude of static force of friction between block A and
block B, Fs = µsN, where N is
reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA +
mB)a,
6 = (20 / 32.2 + 50 / 32.2)a
2.173a = 6
A = 2.76 ft/s^2
To check slipping occurs between block A and block B, consider block A:
P – Ff = mAaA
6 – Ff = 1.71
Ff = 4.29 lb
And also,
N = wA. We know static friction,
Fs = µsN
Fs = 0.4 x 20
Fs = 8lb
Frictional force is less than static friction. Ff < Fs
<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of
block B aB = 2.76 ft/s^2</span>
Answer:
Momentum, p = 5 kg-m/s
Explanation:
The magnitude of the momentum of an object is the product of its mass m and speed v i.e.
p = m v
Mass, m = 3 kg
Velocity, v = 1.5 m/s
So, momentum of this object is given by :
p = 4.5 kg-m/s
or
p = 5 kg-m/s
So, the magnitude of momentum is 5 kg-m/s. Hence, this is the required solution.
