Answer:
the bearing of the line AC will be 154° 51' 48"
Explanation:
given,
bearing of the line AB = 234° 51' 48"
an anticlockwise measure of an angle to the point C measured 80°
to calculate the bearing of AC.
As the bearing of line AB is calculated clockwise from North direction.
the angle is moved anticlockwise now the bearing of AC will be calculated by
= bearing of line AB - 80°
= 234° 51' 48" - 80°
= 154° 51' 48".
= 154.5148
so, the bearing of the line AC will be 154° 51' 48"
The beaker which consists of water at a temperature of 45℃ represents ‘The average kinetic energy of water particles’.
<u>Explanation:
</u>
According to ‘Kinetic Molecular Theory of water’ the average kinetic energy of atoms in a system is directly proportional to its temperature. So, this explains that the kinetic energy in between the inner particles of water which is in the beaker is something which represents the temperature of the water as 45℃. This kinetic energy is produced due to the moment of particles of the matter. (A liquid is the second state of matter which is prepared by tiny particles, such as atoms, inter-molecular bonds between the particles helps these atoms to hold together)
.
Answer:
impulse will be equal to 1.969 N-s
Explanation:
We have given mass of the golf ball m = 60.8 gram = 0.0608 kg
Horizontal speed of the ball is v = 32.4 m/sec
We have to find the impulse of the golf ball
We know that impulse is equal to momentum of the object
So we have to find the momentum of the object
Momentum of the object is equal to product of mass of the object and speed of the object
So
As momentum is equal to impulse so impulse will be equal to 1.969 N-s
C. Making spikes on shoes because it doesn’t cause friction (i think)
Answer:
1 by 3 units
Explanation:
The resistance (R) of a conductor is given by the formula:
R = ρL / A
where L is the length of the conductor, ρ is resistivity and A is the cross sectional area.
Let us assume that the metal bar has a resistivity of ρ.
a) If the leads is attached to the two opposite sides that have dimensions of 1 by 3 units.
The length of the bar would be 13 units and the cross sectional area (A) would be = 1 * 3 = 3 units²
R₁ = ρL / A = ρ(13) / 3 = 13ρ / 3
b) If the leads is attached to the two opposite sides that have dimensions of 3 by 13 units.
The length of the bar would be 1 units and the cross sectional area (A) would be = 3 * 13 = 39 units²
R₂ = ρL / A = ρ(1) / 39 = ρ / 39
c) If the leads is attached to the two opposite sides that have dimensions of 1 by 13 units.
The length of the bar would be 3 units and the cross sectional area (A) would be = 1 * 13 = 13 units²
R₃ = ρL / A = ρ(3) / 13 = 3ρ / 13
Therefore we can see that the largest resistance is gotten If the leads is attached to the two opposite sides that have dimensions of 1 by 3 units