If the 30 N on the rope were pulled straight up, it would offset the force of gravity ( m g = 10 kg * 9.8 N/kg = 98 N) , leaving a net force up from the ground on the sled of 98-30 = 68 N. Since the rope is pulled at the angle of 25o, only part of the force is in the upward direction, (30N)(sin(25) = (30)(.423) = 12.7. So the net force becomes 98 N down offset by 12.7 up or 98-12.7 = 85.3 N. Ah, there it is: C.
Answer:
0.107 m
Explanation:
The magnetic field at the center of a current-carrying loop is given by
where
is the vacuum permeability
I is the current
r is the radius of the loop
In this problem we have
I = 3.40 A is the current in the loop
is the magnetic field at the centre of the loop
So, solving the formula for r we find
Answer:
1.53seconds
Explanation:
Using first equation of motion :
V=U + at
Where final velocity (V) =+8.3m/s
Initial velocity (U) =+4.4m/s
Acceleration (a) = 0.65m/s^2
time(s)=?
V=U + at
+8.3^2 = +4.4 + 0.65 * t
Making t the subject of the formula :
Therefore, t= ( +8.3 - 4.4)/0.65 = 1.53seconds
We can use the ideal gas equation which is expressed as PV = nRT. At a constant volume and number of moles of the gas the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:
T1/P1 = T2/P2
P2 = T2 x P1 / T1
P2 = 273 x 340 / 713
<span>P2 = 130 kPa</span>
The answer is:
B. <span>X: Work is done to the system and temperature increases.
Y: Work is done by the system and temperature decreases.</span>