Answer:
The average speed can be calculated as the quotient between the distance travelled and the time needed to travel that distance.
To go to the school, he travels 2.4 km in 0.6 hours, then here the average speed is:
s = (2.4km)/(0.6 hours) = 4 km/h
To return to his home, he travels 2.4km again, this time in only 0.4 hours, then here the average speed is:
s' = (2.4 km)/(0.4 hours) = 6 km/h.
Now, if we want the total average speed (of going and returning) we have that the total distance traveled is two times the distance between his home and school, and the total time is 0.6 hours plus 0.4 hours, then the average speed is:
S = (2*2.4 km)/(0.6 hours + 0.4 hours)
S = (4.8km)/(1 h) = 4.8 km/h
Answer:
T = 2010 N
Explanation:
m = mass of the uniform beam = 150 kg
Force of gravity acting on the beam at its center is given as
W = mg
W = 150 x 9.8
W = 1470 N
T = Tension force in the wire
θ = angle made by the wire with the horizontal = 47° deg
L = length of the beam
From the figure,
AC = L
BC = L/2
From the figure, using equilibrium of torque about point C
T (AC) Sin47 = W (BC)
T L Sin47 = W (L/2)
T Sin47 = W/2
T Sin47 = 1470
T = 2010 N
New moon or cresant moon i believe
Explanation:
c. if the vector is oriented at 0° from the X -axis.
<u>We are Given:</u>
Mass of the block (m) = 500 grams or 0.5 Kg
Initial velocity of the block (u) = 0 m/s
Distance travelled by the block (s) = 8 m
Time taken to cover 8 m (t)= 2 seconds
Acceleration of the block (a) = a m/s²
<u>Solving for the acceleration:</u>
From the seconds equation of motion:
s = ut + 1/2* (at²)
<em>replacing the variables</em>
8 = (0)(2) + 1/2(a)(2)²
8 = 2a
a = 4 m/s²
Therefore, the acceleration of the block is 4 m/s²
