So there are different ways this could be solved. I'll do try to explain it the way I was taught...
Simon is riding his bike at 12 km/hr relative to the sidewalk, away from where Keesha is.
Simon throws the ball at Keesha, at 5 km/hr.
Keesha sees the ball approaching her at (12-5) = 7 km/hr relative to the ground to her.
Therefore the answer is: 7 km/hr
A resultante das duas forças será zero, já q os sentidos são opostos e sãos iguais em módulo, elas se anulam. Logo, se a força resultante é zero, e F=ma, aceleração também será igual a zero.
G/mL is equivalent to g/cm^3, so we first convert the dimensions into cm:
2.20 cm, 1.35 cm, and 1.25 cm
Then the total volume is: V = lwh = 3.7125 cm^3
To get the density, we divide mass by volume: 2.50 g / 3.7125 cm^3 = 0.6734 g/cm^3 = 0.6734 g/mL
Answer:
the signs of heat and work are; -Q and -W
Explanation:
The first law of thermodynamics is given by; ΔU = Q − W
where;
ΔU is the change in internal energy of a system,
Q is the net heat transfer (the sum of all heat transfer into and out of the system)
W is the net work done (the sum of all work done on or by the system).
Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).
Since work is done by the system, W remains negative.
Thus, the signs of heat and work are; -Q and - W
Answer:
114.44 J
Explanation:
From Hook's Law,
F = ke................. Equation 1
Where F = Force required to stretch the spring, k = spring constant, e = extension.
make k the subject of the equation
k = F/e.............. Equation 2
Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.
Substitute into equation 2
k = 44.5/1.016
k = 43.799 N/m
Work done in stretching the 9 in beyond its natural length
W = 1/2ke²................. Equation 3
Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m
Substitute into equation 3
W = 1/2×43.799×2.286²
W = 114.44 J
