Answer:
0.50m/s
Explanation:
Average velocity is the change in displacement of a body with respect to time.
Velocity = ∆S/∆t
∆S = 100m - 70m
∆S = 30m
∆t = 2min - 1 min
∆t = 1min = 60secs
Substitute the given parameters into the formula for velocity
Velocity = 30m/60s
Velocity = 1/2 m/s
Average Velocity = 0.5m/s
Explanation:
(a) Displacement of an object is the shortest path covered by it.
In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag. She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.
0.4 miles = 0.64 km
displacement = 0.7-0.3+0.64 = 1.04 km
(b) Average velocity = total displacement/total time
t = 15 min = 0.25 hour

Hence, this is the required solution.
Answer:
assume nitrogen is an ideal gas with cv=5R/2
assume argon is an ideal gas with cv=3R/2
n1=4moles
n2=2.5 moles
t1=75°C <em>in kelvin</em> t1=75+273
t1=348K
T2=130°C <em>in kelvin</em> t2=130+273
t2=403K
u=пCVΔT
U(N₂)+U(Argon)=0
<em>putting values:</em>
=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)
<em>by simplifying:</em>
Tfinal=363K
Answer:
U = 1 / r²
Explanation:
In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related
F = - dU / dr
this derivative is a gradient, that is, a directional derivative, so we must have
dU = - F. dr
the esxresion for strength is
F = B / r³
let's replace
∫ dU = - ∫ B / r³ dr
in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product
let's evaluate the integrals
U - Uo = -B (- / 2r² + 1 / 2r₀²)
To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)
U = B / 2r²
we substitute the value of B = 2
U = 1 / r²
Answer:
The magnitude of the average force exerted on the water by the blade is 960 N.
Explanation:
Given that,
The mass of water per second that strikes the blade is, 
Initial speed of the oncoming stream, u = 16 m/s
Final speed of the outgoing water stream, v = -16 m/s
We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :



F = -960 N
So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.