Answer:
The minumum speed the pail must have at its highest point if no water is to spill from it
= 2.64 m/s
Explanation:
Working with the forces acting on the water in the pail at any point.
The weight of water is always directed downwards.
The normal force exerted on the water by the pail is always directed towards the centre of the circle of the circular motion.
And the centripetal force, which keeps the system in its circular motion, is the net force as a result of those two previously mentioned force.
At the highest point of the motion, the top of the vertical circle, the weight and the normal force on the water are both directed downwards.
Net force = W + (normal force)
But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.
At this point of minimum velocity,
Normal force = 0
Net force = W
Net force = centripetal force = (mv²/r)
W = mg
(mv²/r) = mg
r = 0.710 m
g = 9.8 m/s²
v² = gr = 9.8 × 0.71 = 6.958
v = √(6.958) = 2.64 m/s
Hope this Helps!!!
Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.
Hope i helped
Have a good day :)
The final speed of the nickel at the given quantity of heat is determined as 202.1 m/s.
<h3>Final speed of the nickel</h3>
Apply the principle of conservation of energy.
Q = mcΔθ
Q = (18)(0.444)(66 - 20)
Q = 367.63 J
Q = K.E = ¹/₂mv²
2K.E = mv²
v = √(2K.E/m)
where;
v = √(2 x 367.63)/(0.018))
v = 202.1 m/s
Learn more about speed here: brainly.com/question/4931057
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The energy conservation and trigonometry we can find the results for the questions about the movement of the acrobat are;
a) The maximum speed is v = 4.89 m / s
b) The maximum height is h = 1.22 m
The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.
Em = K + U
Let's write the energy in two points.
Starting point. Highest part of the oscillation
Em₀ = U = m g h
Final point. Lower part of the movement
= K = ½ m v²
Energy is conserved.
Emo =
m g h = ½ m v²
v² = 2 gh
Let's use trigonometry to find the height, see attached.
h = L - L cos θ
h = L (1- cos θ)
They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.
h = 3.7 (1- cos 48)
h = 1.22 m
this is the maximum height of the movement.
Let's calculate the velocity.
v = 4.89 m / s
In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;
a) The maximum speed is v = 4.89 m / s
b) The maximum height is h = 1.22 m
Learn more here: brainly.com/question/13010190
Answer:
the magnitude of acceleration will be 1.50m/s^2
Explanation:
To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma
if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.
The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).
you will find that Fnet=15.0N
With that, plug in the values you know to calculate the acceleration of the block:
Fnet=ma
(15.0N)=(10.0kg)a from her you can divide both sides by 10 to isolate a:
1.50=a (and now make sure to label the units of your answer)
a=1.50m/s^2 (which is the typical unit for acceleration)
