From the calculation and the momentum of the body, the velocity is 62 m/s
<h3>What is momentum?</h3>
The term momentum refers to the product of mass and velocity. Now recall that the rate of change of momentum is equal to the impressed force.
Hence;
7440 kg-m/s = 120 kg * v
v= 7440 kg-m/s/120 kg
v = 62 m/s
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Answer:
ω = 0.1 rad/s
v = 0.002 m/s
Explanation:
The angular velcoity of the second hand of the clock can be found by:
ω = θ/t
where,
ω = Angular Speed
θ = Angular Displacement
t = time taken
Now, for one complete revolution of second hand of the clock:
θ = 2π rad
t = 60 s
Therefore,
ω = 2π rad/60 s
<u>ω = 0.1 rad/s</u>
Now, for the linear speed (V):
V = rω
where,
V = Linear Speed of Second Hand = ?
r = radius = length of second hand = 0.02 m
Therefore,
V = (0.02 m)(0.1 rad/s)
<u>V = 0.002 m/s</u>
For rectilinear motions, derived formulas all based on Newton's laws of motion are formulated. The equation for acceleration is
a = (v2-v1)/t, where v2 and v1 is the final and initial velocity of the rocket. We know that at the end of 1.41 s, the rocket comes to a stop. So, v2=0. Then, we can determine v1.
-52.7 = (0-v1)/1.41
v1 = 74.31 m/s
We can use v1 for the formula of the maximum height attained by an object thrown upwards:
Hmax = v1^2/2g = (74.31^2)/(2*9.81) = 281.42 m
The maximum height attained by the model rocket is 281.42 m.
For the amount of time for the whole flight of the model rocket, there are 3 sections to this: time at constant acceleration, time when it lost fuel and reached its maximum height and the time for the free fall.
Time at constant acceleration is given to be 1.41 s. Time when it lost fuel covers the difference of the maximum height and the distance travelled at constant acceleration.
2ax=v2^2-v1^2
2(-52.7)(x) = 0^2-74.31^2
x =52.4 m (distance it covered at constant acceleration)
Then. when it travels upwards only by a force of gravity,
d = v1(t) + 1/2*a*t^2
281.42-52.386 = (0)^2+1/2*(9.81)(t^2)
t = 6.83 s (time when it lost fuel and reached its maximum height)
Lastly, for free falling objects, the equation is
t = √2y/g = √2(281.42)/9.81 = 7.57 s
Therefore, the total time= 1.41+6.83+7.57 = 15.81 s
The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.
The answer is Anorthosite. This is a kind of intrusive igneous rock composed mainly of calcium-rich plagioclase feldspar. All anorthosites located on Earth contain of rough crystals, but some illustrations of the rock reserved from the Moon are outstandingly crystalline.
