The planet MARS is visible without a telescope on many clear nights. The planets JUPITER, MERCURY, VENUS and SATURN are also viewable without the aid of magnification.
D=Vot+1/2at^2
In this case, there is no initial y velocity so the term Vot=0 so d=1/2at^2
acceleration=acceleration due to gravity=-9.8m/s^2
It falls - 22cm or -0.22m
We have - 0.22=1/2(-9.8)t^2
t^2=(-0.44)/(-9.8)
t=sqrt[0.44/9.8]
Answer:
11.95m/s
Explanation:
A moving object has a kinetic energy of 150 J and a momentum of 25.1 kg·m/s.
a) Find the speed of the object. Answer in units of m
K. E =½mv²
150= ½mv²
Multiply both sides by 2
mv² = 300
Divide both sides by v²
m = 300/v² .................. Equation 1
Momentum is the product of mass and velocity
Momentum = mv
25.1 = mv
Divide both sides by v
m = 25.1/v ................ Equation 2
Equate equations 1 and 2
300/v² = 25.1/v
Cross multiply
25.1v² = 300v
Multiply v with both sides
25.1v = 300
Divide both sides by 25.1
v = 300/25.1
V = 11.95m/s
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The answer is 10.5 kg m/s
Impulse (I) is the multiplication of force (F) and time interval (Δt): I = F · Δt
Force (F) is the multiplication of mass (m) and acceleration (a): F = m · a
Acceleration (a) can be expressed as change in velocity (v) divided by time interval (Δt): a = Δv/Δt
So:
a = Δv/Δt ⇒ F = m · a = m · Δv/Δt
F = m · Δv/Δt ⇒ I = m · Δv/Δt · Δt
Since Δt can be cancelled out, impulse can be expressed as:
I = m · Δv = m · (v2 - v1)
It is given:
m = 1.5 kg
v1 = 15 m/s
v2 = 22 m/s
I = 1.5 · (22 - 15) = 1.5 · 7 = 10.5 kgm/s.
The question is incomplete. The complete question is :
In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?
Solution :
Let M.I. of disk A = 
So, M.I. of disk B = 
Angular velocity of A = 
So the kinetic energy of the disk A = 
After coupling, the angular velocity of both the disks will be equal to ω.
Angular momentum will be conserved.
So,
Now,
Therefore, the maximum initial K.E. = 3066.67 J
