Complete Question
An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters ) is the airplane above the ground 13 seconds after takeoff?
Answer:
The height is 
Explanation:
From the question we are told that
The speed at which the plane takes off is 
The angle at which it takes off is 
The time taken is 
The vertical distance traveled is mathematically represented as
Substituting values
Explanation:
Show that the motion of a mass attached to the end of a spring is SHM
Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed
at the a firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface.
If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.
According to "Hook's Law
F = - Kx ---- (1)
Negative sign indicates that the elastic restoring force is opposite to the displacement.
Where K= Spring Constant
If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion
from a to b and then b to a.
According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given by
F = ma ---- (2)
Comparing equation (1) & (2)
ma = -kx
Here k/m is constant term, therefore ,
a = - (Constant)x
or
a a -x
This relation indicates that the acceleration of body attached to the end elastic spring is directly proportional to its displacement. Therefore its motion is Simple Harmonic Motion.
Answer:
The second trumpeter will be playing at frequency = 515 Hz
Explanation: Given that the note sounds lower and they can hear 20 beats in 4.0 s.
Beat frequency = 20/4 = 5 Hz
Beat frequency = F2 - F1
5 = 520 - F1
F1 = 520 - 5
F1 = 515 Hz
Since the note sound lower, the second trumpeter will be playing at 515 Hz frequency
If you have no idea what the voltage is that you're about to measure,
then you should set the meter to the highest range before you connect
it to the two points in the circuit.
Analog meters indicate the measurement by moving a physical needle
across a physical card with physical numbers printed on it. If the unknown
voltage happens to be 100 times the full range to which the meter is set,
then the needle may find itself trying to move to a position that's 100 times
past the highest number on the meter's face. You'll hear a soft 'twang',
followed by a louder 'CLICK'. Then you'll wonder why the meter has no
needle on it, and then you'll walk over to the other side of the room and
pick up the needle off the floor, and then you'll probably put the needle
in your pocket. That will end your voltage measurements for that day,
and certainly for that meter.
Been there.
Done that.
Answer:
5.66 × 10⁻²³ m/s
Explanation:
If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.
Since initial momentum = final momentum,
mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.
My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?
So mv₁ + M × 0 = m × 0 + MV₂
mv₁ = MV₂
V₂ = mv₁/M = 54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s
