Any charged object can<span> exert the force upon other objects ... i think tell me if im right</span>
Answer:
210
Explanation:
A ball rolls horizontally off the cliff at a speed of 30 m/s. It takes 7 seconds for the ball to hit the ground. What is the height of the cliff and the horizontal distance traveled by the ball?
S = (1/2)*9.8 m/s^2 * 7^2 = 240.1 m if the ball is very dense so air resistance, and therefore terminal velocity, can be ignored.
S = v * t = 30 m/s * 7 s = 210 m for the horizontal distance, again assuming negligible air resistance.
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:
a) x₁ = 290.50 feet
, x₂ = 169.74 feet
, b) v_max= 41 mph
Explanation:
For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles
X Axis fr = m a
Y Axis N-W = 0
N = W = mg
The force of friction has the expression
fr = μ N
We replace
μ mg = ma
a = μ g
g = 32 feet / s²
Let's calculate the acceleration for each coefficient and friction
μ a (feet / s2)
0.599 19.168
0.536 17,152
0.480 15.360
0.350 11.200
These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)
v² = v₀² - 2 a x
When the speed stops it is zero
x₁ = v₀² / 2 a₁
Let's reduce speed
v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²
Let's calculate the maximum braking distance
x₁ = 80.667² / (2 11.2)
x₁ = 290.50 feet
The minimum braking distance
x₂ = 80.667² / (2 19.168)
x₂ = 169.74 feet
b) maximum speed to stop at distance x = 155 feet
0 = v₀² - 2 a x
v₀ = √2 a x
We calculate the speed for the two accelerations
v₀₁ = √ (2 11.2 155)
v₀₁ = 58.92 feet / s
v₀₂ = √ (2 19.168 155)
v₀₂ = 77.08 feet / s
To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph
