Answer:
Explanation:
Usando la ley de Hook tenemos:
Solving it for k we have:
Usando la misma ecuación y sabiendo k tenemos:
Espero esto te ayude!
Hey!
First, let's write the problem.
Subtract the numbers, we would do the following operation,
Add 2 to both sides.
This tells us that our final answer would be,
Thanks!
-TetraFish
Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
Answer:
The galaxy is moving away from the observer
Explanation: when a galaxy is moving away from us, the light we percieve from it is "streched". Since the wavelength has an inverse raltionship whith frequency, the longer the wavelength is, the lower the frequency. And lower frequencies correspond to red and infrarred light.
So when we see the light has shifted to the infrarred part of the spectrum, it means the source is traveling away from us, making the light waves we percieve streched and move from visible light to infrarred.
Answer:
a) 20s
b) 500m
Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.
To find time, we apply the UARM formula:
v final = (a x t) + v initial
Replacing the values gives us:
0 = (-10 x t) + 100
-100 = -10t
t = 10s
It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur
So 10s going up and another 10s going down:
10x2 = 20s
b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:
Δy = (1/2)(a)(t^2) + (v initial)(t)
Replacing the values gives us:
Δy = (1/2)(-10)(10^2) + (100)(10)
= (-5)(100) + 1000
= -500 + 1000
= 500 m
Hope this helps, brainliest would be appreciated :)
