<span>EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.</span>
First, determine the mass of the object by dividing its weight on Earth by 9.8 m/s² as shown below,
m = 250 N / 9.8 m/s² = 25.51 kg
Then, multiply the obtained mass by the acceleration due to gravity (g) on Pluto.
W (in Pluto) = (25.51 kg) x (0.61 m/s²) = 15.56 N
Therefore, the object will only weigh 15.56 N.
intraplate earthquake<span> occurs in the interior of a </span>tectonic plate, basically its a lower version of an earth quake but just less damage. The cause of them is the two tectonic plate hitting each other or i should say sliding togather.
Answer:
Resistance, 
Explanation:
Given that,
Voltage of the battery, V = 9 volts
Current produced in the circuit, I = 17 A
We need to find the resistance when shorted by a wire of negligible resistance. It is a case of Ohm's law. The voltage is given by :
So, the resistance in the circuit is 0.529 ohms. Hence, this is the required solution.
The calculated coefficient of kinetic friction is 0.33125.'
The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
given mass of the block=10 kg
spring constant k= 2250 Nm
now according to principal of conservation of energy we observe,
the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.
mgh= μ (mgl) +1/2 kx²
10 x 10 x 3= μ(600) +(1125) (0.09)
μ(600) =300 - 101.25
μ = 198.75÷600
μ =0.33125
The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)
Learn more about kinetic friction here-
brainly.com/question/13754413
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