Answer:
False
Explanation:
It is located in the little dipper whose stars are more faint.
S: 198 m
v=39 m/s
u=0
t=?
a=?
v²=u²+2as
(39)²=(0)²+2(a)(198)
1521=396a
1521/396=a
3.84 m/s^2 = a
Hope I helped :)
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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Answer:
326149.2 KJ
Explanation:
The heat transfer toward and object that suffered an increase in temperature can be calculated using the expression:
Q = m*cv*ΔT
Where m is the mass of the object, cv is the specific heat capacity at constant volume, which basically means the amount of heat necessary for a 1kg of water to increase 1C degree in temperatur, and ΔT is the change in temperature.
A 65000 L swimming pool will have a mass of:
65000L *
= 65000 kg
The specific heat capacity at constant volume of water is equal to 4.1814 KJ/KgC.
We replace the data and get:
Q = m*cv*ΔT = 65000 kg * 4.1814 KJ/KgC * 1.2°C = 326149.2 KJ
