Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
Answer:
0.8 seconds
Explanation:
F=ma
Let x be the seconds the force is applied.
m = 20kg
F = 50 Newtons (kg*m/sec^2)
acceleration, a, is provided for x seconds to increase the speed from 1 m/s to 3 m/s, an increase of 2m/s
Let's calculate the acceleration of the cart:
F=ma
(50 kg*m/s^2) = (20kg)*a
a = 2.5 m/s^2
---
The acceleration is 2.5 m/s^2. The cart increases speed by 2.5 m/s every second.
We want the number of seconds it takes to add 2.0 m/sec to the speed:
(2.5 m/s^2)*x = 2.0 m/s
x = (2.0/2.5) sec
x = 0.8 seconds
Answer:
Heating water to produce steam which drives a turbine
Explanation:
Generation of electricity in coal-burning power plants and nuclear power plants both involve heating water to produce steam which drives a turbine.
Answer:
25 mm = 0 deg C
200 mm = 100 deg C
200 - 25 = 175 = change in thread per 100 deg C
95 - 25 = 70 mm - change in thread from 0 deg C
70 / 175 * 100 = 40 deg C final temperature at 95 mm
<em>1</em><em>.</em><em>259ms^2</em>
Explanation:
since, WORK DONE = FORCE*DISTANCE
AND, FORCE=MASS*ACCELERATION
SO, THE WORK DONE BECOMES=MASS*ACCELERATION*DISTANCE
ACCELERATION=WORK/(MASS*DISTANCE)
AND, WORK=686J
MASS=227kg
DISTANCE=2.4m
THEREFORE, ACCELERATION=686/(227*2.4)
=686/544.8
=1.259ms^2
