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
It’s b because if you’re running at 5 miles per second being a 100kg weight is the fastest
Answer:
The temperature is 2541.799 K
Explanation:
The formula for black body radiation is given by the relation;
Q = eσAT⁴
Where:
Q = Rate of heat transfer 56.6
σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)
A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²
e = emissivity = 0.288
T = Temperature
Therefore, we have;
T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴
T = 2541.799 K
The temperature = 2541.799 K.
The correct answer is:
Work is negative, the environment did work on the object, and the energy of the system decreases.
In fact, the work-energy theorem states that the work done by the system is equal to its variation of kinetic energy:
In this problem, the variation of kinetic energy 
is negative (because the final velocity is less than the initial velocity), so the work is negative, and this means that the environment did work on the object, and its energy decreased.
a) At a position of 2.0m, the Initial energy is
all made up of the potential energy=m*g*hi<span>
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf
The percentage of energy remaining is E=m*g*hi/m*g*hf x 100
and since mass and gravity are constant so it leaves us with
just E=hi/hf
which 1.5/2.0 x100= 75% so we see that we lost 25% of the
energy or 0.25 in fraction
b) Here use the equation vf^2=vi^2+2gd
<span>where g is gravity, vf is the final velocity and vi is the
initial velocity while d is the distance travelled
so in here we are looking for the vi so let us isolate that
variable
we know that at maximum height or peak, the velocity is 0 so
vf is 0
therefore,</span></span>
vi =sqrt(-2gd) <span>
vi =sqrt(-2x-9.81x1.5) </span>
<span>vi =5.4 m/s
<span>c) The energy was converted to heat due to friction with the
air and the ground.</span></span>
