The three inclines have the same height and they are all frictionless, but they have different angles. Which object will have the highest speed at the bottom of the incline
Answer:
the speed of the body must be the same for the three hills regardless of their slope
v = [tex]\sqrt{2gy}[/tex]
Explanation:
To shorten the answer to this exercise, let's use the concept of conservation of energy
starting point. Highest point of the hill
Em₀ = U = m h y
final point. Lower part of the hill
Em_f = K = ½ m v²
as there is no friction the mechanical energy is conserved
Em₀ = Em_f
mg y = ½ m v²
v = [tex]\sqrt{2gy}[/tex]
Let's analyze this result, we see that the speed at the bottom of the hill depends only on the height of the hill, not on the slope, so the speed of the body must be the same for the three hills regardless of their slope
The water side of the wall of a 60-m-long dam is a quarter-circle with a radius of 7 m. Determine the hydrostatic force on the dam and its line of action when the dam is filled to the rim. Take the density of water to be 1000 kg/m3.
Answer:
[tex]26852726.19\ \text{N}[/tex]
[tex]57.52^{\circ}[/tex]
Explanation:
r = Radius of circle = 7 m
w = Width of dam = 60 m
h = Height of the dam will be half the radius = [tex]\dfrac{r}{2}[/tex]
A = Area = [tex]rw[/tex]
V = Volume = [tex]w\dfrac{\pi r^2}{4}[/tex]
Horizontal force is given by
[tex]F_x=\rho ghA\\\Rightarrow F_x=1000\times 9.81\times \dfrac{7}{2}\times 7\times 60\\\Rightarrow F_x=14420700\ \text{N}[/tex]
Vertical force is given by
[tex]F_y=\rho gV\\\Rightarrow F_y=1000\times 9.81\times 60\times \dfrac{\pi 7^2}{4}\\\Rightarrow F_y=22651982.59\ \text{N}[/tex]
Resultant force is
[tex]F=\sqrt{F_x^2+F_y^2}\\\Rightarrow F=\sqrt{14420700^2+22651982.59^2}\\\Rightarrow F=26852726.19\ \text{N}[/tex]
The hydrostatic force on the dam is [tex]26852726.19\ \text{N}[/tex].
The direction is given by
[tex]\theta=\tan^{-1}\dfrac{F_y}{F_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{22651982.59}{14420700}\\\Rightarrow \theta=57.52^{\circ}[/tex]
The line of action is [tex]57.52^{\circ}[/tex].
student builds a simple circuit with a single resistor with resistance R and measures an electric potential difference ΔV across the resistor. Then, the student replaces the resistor with a new one of resistance 2R and keeps the electric potential difference the same. What is the current (I2) through the new circuit in terms of the original current(I1)? 6mks
Answer:
i₂ = ½ i₁
Explanation:
Let's write ohm's law for the first statement
V = i₁ R
i₁ = V / R
tells us that we change the resistance to R '= 2R
let's write ohm's law
V = i₂ R’
indicates that V remains constant
i₂ = V / R '
i₂ = V / 2R
i₂ = ½ V/R
we substitute
i₂ = ½ i₁
On a sunny day, a rooftop solar panel delivers 60 W of power to the house at an emf of 17 V. How much current flows through the panel
Answer:
3.53 amps
Explanation:
Given data
Power= 60W
Voltage= 17V
The expression relating current, power, and voltage is
P= IV
substitute
60= I*17
I= 60/17
I= 3.53 amps
Hence the current that flows is 3.53 amps
How does the size of a wind turbine affect its energy output?
A.)Smaller turbines spin slower.
B.) Larger turbines have a greater storage capacity.
C.) Larger turbines generate more electricity.
D.)Smaller turbines are better for capturing strong winds.
Answer:
Larger tubines generate more electricity.
Explanation:
Larger blades allow the turbine to capture more of the kinetic energy of the wind by moving more air through the rotors. However, larger blades require more space and higher wind speeds to operate. This distance is necessary to avoid interference between turbines, which decreases the power output.
which form of energy is an example of kinetic energy
Answer:
1. realizing of arrow
2. kicking of ball
3. punching the punching bag
The average marathon runner can complete the 42.2-km distance of the marathon in 3 h and 30 min. If the runner's mass is 85 kg, what is the runner's average kinetic energy during the run
Answer:
the runner's average kinetic energy during the run is 476.96 J.
Explanation:
Given;
mass of the runner, m = 85 kg
distance covered by the runner, d = 42.2 km = 42,200 m
time to complete the race, t = 3 hours 30 mins = (3 x 3600s) + (30 x 60s)
= 12,600 s
The speed of the runner, v = d/t
v = 42,200 / 12,600
v = 3.35 m/s
The runner's average kinetic energy during the run is calculated as;
K.E = ¹/₂mv²
K.E = ¹/₂ × 85 × (3.35)²
K.E = 476.96 J
Therefore, the runner's average kinetic energy during the run is 476.96 J.
Two coils have the same number of circular turns and carry the same current. Each rotates in a magnetic field acting perpendicularly to its axis of rotation. Coil 1 has a radius of 4.5 cm and rotates in a 0.21-T field. Coil 2 rotates in a 0.39-T field. Each coil experiences the same maximum torque. What is the radius (in cm) of coil 2
Answer:
Explanation:
Torque acting on a coil in a magnetic field = MBsinθ where M is magnetic moment , B is magnetic field and θ is inclination of the normal to coil with direction of field.
For maximum torque sinθ = 1
Maximum torque = MB
M = NIA where N is no of turns , I is current and A is area of the coil
Maximum torque = NIAB
As maximum torque is same
N₁I₁A₁B₁ = N₂I₂A₂B₂
N₁ = N₂ , I₁ = I₂
A₁B₁ = A₂B₂
π R₁² B₁ = π R₂² B₂
4.5² x .21 = R₂² x .39
R₂² = 10.9
R₂ = 3.3 cm .
The skater lowers her arms as shown in the adjacent
figure decreasing her radius to 0.15 m. Find her new speed.
Answer:
is there more?
Explanation: