Answer:
v = 5.34[m/s]
Explanation:
In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.
Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.
E₁ = mechanical energy at initial state [J]
In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.
In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.
E₂ = mechanical energy at final state [J]
Now we can use the first statement to get the first equation:
where:
W₁₋₂ = work from the state 1 to 2.
where:
h = elevation = 1.5 [m]
g = gravity acceleration = 9.81 [m/s²]
![58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]](https://tex.z-dn.net/?f=58%20%3D%20v%5E%7B2%7D%20%2B29.43%5C%5Cv%5E%7B2%7D%20%3D28.57%5C%5Cv%3D%5Csqrt%7B28.57%7D%5C%5Cv%3D5.34%5Bm%2Fs%5D)
Answer:
Angle: 
Explanation:
<u>Two-Dimension Motion</u>
When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.
Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.
To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is
where 
is the speed of the boy in still water and 
is the angle respect to the shoreline. If the river flows at speed 
, we now set
7.Jupiter is the largest planet in our solar system at nearly 11 times the size of Earth and 317 times its mass.
When we look at Jupiter, we're actually seeing the outermost layer of its clouds.
The Great Red Spot is a storm in Jupiter's southern hemisphere with crimson-colored clouds that spin counterclockwise at wind speeds
8. 58,232 km
The second largest planet in the solar system
Surface. As a gas giant, Saturn doesn't have a true surface. The planet is mostly swirling gases and liquids deeper down.
Saturn's rings are thought to be pieces of comets, asteroids or shattered moons that broke up before they reached the planet,
9. Unlike the other planets of the solar system, Uranus is tilted so far that it essentially orbits the sun on its side, with the axis of its spin nearly pointing at the star.
Uranus' atmosphere is mostly hydrogen and helium, with a small amount of methane and traces of water and ammonia.
As an ice giant, Uranus doesn't have a true surface. The planet is mostly swirling fluids. While a spacecraft would have nowhere to land on Uranus, it wouldn't be able to fly through its atmosphere unscathed either. The extreme pressures and temperatures would destroy a metal spacecraft.
10. 24,622 km
Neptune has an average temperature of -353 Fahrenheit (-214 Celsius).
Neptune's atmosphere is made up mostly of hydrogen and helium with just a little bit of methane.
Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
The potential at a distance can be represented as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
V(r) = ![-[\frac{qr^2}{8 \pi E_0R^3 }]](https://tex.z-dn.net/?f=-%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D)
₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
= -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) = 
V(r) - V(0) = ![[\frac{qr^2}{8 \pi E_0R^3 }]^R](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bqr%5E2%7D%7B8%20%5Cpi%20E_0R%5E3%20%7D%5D%5ER)
V(r) = 
V(r) = 
V(r) 
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
Answer:
-384.22N
Explanation:
From Coulomb's law;
F= Kq1q2/r^2
Where;
K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2
q1 and q2 = magnitudes of the both charges
r= distance of separation
F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2
F= -495.65 × 10^-3/ 1.29 × 10^-3
F= -384.22N
