Initially, the energies are:
[tex]U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\ =K_{i}=\frac{1}{2} m v_{0}^{2}[/tex]
At final point, the energies are:
[tex]U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\ K_{f}=\frac{1}{2} m(0)^{2}=0[/tex]
Using conservation law of energy,
[tex]-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\ -\frac{G M_{e}}{r_{e}}+\frac{v_{0}^{2}}{2} &=-\frac{G M_{e}}{r_{e}+h} \\ \frac{-2 G M_{e}+r_{e} v_{0}^{2}}{2 r_{e}} &=-\frac{G M_{e}}{r_{e}+h} \\ \frac{r_{e}+h}{G M_{e}} &=\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}[/tex]
The equation is further simplified as:
[tex]r_{e}+h &=\left(\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}\right) G M_{e} \\ h &=\frac{2 r_{e} G M_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}-r_{e} \\ &=\frac{2 r_{e} G M_{e}-2 r_{e} G M_{e}+r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}} \\ & h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}[/tex]
A woman shouts at a boy who is underwater what happens to the speed of the sound wave as it moves from the air into the water
Answer:
B. it increases
Explanation:
As shown in the table provided, the speed of sound in water (1493 m/s) is greater than the speed of sound in air (346 m/s).
Answer:
B is the correct answer.
Explanation:
need help ASAP!!!!!!!!!!!
Answer:
The equation says that due to variation in temperature is
delt T = .59 m/s / C = 16 C * .59 m/s = 9.44 m/s
So v = 332 m/s + 9.44 m/s = 341 m/s (to three significant figures)
The moon does not stay at the same distance from the earth.why?
Answer:
The moon does not stay at the same distance of the earth because the ortbit of the moon is slightly elliptical. If earth is not tilted at an angle of 66.5°, there will be no change in the season and the earth will have equal length of days and night.
Explanation:
mark me brainlest
A carnival ride starts at rest and is accelerated from an initial angle of zero to a final angle of 6.3 rad by a rad counterclockwise angular acceleration of 2.0 s2 What is the angular velocity at 6.3 rad?
The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Final angular velocity of the carnival ride
The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;
ωf = ωi + 2αθ
where;
ωf is the final angular velocity of the carnival ride = ?ωi is the initial angular velocity of the carnival ride = 0α is the angular acceleration = 2.0 rad/s²θ is the angular displacement of the carnival ride = 6.3 radωf = 0 + 2(2.0) x 6.3
ωf = 25.2 rad/s
Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Learn more about angular velocity here: https://brainly.com/question/6860269
Answer: 5.0 rad/s
Explanation: Because that’s what khan said so try it out.
A 5kg cart moving to the right with a velocity of 16 m/s collides with a concrete wall and
rebounds with a velocity of 22 m/s. Is the change in momentum of the cart
Explanation:
mass, m = 5kg
initial velocity, u = 16m/s
final velocuty, v = -22m/s
change in momentum, ∆p = ?
∆p = m (v-u)
5(-22-16)
5(38)
∆p = 190kgm/s
check the calculations!
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500
Answer:
8 kV
Explanation:
Here is the complete question
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?
Solution
Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V
So, Q = CV
= 500 × 10⁻⁶ F × 800 V
= 400000 × 10⁻⁶ C
= 0.4 C
Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV
V = Q/C
= 0.4 C/500 × 10⁻⁶ F
= 0.0008 × 10⁶ V
= 800 V
The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)
= 10 × 800 V
= 8000 V
= 8 kV
Will give brainliest!
Describe how heat is moving in the image and label each as Radiation, Conduction, or Convection.
Radiation / Conduction / Convection
Answer:
well in the pot there is conventional heat, the pot itself is giving off conductable heat, and the radiational heat is coming from the stove.
Light containing two different wavelengths passes through a diffraction grating with 1,250 slits/cm. On a screen 17.5 cm from the grating, the third-order maximum of the shorter wavelength falls midway between the central maximum and the first side maximum for the longer wavelength. If the neighboring maxima of the longer wavelength are 8.44 mm apart on the screen, what are the wavelengths in the light
Answer:
[tex]\lambda_s =6.43*10^-4m[/tex]
Explanation:
From the question we are told that:
Diffraction grating [tex]N=1250slits/cm[/tex]
Distance b/w Screen and grating length [tex]d_{sg}=17.5 cm[/tex]
Distance b/w neighboring maxima and Screen [tex]d_{ms}=8.44[/tex]
Generally the equation for grating space is mathematically given by
[tex]d(g)=\frac{1}{N}[/tex]
[tex]d(g)=\frac{100}{1250}[/tex]
[tex]d(g)=0.08[/tex]
Generally the equation for small angle approximation is mathematically given by
[tex]\triangle y=\frac{\lambda d}{L}[/tex]
Therefore for longest wavelength
[tex]\lambda _l=\frac{8.44*10^{-3}*(0.08)}{0.175m}[/tex]
[tex]\lambda _l=3.858*10^{-3}[/tex]
Therefore the third order maximum equation for the shorter wavelength as
[tex]\lambda_s =\frac{1}{6} \lambda_l[/tex]
[tex]\lambda_s =\frac{1}{6} (3.858*10^-^3)[/tex]
[tex]\lambda_s =6.43*10^-4m[/tex]
The wavelengths in the light is given as
[tex]\lambda_s =6.43*10^-4m[/tex]
A step-down transformer has 2500 turns on its primary and 5.0 x 10' tums on its secondary. If the potential difference across the primary is 4850 V, what is
the potential difference across the secondary?
Answer:
I dont know sorry
Explanation:
hehe
Help please. Question about a potential energy.
A fox runs at a speed of 16 m/s and then stops to eat a rabbit. If this all took 120
seconds, what was his acceleration?
Answer:
a = 52s²
Explanation:
How to find acceleration
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.
Solve
We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)
We first need to solve the velocity equation for time (t):
v = u + at
v - u = at
(v - u)/a = t
Plugging in the known values we get,
t = (v - u)/a
t = (16 m/s - 120 m/s) -2/s2
t = -104 m/s / -2 m/s2
t = 52 s
The water pressure to an apartment is increased by the water company. The water enters the apartment through an entrance valve at the front of the apartment. Where will the increase in the static water pressure be greatest when no water is flowing in the system
Answer:
Option C
Explanation:
Options for the question are as follows -
A. At a faucet close to entrance valve
B. At a faucet away from the entrance valve
C. It will be the same at all faucets
D. There will be no increase in the pressure at the faucets
Solution -
The static force will be the same at all faucets and also the area of the faucets be same.
Thus, the pressure created at all faucets will be the same.
Thus, option C is correct
An elevator motor provides 45.0 kW of power while lifting an elevator 35.0 m. If the elevator contains seven passengers each with an average mass of 70.0 kg and it takes 20.0 s to accomplish this task, determine the mass of the elevator.
Find how much work ∆W is done by the motor in lifting the elevator:
P = ∆W / ∆t
where
• P = 45.0 kW = power provided by the motor
• ∆W = work done
• ∆t = 20.0 s = duration of time
Solve for ∆W :
∆W = P ∆t = (45.0 kW) (20.0 s) = 900 kJ
In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is M = (m + 490.0) kg, where m is the mass of the elevator alone. Then
∆W = M g h
where
• g = 9.80 m/s² = acceleration due to gravity
• h = 35.0 m = distance covered by the elevator
Solve for M, then for m :
M = ∆W / (g h) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg
m = M - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg
A man walks 30 m to the west, then 5 m to the east in 45 seconds.
What is his average speed?
6) Which of the following describes a good team member?
A) She is willing to compromise.
B) He is aggressive.
C) She is stubborn.
D) He is conceited.
Answer: A
Explanation:
Because someone who is aggressive, stubborn, or proud of theirselves are more likely to think they're above everyone else and be a bully. However someone who is willing to compromise is better since you can generally make everyone happy that way
HOPE THIS HELPS ^^
The eight plants of the Solar System orbit the Sun in a chaotic random way.
True
False
Answer:
The Solar System has plants? I assume you meant planets. If so, that is false
Explanation:
Which of the following best describes our
atmosphere?
A. envelope of gases that surround Earth
B. a specific range of altitude where plant life flourishes
C. The air, water, and land that form our planet
D. the water vapor in the air surrounding our planet
You are testing a new amusement park roller coaster with an empty car with a mass of 130 kg . One part of the track is a vertical loop with a radius of 12.0 m . At the bottom of the loop (point A) the car has a speed of 25.0 m/s and at the top of the loop (point B) it has speed of 8.00 m/s . Part A As the car rolls from point A to point B, how much work is done by friction
Answer:
work done by friction = 5889 J
Explanation:
We are given;
Mass of car; m = 130 kg
Speed at point A; v1 = 25 m/s
Speed at point B: v2 = 8 m/s
Since radius is 12 m
At point A, distance is; y1 = 12 m
At point B, distance is; y2 = -12 m
Now, formula for work done by all the forces is given by the equation;
Total work;
W_gravity + W_others = K2 - K1
Where W_others is work done by other forces which is equal to work done by friction
Where K2 - K1 is change in kinetic energy.
W_grav is also change in potential energy and is expressed as;
W_grav = mgy1 - mgy2
K2 - K1 = ½m(v1)² - ½m(v2)²
Thus;
mgy1 - mgy2 + W_others = ½m(v1)² - ½m(v2)²
Making W_others the subject;
W_others = ½m(v1)² - ½m(v2)² + mgy2 - mgy1
Plugging in the relevant values;
W_others = (½ × 130 × 25²) - (½ × 130 × 8²) + (130 × 9.8 × -12) - (130 × 9.8 × 12)
W_others = 5889 J
Recall that I earlier said W_others = work done by friction.
Thus, work done by friction = 5889 J
Although planets orbit the Sun in ellipses, all the planetary orbits are fairly close to circular and not very eccentric.
True
False
Answer:
False
Explanation:
The Sun rotates in this same, right-hand-rule direction. All planetary orbits lie in nearly the same plane. All planetary orbits are nearly circular (eccentricity near zero).
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A uniform magnetic field is in the positive z direction. A positively charged particle is moving in the positive x direction through the field. The net force on the particle can be made zero by applying an electric field in what direction
Answer:
We apply an electric field in the negative y direction
Explanation:
Since A uniform magnetic field is in the positive z direction and A positively charged particle is moving in the positive x direction through the field, the magnetic force acting on the positively charged particle is in the positive y direction according to Fleming's right-hand rule.
For the net force on the particle to be zero, we apply an electric field in the negative y direction to create an electric force on the positively charged particle, so as to cancel out the magnetic force.
Review please help.
Answer:
1 and 3
Explanation:
because they are going up from 0
What kind of energy is in a moving skateboard
Answer:
I guess it is kinetic energy
Answer:
kinetic energy because my dog told me
When a 20 kg explosive detonates and sends a 5 kilogram piece traveling to the right at 105 m/s
what is the speed and direction of the other 15 kilogram piece of the explosive!
Answer:
speed: 35m/s
direction: left
Explanation:
Assuming the right side is the positive direction:
before explosion:
P = mv = 0
after explosion:
P' = 15P + 5P
(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')
P' = 0.75mv1' + 0.25mv2'
P' = (15kg)v' + (5kg)(105m/s)
P' = 525kg/m/s + (15kg)v1'
P = P'
525kg/m/s + (15kg)v1' = 0
(15kg)v1' = -525kg/m/s
v1' = -35m/s
speed = |-35| = 35m/s
direction is to the left since the right side is the positive direction.
A scientist measuring the resistivity of a new metal alloy left her ammeter in another lab, but she does have a magnetic field probe. So she creates a 4.5-m-long, 2.0-mm-diameter wire of the material, connects it to a 1.5 V battery, and measures a 3.0 mT magnetic field 1.0 mm from the surface of the wire. What is the material's resistivity
Answer:
[tex]3.49\times 10^{-8}\ \Omega\text{m}[/tex]
Explanation:
r = Radius = [tex]\dfrac{2}{2}=1\ \text{mm}[/tex]
B = Magnetic field = 3 mT
1 mm = Distance from the surface of the wire
V = Voltage
x = Distance from the probe = [tex]r+1=1+1=2\ \text{mm}[/tex]
R = Resistance
L = Length of wire = 4.5 m
Magnetic field is given by
[tex]B=\dfrac{\mu_0I}{2\pi x}\\\Rightarrow I=\dfrac{B2\pi x}{\mu_0}\\\Rightarrow I=\dfrac{3\times 10^{-3}\times 2\times \pi 2\times 10^{-3}}{4\pi 10^{-7}}\\\Rightarrow I=30\ \text{A}[/tex]
Voltage is given by
[tex]V=IR\\\Rightarrow R=\dfrac{V}{I}\\\Rightarrow R=\dfrac{1.5}{30}\\\Rightarrow R=0.05\ \Omega[/tex]
Resistivity is given by
[tex]\rho=\dfrac{RA}{L}\\\Rightarrow \rho=\dfrac{0.05\times \pi (1\times 10^{-3})^2}{4.5}\\\Rightarrow \rho=3.49\times 10^{-8}\ \Omega\text{m}[/tex]
The resistivity of the material is [tex]3.49\times 10^{-8}\ \Omega\text{m}[/tex].
A box having a weight of 8 lb is moving around in a circle of radius rA = 2 ft with a speed of (vA)1 = 5 ft/s while connected to the end of a rope. If the rope is pulled inward with a constant speed of vr = 4 ft/s, determine the speed of the box at the instant rB = 1 ft. How much work is done after pulling in the rope from A to B? Neglect friction and the size of the box
Answer:
W = 1.875 J
Explanation:
For this exercise let's use the relationship between work and kinetic energy
W = ΔK
The kinetic energy of rotational motion is
K₀ = ½ I w²
we can assume that the box is small, so it can be treated as a point object, with moment of inertia
I = m rₐ²
angular and linear velocity are related
v = w r
w = v / r
we substitute in the equation, for point A
K₀ = ½ (m rₐ²) (v / rₐ)²
K₀ = ½ m v²
For the final point B, as the system is isolated the angular momentum is conserved
initial L₀ = Io wo
final L_f = I_f w_f
L₀ = L_f
I₀ w₀ = I_f w_f
(m rₐ²) w₀ = (m [tex]r_{b} ^2[/tex]) w_f
w_f = (rₐ/r_b)² w₀
with this value we find the final kinetic energy
K_f = ½ I_f w_f²
K_f = ½ (m [tex]r_{b}^2[/tex]) ( (rₐ / r_b)² w₀) ²
K_f = ½ m [tex]\frac{r_a^4}{r_b^2} \ w_o^2[/tex]
we substitute in the realcion of work
W = K_f - K₀
W = ½ m [tex]( \( \frac {r_a^2 }{r_b} )^2[/tex] w₀² - ½ m v²
W = ½ m [tex]\frac{r_a^4}{r_b^2} ( \frac{v}{r_a} ) ^2[/tex] - ½ m v²
W = ½ m [tex]\frac{r_a^2}{r_b^2} \ v^2[/tex] - ½ m v2
W = ½ m v² (([tex]( \ (\frac{r_a}{r_b})^2 -1)[/tex]
let's calculate
W = ½ ( [tex]\frac{8}{32}[/tex] ) 5 ((2/1)² -1)
W = 0.625 (3)
W = 1.875 J
In the past, asteroids striking the earth have produced disastrous results. If we discovered an asteroid on a collision course with the earth, we could, in principle, deflect it and avoid an impact by focusing a laser on the surface. Intense surface heating from the laser could cause surface material to be ejected into space at high speed.
Required:
How would this deflect the asteroid?
Answer:
Explained below.
Explanation:
We are told that the surface material is ejected into space at a high speed. This means that it will have a likely high momentum as well.
Now, we can say that the total momentum is conserved because the entire asteroid system behaves like an isolated system.
Also, as the surface material is moving with the high momentum like we established earlier, it will cause the asteroid to move with a speed in an opposite direction which also means deflection in an opposite direction.
Answer:
Explained below.
Explanation:
The material ejected from the surface of the asteroid would have a significant momentum. Since the asteroid and all its material is an isolated system, the ejection would cause an oppositely directed change in momentum of the asteroid, according to the law of conservation of momentum.
The ejected material is analogous to gases expelled from a rocket, and the asteroid is analogous to a rocket.
What is the medium of the wave shown in the photograph?
A. The water
B. Kinetic energy
C. The duck
D. Gravity
Answer:
A. The water
Explanation:
i got it right on A-P-E-X
Which of the following best defines
weather?
A. the expanding or contracting of the atmosphere
B. the measurement of the amount of water vapor in the
atmosphere
C. the condition of the atmosphere at a certain time and
place
Help Resources
D. the average air temperature of a specific region
Answer:
I'd say D
Explanation:
because not all weather happens within the atmosphere, and most weather depends on region (lile if your near the equator or not)
A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward and ideal spring that is attached to a wall , After the block collides with the spring, the spring is compressed a maximum distance of 0.68m . what is the speed of the block when the spring is compressed to only one-half of the maximum distance?
A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall, the maximum speed of the block when the spring is compressed to one-half of the maximum distance is 4.33 m/s
From the conservation of energy; the kinetic energy of the mass is equal to the work done on the spring.
i.e.
[tex]\mathbf{\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2_{max}}[/tex]
Given that:
the mass of the block = 4.0 kg the speed at which it is moving = 5.0 m/scompression of the spring = 0.68 m∴
From the equation above, multiplying both sides with 2, we have:
[tex]\mathbf{mv^2 =kx^2_{max}}[/tex]
Making (k) the subject of the formula;
[tex]\mathbf{k = \dfrac{mv^2}{x^2_{max}}}[/tex]
[tex]\mathbf{k = \dfrac{4 \times 5^2}{0.68^2}}[/tex]
k = 216.26 N/m
However, when compressed to one-half of the maximum distance; the speed is computed as follows:
x = 0.68/2 = 0.34 m
∴
[tex]\mathbf{\dfrac{1}{2}mv_o^2 - \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2}[/tex]
[tex]\mathbf{m(v_o^2 -v^2) =kx^2}[/tex]
[tex]\mathbf{(v_o^2 -v^2) =\dfrac{kx^2}{m}}[/tex]
[tex]\mathbf{(5^2 -v^2) =\dfrac{216.26 \times 0.34^2}{4.0}}[/tex]
25 - v² = 6.25
25 -6.25 = v²
v² = 18.75
[tex]\mathbf{ v= \sqrt{18.75 }}[/tex]
v = 4.33 m/s
Therefore, we can conclude that the speed of the block when the spring is compressed to only one-half of the maximum distance is 4.33 m/s
Learn more about speed here:
https://brainly.com/question/22610586?referrer=searchResults
If you live in Melbourne, Australia, the local magnetic field has a strength of about 4x10-5 T. The magnetic field vector is directed northward, making an angle of 30 deg above the horizontal. An electron in Melbourne is moving parallel to the ground, in the west direction, at a speed of 9x105 m/s. What are the magnitude and direction of the magnetic force on the electron
Answer:
[tex]5.76\times 10^{-18}\ \text{N}[/tex] perpendicular to the velocity and magnetic field
Explanation:
B = Magnetic field = [tex]4\times 10^{-5}\ \text{T}[/tex]
[tex]\theta[/tex] = Angle the magnetic field makes with the horizontal = [tex]30^{\circ}[/tex]
v = Velocity of electron = [tex]9\times 10^5\ \text{m/s}[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ \text{C}[/tex]
Magnetic force is given by
[tex]F=qvB\sin\theta\\\Rightarrow F=1.6\times 10^{-19}\times 9\times 10^5\times 4\times 10^{-5}\sin30^{\circ}\\\Rightarrow F=2.88\times 10^{-18}\ \text{N}[/tex]
The magnitude of the magnetic force is [tex]2.88\times 10^{-18}\ \text{N}[/tex] and the direction is perpendicular to the velocity and magnetic field.
