Answer:
a. [tex]\mathbf{F_{f_2} = 2 F_{f1}}[/tex], [tex]\mathbf { since \ the \ force \ o f \ friction \ is \ represented \ as \ the \ slope \ for \ each \ of \ the \ two \ curves.}[/tex]
Explanation:
From the information given;
Using the work-energy theorem
ΔKE = W = [tex]\mathbf{ F_f \times r}[/tex]
K = [tex]\mathbf{ F_f \times r}[/tex]
∴
[tex]\dfrac{K_1}{K_2} = \dfrac{F_{f1}}{F_{f2}} (\dfrac{r_1}{r_2})[/tex]
Since [tex]K_1 = K_2[/tex] and r_1 = 4, and r_2 = 2 (from the missing diagram which is attached below)
Then;
[tex]1 = \dfrac{F_{f1}}{F_{f2}} (\dfrac{4 \ m}{2 \ m})[/tex]
[tex]\mathbf{F_{f_2} = 2 F_{f1}}[/tex]
A loan of $8.000 accumulates simple interest at an annual interest rate of 5%. After how many years does the value the loan become $9.200?
The most common stars in the universe are red dwarf stars. Red dwarf stars use their hydrogen fuel very slowly. Our universe also contains stars classified as blue supergiants. The table below lists the color and approximate surface temperatures of four stars. Based on the information from the table above, which of the following stars has the greatest absolute brightness?
The Sun
Vega
Bellatrix
Betelgeuse
Answer:
sun
Explanation:
You throw a ball into the air as shown in the diagram. At what point does the ball have the most potential energy?
Answer:
Position X.
Explanation:
To know the correct answer to the question, it is important that we know the definition of potential energy.
Potential energy can be defined as the energy possessed by an object in relation to its position. Mathematically it can be expressed as:
PE = mgh
Where
PE => is the potential energy.
m => is the mass of the object.
g => is the acceleration due to gravity.
h => is the height to which the object is located.
Considering the formula for potential energy (i.e PE = mgh), we can see clearly that the potential energy (PE) is directly proportional to the height (h). This implies that the greater the height, the greater the potential energy and the smaller the height, the smaller the potential energy.
Considering the diagram given above, we can see that the greatest height attained by the ball is at position X. Thus, the ball will have the greatest potential energy at position X.
A skateboarder travels on a horizontal surface with an initial velocity of 4.2 m/s toward the south and a constant acceleration of 2.6 m/s^2 toward the east. Let the x direction be eastward and the y direction be northward, and let the skateboarder be at the origin at t=0.
Required:
a. What is her x position at t=0.60s?
b. What is her y position at t=0.60s?
c. What is her x velocity component at t=0.60s?
d. What is her y velocity component at t=0.60s?
Answer:
Explanation:
Let the skateboarder's movement in x -direction be taken into consideration .
a )
initial velocity in x direction u = 0
acceleration a = 2.6 m /s²
time t = .6 s
displacement in x direction
s = ut + 1/2 a t²
= 0 + .5 x 2.6 x .6²
= .468 m
= 46.80 cm
c )
velocity after .6 s
v = u + at
= 0 + 2.6 x .6
= 1.56 m /s
Let the skateboarder's movement in y -direction be taken into consideration .
b ) initial velocity in y direction u = - 4.2 m /s ( velocity is towards south or - y direction )
acceleration a = 0
time t = .6 s
displacement in y direction
s = ut + 1/2 a t²
= - 4.2 x .6 + 0
= - 2.52 m
2.52 m towards south .
d )
velocity after .6 s
v = u + at
= - 4.2 + 0
= - 4.2 m /s
or 4.2 m /s towards south .
If the kinetic energy of the 40kg box is 784 J, what is the velocity before it strikes the ground?
Answer:
Explanation:
[tex]KE=\frac{1}{2}mv^2[/tex]
[tex]784=\frac{1}{2}(40)v^2[/tex]
[tex]784=20v^2[/tex]
[tex]39.2=v^2[/tex]
[tex]v=6.26m/s[/tex]
Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 3 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension. (Use g = 9.8 m/s2 for the acceleration due to gravity. Round your answers to two decimal places.)
Answer:
Explanation:
Let the tension in ropes be T₁ and T₂ . Ropes are making angle of 52 and 40 degree with the horizontal . The vertical component of tension will add up together to balance the weight of the decoration
T₁ sin52 + T₂ sin40 = 3 x 9.8
.788 T₁ + .643 T₂ = 29.4
The horizontal component of tension will add up to zero because the decoration piece is at rest .
T₁ cos52 + T₂ cos40 = 0
.615 T₁ - .766 T₂ = 0
T₁ = 1.24 T₂
Substituting this value of T₁ in earlier equation , we have
.788 x 1.24 T₂ + .643 T₂ = 29.4
1.62 T₂ = 29.4
T₂ = 18.15 N
T₁ = 1.24 x 18.15 = 22.51 N .
Suppose there is a 3Mbps uplink and 10Mbps downlink between a geostationary satellite and the base station on earth. If the propagation speed is the speed of light (3 * 10^8 m/sec), the packet size is 20Mb, and the distance from the satellite to earth is 36,000 km, what is the uplink propagation delay of the link in milliseconds
Answer:
[tex]120\ \text{ms}[/tex]
Explanation:
Distance between the satellite and Earth = [tex]36000\ \text{km}[/tex]
Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]
Propagation delay is given by distance by the speed of light
[tex]\dfrac{36000\times 10^3}{3\times 10^8}[/tex]
[tex]=0.12\ \text{s}\times 10^3[/tex]
[tex]=120\ \text{ms}[/tex]
The uplink propagation delay of the link is [tex]120\ \text{ms}[/tex].
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Answer:
200
Explanation:
20m/s*10sec=200
g Galileo's telescopes were not of high quality by modern standards. He was able to see the moons of Jupiter, but he never reported seeing features on Mars. Use the small-angle formula to find the angular diameter of Mars when it is closest to Earth. How does that compare with the maximum angular diameter of Jupiter
Answer:
θ₂/ θ₁= 2.58
Explanation:
In this exercise you are asked to compare the angular diameters of Mars and Jupiter. The angular diameter or angle in radians is
θ = D / R
where D is the diameter of the body, the distance from Earth to the body of interest and θ is angle in radians
The different distances are tabulated with respect to the Sun
Sun -Earth 1,496 10¹¹ m
Sun- Mars 2.28 10¹¹ m
Sun - Jupiter 7.78 10 m
The Radii of the planets are
Mars 3.37 10⁶ m
Jupiter 6.99 10⁷ m
let's calculate the angles for each body
a) Mars
θ₁ = 2r / R'
the distance from the ground is
R ’= D_planet - D_earth
R ’= 2.28 10¹¹ - 1.496 10¹¹
R ’= 0.784 10¹¹ m
let's calculate
θ₁ = [tex]\frac{2 \ 3.37 \ 10^6 }{0.784 \ 10^{11}}[/tex]
θ₁ = 8.6 10⁻⁵ radians
b) Jupiter
R ’= 7.78 10¹¹ - 1.496 10¹¹
R ’= 6.284 10¹¹ m
let's calculate
θ₂ = [tex]\frac{2 \ 6.99 \ 10^7}{6.284 \ 10^{11}}[/tex]
θ₂ = 2.22 10⁻⁴ radians
the ratio of the angular diameters is
θ₂/ θ₁ = [tex]\frac{2.22 \ 10^{-4}}{8.6 \ 10^{-5}}[/tex]
θ₂/ θ₁= 2.58
what is calculator program
Answer:
software calculator is a calculator that has been implemented as a computer program, rather than as a physical hardware device. They are among the simpler interactive software tools, and, as such, they: Provide operations for the user to select one at a time.
Answer: The calculator is a compact portable device that performs mathematical calculations. Some calculators also allow easy text editing and programming. It's also a programming software that simulates a portable calculator. Calculator applications help you make basic math calculations without leaving your screen.
The Short Answer:
Earth's tilted axis causes the seasons. Throughout the year, different
parts of Earth receive the Sun's most direct rays. So, when the North
Pole tilts toward the Sun, it's summer in the Northern Hemisphere. And
when the South Pole tilts toward the Sun, it's qvinter in the Northern
Hemisphere.
1. What direction is the Earth tilted in the summer here in
Buffalo?
a) towards the Sun
b) away from the Sun
23.5
Answer:
I cannot fully see the picture, but I'm going to have to tell you to go with towards the sun because it says summer.
In reaching her destination, a backpacker walks with an average velocity of 1.19 m/s, due west. This average velocity results because she hikes for 5.96 km with an average velocity of 3.10 m/s, due west, turns around, and hikes with an average velocity of 0.744 m/s, due east. How far east did she walk
Answer:
x₂ = 4.455 m
Explanation:
The average speed is defined as the displacement traveled between the time interval
v_average = [tex]\frac{\Delta x}{\Delta t}[/tex]
in this case we have a first stop walking west at speed v = 3.10 m / s a distance of x = 5.96 km = 5.96 10³ m
Let's find the time it takes on this tour
v = x / t
t = x / v
t₁ = 5.96 10³ / 3.10
t₁ = 1.9226 10³ s
in a second to walk east at a speed of v₂ = 0.744 m / s
v₂ = x₂ / t₂
t₂ = x₂ / v₂
for full movement
let's assume that the eastward movement is positive
v =[tex]\frac{- x_1 + x_2}{t_1 +t_2}[/tex]
v = [tex]\frac{-x_1 +x_2}{t_1 + \frac{x_2}{t_2} }[/tex]
the only unknown term is the distance to the east. We replace and resolve
1.19 = [tex]\frac{-5.96 \ 10^3 + x_2}{1.92 \ 10^3 + \frac{x_2}{0.744} }[/tex]
1.19 (1.92 10³ + [tex]\frac{x_2}{0.744}[/tex]) = x₂ 1.92 10³ - 5.96 10³
2.2848 10³ + 1.599 x₂ = 1.92 10³ x₂ - 5.96 10³
x₂ (1.92 10³ - 1.599) = 2.2848 10³ + 5.96 10³
x₂ = 8.5448 10³ / 1.918 10³
x₂ = 4.455 m
what is the acceleration of a satellite moving in a circular orbit around the earth of radius 2r
Explanation:
You do the radius times the circumference of the earth
A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground, another ball of mass m is thrown straight downward toward the first ball, also at speed v0. Assume that gravity acts vertically downward on each mass m with a magnitude mg. v0 D b How far above the ground do the balls collide (in terms of only D, v0, and g)?
Answer:
Explanation:
Let the balls collide after time t .
distance covered by falling ball
s₁ = v₀ t + 1/2 g t²
distance covered by rising ball
s₂ = v₀ t - 1/2 g t²
Given ,
s₁ + s₂ = D
D = v₀ t + 1/2 g t² + v₀ t - 1/2 g t²
= 2v₀ t
t = D / 2v₀
s₂ = v₀ t - 1/2 g t²
= v₀ x D / 2v₀ - (1/2) x g x D² / 4v₀²
= D / 2 - gD² / 8 v₀²
A system with a mass of 10 kg, initially moving horizontally with a velocity of 20 m/s, experiences a constant horizontal force of 25 N opposing the direction of motion. As a result, the system comes to rest. Determine the amount of energy transfer by work, in kJ, for this process and the total distance, in m, that the system travels
Answer:
Explanation:
Kinetic energy of the mass of 10 kg
= 1/2 m v² , m is mass and v is velocity .
= .5 x 10 x 20²
= 2000 J
The opposing force stops it . so work done by opposing force will be equal to this energy and it will be negative .
So energy transfer will be - 2000 J .
= 2 kJ .
If distance travelled by mass is d , force 25 N will have a displacement of d . so work done by force of 25 N
= 25 x d
25 d = 2000
d = 80 m .
Hence system travels a distance of 80 m .
A ball is thrown horizontally to the right, from the top of a vertical cliff of height h. A wind blows horizontally to the left, and assume (simplistically) that the effect of the wind is to provide a constant force to the left, equal in magnitude to the weight of the ball. How fast should the ball be thrown so that it lands at the foot of the cliff
Answer:
v = [tex]\sqrt{\frac{y_o \ g}{2} }[/tex]
Explanation:
For this exercise we must use the projectile launch ratios, let's start by finding the time it takes to reach the bottom of the cliff, the initial vertical velocity is zero
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
at the bottom of the cliff y = 0 and as the body is thrown horizontally the initial vertical velocity is zero
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{2y_o/g}[/tex]
this time is the same as the horizontal movement.
Let's use Newton's second law to find the acceleration on this x-axis due to the force of the air
F = m aₓ
they tell us that force is equal to the weight of the body
-mg = maₓ
aₓ = -g
the sign indicates that the acceleration is to the left
we write the kinematics equation
x = x₀ + v₀ₓ t + ½ aₓ t²
They indicate that the final position is the foot of the cliff (x = 0), when it leaves the top it is at x₀ = 0 and has a velocity v₀ₓ = v
we substitute
0 = 0 + v t + ½ (-g) t²
v = ½ g t
we use the drop time
v = ½ g [tex]\sqrt{\frac{2yo}{g} }[/tex]
v = [tex]\sqrt{\frac{y_o \ g}{2} }[/tex]
An athlete swims the length L of a pool in a time t1 and makes the return trip to the starting position in a time t2. If she is swimming initially in the positive x-direction, determine her average velocities symbolically in:
a. the first half of the swim
b. the second half of the swim
c. the round trip.
Answer:
B but I am not sure
Explanation:
3. A stone is thrown vertically upwards from the top of a building 50 m tall with an initial
velocity of 20.0 ms. If the stone just misses the edge of the roof on its return, determine
(a) The time is taken the stone to get to its maximum height.
(b) The maximum height reached by the stone
(c) The time at which the stone return to the point where it was thrown
(d) The velocity of the stone at this instance
(e) The velocity and position of the stone at t = 5 s.
Answer:
13.4436
Explanation:
An air-filled capacitor consists of two parallel plates, each with an area of A , separated by a distance d . A V potential difference is applied to these plates. What is the magnitude of the electric field between the plates
Answer:
E = V / d
Explanation:
In a charged capacitor an electric field is established that goes from the positive to the negative plate, this field is constant,
the potential difference is
D = E d
in this case they do not give the difference in potential V and the distance between the plates d
E = V / d
Acceleration figures for cars usually are given as the number of seconds needed to go from 0.0 to 97 km/h. Convert 97 km/h into m/s.
Answer:
26.9444m/s
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What is the acceleration when a force of 2 N is applied to a ball that has a mass of 0.60 kg?
Answer:
3.333
Explanation:
Acceleration is force divided by mass. So divide the force, 2, by the mass, 0.60, and you will get 3.333. I hope this helps :)
The acceleration of an object is the force divided by time. The acceleration of the ball of 0.60 kg when a force of 2N is applied, is 3.33 m/s².
What is acceleration ?Acceleration of an object is the measure of rate of change in its velocity. Like velocity acceleration is a vector quantity having both magnitude and direction.
Acceleration is defined as the ratio of change in velocity to the change in time. However, according to Newton's second law of motion fore applied on an object is the product of its mass and acceleration.
Hence, F = m a .
Given that, force applied on the ball = 2 N
mass of the ball = 0.60 Kg.
Then acceleration a = force/mass
a = 2 N/ 0.60 Kg = 3.33 m/s²
Therefore, the acceleration of the ball is 3.33 m/s².
Find more on acceleration:
https://brainly.com/question/2303856
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if the forces on an object are balanced the resultant force is equal to zero true false
Answer:
If the forces are balanced, the resultant force is zero. If the forces on an object are unbalanced, this is what happens: a stationary object starts to move in the direction of the resultant force. a moving object changes speed and/or direction in the direction of the resultant force.
Explanation:
Which statement about the sun's energy is correct?
It is entirely re-radiated back into space.
A part of it is destroyed by greenhouse gases.
A part of it is absorbed by atmospheric gases.
It makes Earth too hot for plants and animals to survive.
Answer:
The answer is (A)
Explanation:
We know this because The suns energy is entirely re-radiated back into space.
We know also that the answer is not (D) It makes Earth too hot for plants and animals to survive.
Atmospheric gases are gases located in the Earth's atmosphere
The green house effect is a natural process that warms the Earth's surface
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I took it and got this question right... So i know you will! GOOD LUCK!!
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You've been hired to design the hardware for an ink jet printer. You know that these printers use a deflecting electrode to cause charged ink drops to form letters on a page. The basic mechanism is that uniform ink drops of about 30 microns radius are charged to varying amounts after being sprayed out towards the page at a speed of about 20 m/s. Along the way to the page, they pass into a region between two deflecting plates that are 1.6 cm long. The deflecting plates are 1.0 mm apart and charged to 1500 volts. You measure the distance from the edge of the plates to the paper and find that it is one-half inch. Assuming an uncharged droplet forms the bottom of the letter, how much charge is needed on the droplet to form the top of a letter 3 mm high (11 pt. type)
Answer:
the required charged is 7.06 × 10⁻¹³ C
Explanation:
Given that;
Radius = 30 microns = 30 × 10⁻⁶
Speed v = 20 m/s
length x = 1.6 cm = 0.016 m
spacing d = 1.0 mm = 0.001 m
Voltage V = 1500 V
from the question, the electric field between the plates is uniform and equal to Voltage divided by the distance between the plates.
Electric field E = V/d
E = 1500 V / 0.001 m
E = 1.5 × 10⁶ V/m
Mass of ink drop m = pv
m = 10³ kg/m³ × [tex]\frac{4}{3}[/tex]πr³
m = 1000 kg/m³ × [tex]\frac{4}{3}[/tex]π × (30 × 10⁻⁶)³
m = 1.131 × 10⁻¹⁰ Kg
Time taken to travel t = x / sped
t = 0.016 m / 20 m/s
t = 0.0008 s
From the kinematic equation
to form the top of a letter 3 mm ( 0.003 m )high
y = [tex]\frac{1}{2}[/tex]at²
2y = at²
a = 2y/t²
we substitute
a = (2 × 0.003 m) / (0.0008 s)²
a = 9375 m/s²
Now Force F = Eq = ma
so
q = ma / E
we substitute
q = ( 1.131 × 10⁻¹⁰ Kg × 9375 m/s² ) / ( 1.5 × 10⁶ V/m )
q = 7.06 × 10⁻¹³ C
Therefore, the required charged is 7.06 × 10⁻¹³ C
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a person drops their phone. The phone's mass is 0.115 kilograms and the bridge is 15 meters tall. The instant the they dropped the phone, what was its mechanical energy?
1.725
16.9
200
Two astronauts, each having a mass of 74.3 kg are connected by a 13.1 m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.65 m/s. a. Calculate the magnitude of the initial angular momentum of the system by treating the astronauts as particles.
Answer:
L = 5076.5 kg m² / s
Explanation:
The angular momentum of a particle is given by
L = r xp
L = r m v sin θ
the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1
as we have two bodies
L = 2 r m v
let's find the distance from the center of mass, let's place a reference frame on one of the masses
[tex]x_{cm}[/tex] = [tex]\frac{1}{M} \sum x_{i} m_{i}[/tex]i
x_{cm} = [tex]\frac{1}{m+m} ( 0 + l m)[/tex]
x_{cm} = [tex]\frac{1}{2m} lm[/tex]
x_{cm} = [tex]\frac{1}{2}[/tex]
x_{cm} = 13.1 / 2 = 6.05 m
let's calculate
L = 2 6.05 74.3 5.65
L = 5076.5 kg m² / s
Answer:
5076.5
Explanation:
A skier pushes off the top of a hill with an initial speed of 3.30 m/s. How fast will she be moving after dropping 5.00 meters in elevation if friction is negligible?
Answer:
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Answer:
I think the answer is 63,
First to answer gets brainliest
Answer:
has a charge
Explanation:
Answer:
gain electrons
Explanation:
I looked it up and that was what I found
side note: ions do have a charge
3. Describe how sound signals can be transmitted from a cell phone and received by another cell phone.
Create a labeled diagram that shows this process. (2 points)
Answer:
The sound energy travels through the air into the microphone and makes the diaphragm inside vibrate. The sound energy in your voice makes the air vibrate. Vibrating air carries the sound energy into the phone. The diaphragm in the mouthpiece microphone converts sound energy into electrical energy.
Explanation: Hope this helps! :)