Answer:
the neutron is located in the nucleus of an atom
Answer:
nucleus
Explanation:
Atoms are made up of protons and neutrons located within the nucleus, with electrons in orbitals surrounding the nucleus.
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
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 .
Solids have a definite shape and volume this is because
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
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
I also know the answer because I took the test... FLVS exam 3.09 right?
I took it and got this question right... So i know you will! GOOD LUCK!!
Mark me brainlyest please:)
Question 2 of 10 How many oxygen (O) atoms are in a molecule of C3 H4 03? A. 3 OB. 4 C. 10 ОО D. 1
Answer:
a. 3
Explanation:
Answer:
A. 3
Explanation:
AP3X
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:
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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
pls brainliest
A constant electric field of 5.00 N/C points along the positive x-direction. An electron, initially at rest, moves a distance of 2.00 m in this space. How fast is the electron moving after its 2.00 m journey
Answer:
1.875 x 10⁶ m /s .
Explanation:
Force on electron = E e where E is electric field and e is charge on electron
acceleration generated = Ee / m where m is mass of the electron .
Putting the values
acceleration generated = 5 x 1.6 x 10⁻¹⁹ / 9.1 x 10⁻³¹
= .879 x 10¹² m /s²
v² = u² + 2 as , initial velocity u = 0 , displacement s = 2 m
v² = 0 + 2 x .879 x 10¹² x 2
v = 1.875 x 10⁶ m /s .
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]
Astronomers discover a planet orbiting around a star similar to our sun that is 35 light years away. How fast must a rocket ship go if the round trip is to take no longer than 70 years in time for the astronauts aboard
Answer:
[tex]v = 0.7071c[/tex]
Explanation:
Given
Distance to the planet = 35 light years. So, the entire distance is: 2 * 35 = 70.
[tex]\triangle{x'} = 70[/tex]
[tex]T_0 = 70\ years[/tex] i.e time of travel of the ship.
For the observer on earth, the time is:
[tex]T' = \gamma T_0[/tex]
The required speed so that it does not take more than 70 years is then calculated using:
[tex]\triangle x' = vT'[/tex]
Substitute [tex]T' = \gamma T_0[/tex]
[tex]\triangle x' = v\gamma T_0[/tex]
[tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]
So, we have:
[tex]\triangle x' = \frac{vT_0}{\sqrt{1 - v^2/c^2}}[/tex]
Make v the subject of formula.
Square both sides
[tex]\triangle x'^2 = \frac{v^2T^2_0}{1 - v^2/c^2}[/tex]
Cross Multiply
[tex](1 - \frac{v^2}{c^2}) *\triangle x'^2 = v^2T^2_0[/tex]
Divide both sides by [tex]\triangle x'^2[/tex]
[tex](1 - \frac{v^2}{c^2}) = \frac{v^2T^2_0}{\triangle x'^2}[/tex]
Divide through by [tex]v^2[/tex]
[tex](\frac{1}{v^2} - \frac{v^2}{v^2*c^2}) = \frac{v^2T^2_0}{v^2\triangle x'^2}[/tex]
[tex]\frac{1}{v^2} - \frac{1}{c^2} = \frac{T^2_0}{\triangle x'^2}[/tex]
Make [tex]\frac{1}{v^2}[/tex] the subject
[tex]\frac{1}{v^2} = \frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}[/tex]
Inverse both sides
[tex]v^2 = \frac{1}{\frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}}[/tex]
Take square root of both sides
[tex]v = \sqrt{\frac{1}{\frac{T^2_0}{\triangle x'^2} + \frac{1}{c^2}}}[/tex]
Substitute values for [tex]T_0[/tex] and [tex]\triangle x[/tex]
[tex]v = \sqrt{\frac{1}{\frac{70^2}{(70c)^2} + \frac{1}{c^2}}}[/tex]
[tex]v = \sqrt{\frac{1}{\frac{70^2}{70^2*c^2} + \frac{1}{c^2}}}[/tex]
[tex]v = \sqrt{\frac{1}{\frac{1}{c^2} + \frac{1}{c^2}}}[/tex]
[tex]v = \sqrt{\frac{1}{\frac{2}{c^2}}}[/tex]
[tex]v = \sqrt{\frac{c^2}{2}}[/tex]
[tex]v = c\sqrt{\frac{1}{2}}[/tex]
[tex]v = c * 0.7071[/tex]
[tex]v = 0.7071c[/tex]
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].
Question: How did NASA use Newton's Laws to land the Perseverance lander,
safely
on Mars?
Please help asap.
Answer:
If the thrust is increased, the aircraft accelerates and the velocity increases. This is the second part sited in Newton's first law; a net external force changes the velocity of the object. The drag of the aircraft depends on the square of the velocity. So the drag increases with increased velocity.
High-voltage power lines are a familiar sight throughout the country. The aluminum wire used for some of these lines has a cross-sectional area of 4.8 x 10-4 m2. What is the resistance of 14 kilometers of this wire
Answer:
Explanation:
For resistance of a wire , the formula is as follows
R = ρ L / S
where ρ is specific resistance , L is length and S is cross sectional area
Given L = 14 000 m ,
S = 4.8 x 10⁻⁴ m²
specific resistance of aluminum = 2.8 x 10⁻⁸ ohm-meter
Putting the values in the formula
R = 2.8 x 10⁻⁸ x 14 x 10³ / (4.8 x 10⁻⁴ )
R = 0.8167 ohm .
= .82 ohm .
Giving 50 points exactly 50 point and making brainles and if answered wrong I will report and get you kicked out
Draw a picture showing how Heat is added, released, and transferred from one object to another. Also, draw a picture explaining how sublimation or deposition works. please you can draw in your book and take a picture and post it
Answer:
i hope this helps i did'nt quiet understand the secound one. I hope youcan see my picture well .
Explanation:
sincerily, MEMC3891
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:
eeeeeeeeeeeeeeeeeeeee
When a wave hits an object,energy from the wave is both absorbed and reflected off the object
A. True
B. False
P.S pls help
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:
I need help please will mark brainliest
Answer:
200
Explanation:
20m/s*10sec=200
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
Write any two uses of plane mirrors?
Answer:
Uses of plane mirrorsThey are used in periscopes, for signalling, in kaleidoscopes, to see round dangerous bends, in meters, as mirror tiles, in a sextant, in an overhead projector, an SLR camera, car wing mirrors, in microscopes and as reflecting number plates to mention only some!
Explanation:
Hope it is helpful....
Answer:
two uses are:
they are using for looking glassthey are used to make periscope3. 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! :)
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:
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 .
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]
If the gravitational constant is extremely weak, how is the force of gravity on earth so strong?
list 5 types of food that should be consumed daily in a healthy diet.Give an example of each type.
Answer:
vegetables and legumes or beans
fruit
lean meats and poultry, fish, eggs, tofu, nuts and seeds, legumes or beans
grain (cereal) foods, mostly wholegrain or high cereal fibre varieties
milk, yoghurt, cheese or alternatives, mostly reduced fat.
Explanation:
Foods are grouped together because they provide similar amounts of key nutrients. For example, key nutrients of the milk, yoghurt, cheese and alternatives group include calcium and protein, while the fruit group is a good source of vitamins, especially vitamin C.
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
How can you describe the orbital period (planetary year) of each planet?
Answer:
How can you describe the orbital period (planetary year) of each planet?
A year is defined as the time it takes a planet to complete one revolution of the Sun, for Earth this is just over 365 days. This is also known as the orbital period. Unsurprisingly the the length of each planet's year correlates with its distance from the Sun.
Explanation:
Answer: The orbital period is how long it takes a plane to fulfill its "revolution"(365 days)
Explanation:
How do we know that an object has accelerated?