Answer:
If the force is 10N and mass is 5000G, then a=2m/s2
Explanation:
F=10N
m=5000
g=5kg
a=10N/ 5kg
and finally; a= 2m/s2
When there is a temperature inversion, you would expect to experience Group of answer choices clouds with extensive vertical development above an inversion aloft. good visibility in the lower levels of the atmosphere and poor visibility above an inversion aloft. an increase in temperature as altitude increases.
Temperature inversion leads to an increase in temperature as altitude increases.
The term temperature inversion refers to a situation in which a layer of warm air lies over a layer of cool air. This is also referred to as thermal inversion. This occurs when the air below to loose heat rapidly.
One of the effects of temperature inversion is reduction in visibility. So, thermal inversion leads to an increase in temperature as altitude increases.
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When making a left turn, drive into the intersection, make the turn smoothly and without strain on the engine, braking to about ... to .........mph and stay on the brake until approximately half way into the intersection.
Answer:
whats the question
Explanation:
I'LL MARK YOU BRAINLIST!!
ali starts moving his car from rest and accelerates it at 5 m/s^2 over 8 seconds. what is the velocity of ali's car reached by the end of these 8 seconds ?
Explanation:
u=0, v=?, a=5m/s², t=8sec
So, by the formula,
v=u+at
v=0+5×8
v=40m/s.
hope this helps you .
consider two vectors A and B Where A-B and |A+B|=|A-2B| vector B is directed in the positive x_direction and A makes an angle of 60°above the positive x_axis find magnitude of the two vectors
Answer:
It should be mentioned that an efficient way to work this vector addition problem is with the cosine law for general triangles (and since
a
,
b
and
r
from an isosecles triangle, the angles are easy to figure). However, in the interest of reinforcing the usual systematic approach to vector addition, we note that the angle
b
makes with the +x axis is 30
0
+105
0
=135
0
(a) The x component of
r
x
=(10.0m)cos30
0
+(10.0m)cos135
0
=1.59m
(b) the y component of
r
is r
y
=(10.0m)sin135
0
= 12.1 m.
(c) the magnitude of
r
is r = ∣
r
∣=
(1.59m)
2
+(12.1m)
2
=12.2m
(d) The angle between
r
and the +x direction is tan
−1
[(12.1m)/(1.59m)]=82.5
0
Explanation:
If a statue is made out of wood, which of these words correctly describes the statue?
OA.
wooded
ОВ.
woody
OC.
non-wood
OD.
wooden
Answer:
Od- Wooden
Explanation:
It would not make sense if it was non-wood because that make it no a wood base product, OC is not right because its not a a plant that produces wood as its structural tissue and thus has a hard stem. An OA justT o take or get a supply of wood. so its Od
When riding your bike, you come up to a small child who jumps in front of you. You slam on the brakes, and your bike comes to a sliding stop. Which force greatly increases when you lock the brakes and slide
Answer: friction :)
Explanation:
I took the test and got 100%
negatively charged particles of radiation emitted from the decay of radioactive substances are known as
Answer: beta particles
Explanation:
that's the answer
Krishna wants to measure the mass and volume of a key. Which tools should she use?
a balance and a beaker of water
a balance and a meter stick
a beaker of water and a ruler
a ruler and a meter stick
Answer:
a balance and a beaker of water
Explanation:
The mass can easily be measured with balance. If you drop the key in a beaker of water, the water inside the beaker will increase and this increases the volume of water that will be equal to volume of key.
The drone can fly for 25 minutes before the battery needs recharging. the power output of the battery is 65.0 watts. calculate the maximum energy stored by the battery
help please!! needed asap!!
Answer:
stored Energy = Power x time = 65 x 25x 60 = 105.6 Kjoules
The object in this diagram has a mass of 2 kg.
According to this diagram, what is the acceleration of this object?
A. 0 m/s2
B. 4 m/s2
C. 7 m/s2
D. 11 m/s2
Answer:
[tex]4 ms^{-2}[/tex]
Explanation:
Normal force - should be the reaction from the surface it's on - nullifies the weight of the object. At this point the only force is of [tex]22-14 = 8N[/tex] from whoever is pushing the square and the friction, towards the right. At this point we divide the force by the mass of the object to obtain an acceleration of [tex]8/2 = 4 ms^{-2}[/tex]
please solve due is today....asap please
Answer:
a)object is in uniform motion
Explanation:
velocity is constant.and a=0
body is in uniform motion.
BRAINLIST PLS
choose all the answers that apply.
Major areas of science include
____.
A.basic science
B.physical science
C.life science
D.social science
E. ethical science
Answer:
physical science
Explanation:
i guess??
An object with a mass of 90 grams, moving at a constant velocity of 6 meters per second, has __________. A. A momentum of 540 gram meters per second B. An acceleration of 10 meters per second squared C. A momentum of 90 gram meters per second D. An acceleration of 6 meters per second squared Please select the best answer from the choices provided. A B C D.
Answer: 540 g m/s.
Explanation:
Answer: A.
A momentum of 540 gram meters per second!
Explanation: On Edge!!
What average force magnitude is required to stop a 15,000 kg train in 12.1 s if the train is traveling at 86 km/hr
Answer:
Explanation:
Givens
m = 15000 kg
t = 12.1 seconds
vi = 86 km / hr
vf =0
Formula
F = m*a
a = (vf - vi)/ t
vi has to be converted to m/s
86 km/hr [1000 m / 1 km] * [1 hour / 3600 seconds]
86 * 1000 / 3600 = 23.89 m/s
Solution
a = (vf - vi) / t
a = (0 - 23.89)/12.1
a = - 1.97 m/s^2 The minus means that the trains is slowing down.
F = 15000 kg * - 1.97 m/s^2
F = -29615.7 Newtons.
A ____ is a region where jets of gas from young stars impact and heat the gas surrounding the young star.
a ball is thrown with a speed of 100 m/s at a height of 150 m take g equals 9.8 m per second calculate the time of flight,angle of projection,range pls help I don't know how to solve it
Hi there!
Since the ball is thrown with an initial HORIZONTAL velocity, we can treat this as a free-fall situation since the horizontal motion does NOT impact the ball's vertical motion.
We can use the derived kinematic equation:
[tex]t = \sqrt{\frac{2h}{g}}[/tex]
Plug in the given values:
[tex]t = \sqrt{\frac{2(150)}{9.8}} = \large\boxed{5.53 s}}[/tex]
The angle of projection is 0° because there is no vertical component to the velocity as the initial velocity is purely horizontal.
Range:
We can use the equation:
dₓ = vₓt
displacement (x direction) = velocity (x direction) · time
Used the solved-for time and given velocity:
dₓ = 100 · 5.53 = 553 m
What would the kinetic energy of a 20kg person running at a velocity of 2.5m/s?
Answer:
62.5 JExplanation:
The kinetic energy of an object can be found by using the formula
[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]
m is the mass in kg
v is the velocity in m/s
From the question
m = 20 kg
v = 2.5 m/s
From the question we have
[tex]k = \frac{1}{2} \times 20 \times {2.5}^{2} \\ = 10 \times 6.25 \\ = 62.5\: \: \: \: \: \: \: \: \: \: [/tex]
We have the final answer as
62.5 JHope this helps you
Choose all of the following that are compounds.
[]Air
[]Ink
[]Sodium (Na)
[]Mud
[]Hydrochloric acid (HCI)
[]Ocean water
[]carbon dioxide (CO2)
[]Hydrogen gas (H2)
[]water (H20)
Answer:
Ink
mud
hcl
ocean water
co2
h20
Please help me calculate this question along with the equation
Hi there!
We can use the following:
U (Gravitational Potential Energy) = mgh
m = mass of object (kg)
g = acceleration due to gravity OR gravitational field strength
h = height, or displacement (m)
ΔU = Uf - Ui, so:
Uf = 3 × 1.5 × 10 = 45 J
Ui = 3 × 0 × 10 = 0
ΔU = Uf - Ui = 45 - 0 = 45 J
2. Summarize how to use a spring scale to measure a pull and how to use it to apply a push with
a specific force.
If the spring constant is known, the extension or compression of the spring can be used to calculate the applied force (pull or push) on the spring.
According to Hook's law, the force applied to an elastic material is directly proportional to the extension of the material.
The force applied;
F = kx
where;
k is the springThe applied force can be in form of push or pull. A push force can result in compression of the spring while a pull force can result in extension of the spring.
If the spring constant is known, the extension or compression of the spring can be used to calculate the applied force (pull or push) on the spring.
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How do you find the magnitude of the air inside a balloon?
Answer:
This demonstration is often done following a discussion of the ideal gas equation of state, PV=nRT.
We begin by weighing a balloon, then blowing it up and weighing it again. In the photo shown on right, the mass indication increased from 3.4 to 3.5 grams. At this point, it is important to note that the scale measures force, even though it reports a conclusion about mass based on the force measurement.
One assumption made in reaching the conclusion is that the buoyant force on the object being weighed is negligible. In the case of the balloon, this is incorrect. The buoyant force on this balloon is equal to the weight of the air displaced.
Since the volume of air inside the balloon is essentially the same as the volume of air displaced, we should expect that the buoyant force would support the weight of the air inside the balloon: The reported mass should not go up at all, because the force required of the scale should not change.
The increase in reported mass of .1 gram is attributed to the higher density of the air inside the balloon: The tension in the balloon compresses the air inside, as attested by the pressure required to blow the balloon up. Evidently, for this experiment, the pressure inside is greater than atmospheric by about 2%.
In the picture at right, the balloon is being pressed into a pan of liquid nitrogen. (The pan is the styrofoam lid of a small lunch box.) The balloon floats lightly on the liquid nitrogen unless pressed down. Pressing down places more surface area in contact with the cold nitrogen and speeds the demonstration. It is interesting to note the buoyant force by this liquified constituent of air.
The balloon shrinks dramatically, as indicated below. When left in contact with the liquid nitrogen long enough (perhaps 5 minutes) the oxygen inside the balloon liquifies, and then the nitrogen liquefies also. Close observation of the photo at the upper left corner of the pan shows some liquid nitrogen bubbles may forming above the dark spot in the center of the pan. One can also make out a faint line at the upper left corner of the pan which is the liquid nitrogen surface. The balloon still floats, riding rather high on that surface. Evidently, some of the balloon contents remain in the gas phase, making the mass of the balloon less than the mass of the displaced liquid nitrogen.
Next, we take the shrunken balloon and place it back on the scale, as above. In this instance, the reported mass is 8.7 grams, an increase of 5.2 grams.
A look at the figure on the right shows a faint line near the bottom of the cold balloon. Above that line, the balloon contains gas; below the liquid. That line represents the top surface of the liquid air inside the balloon. With this evidence, the easy thing to say would be, "Of course, liquids are heavier than gases," but that would be incorrect. We assert that the amount of air inside the balloon has not changed and that the mass of that air is not dependent on temperature.
If these assertions are true, then the force of gravity on the balloon has not changed. The scale reading is determined by the force which it must exert on the balloon in order to keep it stationary. Evidently, the required force is larger when the balloon is shrunken. The reason is that the buoyant force (upward) has decreased to practically zero, leaving the scale alone to balance the downward force by gravity.
From the data, we can say that the change in the buoyant force is equal to the weight associated with the apparent change in mass. The weight of 5.2 grams is about .052 newtons. The buoyant force is less now because the balloon displaces less air. If we could measure the change in volume of the balloon as DV, then the buoyant force would be (r g DV) upwards, where r is the density of air that was displaced by the balloon, and g is the gravitational field strength, 9.8 Newton/kg.
Note that the .052 newton force is not the weight of the air inside the balloon. Rather, it is the weight of the air that was displaced by the balloon. If we ignore the compression of air inside the balloon, the two numbers are the same. However, the two samples are completely different.
We can estimate the volume of the balloon by assuming that the hand in the photograph is about .1meters across. For purposes of estimation, we say that the volume shrank to almost zero when the balloon was cold so that the change in volume was nearly equal to the original volume. Plugging in numbers gives fair agreement with the book value of 1kg/cubic meter for the density of air.
The value for the density of air is secondary to two main features of this demonstration:
Large changes in temperature produce the large changes in volume that are indicated by the ideal gas equation.
The mass of air in a volume equal to the volume of a balloon can be determined provided that the buoyant force is understood.
In your response be very specific and break down each item on the list, then state if they would be classified the same or differently.
Consider the definition for each as you respond.
• atoms
• elements
• compounds
• molecules
• matter
CASE CLEANING:The plastic case that houses the PC components can be cleaned with lint-free cloth that has been slightly dampened with water. For stubborn stains, add a little household detergent to the cloth. Use vacuum around each of the holes, vents, and crevices on the computer.On the other hand, cleaning the keyboard needs to disconnect first the computer and peripheral from their power sources. Use a vacuum to remove any dust from the slots and holes on the keyboard.To blow the dust and debris stored in a local office supply use compressed air. Do you agree with the process presented above? MAKE A CLAIM AND WRITE A STATEMENT OR CONCLUSION THAT RESPONDS TO THE ORIGINAL QUESTION OR PROBLEM.
Answer:
She took it as priority and got to work
A tennis ball thrown with a velocity of 25m/s[Fwd] in the horizontal direction from the top of an 80m high building.
1. How long does it take the tennis ball to reach the ground?
2. How far is the range (how far from the base of the building will the ball land) ?
3. What is the final velocity?
Even answering one will help me out and i will give brainlest!
Answer:
Explanation:
Ignoring air resistance
1) from vertical rest
s = ½at²
t = √(2s/a)
t = √(2(80)/9.8)
t = 4.040610...
t = 4.0 s
2) d = vt
d = 25(4.0)
d = 100 m
3) v² = u² + 2as
v² = 25² + 2(9.8)(80)
v² = 2193
v = 46.8294...
v = 47 m/s
θ = - 90 + arcsin(25/46.8294)
θ = -57.7339...
θ = 58° below the horizontal
What time is the lunar eclipse tonight in california?
please no irrelevant answers!
Light enters a glass block at an angle of incidence of 46°. The light refracts at an angle of refraction of 26°. What is the refractive index of the glass?
A 0.57
B 0.61
C 1.64
D 1.77
Answer:
Choice C: Approximately [tex]1.64[/tex].
(Assuming that light entered into this glass block from air.)
Explanation:
Let [tex]n_{1}[/tex] denote the refractive index of the first medium (in this example, air.) Let [tex]\theta_{1}[/tex] denote the angle of incidence.
Let [tex]n_{2}[/tex] denote the refractive index of the second medium (in this example, the glass block.) Let [tex]\theta_{2}[/tex] denote the angle of refraction.
By Snell's Law:
[tex]n_{1} \, \sin(\theta_{1}) = n_{2}\, \sin(\theta_{2})[/tex].
Rearrange the equation to find an expression for [tex]n_{2}[/tex], the refractive index of the second medium (the glass block.)
[tex]\begin{aligned}n_{2} = \left(\frac{\sin(\theta_{1})}{\sin(\theta_{2})}\right)\, n_{1}\end{aligned}[/tex].
The refractive index of air is approximately [tex]1.00[/tex]. Substitute in the values and solve for [tex]n_{2}[/tex], the refractive index of the glass block:
[tex]\begin{aligned}n_{2} &= \left(\frac{\sin(\theta_{1})}{\sin(\theta_{2})}\right)\, n_{1} \\ &= \left(\frac{\sin(46^{\circ})}{\sin(26^{\circ})}\right)\times 1.00 \\ &\approx 1.64\end{aligned}[/tex].
A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller diagram below.
1) Find the total energy for the roller coaster at the initial point.
2) Find the potential energy at point A using the PE formula.
3) Use the conservation of energy to find the kinetic energy (KE) at point B.
4) Find the potential energy at point C.
5) Use the conservation of energy to find the Kinetic Energy (KE) of the roller coaster at point C.
6) Use the Kinetic Energy from C, find velocity of the roller coaster at point C.
For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:
1) The total energy for the roller coaster at the initial point is 19820 J
2) The potential energy at point A is 19620 J
3) The kinetic energy at point B is 10010 J
4) The potential energy at point C is zero
5) The kinetic energy at point C is 19820 J
6) The velocity of the roller coaster at point C is 19.91 m/s
1) The total energy for the roller coaster at the initial point can be found as follows:
[tex] E_{t} = KE_{i} + PE_{i} [/tex]
Where:
KE: is the kinetic energy = (1/2)mv₀²
m: is the mass of the roller coaster = 100 kg
v₀: is the initial velocity = 2 m/s
PE: is the potential energy = mgh
g: is the acceleration due to gravity = 9.81 m/s²
h: is the height = 20 m
The total energy is:
[tex] E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J [/tex]
Hence, the total energy for the roller coaster at the initial point is 19820 J.
2) The potential energy at point A is:
[tex] PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J [/tex]
Then, the potential energy at point A is 19620 J.
3) The kinetic energy at point B is the following:
[tex] KE_{A} + PE_{A} = KE_{B} + PE_{B} [/tex]
[tex] KE_{B} = KE_{A} + PE_{A} - PE_{B} [/tex]
Since
[tex] KE_{A} + PE_{A} = KE_{i} + PE_{i} [/tex]
we have:
[tex] KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J [/tex]
Hence, the kinetic energy at point B is 10010 J.
4) The potential energy at point C is zero because h = 0 meters.
[tex] PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J [/tex]
5) The kinetic energy of the roller coaster at point C is:
[tex] KE_{i} + PE_{i} = KE_{C} + PE_{C} [/tex]
[tex] KE_{C} = KE_{i} + PE_{i} = 19820 J [/tex]
Therefore, the kinetic energy at point C is 19820 J.
6) The velocity of the roller coaster at point C is given by:
[tex] KE_{C} = \frac{1}{2}mv_{C}^{2} [/tex]
[tex] v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s [/tex]
Hence, the velocity of the roller coaster at point C is 19.91 m/s.
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What might be an example of newtons second law of motion
Answer:
Pushing a car and a truck
Explanation:
how long does it take saturn to revolve around the sun
Answer: about 29.4 Earth years
Orbit and Rotation
One day on Saturn takes only 10.7 hours (the time it takes for Saturn to rotate or spin around once), and Saturn makes a complete orbit around the Sun (a year in Saturnian time) in about 29.4 Earth years (10,756 Earth days).
What is the potential energy of a 50kg car on top of a 600m hill?
Answer:
294,000 JExplanation:
The potential energy of a body can be found by using the formula
PE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 9.8 m/s²
From the question we have
PE = 50 × 9.8 × 600 = 294,000
We have the final answer as
294,000 JHope this helps you