Answer:As the temperature increases, the helium in the ballon expands.
Explanation:
I took the quiz already
The answer is C) As the temperature increases, the helium in the balloons expands.
Have a blessed day!
If someone is attempting to focus on multiple tasks at one time, it could be said that they have divided attention.
Please select the best answer from the choices provided
T
F
If someone is attempting to focus on multiple tasks at one time, it could be said that they have divided attention. the statement is true.
What is attention?The capacity for active processing of a particular piece of environmental information while tuning out other details is known as attention. In order to make sense of the world, we need to be able to properly manage the attentional resources we have at our disposal because attention is constrained in terms of both capacity and duration.
What is divided attention?Our brain's capacity to attend to two separate stimuli at once and respond to the various demands of our environment is known as divided attention. Divided attention, a type of simultaneous attention, enables us to successfully process several information sources and do multiple tasks at once. It is crucial to have these cognitive abilities since they help us live more effectively every day.
When someone is attempting to focus on multiple tasks at one time, his brain attends more than one stimuli at the same time and respond to the more than one demands of his surroundings. So, it could be said that they have divided attention.
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BRAINLIEST IF UR RIGHT!!!
Answer:cart 3
Explanation:
What is White light? Give your answer in one word.
Answer:
Colorless light
Explanation:
An example is day light.
(Score for Question 3: ___ of 8 points)
A 75 kg barrel is pulled up by a rope. The barrel accelerates at 3".4 m"/s^2. Find the force exerted on the barrel by the rope. Show all your work.
Answer:
Type your answer here.
Answer:
255N
Explanation:
Given parameters:
Mass of barrel = 75kg
Acceleration = 3.4m/s²
Unknown:
Force exerted = ?
Solution:
Force is the product of mass and acceleration according to Newton's second law of motion.
Force = mass x acceleration
Now insert given parameters and solve,
Force = 75 x 3.4 = 255N
The net force on the barrel is the sum of the pulling force (magnitude p, acting upward) and the barrel's weight (mag. w, opposing p ). So by Newton's second law,
∑ F = p - w = m a
p = w + m a
p = m g + m a
p = m (g + a)
p = (75 kg) (9.80 m/s² + 3.4 m/s²)
p = 990 N
The two blocks are attached to each other by a massless string that is wrapped
around a frictionless pulley as shown in figure-2. When the bottom 4.00 kg block is
pulled to the left by the constant force P~ = 45 N, the top 2.00 kg block slides across
it to the right. Assume that the coefficient of kinetic friction between all surfaces
is μk = 0.400
Answer:
a) The complete free body diagram of the system is presented in the image attached below according to the Newton's Law.
b) The acceleration of the top block is 0.962 meters per square second in the -x direction and the acceleration of the bottom block is 0.962 meters per square second in the -x direction.
c) The tension on the rope is 9.770 newtons.
Explanation:
a) The complete free body diagram of the system is presented in the image attached below according to the Newton's Law.
b) At first we have to construct corresponding equations of equilibrium for each mass:
Mass A
[tex]\Sigma F_{x} = T - \mu_{k}\cdot N_{A} = m_{A}\cdot a_{A}[/tex] (1)
[tex]\Sigma F_{y} = N_{A}-m_{A}\cdot g = 0[/tex] (2)
Mass B
[tex]\Sigma F_{x} = -P+\mu_{k}\cdot N_{A}+\mu_{k}\cdot N_{B}+T= -m_{B}\cdot a_{A}[/tex] (3)
[tex]\Sigma F_{y} = N_{B}-N_{A}-m_{B}\cdot g= 0[/tex] (4)
Where:
[tex]P[/tex] - Force exerted on the bottom block, measured in newtons.
[tex]N_{A}[/tex], [tex]N_{B}[/tex] - Normal forces, measured in newtons.
[tex]m_{A}[/tex], [tex]m_{B}[/tex] - Masses of the top and bottom blocks, measured in kilograms.
[tex]\mu_{k}[/tex] - Kinetic coeffcient of friction, dimensionless.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]a_{A}[/tex], [tex]a_{B}[/tex] - Net accelerations of the top and bottom blocks, measured in meters per square second.
From (2):
[tex]N_{A} = m_{A}\cdot g[/tex] (5)
(2) in (4):
[tex]N_{B}=(m_{A}+m_{B})\cdot g[/tex] (6)
(5) in (1):
[tex]T = m_{A}\cdot a_{A}+\mu_{k}\cdot m_{A}\cdot g[/tex] (7)
(6) in (3):
[tex]T = -m_{B}\cdot a_{A}+P-\mu_{k}\cdot m_{A}\cdot g-\mu_{k}\cdot (m_{B}+m_{A})\cdot g[/tex]
[tex]T = -m_{B}\cdot a_{A}+P -\mu_{k}\cdot (2\cdot m_{A}+m_{B})\cdot g[/tex] (8)
And we obtain an expression for the magnitude of the acceleration by eliminating tension:
[tex]m_{A}\cdot a_{A} +\mu_{k}\cdot m_{A}\cdot g = -m_{B}\cdot a_{A}+P-\mu_{k}\cdot (2\cdot m_{A}+m_{B})\cdot g[/tex]
[tex](m_{A}+m_{B})\cdot a_{A} = P-\mu_{k}\cdot (3\cdot m_{A}+m_{B})\cdot g[/tex]
[tex]a_{A} = \frac{P-\mu_{k}\cdot (3\cdot m_{A}+m_{B})\cdot g}{m_{A}+m_{B}}[/tex]
If we know that [tex]P = 45\,N[/tex], [tex]m_{A} = 2\,kg[/tex], [tex]m_{B} = 4\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\mu_{k} = 0.4[/tex], then the magnitude of the acceleration is:
[tex]a_{A} = \frac{45\,N-(0.4)\cdot [3\cdot (2\,kg)+4\,kg]\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{2\,kg+4\,kg}[/tex]
[tex]a_{A} = 0.962\,\frac{m}{s^{2}}[/tex]
The acceleration of the top block is 0.962 meters per square second in the -x direction and the acceleration of the bottom block is 0.962 meters per square second in the -x direction.
c) By means of (7) and knowing that [tex]m_{A} = 2\,kg[/tex], [tex]a_{A} = 0.962\,\frac{m}{s^{2}}[/tex], [tex]\mu_{k} = 0.4[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], we find that the tension on the rope is:
[tex]T = (2\,kg)\cdot \left(0.962\,\frac{m}{s^{2}} \right)+(0.4)\cdot (2\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]T = 9.770\,N[/tex]
The tension on the rope is 9.770 newtons.
5
b. What is the molecular shape of the molecule? (3 points)
Answer:
tetrahedral shape.
Explanation:
Answer:
Molecular Geometries. The VSEPR theory describes five main shapes of simple molecules: linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. Apply the VSEPR model to determine the geometry of molecules where the central atom contains one or more lone pairs of electrons.
When you use a match to light a candle _______________________________ transformations of energy
Answer:
thermal or heat energy
Which organ completes the majority of chemical digestion?
Answer:
The majority of chemical digestion occurs in the small intestine. Digested chyme from the stomach passes through the pylorus and into the duodenum.
You are driving on the highway at a speed of 40 m/s (which is over the speed limit) when you notice a cop in front of you. To avoid a ticket, you press on the brake and slow to a speed of 30 m/s over the course of 5 seconds. What is the acceleration of the car? SHOW WORK OR GET REPORTED(have 5 accounts)
What is your car's initial velocity?
What is your car's final velocity?
How long does it take the car to slow down?
Write the equation you will use to solve this problem.
What is the acceleration of your vehicle?
+ 2.0 m/s^2
- 2.0 m/s^2
+ 8.0 m/s^2
- 6.0 m/s^2
Answer:
What is the acceleration of your vehicle?
+ 2.0 m/s^2 ✔
- 2.0 m/s^2
+ 8.0 m/s^2
- 6.0 m/s^2
Fill in the blanks to complete each statement about metamorphic rocks.
Rocks beneath the surface are forced toward the
. This increases pressure and heat.
pockets rise, increasing heat. Temperature and pressure cause
to change.
Answer:
1) MANTLE 2) MAGMA 3) MINERALS
Explanation:
I just did the assignment and got it correct hope this helps
Rocks beneath the surface are forced toward the mantle. This increases pressure and heat. Magma pockets rise, increasing heat. Temperature and pressure cause to change minerals
What is Metamorphic rock ?A Metamorphic rock is a type of rock which has been changed by extreme heat and pressure
Subduction is the process in which a dense tectonic plate slips beneath a more buoyant one . Hence ,Metamorphic rock and minerals of lithosphere transported to the mantle .
hence , Rocks beneath the surface are forced toward the Mantle
The reduction in overlying pressure , decompression enables the mantle rock and form magma
hence ,magma pockets rise, increasing heat.
When pressure and temperature change , chemical reactions occur to cause the mineral in the Metamorphic rock to change to an assemblage that is stable at the new pressure and temperature conditions.
Temperature and pressure cause to change minerals
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A wheelbarrow is an example of which simple machine?
O pulley
O lever
O screw
wheel and axel
What is a difference between an object's speed and velocity?
Answer:
.
Explanation:
Speed, being a scalar quantity, is the rate at which an object covers distance. ... On the other hand, velocity is a vector quantity; it is direction-aware. Velocity is the rate at which the position changes. The average velocity is the displacement or position change (a vector quantity) per time ratio.
Answer:
velocity is the magnitude and the direction and speed is the distance something traveled in a certain amount of time
SOMETHING:
hope this helps
A hockey puck attached to a horizontal spring oscillates on a frictionless, horizontal surface. The spring has force constant 4.50 N/m and the oscillation period is 1.20 s. What is the mass of the puck
Answer:
m = 0.164 kg
Explanation:
T (period)
k (force/spring constant)
m (mass)
T = 2*Pi*sqrt(m/k)
T/(2*Pi) = sqrt(m)/sqrt(k)
(T/(2*Pi))*sqrt(k) = sqrt(m)
m = ((T/(2*Pi))*sqrt(k))^2
m = 4.5*((1.2/(2*Pi)))^2
m = 0.1641403175
1 point
You throw a ball up in the air with a velocity of 30 m/s. How high does it
go?
Answer:
Explanation:
2as=vf^2-Vi^2
vf=30 m/s
vi= 0 m/s
a=g=9.8 m/s^2
s=vf^2-Vi^2/2a
s=(30)²-(0)²/2*9.8
s=900-0/19.6
s=45.9=46 m
what problems might it cause if a company tried to recycle materials without sorting them first? at least 2 or 3 sentences please!!!!
Answer:
They would probably not be looking good.. like if u don't sort it out they won't be shaped well and there are many types of plastic and if we combine them together they won't give the correct form..
hope this helps :)
Why sun is known as main or ultimate
source of energy?( please in short answer)
Answer:
because of the heats came from to ultraviolet trnasform and i thankyou
Determine the velocity (in m/s) of the object during the last six seconds. Include a numerical answer accurate to the second decimal place. If negative, include the - sign
Answer:
0.33 m/s
Explanation:
The graph given is that of distance against time. The slope of such a graph gives the velocity of the object while travelling.
In this graph the change in distance for the last 6 seconds is given by;
The distance is : 10-8 = 2 m
The time in seconds is: 6 s
The velocity = 2/6 = 1/3 m/s
The velocity expressed in decimal form is : 0.33 m/s
fun facts are,
Butterflies taste with their hind feet.
A group of crows is called murder.
Tomato sauce was sold in the 1800's as medicine.
A donkey will sink in quicksand, but a mule won't.
Lakes Can Explode. ...
Earth Used to Be Purple. ...
60 Tons of Cosmic Dust Fall to Earth Daily. ...
The Oceans Hold $771 Trillion Worth of Gold. ...
Earth Once Had Two Moons. ...
Earth Might Still Have Two Moons. ...
The Greatest Vertical Drop Is in Canada. ...
The Hottest Spot in the World Is in California.
Answer:
that is cool and i have one interesting fact
Explanation: North Korea and Cuba are the only places you can't buy Coca-Cola
Please help with this question
Answer:
I believe the answer is 5718.75. Respond if it is wrong please.
Explanation:
I used a calculator.
How do the forces on a skydiver affect the velocity of the skydiver as they fall?
Hellp
Answer:
Explanation:
As a skydiver falls, he accelerates downwards, gaining speed with each second. The increase in speed is accompanied by an increase in air resistance. This force of air resistance counters the force of gravity.
When a skydiver falls, he constantly accelerates. Therby, falling faster and faster. But also he is slowed down a bit due to the force of air resistance.
The velocity of a Bus is reduced uniformly from 15 m/s to 7 m/s while traveling a distance of 90 m.
i. Compute the acceleration.
ii. How much further will the Bus travel before coming to rest, provided the acceleration remain constant?
Answer:
i) a = 0.977 [m/s²]
ii) x = 115.06 [m]
Explanation:
In order to determine the acceleration, we must use the following equation of kinematics.
[tex]v_{f} ^{2} =v_{o} ^{2} -2*a*x[/tex]
where:
Vf = final velocity = 7 [m/s]
Vo = initial velocity = 15 [m/s]
a = acceleration [m/s²]
x = displaciment = 90 [m]
Now replacing:
[tex](7)^{2} =(15)^{2} -2*a*90\\2*a*90 = 15^{2} - 7^{2} \\180*a=176\\a=0.977[m/s^{2}][/tex]
When the bus coming to rest.
[tex]v_{f}^{2} =v_{o} ^{2} -2*a*x\\0 = 15^{2} -2*0.977*x\\x = 115.06[m][/tex]
5) A train travels 120 km in 90 minutes (min). What is its average speed?
Answer:
1.33 km per minute
Explanation:
120 km divided by 90 is 1.33 km
Two protons are on either side of an electron as shown below:
Diagram
The electron is 30 µm away from the proton on its left and 10 µm away from the proton on its right. What is the magnitude and direction of the net electric force acting on the electron? Note that
P=1.6×10(-19) c=-e
Select one:
a. 2.0×10−18N to the right
b. 7.0×10−24N to the right
c. 2.0×10−18N to the left
d. −9.59×10−24N to the right
e. 7.0×10−24N to the
Answer:
Choice a. approximately [tex]2 \times 10^{-18}\; \rm N[/tex] to the right.
Explanation:
Look up the Coulomb constant, [tex]k[/tex]:
[tex]k \approx 8.987552 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex].
Coulomb's Law may be used to find the magnitude of the electric force between the electron and each proton.
Let [tex]q_1[/tex] and [tex]q_2[/tex] denote the magnitude of electric charge on two point charges. Let [tex]r[/tex] denote the distance between these two charges.
By Coulomb's Law, the magnitude of the force between these two point charges would be:
[tex]\displaystyle F = \frac{k\, q_1\, q_2}{r^2}[/tex].
The unit of distance in the Coulomb's constant here[tex]k \approx 8.987552 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex] is meters. However, the distance between the electron and each proton is given in micrometers, [tex]\rm \mu m[/tex]. Note that [tex]1\; \rm \mu m = 10^{-6}\; \rm m[/tex].
Convert these units to meters:
[tex]30 \; \rm \mu m = 30 \times 10^{-6}\; \rm m[/tex].
[tex]10 \; \rm \mu m = 10 \times 10^{-6}\; \rm m[/tex].
The question states that the magnitude of electric charge on each proton and each electron is [tex]1.6\times 10^{-19}\; \rm C[/tex].
Using these information, find the magnitude of the electric force between the electron and each proton:
Between the electron and the proton on the left-hand side of the electron: [tex]\begin{aligned}F &= \frac{k\, q_1\, q_2}{r^2}\\ &\approx \frac{\left(8.987552\times 10^{9}\; \rm N \cdot m^2\cdot C^{-2}\right) \times \left(1.6 \times 10^{-19}\; \rm C\right) \times \left(1.6 \times 10^{-19}\; \rm C\right)}{\left(30 \times 10^{-6}\; \rm m\right)} \\ &\approx 0.26 \times 10^{-18}\; \rm N\end{aligned}[/tex].Between the electron and the proton on the right-hand side of the electron:[tex]\begin{aligned}F &= \frac{k\, q_1\, q_2}{r^2}\\ &\approx \frac{\left(8.987552\times 10^{9}\; \rm N \cdot m^2\cdot C^{-2}\right) \times \left(1.6 \times 10^{-19}\; \rm C\right) \times \left(1.6 \times 10^{-19}\; \rm C\right)}{\left(10 \times 10^{-6}\; \rm m\right)} \\ &\approx 2.56 \times 10^{-18}\; \rm N\end{aligned}[/tex].The two protons in this question will both attract the electron. Therefore, the force between the electron and the proton on the left-hand side of the electron would point to the left. Similarly, the force between the electron the proton on the right-hand side of the electron would point to the right.
The magnitude of the net electric force on the electron would be:
[tex]2.56 \times 10^{-18}\; \rm N - 0.26\times 10^{-18}\;\rm N \approx 2\times 10^{-18}\; \rm N[/tex].
These two forces act along the same line in opposite direction. Therefore, the resultant force of these two force would be in the direction of the larger force of the two.
In this question, the electric force between the electron and the proton on the right-hand side of the electron is larger. Hence, the net electric force of the protons on the electron should point to the right (towards the proton that is closer to the electron.)
1. Newton's third law states that any action will have a(n) and reaction O A. Equal and opposite O B. Greater and opposite C. Equal and similar D. Equal and different
Answer: Equal and Opposite
Explanation:
What constant acceleration (in ft/s2) is required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds
Answer:
6.746 ft/s^2
Explanation:
v(t)=50
v(0)=27
t=5/3600 = 1/720 hours
v(t)-v(0)= a(t-0)
50-27= a(1/720)
a= 23*720= 16560 mi/h^2
16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2
The acceleration required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds is 6.75 ft/s².
Acceleration is given by:
[tex] a = \frac{\Delta v}{\Delta t} = \frac{v_{f} - v_{i}}{t_{f} - t_{i}} [/tex]
Where:
[tex] v_{f} [/tex]: is the final speed of the car = 50 mi/h
[tex] v_{i}[/tex]: is the initial speed of the car = 27 mi/h
[tex] \Delta t[/tex]: is the change in time = 5 seconds
The constant acceleration required is:
[tex] a = \frac{v_{f} - v_{i}}{\Delta t} [/tex]
[tex] a = \frac{50 \frac{mi}{h}*\frac{1 h}{3600 s}*\frac{5280 ft}{1 mi} - 27 \frac{mi}{h}*\frac{1 h}{3600 s}*\frac{5280 ft}{1 mi}}{5 s} [/tex]
[tex] a = 6.75 ft/s^{2} [/tex]
Therefore, the constant acceleration required is 6.75 ft/s².
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Determine the mass of an object if it's moving at 5 m/s and has a momentum of 50 kg*m/s
A girl swings a 0.250 kg rock attached to a taut string in a circle around her head. Her hand holds the end of the string above her head, and the string angles down slightly (11.9° below the horizontal.). The string is massless and 0.75m long.
A coordinate system lies with its origin a the location where the string comes out of the girl's hand. The positive z-axis points vertically. the projection onto the horizontal plane is such that the string makes an angle of 34.6° with the x-axis.
At a certain instant, the rock makes 2.50 revolutions per second (rev/s) in a counterclockwise direction as seen from above.
What is the tangential velocity vector at this instant?
Complete Question
The diagram of with this question is shown on the first uploaded image
Answer:
The value is [tex]v = -6.543 \^ i + 9.47 \^ j + 0 \^ k[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m = 0.250 \ kg[/tex]
The length of the string is [tex]L = 0.75 \ m[/tex]
The angle the string makes horizontal is [tex]\theta = 11.9^o[/tex]
The angle which the projection of the string onto the xy -plane makes with the positive x-axis is [tex]\phi = 34.6^o[/tex]
The angular velocity of the rock is [tex]w = 2.50 rev/s = 2.50 * 2\pi =15.7 \ rad/s[/tex]
Generally the radius of the circle made by the length of the string is mathematically represented as
[tex]r = L cos(\theta )[/tex]
=> [tex]r = 0.75 cos(11.9 )[/tex]
=> [tex]r = 0.734 \ m[/tex]
Generally the resultant tangential velocity is mathematically represented as
[tex]v__{R}} = w * r[/tex]
=> [tex]v__{R}} = 15.7 *0.734[/tex]
=> [tex]v__{R}} = 11.5 \ m/s[/tex]
Generally the tangential velocity along the x-axis is
[tex]v_x = -v__{R}} * sin(\phi)[/tex]
=> [tex]v_x =- 11.5 * sin(34.6)[/tex]
=> [tex]v_x = -6.543 \ m/s[/tex]
The negative sign show that the velocity is directed toward the negative x-axis
Generally the tangential velocity along the y-axis is
[tex]v_y = v__{R}} * cos(\phi)[/tex]
=> [tex]v_y = 11.5 * cos(34.6)[/tex]
=> [tex]v_y = 9.47 \ m/s[/tex]
Generally the tangential velocity along the y-axis is
[tex]v_z = v__{R}} * cos(90)[/tex]
=> [tex]v_z = 0 \ m/s[/tex]
Generally the tangential velocity at that instant is mathematically represented as
[tex]v = -6.543 \^ i + 9.47 \^ j + 0 \^ k[/tex]
The standard kilogram is a platinum-iridium cylinder 39.0 mm in height and 39.0 mm in diameter. What is the density of the material?
Density is mass per unit of volume, usually measured in g/cm³. So first determine the volume of the cylinder:
v = π r ² h = π ((39.0 mm) / 2)² (39.0 mm)
v = 59,319/4 π mm³ ≈ 46,589 mm³ ≈ 46.589 cm³
Then the density is
ρ = m / v
ρ = (1 kg) / v = (1000 g) / v
ρ ≈ 21.464 g/cm³
Or, if you want to preserve the given units, that's
ρ ≈ 0.00002146 kg/mm³
7. How many forces are acting on this object? *
Fnorm
Ffrict
Fapp
Fgray
Å
Answer:
4
Explanation:
As you can see in the free body diagram there are 4 forces acting on the body.
A horizontal force is used to pull a 5.0 kilogram cart at a constant speed of 5.0 m/s across the floor. The force of friction between the cart and the floor is 10.0 newtons. What is the magnitude of the horizontal force along the handle of the cart?
5 N
10 N
15 N
20 N
Answer:10N
Explanation: