Answer:
the average kinetic energy of the 10 molecules is 10 J.
Explanation:
Given;
energy on one molecule in motion, E = 100 J
number of molecules, n = 10
(A) The average kinetic energy of the 10 molecules
since the remaining 9 molecules are at rest, their kinetic energy = 0
[tex]E_{Avg} = \frac{E_1 + E_9}{10} \\\\E_{Avg} = \frac{100J+ 0}{10} \\\\E_{Avg} = \frac{100J}{10} \\\\E_{Avg} = 10J[/tex]
Therefore, the average kinetic energy of the 10 molecules is 10 J.
What do you call the height of a wave?
a. wavelength
b. frequency
c. amplitude
d. resonance
Answer:
amplitude is the answer
Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.
How much kinetic energy does the rake gain?
Answer:
10 J.
Explanation:
Given that,
Net force acting on the rake, F = 2 N
Distance moved by the rake, d = 5 m
We need to find the kinetic energy gained by the rake. We know that,
Kinetic energy = work done
So,
K = F×d
K = 2 N × 5 m
K = 10 J
So, 10 J of kinetic energy is gained by the rake.
Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.
How much kinetic energy does the rake gain?
Answer: 10 J
You are walking on a moving walkway in the airport. The length of the walkway is 59.1 m. If your velocity relative to the walkway is 2.35 m/s, and the walkway moves with a velocity of 1.85 m/s, how long will it take you to reach the other end of the walkway
Answer:
14.1seconds approx
Explanation:
Given data
Distance= 59.1m
Your velocity= 2.35m/s
Walkway velocity= 1.85m/s
Total velocity= 2.35+1.85= 4.2m/s
We know that
Velocity= distance/time
time= distance/velocity
substitute
time= 59.1/4.2
time= 14.07
time=14.1seconds approx
Hence the time is 14.1seconds approx
Two loudspeakers are about 10 mm apart in the front of a large classroom. If either speaker plays a pure tone at a single frequency of 400 HzHz, the loudness seems pretty even as you wander around the room, and gradually decreases in volume as you move farther from the speaker. If both speakers then play the same tone together, what do you hear as you wander around the room
Answer:
I hear points of low volume sound and points of high volume of sound.
Explanation:
This is because, since the two sources of sound have the same frequency and are separated by a distance, d = 10 mm, there would be successive points of constructive and destructive interference.
Since their frequencies are similar, we should have beats of high and low frequency.
So, at points of low frequency, the amplitude of the wave is smallest and there is destructive interference. The frequency at this point is the difference between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, f - f' = 400 Hz - 400 Hz = 0 Hz. So, the volume of the sound is low(zero) at these points.
Also, at points of high frequency, the amplitude of the wave is highest and there is constructive interference. The frequency at this point is the sum between the frequencies from both speakers. Since the frequency from both speakers is 400 Hz, we have, (f + f') = 400 Hz + 400 Hz = 800 Hz. So, the volume of the sound is high at these points.
So, as you wander around the room, I should hear points of high and low sound across the room.
Question 18 of 25
Which type of reaction is shown in this energy diagram?
Energy
Products
Activation
Energy
Reoctants
to
ti
Time
A. Endothermic, because the products are lower in energy
B. Exothermic, because the reactants are lower in energy
C. Endothermic, because the reactants are lower in energy
D. Exothermic, because the products are lower in energy
Answer:
Endothermic, because the reactants are lower in energy (C)
Explanation:
From the graph, you can see the energy of the products is higher than the energy of the reactants. If you recall that when the enthalpy change Eproducts is gretater than Ereactants, the reaction is said to be endothermic.
The planet Mars is host to five functioning spacecraft, three in orbit about the planet and two on the surface of the planet. Thanks to those spacecraft, we know that the planet Mars has a mass that is 0.11 times that of Earth and a radius that is 0.53 times that of Earth. The acceleration of an object in free-fall near the surface of Mars is most nearly what in terms of the local value of g on Earth
Answer:
0.392
Explanation:
Mm = 0.11Me
Rm = 0.53Re
g = GM / r^2
G = 6.67 * 10^-11
gmars = (G * 0.11Me) / (0.53Re)^2
Recall:
gearth = GMe /Re^2
Hence, gmars in terms of gearth equals
gmars = gearth * (0.11 / 0.53^2)
gmars = gearth * 0.3915984
gmars = 0.392gearth
A ball is dropped from a building taking 3sec to fall to the ground. Calculate:
Speed when falling to the ground
The height of the building
Answer:
Vf = 29.4 m/s
h = 44.1 m
Explanation:
Data:
Initial Velocity (Vo) = 0 m/sGravity (g) = 9.8 m/s²Time (t) = 3 sFinal Velocity (Vf) = ?Height (h) = ?==================================================================
Final Velocity
Use formula:
Vf = g * tReplace:
Vf = 9.8 m/s² * 3sMultiply:
Vf = 29.4 m/s==================================================================
Height
Use formula:
[tex]\boxed{h=\frac{g*(t)^{2}}{2}}[/tex]Replace:
[tex]\boxed{h=\frac{9.8\frac{m}{s^{2}}*(3s)^{2}}{2}}[/tex]Multiply time squared:
[tex]\boxed{h=\frac{9.8\frac{m}{s^{2}}*9s^{2}}{2}}[/tex]Simplify the s², and multiply in the numerator:
[tex]\boxed{h=\frac{88.2m}{2}}[/tex]It divides:
[tex]\boxed{h=44.1\ m}[/tex]What is the velocity when falling to the ground?
The final velocity is 29.4 meters per seconds.
How high is the building?
The height of the building is 44.1 meters.
What makes electromagnets useful for sorting metals in recycling centers?
O A. The current can be turned on to pick up items containing all
metals and turned off to drop them.
O B. The current can be turned off to pick up items containing all
metals and turned off to drop them.
O C. The current can be turned on to pick up items containing iron and
turned off to drop them.
D. The current can be turned off to pick up items containing iron and
turned on to drop them.
C
It is right because I took this and I got this answer correct
5.0 L/s water flows through a horizontal pipe that narrows smoothly from 10.0 cm diameter to 5.0 cm diameter. A pressure gauge in the narrow section reads 50 kPa. What is the reading of the pressure gauge in the wide section
Solution :
The volume rate of flow is given by : R = 5.0 L/s
[tex]$ = 5.0 \times 10^{-3} \ m^3/s$[/tex]
The radius of the pipe, [tex]$r_1= 5 \times 10^{-2} \ m$[/tex]
∴ [tex]$ 5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]
then, [tex]$v_1 = \frac{5.0 \times 10^{-3}}{(3.14)(5 \times 10^{-2})^2}$[/tex]
= 0.637 meter per second
Then the speed of the water at wider section,
[tex]$R=A_1v_1$[/tex]
Similarly, the speed of water at narrow pipe.
The radius of the [tex]$r_2 = 2.5 \times 10^{-2}$[/tex] m
[tex]$5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]
then, [tex]$v_2 = \frac{5.0 \times 10^{-3}}{(3.14)(2.5 \times 10^{-2})^2}$[/tex]
= 2.55 meter per sec
Now from Bernoulli's theorem,
[tex]$P_1 + \frac{1}{2} \rho v_1^2 =P_2 + \frac{1}{2} \rho v_2^2 $[/tex]
[tex]$P_1 = P_2 + \frac{1}{2} \rho (v_2^2 - v_1^2)$[/tex]
[tex]$= 50 \kPa + (0.5)(10^3)[(2.55)^2-(0.637)^2]$[/tex]
= 50 kPa + 3.05 kPa
= 53.05 kPa
or 53000 Pa
This question involves the concepts of Bernoulli's Theorem and Volumetric Flowrate.
The pressure reading in the wide section is "53.05 KPa".
First, we will use the volumetric flow rate to find the velocities of the water at wide and narrow sections.
[tex]V = A_1v_1[/tex]
where,
V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s
r₁ = radius of narrow section = 5 cm/2 = 2.5 cm = 0.025 m
A₁ = Area of narrow section = πr₁² = π(0.025 m)²
v₁ = velocity at narrow section = ?
Therefore,
[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.025\ m)^2](v_1)\\\\v_1=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.025\ m)^2}\\\\v_1=2.55\ m/s\\[/tex]
Similarly,
[tex]V = A_2v_2[/tex]
where,
V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s
r₂ = radius of wide section = 10 cm/2 = 5 cm = 0.05 m
A₂ = Area of wide section = πr₁² = π(0.05 m)²
v₂ = velocity at wide section = ?
Therefore,
[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.05\ m)^2](v_2)\\\\v_2=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.05\ m)^2}\\\\v_2=0.64\ m/s\\[/tex]
Now, we will use Bernoulli's Theorem to find out the pressure wide section.
[tex]P_1 + \frac{1}{2}\rho v_1^2=P_2 + \frac{1}{2}\rho v_2^2[/tex]
where,
[tex]\rho[/tex] = density of water = 1000 kg/m³
P₁ = pressure in narrow section = 50 KPa = 50000 Pa
P₂ = pressure in wide section = ?
Therefore,
[tex]50000\ Pa + \frac{1}{2}(1000\ kg/m^3)(2.55\ m/s)^2=P_2 + \frac{1}{2}(1000\ kg/m^3)(0.64\ m/s)^2[/tex]
P₂ = 50000 Pa + 3251.25 Pa - 204.8 Pa
P₂ = 53046.45 Pa = 53.05 KPa
Learn more about Bernoulli's Theorem here:
https://brainly.com/question/13098748?referrer=searchResults
The attached picture shows Bernoulli's Theorem.
A plane wishes to fly due north to an airport which is 205 km away. The plane can fly at a speed in still air of 220 km/h. A wind of 43 km/h blows from east to west.
a. In which direction,relative to north, should the plane head to reach it’s destination?
b. How long does this take?
Answer:
nique ta mama
Explanation:
How does Doppler ultrasound technology differ from ultrasound technology
that does not use the Doppler effect?
A. Doppler ultrasound collects data from moving objects.
B. Other ultrasound technology creates images, but Doppler
ultrasound does not.
C. Doppler ultrasound creates images, but other ultrasound
technology does not.
D. Doppler ultrasound is based on absorption of sound, and other
ultrasound technology is based on reflection.
Answer:
A. Doppler ultrasound collects data from moving objects
Explanation:
Did the test !!
Answer:A. Doppler ultrasound collects data from moving objects.
Explanation: just got it right
on my test
Iron has a specific heat of o.45 J/g °C. Removing -1.16 E 2 J of energy lowered the temperature of iron from 89 °C to 26.41 °C. What was the mass of the iron?
Answer:
m = 4.11 grams
Explanation:
Given that,
The specific heat of Iron, c = 0.45 J/g°C
Heat removed, [tex]Q=1.16\times 10^2\ J[/tex]
Initial temperature, [tex]T_i=89^{\circ} C[/tex]
Final temperature, [tex]T_f=26.41^{\circ} C[/tex]
We need to find the mass of the iron. We know that the heat removed in terms of specific heat is given by :
[tex]Q=mc\Delta T\\\\m=\dfrac{Q}{c\Delta T}\\\\m=\dfrac{-1.16\times 10^2}{0.45\times (26.41-89)}\\\\m=4.11\ g[/tex]
So, the mass of the iron is 4.11 grams.
Hearing rattles from a snake, you make two rapid displacements of magnitude 1.8 m and 2.4m. Draw sketches, roughly to scale, to show how your two displacements might add to give the following resultant of magnitudes.
a. 4.2 m
b. 0.6 m
c. 3.2 m
Answer:
The answer is a 4.2m
Explanation:
Given data
Please see attached the rough drawing for your reference.
From the drawing, you ran 18m west and 2.4m south
The displacement is
= 1.8+2.4
=4.2m
Q4. What is the speed of light in a block of glass that
has an Index of refraction of 1.11?
A certain brand of hotdog cooker works by applying a potential difference of 144 V across opposite ends of a hot dog and allowing it to cook by means of the thermal energy produced. The current is 10.5 A, and the energy required to cook one hot dog is 60.6 kJ. If the rate at which energy is supplied is unchanged, how long will it take to cook three hot dogs simultaneously
Answer:
120.237 seconds
Explanation:
Given that:
V = 144 V
I = 10.5 A
H = 60.6 kJ
Using the formula:
H = I²RT
From H = I²RT; making T the subject, we have:
[tex]T = \dfrac{H}{I^2R}[/tex]
where;
[tex]R = \dfrac{V}{I}[/tex]
∴
[tex]T = \dfrac{H}{V \times I}[/tex]
[tex]T = \dfrac{60.6 \times 10^3 }{144 \times 10.5}[/tex]
T = 40.079
[tex]T_{neq} = 3T[/tex]
[tex]T_{neq} =3 \times 40.079[/tex]
[tex]\mathbf{T_{neq} =120.237 \ sec}[/tex]
2. A force of 10 N is applied to an object which accelerates at a rate of 2m/s2. What is the mass
of the object?
(10 Points)
10 kg
5 kg
2 kg
20 kg
kinetic energy portfolio in part 2 the independent changes to----?
what are the precautions to be taken while performing a rectangular glass prism experiment
Answer:
PRECAUTIONS
-The refracting faces of the glass prism should be smooth, transparent and without any air bubble or broken edge. ...
-Use a sharp pencil to draw boundary of the prism and rays of light.
-The alpins should have sharp tip and should be fixed exactly vertical to the plane of the paper.
Explanation:
Please give thanks to all my answers and please mark as brilliant and please follow me
What simple machine can best be described as "a simple machine that uses an inclined plane wrapped around a rod"?
a wedge
a screw
a wheel and axle
a lever
Answer:
It is a screw.
Explanation:
On a scale of 1-10 how much do you care of what people think of you?
Answer:
3
Explanation:
my family i hope thinks of me. And I don't have friends for them to think of me.
A spacecraft and a staellite are at diametrically opposite position in the same circular orbit of altitude 500 km above the earth. As it passes through point A, the spacecraft fires its engine for a short interval of time to increase its speed and enter an elliptical orbit. Knowing that the spacecraft returns to A at the same time the satellite reaches A after completing one and a half orbits, determine (a) the increase in speed required, (b) the periodic time for the elliptic orbit
Answer:
Hello the diagram related to your question is attached below
answer: a) 851 m/s
b) 8506.1 secs
Explanation:
calculate the periodic time of the satellite using the equation below
t = [tex]\frac{2\pi }{R} \sqrt{\frac{(R+h)^{3} }{g} }[/tex] -- ( 1 )
where ; R = 6370 km
h = 500 km
g = 9.81 m/s^2
input given values into equation 1
t = 5670.75 secs
next calculate the periodic time taken by the space craft
a) determine the increase in speed
V = v - [tex]\sqrt{\frac{gR^2}{R + h} }[/tex]
where ; v = 8463 m/s , R = 6370 km, h = 500 km
V = 851 m/s
b) Determine the periodic time for the elliptic orbit
τ = [tex]\frac{3t}{2}[/tex]
= [tex]\frac{3*5670.76}{2}[/tex] = 8506.1 secs
attached below is the remaining part of the detailed solution
A pendulum has a period of 6.98s. Calculate the length of the pendulum. Use
9.8m/s^2 for gravity. *
Answer:
Length, l = 0.126 meters.
Explanation:
Given the following data;
Period = 6.98s
Acceleration due to gravity, g = 9.8m/s²
To find the length, l;
[tex] Period, T = 2 \pi \sqrt {lg} [/tex]
Substituting into the equation, we have;
[tex] 6.98 = 2*3.142 \sqrt {l*9.8} [/tex]
[tex] 6.98 = 6.284 \sqrt {9.8l} [/tex]
[tex] \frac {6.98}{6.284} = \sqrt {9.8l} [/tex]
[tex] 1.1108 = \sqrt {9.8l} [/tex]
Taking the square of both sides
[tex] 1.1108^{2} = 9.8l [/tex]
[tex] 1.2339 = 9.8l [/tex]
[tex] l = \frac {1.2339}{9.8} [/tex]
Length, l = 0.126m.
A car accelerates from zero to a speed of 110
km/hr in 15 seconds. What is the car's rate of
acceleration?
The car's rate of acceleration : a = 2.04 m/s²
Further explanationGiven
speed = 110 km/hr
time = 15 s
Required
The acceleration
Solution
110 km/hr⇒30.56 m/s
Acceleration is the change in velocity over time
a = Δv : Δt
Input the value :
a = 30.56 m/s : 15 s
a = 2.04 m/s²
Thomas knows that many machines transform electrical energy into other forms of energy
Answer:
Only the car transforms electrical energy into more than one form of energy.
Explanation:
The motion of the car is mechanical energy but it can also transform into another energy witch is electrical energy
A building inspector standing on the top floor of a building wishes to determine the depth of the elevator shaft. They drop a rock from rest and hear it hit bottom after 2.56 as. (a) How far (in m) is it from where they drop the rock to the bottom of the shaft
Answer:
d = 29.89 m
Explanation:
To solve this, we need to separate this problem in two parts.
One part would be the the time taken by the rock to actually hit the bottom, and the other part would be the time taken by the sound to reach the inspector.
Joining these two times we have:
t = t₁ + t₂ (1)
This time is 2.56 s.
Now, as we are asked to determine the distance from the top floor to the bottom, and we have two times taken in different ways, one by sound and the other the actual, we can say the same thing on distance, we need a distance relationed to the time taken by rock to hit the bottom, and the other distance relationet to the time taken by sound to reach the inspector.
Doing this we have that the distance traveled by the rock is:
y₁ = gt²/2
y₁ = 9.8t²/2 = 4.9t₁² (2)
Now, the distance traveled by sound would be:
y₂ = v * t₂ = 336t₂ (3)
Remember that the speed of the sound is 336 m/s
From this last expression (3), we can actually write t₂ in function of t₁, using (1):
2.56 = t₁ + t₂
t₂ = 2.56 - t₁ (4)
Replacing (4) in (3):
y₂ = 336(2.56 - t₁) (5)
Now that we have y₁ and y₂, we can equal (2) and (5), both expressions to get the value of t₁, and then, calculate the distance:
4.9t₁² = 336(2.56 - t₁)
4.9t₁² = 860.16 - 336t₁
4.9t₁² + 336t₁ - 860.16 = 0
Using the quadractic formula, we can calculate t₁:
t₁ = -336 ±√(336)² + 4*4.9*860.16 / (2*4.9)
t₁ = -336 ±√129,7555.136 / 9.8
t₁ = -336 ± 360.21 / 9.8 Using only the positive value we have:
t₁ = 2.47 s
This means that the rock hits the bottom in 2.47 s, and the remaining 0.09 s belongs to the time taken by sound. (2.47 + 0.09 = 2.56 s)
With this, we can calculate the distance of the rock using expression (2):
y₁ = 4.9 * (2.47)²
y₁ = 29.89 mHope this helps
How fast would an object have to travel on the surface of Jupiter at the equator to keep up with the Sun (that is, so the Sun would appear to remain in the same position in the sky)? Use the facts that the radius of Jupiter is approximately 44,360 miles and its revolution is approximately 10 hours.
Answer:
27872.2 miles per hour
Explanation:
Given that :
Radius of Jupiter is approximately = 44,360 miles
Revolution is 10 hours ;
Jupiter makes one revolution in 10 hours :
Using the relation to obtain the velocity :
V = re
r = radius
w = 2π/T
Hence,
V = r * 2π/ T
V =44360 * 2 * π/10
V = 88720 * π/10
V = 278722.10 / 10
V = 27872.210
V = 27872.2 miles per hour
The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would....
A Increase
B Decrease
C Stay the Same
D Vary with day and night
Answer:The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would..
Explanation: It would decrease.
The Moon has a smaller mass than the Earth. If you were to travel to the moon your weight would decrease because the acceleration due by gravity on the moon is less than the acceleration due to gravity on the earth, therefore the correct answer is option C.
What is gravity?It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another.
In comparison to the Earth, the Moon is less massive. Your weight would drop if you traveled to the moon because the acceleration caused by gravity there is lower than that caused by gravity here on Earth.
As a result of the less gravity on the moon, the weight would decrease.
Thus, the correct answer is option C.
To learn more about gravity from here, refer to the link;
brainly.com/question/4014727
#SPJ2
a. Use the graph and the element made in question 2 to determine the mass of the star.
A body moves on a coordinate line such that it has a position sequalsf(t)equalstsquaredminus4tplus3on the interval 0less than or equalstless than or equals7,with s in meters and t in seconds.a. Find the body's displacement and average velocity for the given time interval.b. Find the body's speed and acceleration at the endpoints of the interval.c. When, if ever, during the interval does the body change direction?
Answer:
A) Δf = - 49 m, B) v (7) = -56 m / s, a = - 8 m / s², C) t = 0.866 s
Explanation:
A) In this exercise ask to find the displacement and the average velocity, give the function of the movement
f (t) = - 4t² +3
and the range of motion 0≤ t ≤ 7
the displacement is
for t = 0
f (0) = 3
for t = 7 s
f (7) = - 4 7² +3
f (7) = -46 m
the total displacement is
Δf = f (7) - f (0)
Δf = -46 - 3
Δf = - 49 m
the average speed is defined as the displacement between the time interval
v = Df / Dt
v = -49 / 7
v = - 7 m / s
B) the speed and acceleration of the end points of the motion
the speed of defined by
v = [tex]\frac{dx}{dt}[/tex]
in this case
v = [tex]\frac{df}{dt}[/tex]
v = -8t
let's calculate
v (7) = -8 7
v (7) = -56 m / s
acceleration is defined by
a = [tex]\frac{dv}{dt}[/tex]
a = - 8 m / s²
acceleration is constant throughout the movement
C) the point where the direction changes.
This point is a point where the position goes from positive to negative, the point f = 0
0 = -4t² +3
t = √¾
t = 0.866 s
NEED HELP WITH THIS PLEASE
Answer:
A
Explanation: