Answer:
a) v = 3.116 m / s, b) μ = 1.65 10⁻²
Explanation:
a) to find the velocity of the set, let's define a system formed by the person and the sled, so that the forces during the collision are internal and the moment is conserved
initial instant. Before the crash
p₀ = M v₀
final instant. After the crash
p_f = (M + m) v
the moment is preserved
M v₀ = (M + m) v
v = [tex]\frac{M}{M+m} \ v_o[/tex]
let's calculate
v = [tex]\frac{41.6}{41.6 + 14.6} \ 4.21[/tex]
v = 3.116 m / s
b) for this part let's use the relationship between work and kinetic energy
W = ΔK
as the body has its final kinetic energy is zero
the work of the friction forces is
W = - fr x
the negative sign is because the friction forces always oppose the movement
let's write Newton's second law
Y axis
N - W_sled -W_person = 0
N = mg + M g
N = (m + M) g
X axis
fr = ma
the friction force has the expression
fr = μ N
fr = μ g (m + M)
we substitute
- μg (m + M) x = 0- ½ (m + M) v²
μ = [tex]\frac{1}{2} \ \frac{v^2 }{g \ x }[/tex]
let's calculate
μ = [tex]\frac{1}{2} \ \frac{3.116^2}{9.8 \ 30.0}[/tex]
μ = 0.0165
μ = 1.65 10⁻²
Which two chemical equations show double-replacement reactions?
A. C+02 - CO2
B. 2Li + CaCl2 - 2LiCl + Ca
I C. Ca(OH)2 + H2S04 - CaSO4 + 2H20
D. Na2CO3 + H2S - H2CO3 + Na2S
The two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.
What is double replacement reaction?A double replacement reaction have two ionic compounds that are exchanging anions or cations.
From the given options, we can choose the following based on their exchange of anions or cations.
Ca(OH)2 + H2S04 - CaSO4 + 2H20Na2CO3 + H2S - H2CO3 + Na2SThus, the two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.
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a Ferris wheel with a diameter of 35 m starts from rest and achieves its maximum operational tangential speed of 2.3 m/s in a time of 15 s. what is the magnitude of the wheels angular acceleration?
b. what is the magnitude of the tangential acceleration after the maximum operational speed is reached?
The following statements address the science behind the pulley system illustrated:
A. The pulleys increase the entropy of the system.
B. The force applied to the rope is less than the force needed to lift the object.
C. The pulleys help generate as much energy as possible.
D. The pulleys multiply energy input, resulting in more energy output.
E. The pulleys generate no thermal energy.
Which of these statements is/are true?
i. Statements A and B
ii. Statements D and E
iii. Only statement C
iv. All of the statements
Answer:
i. Statements A and B
Explanation:
Sana nakatulong
A 85-W lamp is connected to 100 V. What is the resistance of the lamp?
What is an example of a combined approach in psychology?
Answer:
The third approach to cross-cultural studies of personality is the combined approach, which serves as a bridge between Western and indigenous psychology as a way of understanding both universal and cultural variations in personality
Explanation:
Answer:
The third approach to cross-cultural studies of personality is the combined approach, which serves as a bridge between Western and indigenous psychology as a way of understanding both universal and cultural variations in personality
In which type of circuit does charge move in only one direction?
A. A D.C CIRCUIT
B. AN A.C CIRCUIT
C. A COMBINED CIRCUIT
D. A PARALLEL CIRCUIT
Hi please zoom in to see it clearly, uh you don’t have to answer them all but it would be nice !!! (no links please)
Answer:
3. (a) - is the answer most likely
4: (a)
5 (d or c)
6 (b)
a highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of:
A car making this turn is pulled downward by its own weight, and pushed up by the road at an angle of 45°, so by Newton's second law,
• the net horizontal force on the car is
∑ F = N cos(45°) = m a = m v ² / R
• the net vertical force on the car is
∑ F = N sin(45°) - m g = 0
where
• N = magnitude of the normal force
• m = mass of the car
• a = v ² / R = centripetal acceleration of the car
• v = tangential speed of the car
• R = 100 m = radius of curvature
• g = 9.8 m/s² = acceleration due to gravity
From the net vertical force equation, we get
N = m g / sin(45°)
and substituting this into the net horizontal force equation and solving for v gives
(m g / sin(45°)) cos(45°) = m v ² / R
v = √(R g cos(45°) / sin(45°)) ≈ 31 m/s
We have that A highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of
V=32m/s
From the question we are told
a highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of:
Generally the equation for the Velocity is mathematically given as
[tex]V=\sqrt{rgtan\theta}[/tex]
Therefore
[tex]V=\sqrt{rgtan\theta}\\\\V=\sqrt{100*9.8*tan45}\\\\V=32m/s[/tex]
Therefore
A highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of
V=32m/s
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I need help with this review question. I’ll give extra points.
Answer:
1.9m/s²
Explanation:
Use the equation v=u+at, where v is the final speed, u is the initial speed, a is the acceleration and t is the time.
v=u+at
15.3=0+a(8)
a=15.3/8
a= 1.9125 m/s²
A uniform sphere has a moment of inertia that is (2/5)MR2. A sphere of uniform density, with mass 29 kg and radius 0.5 m is located at the origin, and rotates around an axis parallel with the x axis. If you stand somewhere on the x axis and look toward the origin at the sphere, the sphere spins counterclockwise. One complete revolution takes 0.5 seconds. What is the rotational angular momentum of the sphere
Answer:
[tex]36.44\ \text{kg m/s}\hat{i}[/tex]
Explanation:
I = Moment of inertia of sphere = [tex]\dfrac{2}{5}MR^2[/tex]
M = Mass of sphere = 29 kg
R = Radius of sphere = 0.5 m
T = Time taken for one revolution = 0.5 s
[tex]\omega[/tex] = Angular velocity = [tex]\dfrac{2\pi}{T}[/tex]
[tex]L=I\omega\\\Rightarrow L=\dfrac{2}{5}MR^2\dfrac{2\pi}{T}\\\Rightarrow L=\dfrac{4MR^2\pi}{5T}\\\Rightarrow L=\dfrac{4\times 29\times 0.5^2\pi}{5\times 0.5}\\\Rightarrow L=36.44\ \text{kg m/s}[/tex]
The rotational angular momentum of the sphere is [tex]36.44\ \text{kg m/s}\hat{i}[/tex].
Please help me!
8. Give an example of a poor blackbody radiator and explain why it is not a good blackbody radiator.
9. Does a blackbody radiator emit light waves? Explain.
Answer:
A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").
A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).
a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.
b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.
Answer:
A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").
A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).
a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.
b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.
Explanation:
A box having a weight of 8 lb is moving around in a circle of radius rA = 2 ft with a speed of (vA)1 = 5 ft/s while connected to the end of a rope. If the rope is pulled inward with a constant speed of vr = 4 ft/s, determine the speed of the box at the instant rB = 1 ft. How much work is done after pulling in the rope from A to B? Neglect friction and the size of the box
Answer:
W = 1.875 J
Explanation:
For this exercise let's use the relationship between work and kinetic energy
W = ΔK
The kinetic energy of rotational motion is
K₀ = ½ I w²
we can assume that the box is small, so it can be treated as a point object, with moment of inertia
I = m rₐ²
angular and linear velocity are related
v = w r
w = v / r
we substitute in the equation, for point A
K₀ = ½ (m rₐ²) (v / rₐ)²
K₀ = ½ m v²
For the final point B, as the system is isolated the angular momentum is conserved
initial L₀ = Io wo
final L_f = I_f w_f
L₀ = L_f
I₀ w₀ = I_f w_f
(m rₐ²) w₀ = (m [tex]r_{b} ^2[/tex]) w_f
w_f = (rₐ/r_b)² w₀
with this value we find the final kinetic energy
K_f = ½ I_f w_f²
K_f = ½ (m [tex]r_{b}^2[/tex]) ( (rₐ / r_b)² w₀) ²
K_f = ½ m [tex]\frac{r_a^4}{r_b^2} \ w_o^2[/tex]
we substitute in the realcion of work
W = K_f - K₀
W = ½ m [tex]( \( \frac {r_a^2 }{r_b} )^2[/tex] w₀² - ½ m v²
W = ½ m [tex]\frac{r_a^4}{r_b^2} ( \frac{v}{r_a} ) ^2[/tex] - ½ m v²
W = ½ m [tex]\frac{r_a^2}{r_b^2} \ v^2[/tex] - ½ m v2
W = ½ m v² (([tex]( \ (\frac{r_a}{r_b})^2 -1)[/tex]
let's calculate
W = ½ ( [tex]\frac{8}{32}[/tex] ) 5 ((2/1)² -1)
W = 0.625 (3)
W = 1.875 J
Becoming informed about economics helps a person understand the reasons a command economy is ideal. role of government in regulating production. why consumers receive tax revenue. reasons an economy must always be completely regulated. Mark this and return
Answer:
Role of government in regulating production
Explanation:
The role of government in regulating show , provides the legal and social framework, uphold competition, provides public goods and services.
What is the role of economics in the community?The community's role in conserving and enhancing common-property resources is well known.
In extra, its role in helping market growth by its power to execute trade agreements among transacting parties belonging to the community network is stressed.
Thus, it provides the legal and social framework, maintains competition, and provides public goods and services.
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When a 20 kg explosive detonates and sends a 5 kilogram piece traveling to the right at 105 m/s
what is the speed and direction of the other 15 kilogram piece of the explosive!
Answer:
speed: 35m/s
direction: left
Explanation:
Assuming the right side is the positive direction:
before explosion:
P = mv = 0
after explosion:
P' = 15P + 5P
(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')
P' = 0.75mv1' + 0.25mv2'
P' = (15kg)v' + (5kg)(105m/s)
P' = 525kg/m/s + (15kg)v1'
P = P'
525kg/m/s + (15kg)v1' = 0
(15kg)v1' = -525kg/m/s
v1' = -35m/s
speed = |-35| = 35m/s
direction is to the left since the right side is the positive direction.
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Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit
Answer:
0.03 A
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12 V
Resistor (R) = 470 Ω
Current (I) =?
From ohm's law, the voltage, current and resistor are related by the following formula:
Voltage = current × resistor
V = IR
With the above formula, we can obtain the current in the circuit as follow:
Voltage (V) = 12 V
Resistor (R) = 470 Ω
Current (I) =?
V = IR
12 = I × 470
Divide both side by 470
I = 12 / 470
I = 0.03 A
Thus, the current in the circuit is 0.03 A
Answer:
0.03 A
Explanation:
Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit
0.03 A
Review please help.
Answer:
1 and 3
Explanation:
because they are going up from 0
A carnival ride starts at rest and is accelerated from an initial angle of zero to a final angle of 6.3 rad by a rad counterclockwise angular acceleration of 2.0 s2 What is the angular velocity at 6.3 rad?
The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Final angular velocity of the carnival ride
The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;
ωf = ωi + 2αθ
where;
ωf is the final angular velocity of the carnival ride = ?ωi is the initial angular velocity of the carnival ride = 0α is the angular acceleration = 2.0 rad/s²θ is the angular displacement of the carnival ride = 6.3 radωf = 0 + 2(2.0) x 6.3
ωf = 25.2 rad/s
Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
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Answer: 5.0 rad/s
Explanation: Because that’s what khan said so try it out.
The moon does not stay at the same distance from the earth.why?
Answer:
The moon does not stay at the same distance of the earth because the ortbit of the moon is slightly elliptical. If earth is not tilted at an angle of 66.5°, there will be no change in the season and the earth will have equal length of days and night.
Explanation:
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Batteries are not perfect. They can't deliver infinite current. As the current load on a battery gets larger, the voltage output gets smaller.
a. True
b. False
A woman shouts at a boy who is underwater what happens to the speed of the sound wave as it moves from the air into the water
Answer:
B. it increases
Explanation:
As shown in the table provided, the speed of sound in water (1493 m/s) is greater than the speed of sound in air (346 m/s).
Answer:
B is the correct answer.
Explanation:
If each Coulomb of charge is given 20 Joules of energy, what is the voltage of the battery?
A. 20 V
B. 5 V
C. 10 V
D. Not enough info
Answer:
Explanation:
V = J/C
V = 20/1
= 20 v
Option A is the correct answer
A van has a weight of 4000 lb and center of gravity at Gv. It carries a fixed 900 lb load which has a center of gravity at Gl. If the van is traveling at 40 ft/s, determine the distance it skids before stopping. The brakes cause all the wheels to lock or skid. The coefficient of kinetic friction between the wheels and the pavement is . Assume that the two rear wheels are one normal, NB, and the two front wheels are one normal, NA.
Answer:
x = 25 / μ [ ft]
Explanation:
To solve this exercise we can use Newton's second law.
Let's set a reference system where the x axis is parallel to the road
Y axis
N_B + N_A - W_van - W_load = 0
N_B + N_A = W_van + W_load
X axis
fr = ma
a = fr / m
the total mass is
m = (W_van + W_load) / g
the friction force has the expression
fr = μ N_{total}
fr = μy (W_van + W_load)
we substitute
a = μ (W_van + W_load) [tex]\frac{g}{W_van + W_load}[/tex]
a = μ g
taking the acceleration let's use the kinematic relations where the final velocity is zero
v² = v₀² - 2 a x
0 = v₀² -2a x
x = [tex]\frac{v_o^2}{2a}[/tex]
x = [tex]\frac{v_o^2}{2 \mu g}[/tex]
x = [tex]\frac{40^2}{2 \ 32 \ \mu}[/tex]
x = 25 / μ [ ft]
Example of the center of the gravity
Answer:
The example of the center of the gravity is the middle of a seesaw
Explanation:
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The Brazilian rain forest is an area with significant biodiversity. As the rain forest is replaced with agricultural land, it is reasonable to predict a reduction in -
Answers-
A: consumption of solar energy.
B: sustainability over time.
C: precipitation levels.
D: average daily temperature.
Answer:
Bb
Explanation:
g A thin-walled hollow cylinder and a solid cylinder, both have same mass 2.0 kg and radius 20 cm, start rolling down from rest at the top of an incline plane. The height of top of the incline plane is 1.2 m. Find translational speed of each cylinder upon reaching the bottom and determine which cylinder has the greatest translational speed upon reaching the bottom. Moment of inertia of hollow cylinder about its axis passing through the center is mr2 and for solid cylinder mr2/2
Answer:
a. i. 3.43 m/s ii. 2.8 m/s
b. The thin-walled cylinder
Explanation:
a. Find translational speed of each cylinder upon reaching the bottom
The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy
So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.
i. translational speed of thin-walled cylinder upon reaching the bottom
So, For the thin-walled cylinder, I = mr², we find its translational velocity, v
So, mgh = 1/2mv² + 1/2Iω²
mgh = 1/2mv² + 1/2(mr²)(v/r)²
mgh = 1/2mv² + 1/2mv²
mgh = mv²
v² = gh
v = √gh
v = √(9.8 m/s² × 1.2 m)
v = √(11.76 m²/s²)
v = 3.43 m/s
ii. translational speed of solid cylinder upon reaching the bottom
So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'
So, mgh = 1/2mv'² + 1/2Iω²
mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²
mgh = 1/2mv'² + mv'²
mgh = 3mv'²/2
v'² = 2gh/3
v' = √(2gh/3)
v' = √(2 × 9.8 m/s² × 1.2 m/3)
v' = √(23.52 m²/s²/3)
v' = √(7.84 m²/s²)
v' = 2.8 m/s
b. Determine which cylinder has the greatest translational speed upon reaching the bottom.
Since v = 3.43 m/s > v'= 2.8 m/s,
the thin-walled cylinder has the greatest translational speed upon reaching the bottom.
What kind of energy is in a moving skateboard
Answer:
I guess it is kinetic energy
Answer:
kinetic energy because my dog told me
A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1490 Hz. The bird-watcher, however, hears a frequency of 1505 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound
Answer:
The speed of the bird is 1.00% of the speed of sound.
Explanation:
The speed of the bird can be found by using the Doppler equation:
[tex] f = f_{0}(\frac{v - v_{r}}{v - v_{s}}) [/tex]
Where:
v: is the speed of sound = 343 m/s
f₀: is the frequency emitted = 1490 Hz
f: is the frequency observed = 1505 Hz
[tex]v_{r}[/tex]: is the speed of the receiver = 0 (it is stationary)
[tex]v_{s}[/tex]: is the speed of the source =?
The minus sign of [tex]v_{s}[/tex] is because the source is moving towards the receiver.
By solving the above equation for [tex]v_{s}[/tex] we have:
[tex] v_{s} = v - \frac{f_{0}*v}{f} = 343 - \frac{1490*343}{1505} = 3.42 m/s [/tex]
The above speed in terms of the speed of sound is:
[tex]\% v_{s} = \frac{3.42}{343}\times 100 = 1.00 \%[/tex]
Therefore, the speed of the bird is 1.00% of the speed of sound.
I hope it helps you!
Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10−2 cm , mounted coaxially within a cylindrical anode with a radius of 0.5580 cm . The potential difference between the anode and cathode is 400 V . An electron leaves the surface of the cathode with zero initial speed (vinitial=0). Find its speed vfinal when it strikes the anode.
Answer:
The final speed will be "[tex]1.185\times 10^7 \ m/sec[/tex]".
Explanation:
The given values are:
Potential difference,
Δv = 400 v
Radius,
r = 0.5580 cm
As we know,
⇒ [tex]W=e \Delta v[/tex]
and,
⇒ [tex]\frac{1}{2}mv^2=e \Delta v[/tex]
then,
⇒ [tex]v=\sqrt{\frac{2e \Delta v}{m} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 400}{9.11\times 10^{-31}} }[/tex]
⇒ [tex]=\sqrt{\frac{1.6\times 10^{-19}\times 800}{9.11\times 10^{-31}}}[/tex]
⇒ [tex]=1.185\times 10^7 \ m/sec[/tex]
A proton is moved so that its electric potential energy increases from 4.0 × 10-14 J to 9.0 × 10-14 J. The magnitude of the charge on a proton is 1.602 × 10-19 C.
What is the electric potential difference through which the proton moved?
2.5 × 105 V
3.1 × 105 V
5.6 × 105 V
8.1 × 105 V
Answer:
B. 3.1 × 10^5 V
Explanation:
Answer:
B
Explanation:
e2021
