Given,
The radius of the wheel, r=0.300 m
The mass of the wheel, m=1.47 kg
The initial angular frequency of the wheel, f=4.00 rev/s
The time it takes for the wheel to come to a stop, t=56.4 s
The final angular velocity of the wheel, ω=0 rad/s
The initial angular velocity of the wheel is
[tex]\begin{gathered} \omega_0=2\pi f \\ =2\pi\times4.00 \\ =25.13\text{ rad/s} \end{gathered}[/tex]From the equation of motion,
[tex]\omega=\omega_0+\alpha t[/tex]Where α is the angular acceleration of the wheel.
On substituting the known values,
[tex]\begin{gathered} 0=25.13+\alpha\times56.4 \\ \Rightarrow\alpha=\frac{-25.13}{56.4} \\ =-0.45\text{ rad/s}^2 \end{gathered}[/tex]The moment of inertia of the wheel whose mass is concentrated on the rim is given by,
[tex]I=mr^2^{}[/tex]The torque applied by the friction is given by,
[tex]\begin{gathered} \tau=I\alpha \\ =mr^2\alpha \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} \tau=1.47\times0.300^2\times-0.45 \\ =-0.06\text{ Nm} \end{gathered}[/tex]The negative sign of the torque indicates that the torque is applied to the wheel in the direction opposite to the direction of rotation of the wheel.
Thus the magnitude average torque due to frictional force is 0.06 Nm
Number 1. Part b: what are the final kinetic energy of the system
Given that there is a cart of mass, m = 0.12 kg moving with initial speed of, u1 = 0.45 m/s and it collides with another cart of mass, m = 0.12 kg with initial speed, u2 = 0 m/s
We have to find the initial and final kinetic energy.
(a) Initial kinetic energy,
[tex]\begin{gathered} K\mathrm{}E.1=\frac{1}{2}mv^2 \\ =\frac{1}{2}\times0.12\times(0.45)^2 \\ =0.012\text{ J} \end{gathered}[/tex]According to the conservation of linear momentum,
[tex]mu1+mu2=2mv[/tex]Here, v is the final speed.
[tex]\begin{gathered} 0.12\times0.45=2\times0.12\times v \\ v=\frac{0.45}{2} \\ =0.225\text{ m/s} \end{gathered}[/tex]Here, the final speed is 0.225 m/s.
(b) The formula to find kinetic energy is
[tex]K\mathrm{}E\mathrm{}=\frac{1}{2}(2m)v^2[/tex]Substituting the values, we get
[tex]\begin{gathered} K\mathrm{}E\mathrm{}=0.12\times(0.225)^2 \\ =6.075\times10^{-3}\text{ J} \end{gathered}[/tex]Hence the kinetic energy is 6.075 x 10^(-3) J.
A 0.0012 C amount of charge is produced on a Van der Graaf with 1,093 J of energy. A spark of this charge is produced in 0.005 s. What is the resistance of the air gap of the spark? (Voltage can be calculated by using energy divided by charge.)
Given,
The charge generated, q=0.0012 C
The energy of the Van der Graaf generator, E=1093 J
The time it takes for the generator to produce the given amount of charge, t=.005 s
The current is given by the time rate of the flow of charges. That is,
[tex]I=\frac{q}{t}\text{ }\rightarrow\text{ (i)}[/tex]The voltage is calculated using the formula,
[tex]V=\frac{E}{q}\text{ }\rightarrow\text{ (ii)}[/tex]From Ohm's law, the voltage is given by,
[tex]\begin{gathered} V=IR \\ R=\frac{V}{I}\text{ }\rightarrow\text{ (iii)} \end{gathered}[/tex]On substituting equation (i) and (ii) in equation (iii),
[tex]\begin{gathered} R=\frac{\frac{E}{q}}{\frac{q}{t}} \\ =\frac{Et}{q^2} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} R=\frac{1093\times0.005}{0.0012^2} \\ =3.8\times10^6\text{ }\Omega \end{gathered}[/tex]Thus the resistance of the air gap is 3.8×10⁶ Ω
It takes 225 kJ of work to accelerate a car from 20.1 m/s to 28.1 m/s. What is the car's mass?
Answer:
THE REMAINIG WILL BE 75
Explanation:
HOPE IT HELPS YOU
A tensile load of 190 kN is applied to a round metal bar with a diameter of 16mm and a gage length Of 50mm. Under this load the bar elastically deforms so that the gage length increases to 50.1349 mm and the diameter decreases to 15.99 mm. Determine the modulus of elasticity and Poisson s ratio for this metal.
The elasticity and Poisson's ratio for this metal is 0.232.
What is ratio?
The realation between two numbers which shows how much bigger one quantity is than another.
Sol-
As per the given question
P=190KN
d=16 mm
Lo=50mm
X=50.1349-50=0.1349mm
Y=15.99-16=-0.01mm
The formula-
E=ó/€
Ó=P/A
A=r/4 d^2 =π/4(16)^2=201.062 mm
ó={190(1000)}201.062=944.982 Mpa
E=944.982/0.002698=350.253 GPa
€y=-0.000625
v=0.232(answer)
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Steve takes his car out for a joy ride and travels 400 meters north. He then travels 100 meters east and picks up his buddy Frank. They then stop at a 7-11 which is 200 meters south from Frank’s house. If the total trip takes 10 minutes, determine the average velocity of Steve’s car. (Draw a picture!) NEED HELPP ASAPPPP !!!!
The average velocity of Steve's car is 1.17m/s.
How to calculate average velocity?Average velocity is the change in position or displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs.
It can be calculated by dividing the total distance of a moving body by the time taken.
According to this question, Steve takes his car out for a joy ride and travels 400 meters north. He then travels 100 meters east and picks up his buddy Frank. They then stop at a 7-11 which is 200 meters south from Frank’s house.
The total distance traveled by Steve is 400m + 100m + 200m = 700m.
Average velocity = 700m ÷ 600s
Average velocity = 1.17m/s
Therefore, 1.17m/s is the average velocity of the car.
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Three blocks are sitting on a horizontal, frictionless table. They
are pushed from the left by an applied force F = 10 N, as
shown. How much force does block 3 exert on block 2, if
m₂ = 1 kg, m₂ = 2 kg, m₂ = 3 kg ?
a) 3 N
b) 5 N
c) 6 N
d) 8 N
Newton's laws of motion, which are composed of three basic principles, describe the interaction between an object's motion and the forces acting on it.
These laws can be summarized as follows: A body remains at rest or in motion in a straight line at a constant speed unless moved upon by a force.
The first law states that an object's motion cannot be changed until a force acts on it.
The second law states that an object's force is calculated by dividing its mass by its acceleration.
The third law states that when two objects come into contact, they apply pressures that are equal in size and direction to one another.
Let a represent the system's acceleration.
T1 = M1a .....(1)
T2 − T1 = M2a ....(2)
F2 − T2 = M3a ......(3)
When we combine (1), (2), and (3), we obtain
(M1 + M2 + M3)a = F
or (1 + 2 + 3) a= 6 a = 1 m/s²
Now , T2 = (M1 + M2)a = (1 + 2)(1) = 3N
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How to do calculate this the error? What are the rules?(Mark scheme Answer is B)
Here,
potential difference(V)= (2.00±0.02) V;
current (I)=( 5.3 ± 0.1) mA;
Resistance (X) will be given by
[tex]\begin{gathered} X=\frac{V}{I}; \\ X=\text{ }\frac{2.00±0.02}{5.3±0.1}\text{ }\times1000\begin{cases}V={(2.00±0.02)} \\ I=({5.3±0.1\text{ \rparen mA= \lparen5.3}\pm0.1)\times10^{-3\text{ }}}A\end{cases} \\ \\ \therefore X=(\text{ 377.358 }\pm0.001) \\ \text{ } \\ \end{gathered}[/tex]A football player kicks a football off a tee with a speed of 16 m/s at an angle of 63°. How far is the ball from the football player when it lands? How much farther would the ball go if he kicked it with the same speed, but at a 45° angle? Which ball will land first: the ball kicked at 16 m/s and at a 63° angle, or one kicked at 9 m/s and at a 45° angle?
A football player kicks a football off a tee with a speed of 16 m/s at an angle of 63°
The horizontal and vertical speed of the ball is given by
[tex]\begin{gathered} v_x=v\cos (\theta) \\ v_y=v\sin (\theta) \end{gathered}[/tex]Where v = 16 m/s and θ = 63°
[tex]\begin{gathered} v_x=16\cos (63\degree)=7.26\; \frac{m}{s} \\ v_y=16\sin (63\degree)=14.26\; \frac{m}{s} \end{gathered}[/tex]How far is the ball from the football player when it lands?
The range of the ball is given by
[tex]x=v_x\times t[/tex]Where t is the time the ball remains in the air.
The time (t) can be found as
[tex]y=v_yt+\frac{1}{2}at^2[/tex]y = 0 when the ball is in the air.
The acceleration is due to gravity (-9.8 m/s²)
[tex]\begin{gathered} 0=14.26t+\frac{1}{2}(-9.8)t^2 \\ 0=14.26t-4.9t^2 \\ 0=t(14.26-4.9t) \\ 0=14.26-4.9t \\ 4.9t=14.26 \\ t=\frac{14.26}{4.9} \\ t=2.91\; s \end{gathered}[/tex]Finally, the range is
[tex]x=v_x\times t=7.26\times2.91=21.13\; m[/tex]Therefore, the ball will land 21.13 meters far from the football player.
How much farther would the ball go if he kicked it with the same speed, but at a 45° angle?
We need to repeat the above calculations
The horizontal and vertical speed of the ball is given by
[tex]\begin{gathered} v_x=v\cos (\theta)=16\cos (45\degree)=11.31\; \frac{m}{s} \\ v_y=v\sin (\theta)=16\sin (45\degree)=11.31\; \frac{m}{s} \end{gathered}[/tex]The time (t) is given by
[tex]\begin{gathered} y=v_yt+\frac{1}{2}at^2 \\ 0=11.31_{}t+\frac{1}{2}(-9.8)t^2 \\ 0=11.31_{}t-4.9t^2 \\ 0=11.31_{}-4.9t \\ 4.9t=11.31_{} \\ t=\frac{11.31_{}}{4.9} \\ t=2.31\; s \end{gathered}[/tex]Finally, the range is
[tex]x=v_x\times t=11.31\times2.31=26.13\; m[/tex]Therefore, the ball will land 26.13 meters far from the football player.
Which ball will land first: the ball kicked at 16 m/s and at a 63° angle, or one kicked at 9 m/s and at a 45° angle?
The ball kicked at 16 m/s and at a 63° angle takes 2.91 s to land.
The ball kicked at 9 m/s and at a 45° angle will take
[tex]v_y=9\sin (45\degree)=6.36\; \frac{m}{s}[/tex][tex]\begin{gathered} y=v_yt+\frac{1}{2}at^2 \\ 0=6.36t+\frac{1}{2}(-9.8)t^2 \\ 0=6.36t-4.9t^2 \\ 0=6.36-4.9t \\ 4.9t=6.36 \\ t=\frac{6.36}{4.9} \\ t=1.30\; s \end{gathered}[/tex]So, the ball kicked at 9 m/s and at a 45° angle takes 1.30 s to land.
Therefore, the ball kicked at 9 m/s and at a 45° angle will land first.
You throw an object up with an initial velocity of Voy = 11 m/s from a height of y = 25 m. How long, in seconds, does it take for the object to reach the ground? What is the objects final velocity, in meters per second, as it impacts the ground? Find the time, in seconds, if you instead threw the object DOWN with the same velocity Voy
Calculate the ball's greatest height using the vertical motion model, h = -16t2 + vt + s, where v is the beginning velocity in feet/second and s is the height in feet.
What does a ball being thrown upwards accelerate to?A ball is thrown into the air, where it gradually loses speed until it abruptly comes to a rest at the peak of the motion. The body is travelling upward against gravity at the top, hence the acceleration there is 9.8 ms2. For example, g=9.8 ms2 is the formula for the acceleration caused by gravity.
Only at the greatest point of a body being hurled vertically upwards would velocity be zero because of the constant downward acceleration brought on by the gravitational force. As a result, velocity is zero throughout the rest of the motion.
A ball is originally moving upward when it is tossed into the air, for instance.
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5. Was you hypothesis supported by the data collected for question 2? Why or why not ?
Answer:
Your hypothesis was not supported by the data collected
Explanation:
On question 2 you said that the point with the greatest kinetic energy would be point 5. However, when you analyze the data, the point with the greatest velocity was point 2 which means that this is the point with the greatest kinetic energy.
Therefore, your hypothesis was not supported by the data collected because based on the picture, it is hard to say where is the lowest point of the roller coaster.
Determine the net force (in N) necessary to give an object with a mass of 2.10 kg an acceleration of 5.20 m/s².N
Newton's second law states that the force is equal to the rate of change of momentum; for a constant mass, force equals mass times acceleration. In mathematical terms this means that:
[tex]F=ma[/tex]where m is the mass and a is the acceleration.
In this case the mass is 2.10 kg and the acceleration is 5.20 m/s²; plugging these values in Newton's second law we have that:
[tex]\begin{gathered} F=(2.10)(5.20) \\ F=10.92 \end{gathered}[/tex]Therefore, the force needed is 10.92 N
Given two 2.00μC charges on the horizontal axis are positioned at x=0.8m, and the
other at x=-0.8m, and a test charge q = 1.28x10-18C at the origin.
(a) What is the net force exerted on q by the two 2.00μC charges? [5]
(b) What is the electric field at the origin due to 2.00μC charges? [5]
(c) What is the electric potential at the origin due to the two 2.00μC charges
Answer:Question 1
Given q1=2µC
q2=2µC
q= 1.2×10^-18C at origin
Net force exerted by two charges on q
F_1= force on q due to q1
F_2= force on q due to q2
F_net= F_(1-) F_2
= (Kqq_1)/r^2 - (kqq_2)/r^2 Then q_1=q_2=〖2×10〗^(-6)
F_net=0N
b) The electric field at charge q
E_net= E_1- E_2
= (kq_1)/r^2 - (kq_2)/r^2
Then q_1=q_2
E)_net= 0 N/C
c) The electric potential at origin due to two charge
V_net= V_(1 )- V_2
= (kq_1)/r - (kq_2)/r
Then q1= q2
V_net= 0 V
Explanation:
Suppose you walk 16 m straight east and then 22 m straight south. At what angle, in degrees South of East, is a line connecting your starting point to your final position?
53.9 angle, in degrees South of East and 27.20 m is a line connecting your starting point to your final position.
What is initial position and final position?The distance in decimeters between the starting point and the ending position is in a straight line. The distance between an object's original position and its ultimate position is known as displacement.
Briefing:You walk 16 m straight east and then 22 m straight south. This forms a right angled triangle with a horizontal distance of 16 m, a vertical distance of 22 m and the hypotenuse is the distance between the ending and starting point. Let x represent the distance between the ending and starting point. Using Pythagoras theorem:
x² = 16² + 22²
x² = 256 + 484
x² = 740
Taking square root of both sides:
√x² = √740
x = √740
x = 27.20 m = distance between the ending and starting point.
Now use trigonometry:
sinθ=B/R
sinθ=22/27.20
sinθ= 0.808
θ = 53.9 degree. This is your angle.
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A net force of 43.1 N causes a mass to accelerate at a rate of 0.2 m/s2. Determine the mass in kilograms.
The mass of the an object caused to accelerate at 0.2m/s² by a 43.1 Newton force is 215.5 kilograms.
What is the mass of the object?A force is simply referred to as either a push or pull of an object resulting from the object's interaction with another object.
Force, according to Newton's Second Law is expressed as;
F = m × a
Where a is acceleration and m is the mass.
Given the data in the question;
Force applied F = 43.1N = 43.1kgm/s²Acceleration a = 0.2m/s²Mass m = ?Plug the given values into the formula above and solve for m.
F = m × a
43.1kgm/s² = m × 0.2m/s²
m = 43.1kgm/s² / 0.2m/s²
Mass m = 215.5kg
Therefore, the mass of the object is 215.5 kilograms.
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A resistor uses energy at a rate of 2.50W when there is a current of 4.00A passing through it. What must be the potential difference across the resistor?1.25V1.50V0.625V10.0V
As we know
[tex]P=\text{ V}\times I;[/tex]Where,
P= electric power= 2.50 W
V= potential difference= ?
I= current = 4.00A
Using above formula we get,
[tex]\begin{gathered} P=V\times I; \\ \therefore2.50=\text{ V}\times4; \\ V=\text{ }\frac{2.50}{4}=\text{ 0.625V} \end{gathered}[/tex]Final answer is :- 0.625 V
n Fig. P9.82, the cylinder Figure P9.82 and pulley turn without friction about stationary horizontal axles that pass through their centers. A light rope is Pulley wrapped around the cylinder, passes over the pulley, and has a 3.00 kg box Cylinder Box suspended from its free end. There is no slipping between the rope and the pulley surface. The uniform cylinder has mass 5.00 kg and radius 40.0 cm. The pulley is a uniform disk with mass 2.00 kg and radius 20.0 cm. The box is released from rest and descends as the rope unwraps from the cylinder. Find the speed of the box when it has fallen 2.50 m.
The speed of the box when it has fallen 2.50 m is 4.22 m/s.
What is the speed of the box?
The speed of the box when it has fallen through the given height is calculated as follows;
Apply the principle of conservation of energy to determine the speed of the box.
ΔK.E = ΔP.E
K.Ef - K.Ei = mg(hf - hi)
K.Ef - 0 = mg(hf - 0)
K.Ef = mghf
where;
K.E is the final kinetic energy = rotational + translational kinetic energyhf is the final height of the box¹/₂mv² + ¹/₂I_pω² + ¹/₂I_cω²= mghf
¹/₂mv² + ¹/₂(I_p + I_c)ω² = mghf
where;
I_p is moment of inertia of the pulleyI_c is the moment of inertia of the cylinderω is the angular speed of the boxm is mass of the boxI_p = ¹/₂MR²
where;
M is mass of the pulleyR is the radius of the pulleyI_p = ¹/₂(2)(0.2)² = 0.04 kgm²
I_c = MR²
I_c = (5)(0.4)²
I_c = 0.8 kgm²
¹/₂mv² + ¹/₂(I_p + I_c)(v/r)² = mghf
¹/₂mv² + ¹/₂r²(I_p + I_c)v² = mghf
¹/₂v²[m + 1/r²(I_p + I_c)] = mghf
v²[m + 1/r²(I_p + I_c)] = 2mghf
v² [3 + 1/0.4²(0.04 + 0.8)] = 2(3)(9.8)(2.5)
v² [3 + 1/0.4²(0.04 + 0.8)] = 2(3)(9.8)(2.5)
8.25v² = 147
v² = 147/8.25
v² = 17.8
v = √17.8
v = 4.22 m/s
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Information givenknown: mass of Christine=60 kgmass of cart= 22 kg mass of hailey=69The two girls on the cart to the left pushing off of Conner take .3833 s to reach a distance of 0.3m. Conner reaches the same distance in 0.2333s. What is the mass of Conner?
We will have the following:
We use conservation of momentum to solve, that is:
[tex]\begin{gathered} (60kg+22kg+69kg)(0.3m/0.3833s)=(m+22kg)(0.3m/0.233s) \\ \\ \Rightarrow118.1841899kg\ast m/s=(m+22kg)(\frac{300}{233}m/s) \\ \\ \Rightarrow m+22kg=91.78968976kg\Rightarrow m=69.78968976... \\ \\ \Rightarrow m\approx69.8 \end{gathered}[/tex]So, Conner's mass is approximately 69.8 kg.
An ant can crawl 72 inches every 6 minutes. The table below shows the distance the ant can travel for different amounts of time at this rate.According to the table, if the ant moves at a constant speed, what is its speed? (Remember that speed is a unit rate).
Given,
The ant crawls 72 inches every 6 minutes.
And the table of the time and distance covered by the ant.
From the table, the values of distance are,
d₁=6 inches
d₂=12 inches
d₃=18 inches
And the values of corresponding time are,
t₁=0.5 min
t₂=1 min
t₃=1.5 min
The speed of an object is given by the ratio of the distance covered by the object to the time it takes the object to cover the distance.
Thus the speed of the ant is given by,
[tex]v=\frac{d_1}{t_1}=\frac{d_2}{t_2}=\frac{d_3}{t_3}[/tex]On substituting the known values,
[tex]\begin{gathered} v=\frac{6}{0.5}=\frac{12}{1}=\frac{18}{1.5} \\ =12\text{ imches/min} \end{gathered}[/tex]Thus the speed of the ant is 12 inches/min
The tortoise and the hare A tortoise and a hare run to the East. The hare knows that it is faster, so it gives the tortoise a 30-meter head start. The tortoise is moving east at 1 m/s and the hare is moving east at 4 m/s.
The time taken for the hare to catch up with the tortoise is 10 seconds.
What is the time taken for the hare to catch the tortoise?
The time taken for the hare to catch up with the tortoise is calculated by applying the principle of relative velocity as shown below.
Distance travelled by hare + Distance travelled by tortoise = Total distance
V₁t + V₂t = d
where;
V₁ is the velocity of hareV₂ is the velocity of tortoiset is the time taken for them to meetd is the distance between themSince they are moving in the same direction, the relative velocity becomes
V₁t - V₂t = d
(V₁ - V₂)t = d
(4 - 1)t = 30
3t = 30
t = 30/3
t = 10 seconds
Thus, the time taken for the hare to catch up with the tortoise is 10 seconds.
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The complete question is below:
A tortoise and a hare run to the East. The hare knows that it is faster, so it gives the tortoise a 30-meter head start. The tortoise is moving east at 1 m/s and the hare is moving east at 4 m/s. At what time does the hare catch up with the tortoise?
A 12 N force acts at a 25 degrees and an 8 N force acts at 65 degrees. Determine the magnitude and direction (include angle) of the resultant . Scale is 1cm = 1N
NEEDDD HELLPP ASAPPPPP
A 12 N force acts at a 25-degree angle, while an 8 N force acts at a 65-degree angle. The magnitude and direction (including angle) of the resultant force are 21.43.
Consider the formula for a force, F=M×Cos∝
As M is a Mass of an object and, ∝ Is the angle at which force is acting on an object.
[tex]F_{1}[/tex]=M × Cos ∝
M=12N, and Cos∝= Cos25 = 0.906
[tex]F_{1}[/tex] = 12× 0.906
∴ [tex]F_{1}[/tex]= 10.87N
For [tex]F_{2}[/tex] = 25×Cos65
= 10.56N
According to Superposition resulting force is the Sum of total forces.
[tex]F_{RESULTANT} = F_{1} + F_{2}[/tex]
= 10.87+ 10.56
=21.43N
∴The resultant force is 21.43.N
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In a physics lab a student discovers that the magnitude of the magnetic field in a specific location near a long wire is 21.919 microTesla. If the wire carries a current of 35.483 A, what is the distance from the wire to that location ?
We will have the following:
First, we have that the permeability of free space is:
[tex]\mu_0=4\pi\ast10^{-7}Tm/A[/tex]Then:
[tex]\begin{gathered} B=\frac{\mu_0I}{2r\pi}\Rightarrow r=\frac{\mu_0I}{2B\pi} \\ \\ \Rightarrow r=\frac{(4\pi\ast10^{-7}Tm/A)(35.483A)}{2\pi(2.1919\ast10^{-5}T)}\Rightarrow r=0.3237647703...m \\ \\ \Rightarrow r\approx0.32m \end{gathered}[/tex]So, the distance is approximately 0.32 m.
Calculate the depth in the ocean at which the pressure is three times the atmospheric pressure
ANSWER:
20.17 meters
STEP-BY-STEP EXPLANATION:
Given:
Pressure = Po = 1.013x10^5 Pa
Pressure at depth = P = 3Po
Density of sea water = 1025 kg/m^3
We can calculate the depth as follows:
[tex]\begin{gathered} P=P_o+d\cdot g\cdot h \\ \text{ we solve for h} \\ dgh=P-P_o \\ h=\frac{P-P_o}{d\cdot g} \end{gathered}[/tex]We replacing and calculate the depth:
[tex]\begin{gathered} h=\frac{3\cdot P_o-P_o}{1025\cdot9.8}=\frac{2P_o}{10045}=\frac{2\cdot1.013\cdot10^5}{10045} \\ h=20.17\text{ m} \end{gathered}[/tex]Therefore, the depth is equal to 20.17 meters
What is the “time constant” for a capacitor, and why is it important?
We know that in a RC circuit the voltage in the capacitor when is charging is given by:
[tex]V_C(t)=V_0(1-e^{-\frac{t}{RC}})[/tex]when this is happening the voltage in the resistor is given by:
[tex]V_R(t)=V_0e^{-\frac{t}{RC}}[/tex]In both equations V0 denotes the voltage given by the source, R is the resistance of the resistor and C is the capacitance of the capacitor.
We notice that in both expressions the product RC appear, this product is what we call the time constant of the capacitor; and it is important since it determines the time intervals in which the voltage, charges and currents chage in a RC circuit. This means that while the capacitor is charging or discharging the variables mentioned will always have the time constant in their expressions.
this is a 2 part question28) A 1100-kg car coasts on a horizontal road with a speed of 19 m/s. After crossing anunpaved, sandy stretch of road 32 m long, its speed decreases to 12 m/s. (a) Was the net work done on the car positive, negative, or zero? Explain. (b) Find the magnitude of the average net force on the car in the sandy section.
ANSWER:
(a) negative
(b) -120.3 N
STEP-BY-STEP EXPLANATION:
(a)
The net work done on the car is negative because its speed decreases from 19 m/s to 12 m/s
(b)
The acceleration is found from the equationof motion that:
[tex]v^2-u^2=2\cdot a\cdot s[/tex]Replacing and solving for a:
[tex]\begin{gathered} 12-19=2\cdot a\cdot32 \\ -7=64a \\ a=-\frac{7}{64}\frac{m}{s^2} \end{gathered}[/tex]Therefore, the force would be:
[tex]\begin{gathered} F=m\cdot a \\ F=1100\cdot-\frac{7}{64} \\ F=-120.3\text{ N} \end{gathered}[/tex]what is constant angular speed
Answer:
in optical storage, constant angular velocity (CAV) is a qualifier for the rated speed of any disc containing information, and may also be applied to the writing speed of recordable discs. A drive or disc operating in CAV mode maintains a constant angular velocity, contrasted with a constant linear velocity (CLV)
describe how moral relativism was influenced by einstein theories of relativity and subsequently the trend toward the idea there are no absolutes?
The special and general theories of relativity and Albert Einstein's audacious theory that light is a particle are his most famous works as a physicist and Nobel winner. The most well-known scientist of the 20th century is perhaps him.
In March 1879, he was born in Ulm, Württemberg. He had a great interest in nature and the capacity to comprehend challenging mathematical ideas even as a young man in Munich. He had an unremarkable high school experience, doing exceptionally well in arithmetic but completely failing the classics, which were then thought to be crucial for anybody planning to attend college. He detested school's dreary regimentation and uncreative atmosphere.
The second study established a lot of information regarding the nature of molecules and explained Brownian motion, which is the random jostling of molecules floating in a fluid. 16 years later, this study helped him win the physics Nobel Prize.
However, his third work, "On the Electrodynamics of Moving Bodies," was left out of the award's wording. The third article was the one that would have the biggest impact on contemporary physics. It included Einstein's Special Theory of Relativity, which greatly simplified how we think about how radiation, like light, interacts with matter. Speaking about one body moving and another being motionless has no real significance, according to Einstein. Only in connection to one other can bodies be conceived of as moving;
This specifically implies that, regardless of the frame of reference, electromagnetic radiation's (such as light's) speed remains constant. Even well-known scientists struggled to comprehend this theory because of Einstein's insightful and audacious viewpoint. But over time, when the predictions made by his theory were repeatedly verified, the Special Theory of Relativity finally transformed how scientists thought about matter, space, time, and everything that interacts with them.
To know more about Albert Einstein refer to https://brainly.com/question/12962864
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mick took his friend out to dinner the bill was $40 but he applied a coupon then the total price was $33 what was the % off?
ANSWER
17.5 %
EXPLANATION
We have to find the percent change, given that the initial price was $40 and the final price was $33.
[tex]\text{ \% }change=\frac{final.price-initial.price}{initial.price}\times100[/tex][tex]\text{\% }change=\frac{33-40}{40}\times100=\frac{-7}{40}\times100=-0.175\times100=-17.5\text{\%}[/tex]The coupon was for 17.5% off
A superelastic collision is one in which1) kinetic energy before the collision equals kinetic energy after the collision.2) kinetic energy after the collision is zero.3) kinetic energy before the collision is less than kinetic energy after the collision.4) kinetic energy before the collision is greater than kinetic energy after the collision
A superelastic collision is the the collision in which the potential energy of the system is converted into the kinetic energy such the kinetic energy of the system after the collision is more than the kinetic energy of the system before the collision.
Hence, 3rd option is the correct answer.
Diffraction is:A.the difference in density of the compression and rarefaction parts of a sound wave.B.the change of frequency heard by an observer when sound waves come from a moving source.C.when waves suddenly appear in a medium without a source.D.the apparent bending of sound waves around obstacles.
Diffraction is a phenomenon that occurs when a wave hits an object or it passes through a small gap.
The propagation of the wave will have a circular pattern after the diffraction effect:
Therefore the correct option is D.
I just started a new lesson and have this study guide, but the material is unfamiliar to me and
2.79 eV
Explanation:Given
The wavelength, λ = 445 nm
The planck's constant, h = 6.626 x 10^-34 m² kg/s
The speed of light, c = 3 x 10^8 m/s
The energy of the photon is calculated as:
[tex]\begin{gathered} E=\frac{hc}{\lambda} \\ \\ E=\frac{6.626\times10^{-34}\times3\times10^8}{445\times10^{-9}} \\ \\ E=\frac{6.626\times3}{445}\times10^{-34+8+9} \\ \\ E=0.0447\times10^{-17} \\ \\ E=4.47\times10^{-2}\times10^{-17} \\ \\ E=4.47\times10^{-19}J \end{gathered}[/tex]But, 1 eV = 1.6 x 10^-19J
[tex]\begin{gathered} E=\frac{4.47\times10^{-19}}{1.6\times10^{-19}} \\ \\ E=2.79eV \end{gathered}[/tex]