Answer:
y = 4.666 10⁻² m
Explanation:
The constructive interference experiment for the double slit
d sin sin θ = m λ
Let's use trigonometry to find a sine relationship.
Tan θ = y / L
tan θ = sin θ/ cos θ
in these experiments the angles are very small
tan θ = sin θ
sin θ = y / L
[tex]d \frac{y}{L}[/tex] = m λ
y = [tex]\frac{ m \lambda \ L}{d}[/tex]
we replace the values
y = 3 560 10⁻⁹ 3.0 / 0.108 10⁻³
y = 4.666 10⁻² m
Which types of reactions are also redox reactions?
Answer:
Your ans is B)
Explanation:
Hope its right
if wrong firgive me
A projectile is thrown with a velocity of 40 m/s, 45 degrees above the +x-axis. Determine its initial x-velocity and its initial y-velocity.
x-component is V·cos(angle)
y-component is V·sin(angle)
When the angle is 45°, its sin and cos are both 1/2·√2 .
Vx = 20·√2 = 28.28 m/s
Vy = 20·√2 = 28.28 m/s
An elevator motor in a high-rise building can do 3500 kJ of work in 5 min. Find the power developed by the motor. Explain if you can plz
Answer:
P = 11666.6 W
Explanation:
Given that,
Work done by the motor, W = 3500 kJ
Time, t = 5 min = 300 s
We need to find the power developed by the motor. Power developed is given by :
[tex]P=\dfrac{E}{t}\\\\P=\dfrac{3500\times 10^3}{300}\\\\P=11666.7\ W[/tex]
So, the required power is 11666.6 W.
A uniform 140 g rod with length 57 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 30 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 11 cm on each sides of the center, at which time the system rotates at an angular speed of 23 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.
Answer:
The correct answer is "12 rad/s"
Explanation:
The given values are,
Mass of rod,
M = 140 g
i.e.,
= 0.14 kg
Length,
L = 57 cm
i.e.,
= 0.57 m
Mass of beads,
M = 30 g
i.e.,
= 0.03 kg
Angular speed,
r = 11 cm
i.e.,
= 0.11 m
Now,
The inertia of rods will be:
= [tex]\frac{1}{12}ML ^2[/tex]
On substituting the values, we get
= [tex]\frac{1}{12}\times 0.14\times (0.57)^2[/tex]
= [tex]0.0037905 \ kg-m^2[/tex]
The inertia of beads will be:
= [tex]mr^2[/tex]
On substituting the values, we get
= [tex]0.03\times (0.11)^2[/tex]
= [tex]0.000726 \ kg-m^2[/tex]
The total inertia will be:
= [tex]Inertia \ of \ rods+Inertia \ of \ beads[/tex]
= [tex]0.0037905 + 0.000726[/tex]
= [tex]0.0045165 \ kg-m^2[/tex]
now,
The angular speed of the system will be:
⇒ [tex]L_1w_1=L_2w_2[/tex]
On substituting the values in the above equation, we get
⇒ [tex]0.0045165\times 23 = (0.0037905 + (0.03\times 0.285^2)\times 2 )\times w_2[/tex]
⇒ [tex]0.1038795 = 0.0037905 + (0.00243675\times 2 )\times w_2[/tex]
⇒ [tex]w_2 = 12 \ rad/s[/tex]
how much force is needed to accelerate a 2500kg car at a rate of 4m/s2
Answer:
10,000 N
Explanation:
Using Newton's Second Law of Motion, Force = Mass*Acceleration.
Plugin 2500 kg for the mass and 4 m/s^2 for the acceleration.
Force = 2500 kg*4 m/s^2 = 10000 Newtons
(SI unit of force is Newtons)
A student sets up a standing wave of wavelength 6 meters in a coiled spring by moving his hand up and down twice each second. What is the velocity of the wave?
Answer:
Velocity = 12 m/s
Explanation:
Given the following data;
Wavelength = 6 meters
Period = 0.5 seconds. This is due to the fact that the student is moving his hand up and down twice each second.
To find the velocity;
Velocity = wavelength * frequency
But, frequency = 1/period
Frequency = 1/0.5
Frequency = 2
Substituting the values into the velocity formula, we have;
Velocity = 6 * 2
Velocity = 12 m/s
Therefore, the velocity of the wave is 12 meters per seconds.
faraday's law states that the potential difference can be increased by... (choose all that apply)
1. using static electricity instead of current electricity
2. increase the number of coils, increase the strength of the magnet
3. using a bunch of small magnets
4. increasing the time used to power up an electromagnet
5. increase the speed of the magnet
Id say its the last option of the fourth option
Explanation:
Answer:
1:increase the number of coils, increase the strength of the magnet
2:increase the speed of the magnet
3: increasing the time used to power up an electromagnet
Explanation:
1). Increasing the number of turns of wire in the coil – By increasing the amount of individual conductors cutting through the magnetic field, the amount of induced emf produced will be the sum of all the individual loops of the coil, so if there are 20 turns in the coil there will be 20 times more induced emf than in one piece of wire.
2). Increasing the speed of the relative motion between the coil and the magnet – If the same coil of wire passed through the same magnetic field but its speed or velocity is increased, the wire will cut the lines of flux at a faster rate so more induced emf would be produced.
3). Increasing the strength of the magnetic field – If the same coil of wire is moved at the same speed through a stronger magnetic field, there will be more emf produced because there are more lines of force to cut.
A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 11.5 pC. The inner cylinder has a radius of 0.550 mm, the outer one has a radius of 7.20 mm, and the length of each cylinder is 13.0 cm.
(a) What is the capacitance?
(b) What applied potential difference is necessary to produce these charges on the cylinders?
Answer:
A)2.811 × 10^-12 F
B)4.09V
Explanation:
(a) What is the capacitance?
Capacitance of coaxial cylinder can be determined using below expresion
C= (2π ε0 L)/ Ln[ rb/ra]
Where
ε0= permittivity of free space= 8.85×10^−12 Fm^-1
L= length of each cylinder =13.0 cm= 13×10^-2m
rb= radius of outer cylinder= 7.20 mm= 7.20×10^-3m
ra=radius of inner cylinder=0.550 mm= 0.550×10^-3 m
If we substitute the values we have,
C= (2π ε0 L)/ Ln[ rb/ra]
C= ( 2 × π × 8.85×10^−12 ×13×10^-2) / Ln[
7.20×10^-3/0.550×10^-3]
C=( 7.2288×10^-12 )/2.5729
C=2.811 × 10^-12 F
B) (b) What applied potential difference is necessary to produce these charges on the cylinders?
Vba= Q/C
Where Vba=potential difference
Q= charge on each = 11.5 pC. = 11.5×10^-12
C= capacitance= 2.811 × 10^-12 F
If we substitute the values we have
Vba=(11.5×10^-12)/2.811 × 10^-12
= 4.09V
A cart moves with negligible friction or air resistance along a roller coaster track. The cart starts from rest at the top of a hill of unknown height. It then goes down that first hill and up another circularly shaped hill of height 10 meters and radius 15 meters. The ride is designed so that the cart just barely loses contact with the track at the top of the second hill (10 meters above ground level). What is the height of the starting hill relative to ground
Answer:
hinit = 17.5 m
Explanation:
Assuming no friction present, the mechanical energy must be conserved, which means that at any point of the trajectory, the sum of the gravitational potential energy and the kinetic energy must keep the same.At the top of the hill, since it starts from rest, all the energy must be potential, and we can express it as follows:[tex]E_{o} = U_{o} = m*g*h_{init} (1)[/tex]
When the car arrives to the top of the second hill, as we know that it is lower than the first one, the energy of the car, must be part gravitational potential energy, and part kinetic energy.We can express this final energy as follows:[tex]E_{f} = U_{f} + K_{f} = m*g* h_{2} + \frac{1}{2} *m*v_{f} ^{2} (2)[/tex]
In order to find hinit, we need to make (1) equal to (2), and solve for it.In (2) we have the value of h₂ (10 m), but we still need the value of the speed at the top of the second hill, vf.Now, when the car is at the top of the hill, there are two forces acting on it, in opposite directions: the normal force (upward) and the weight (downward).We know also that there is a force that keeps the car along the circular track, which is the centripetal force.This force is just the net downward force acting on the car (it's vertical at the top), and is just the difference between the weight and the normal force.If the cart just barely loses contact with the track at the top of the second hill, this means that at that point the normal force becomes zero.So, the centripetal force must be equal to the weight.The centripetal force can be expressed as follows:[tex]F_{c} = m*\frac{v_{f} ^{2}}{R} (3)[/tex]
We have just said that (3) must be equal to the weight:[tex]F_{c} = m*\frac{v_{f} ^{2}}{R} = m*g (4)[/tex]
Simplifying, and rearranging, we can solve for vf², as follows:[tex]v_{f}^{2} = R*g (5)[/tex]
Replacing (5) in (2), simplifying and rearranging in (1) and (2) we finally have:[tex]h_{init} = h_{2} + \frac{1}{2} R = 10m + 7.5 m = 17.5 m (6)[/tex]
PLEASE HELP ME WITH ONE QUESTION
If 1000 J of heat energy is lost from a piece of hot metal which is placed into a cup of cool water, how much energy is gained by the water?
Answer:
1000J
Explanation:
According to the law of calorimetry;
Heat lost by cold substance = Heat gained by hot substance
If 1000 J of heat energy is lost from a piece of hot metal,
heat lost by hot substance = 1000J
Since we are to determine how much energy is gained by the water, using the law
Heat lost by hot metal = heat gained by the water
Hence heat gained by the water = 1000J