Answer:
Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists. When spelled out, it means voltage = current x resistance, or volts = amps x ohms, or V = A x Ω.
ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attached at the 16.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick
Answer:
0.114 kg or 114 g
Explanation:
From the diagram attaches,
Taking the moment about the fulcrum,
sum of clockwise moment = sum of anticlockwise moment.
Wd = W'd'
Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.
make W' the subject of the equation
W' = Wd/d'................ Equation 1
Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm
Substitute these values into equation 1
W' = 0.5047(23.2)/10.5
W' = 1.115 N.
But,
m' = W'/g
m' = 1.115/9.8
m' = 0.114 kg
m' = 114 g
Work out the velocity v at the end of a rollercoaster ride (0). (rearrange the equation for KE to make velocity v the subject)
KE=1/2mv^2
Explanation:
If the kinetic energy of an object is given and we need to find its velocity of motion, then we can find it by using the formula of kinetic energy as :
[tex]K=\dfrac{1}{2}mv^2[/tex]
m is mass of the object
We can rearrange the above equation such that,
[tex]v=\sqrt{\dfrac{2K}{m}}[/tex]
Hence, this is the velocity at the end of a rollercoaster ride.
A rod of mass M = 154 g and length L = 35 cm can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 11 g, moving with speed V = 9 m/s, strikes the rod at angle θ = 29° a distance D = L/3 from the end and sticks to the rod after the collision.Calculate the rotational kinetic energy, in joules, of the system after the collision.
Answer:
Explanation:
moment of inertia of the rod = 1/3 mL² , m is mass and L is length of rod.
1/3 x .154 x .35²
= .00629
moment of inertia of putty about the axis of rotation
= m d² , m is mass of putty and d is distance fro axis
= .011 x( .35 / 3 )²
= .00015
Total moment of inertia I = .00644 kgm²
angular momentum of putty about the axis of rotation
= mvRsinθ
m is mass , v is velocity , R is distance where it strikes the rod and θ is angle with the rod at which putty strikes
= .011 x 9 x .35 / 3 x sin 29
= .0056
Applying conservation of angular momentum
angular momentum of putty = angular momentum of system after of collision
.0056 = .00644 ω where ω is angular velocity of the rod after collision
ω = .87 rad /s .
Rotational energy
= 1/2 I ω²
I is total moment of inertia
= .5 x .00644 x .87²
= 2.44 x 10⁻³ J .
A hydraulic lift is made by sealing an ideal fluid inside a container with an input piston of cross-sectional area 0.004 m2 , and an output piston of cross-sectional area 1.2 m2 . The pistons can slide up or down without friction while keeping the fluid sealed inside. What is the maximum weight that can be lifted when a force of 60 N is applied to the input piston
Answer:
Maximum weight that can be lifted = 18,000 N
Explanation:
Given:
Cross-sectional area of input (A1) = 0.004 m²
Cross-sectional area of the output (A2) = 1.2 m ²
Force (F) = 60 N
Computation:
Pressure on input piston (P1) = F / A1
Assume,
Maximum weight lifted by piston = W
Pressure on output piston (P2) = W / A2
We, know that
P1 = P2
[F / A1] = [W / A2]
[60 / 0.004] = [W / 1.2]
150,00 = W / 1.2
Weight = 18,000 N
Maximum weight that can be lifted = 18,000 N
Question 10
Air with a density of 1.20 kg/m3 flows through a 75.0 cm diameter pipe with a velocity of 2.00 m/s. What is the mass flow rate?
Answer:
75.0 cm
Explanation:
becouse i don,t no the right answer
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed
Complete Question
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed?
A It should stay the same
B It should be quadrupled.
C It should be quintupled
D It should be doubled.
E It should be tripled
Answer:
Option D is the correct option
Explanation:
Generally electric field is mathematically represented as
[tex]E = \frac{\sigma}{\epsilon_o}[/tex]
Where [tex]\sigma[/tex] is the charge per unit area (Charge density )
From the question we are told that [tex]\sigma[/tex] is doubled hence the
[tex]E = \frac{2 \sigma }{\epsilon_o}[/tex]
Looking the equation above we see that the value of the electric field will also double given that it is directly proportional to the charge density
student conducted an experiment and find the density of an ICEBERGE. A students than recorded the following readings. Mass 425 25 g Volume 405 15 mL What experimental value should be quoted for the density of the ICEBERG? Compare your answer with the density of water, which is 3 1.00 10 kg . Show any calculations necessary to justify your answer
Complete Question
The complete question is shown on the first uploaded image
Answer:
The experimental value of density is [tex]\rho = 1.05*10^{3} \ kg/m^3 \pm 101 \ kg/m^3[/tex]
Comparing it with the value of density of water ([tex]1.0*10^{3} \ kg/m^3[/tex]) we can see that the density of ice is greater
Explanation:
From the question we are told
The mass is [tex]M = (425 \pm 25) \ g =(0.425 \pm 0.025) \ kg[/tex]
The volume is [tex]V = (405 \pm 15 ) \ mL = (0.000405 \pm 1.5*10^{-5}) \ m^3[/tex]
The experimental value of density is mathematically evaluated as
[tex]\rho = \frac{M}{V}[/tex]
[tex]\rho = \frac{0.425}{0.000405}[/tex]
[tex]\rho = 1.05 *10^{3} \ kg/m^3[/tex]
The possible error in this experimental value of density is mathematically evaluated as
[tex]\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} +\frac{\Delta V}{V}[/tex]
substituting value
[tex]\frac{\Delta \rho}{1.05*10^{3}} = \frac{0.025}{0.425} +\frac{1.5*10^{-5}}{0.000405}[/tex]
[tex]\Delta \rho = 101 \ kgm^{-3}[/tex]
Thus the experimental value of density is
[tex]\rho = 1.05*10^{3} \ kg/m^3 \pm 101 \ kg/m^3[/tex]
Light bulb 1 operates with a filament temperature of 3000 K, whereas light bulb 2 has a filament temperature of 2000 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.
Answer:
A₁/A₂ = 0.44
Explanation:
The emissive power of the bulb is given by the formula:
P = σεAT⁴
where,
P = Emissive Power
σ = Stefan-Boltzman constant
ε = Emissivity
A = Surface Area
T = Absolute Temperature of Surface
FOR BULB 1:
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₁T₁⁴ ----------- equation 1
where,
A₁ = Surface Area of Bulb 1
T₁ = Temperature of Bulb 1 = 3000 k
FOR BULB 2:
Since, emissivity and emissive power are constant.
Therefore,
P = σεA₂T₂⁴ ----------- equation 2
where,
A₂ = Surface Area of Bulb 2
T₂ = Temperature of Bulb 1 = 2000 k
Dividing equation 1 by equation 2, we get:
P/P = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁(3000)²/A₂(2000)²
A₁/A₂ = (2000)²/(3000)²
A₁/A₂ = 0.44
Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (mb). Note how the air pressure changes as you move StationB towards the center of the high-pressure system.
A. What do you notice?
B. Why do you think this is called a high-pressure system?
Answer:
a) When moving towards a high pressure center the pressure values increase in the equipment
b) This area is called high prison since the weight of the atmosphere on top is maximum
Explanation:
A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values 85.5 kPa.
When moving towards a high pressure center the pressure values increase in the equipment
B) This area is called high prison since the weight of the atmosphere on top is maximum
in general they are areas of good weather
A type of friction that occurs when air pushes against a moving object causing it to negatively accelerate
a) surface area
b) air resistance
c) descent velocity
d) gravity
Answer:
Air resistance
Answer B is correct
Explanation:
The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.
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Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven, Wtoaster , which is used for 5.40 minutes , and then calculate the amount of energy that an 11.0 W compact fluorescent light (CFL) bulb, Wlight , uses when left on for 10.50 hours .
Energy = (power) x (time)
-- For the toaster:
Power = 1.4 kW = 1,400 watts
Time = 5.4 minutes = 324 seconds
Energy = (1,400 W) x (324 s) = 453,600 Joules
-- For the CFL bulb:
Power = 11 watts
Time = 10.5 hours = 37,800 seconds
Energy = (11 W) x (37,800 s) = 415,800 Joules
-- The toaster uses energy at 127 times the rate of the CFL bulb.
-- The CFL bulb uses energy at 0.0079 times the rate of the toaster.
-- The toaster is used for 0.0086 times as long as the CFL bulb.
-- The CFL bulb is used for 116.7 times as long as the toaster.
-- The toaster uses 9.1% more energy than the CFL bulb.
-- The CFL bulb uses 8.3% less energy than the toaster.
A certain metal with work function of Φ = 1.7 eV is illuminated in vacuum by 1.4 x 10-6 W of light with a wavelength of λ = 600 nm. 1)How many photons per second, N, are incident on the metal? N = photons per second Submit 2)What is KEmax, the maximum kinetic energy of the electron that is emitted from the metal? KEmax = eV
Answer:
1) n = 4.47*10^12 photons
2) K = 0.25 eV
Explanation:
This is a problem about the photoelectric effect.
1) In order to calculate the number of photons that impact the metal, you take into account the power of the light, which is given by:
[tex]P=\frac{E}{t}=1.4*10^{-6}\frac{J}{s}[/tex] (1)
Furthermore you calculate the energy of a photon with a wavelength of 600nm, by using the following formula:
[tex]E_p=h\frac{c}{\lambda}[/tex] (2)
c: speed of light = 3*10^8 m/s
h: Planck's constant = 6.626*10^-34 Js
λ: wavelength = 600*10^-9 m
You replace the values of the parameters in the equation (2):
[tex]E_p=(6.626*10^{-34}Js)\frac{3*10^8m/s}{600*10^{-9}m}=3.131*10^{-19}J[/tex]
Next, you calculate the quotient between the power of the light (equation (1)) and the energy of the photon:
[tex]n=\frac{P}{E_p}=\frac{1.4*10^{-6}J/s}{3.131*10^{-19}J}=4.47*10^{12}photons[/tex]
The number of photons is 4.47*10^12
2) The kinetic energy of the electrons emitted by the metal is given by the following formula:
[tex]K=E_p-\Phi[/tex] (3)
Ep: energy of the photons
Φ: work function of the metal = 1.7 eV
You first convert the energy of the photons to eV:
[tex]E_p=3.131*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.954eV[/tex]
You replace in the equation (3):
[tex]K=1.95eV-1.7eV=0.25eV[/tex]
The kinetic energy of the electrons emitted by the metal is 0.25 eV
(1). The Number of photons per second is,[tex]4.23*10^{12}[/tex]
(2). The maximum kinetic energy of the electron is 0.37eV.
(1). The power of light is given as,
[tex]P=1.4*10^{-6}W[/tex]
Energy is given as,
[tex]E=\frac{hc}{\lambda} =\frac{6.626*10^{-34}*3*10^{8} }{600*10^{-9} } \\\\E=3.313*10^{-19} Joule\\\\E=\frac{3.313*10^{-19}}{1.6*10^{-19} }=2.07eV[/tex]
Number of photons per second is,
[tex]N=\frac{P}{E}=\frac{1.4*10^{-6} }{3.313*10^{-19} } =4.23*10^{12}[/tex]
(2). the maximum kinetic energy of the electron is,
[tex]K.E=E-\phi[/tex]
Where [tex]\phi[/tex] is work function.
[tex]K.E=2.07-1.7=0.37eV[/tex]
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Electrical current is defined as _____.
the capacity to store charge
the flow of electric charge per unit time
the amount of stored electric energy
the voltage of the battery
Electrical current is defined as the flow of electric charge per unit time.
A uniform thin spherical shell of mass M=2kg and radius R=0.23m is given an initial angular speed w=18.3rad/s when it is at the bottom of an inclined plane of height h=3.5m, as shown in the figure. The spherical shell rolls without slipping. Find wif the shell comes to rest at the top of the inclined plane. (Take g-9.81 m/s2, Ispherical shell = 2/3 MR2 ).Express your answer using one decimal place.
Answer:
47.8rad/s
Explanation:
For energy to be conserved.
The potential energy sustain by the object would be equal to K.E
P.E = m× g× h = 2 × 9.81× 3.5= 68.67J
Now K.E = 1/2 × I × (w1^2 - w0^2)
I = 2/3 × M × R2
= 2/3 × 2 × (0.23)^2= 0.0705
Hence
W1 = final angular velocity
Wo = initial angular velocity
From P.E = K.E we have;
68.67J = 1/2 × 0.0705 × (w1^2 - w0^2)
(w1^2 - w0^2) = 1948.09
W1^2 = 1948.09 + (18.3^2)
W1^2=2282.98
W1 = √2282.98
=47.78rad/s
= 47.8rad/s to 1 decimal place.
a) Write the names of the materials used in the ohm law according to the Figure 1?
b) If the voltage of a circuit is 12 V and the resistance is 40 , What is the generated power?
Answer:
a. i. conducting wire
ii high-pass and low-pass filters
iii. Cobra-4 Xpert-link
iii. voltage source
b. Power generated is 3.6 W.
Explanation:
Ohm's law state that the current passing through a metallic conductor, e.g wire is directly proportional to the potential difference across its ends, provided temperature is constant.
i.e V = IR
i. conducting wire
ii high-pass and low-pass filters
iii. Cobra-4 Xpert-link
iii. voltage source
b. Given that; V = 12 V and R = 40 Ohm's.
P = IV
From Ohm's law, I = [tex]\frac{V}{R}[/tex]
So that;
P = [tex]\frac{V^{2} }{R}[/tex]
= [tex]\frac{12^{2} }{40}[/tex]
= [tex]\frac{144}{40}[/tex]
= 3.6 W
The power is 3.6 W.
We say that the displacement of a particle is a vector quantity. Our best justification for this assertion is: A. a displacement is obviously not a scalar. B. displacement can be specified by a magnitude and a direction. C. operating with displacements according to the rules for manipulating vectors leads to results in agreement with experiments. D. displacement can be specified by three numbers. E. displacement is associated by motion.
Answer:
Option B - displacement can be specified by a magnitude and a direction.
Explanation:
A Vector quantity is defined as a physical quantity characterized by the presence of both magnitude as well as direction. Examples include displacement, force, torque, momentum, acceleration, velocity e.t.c
Whereas a scalar quantity is defined as a physical quantity which is specified with the magnitude or size alone. Examples include length, speed, work, mass, density, etc.
Displacement is the difference between the initial position and the final position of a body. Displacement is a vector quantity and not a scalar quantity because it can be described by using both magnitude as well as direction.
Looking at the options, the only one that truly justifies this definition is option B.
A sphere of diameter 6.0cm is moulded into a thin uniform wire of diameter 0.2mm. Calculate the length of the wire in metres (Take π = 22/7) *
Answer:
2025m
Explanation:
Since all materials of the sphere is made to a cylindrical wire, it implies the volume of the sphere material is same as that of the cylinder. This is expressed mathematically thus.
Volume of Sphere= volume of cylinder
4/3 ×π×R^3= π× r2× L
4/3 ×R^3= r^2×L
Hence
L = 3/4 × R^3/ r^2
But R = 6.0/2 = 3.0cm{ Diameter is twice raduis}
r= 0.2/2 = 0.1mm=>0.01cm{ Diameter is twice raduis and unit converted by dividing by 10 since 10mm = 1cm}
Substituting R and r into the expression for L, we have :
L = 3/4 × 3^3/ 0.01^2= 0.75 ×27/0.0001 = 202500cm
202500/100= 2025m{ we divide by 100 because 100cm=1m}
A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes the point x=385 with a speed of 45.0 m/s at t=20.0 s. Find the average velocity and the average acceleration between t=3.0 s and 20.0 s.
Answer:
Average velocity v = 21.18 m/s
Average acceleration a = 2 m/s^2
Explanation:
Average speed equals the total distance travelled divided by the total time taken.
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
Average acceleration equals the change in velocity divided by change in time.
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
Where;
v1 and v2 are velocities at time t1 and t2 respectively.
And x1 and x2 are positions at time t1 and t2 respectively.
Given;
t1 = 3.0s
t2 = 20.0s
v1 = 11 m/s
v2 = 45 m/s
x1 = 25 m
x2 = 385 m
Substituting the values;
Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)
v = (385-25)/(20-3)
v = 21.18 m/s
Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)
a = (45-11)/(20-3)
a = 2 m/s^2
A certain type of laser emits light that has a frequency of 4.6 x 1014 Hz. The light, however, occurs as a series of short pulses, each lasting for a time of 3.1 x 10-11s. The light enters a pool of water. The frequency of the light remains the same, but the speed of light slows down to 2.3 x 108 m/s. In the water, how many wavelengths are in one pulse
Answer:
14,260
Explanation:
Relevant data provided for computing the wavelengths are in one pulse is here below:-
The number of wavelengths in Ls = [tex]4.6\times 10_1_4[/tex]
Therefore the Number of in time = Δt = [tex]3.1\times 10_-_1_1[/tex]
The number of wavelengths are in one pulse is shown below:-
[tex]Number\ of\ wavelengths = \triangle t\times f[/tex]
[tex]= 3.1\times 10_-_1_1\times 4.6\times 10_1_4[/tex]
= 14,260
Therefore for computing the number of wavelengths are in one pulse we simply applied the above formula.
A force of 640 newtons stretches a spring 4 meters. A mass of 40 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 6 m/s. Find the equation of motion.
A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible. Part A What is the speed of the rock when the string passes through the vertical position
Answer:
v = 3.33 m/s
Explanation:
In the position of 45 degrees, all the energy of the rock is gravitational, then we have:
E = m*g*L*cos(angle)
and in the vertical position of the string, all the energy is kinetic, so we have:
E = m*v^2/2
If there is no dissipation, both energies are equal, so we have:
m*g*L*cos(45) = m*v^2/2
9.81 * 0.8 * 0.7071 * 2 = v^2
v^2 = 11.0986
v = 3.33 m/s
Determined to test the law of gravity for himself, a student walks off a skyscraper 180 m high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student.
a) Superman leaves the roof with an initial velocity that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. What must the value of the initial velocity be so that Superman catches the student just before they reach the ground?
b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a).
c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?
Answer:
a) v₀ = - 164.62 m / s , c) y = 122.5 m
Explanation:
We can solve this exercise using the free fall kinematic relations.
We put our reference system on the floor, so the height of the skyscraper is y₀ = 180m and the floor level is y = 0 m
For the boy
y = y₀ + v₀ t - ½ g t²
with free fall its initial speed is zero
y = ½ g t2
For superman
y = y₀ + v₀ (t-5) - ½ g (t-5)²
how superman grabs the lot just before hitting the ground
we look for the time it takes the boy down
t = √ (2 y₀ / g)
t = √ (2 180 / 9,8)
t = 6.06 s
in the equation for superman, we clear the volume and calculate
v₀ (t-5) = -y₀ + ½ g (t-5)²
v₀ (6.06 -5) = -180 + ½ 9.8 (6.06 -5)²
v₀ 1.06 = -174.49
v₀ = - 174.49 / 1.06
v₀ = - 164.62 m / s
the negative sign indicates that the initial speed is down
b) to graph the position of the two we use the table
t (s) Y_boy (m) Y_superman (m)
0 180 180
1 175.1 180
5 57.5 180
6 3.6 10.18
see attachment for the two curves
c) calculate the height that falls a lot in the 5 seconds (t = 5)
y = -1/2 g t²
y = ½ 9.8 5²
y = 122.5 m
for this height superman has not yet left the skyscraper, so the boy hits the ground
Astronauts are testing the gravity on a new planet. A rock is dropped between two photogates that are 0.5 meters apart. The first photogate reads a velocity of 1.2 m/s and the the second photogate reads a velocity of 4.3 m/s . What is the acceleration of gravity on this new planet?
Answer:
a = 17 m / s²
Explanation:
For this experiment that the astronauts are carrying out, the value of the relation of gravity is cosecant, therefore we can use the kinematic relations
v² = v₀² + 2a y
They indicate the initial speed 1.2 m / s the final speed 4.3 m / s and the distance remembered 0.5 m
we clear
a = (v² - v₀²) / 2y
we calculate
a = (4.3² -1.2²) / 2 0.5
a = 17 m / s²
this is the gravity of the new planet
what is the speed of light in quartz
Answer:
1.95 x 10^8 m/s.
Explanation:
Answer:
the answer is 1.95 x 10^8 m/s
Explanation:
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
Answer:
rod end A is strongly attracted towards the balls
rod end B is weakly repelled by the ball as it is at a greater distance
Explanation:
When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.
Therefore rod end A is strongly attracted towards the balls and
rod end B is weakly repelled by the ball as it is at a greater distance
A bus travelling at a speed of 40 kmph reaches its destination in 8 minutes and 15 seconds. How far is the destination? a. 5.43 km b. 5.44 km c. 5.50 km d. 9.06 km
Answer:
c. 5.50 km
Explanation:
8 min * 1h/(60 min) = 8/60 = 2/15 h
15 sec* 1 min/60 sec = 1/4 min * 1h/(60 min) = 1/240 h
8 min 15 sec = (2/15+1/240)h
40 km/h *(2/15 +1/240)h =5.50 km
Answer: 5.50 km
Explanation:
A low C (f = 65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and its mass density is 5.0 g/m2, determine the tension of the wire.
Answer:
Tension of the wire(T) = 169 N
Explanation:
Given:
f = 65Hz
Length of the piano wire (L) = 2 m
Mass density = 5.0 g/m² = 0.005 kg/m²
Find:
Tension of the wire(T)
Computation:
f = v / λ
65 = v / 2L
65 = v /(2)(2)
v = 260 m/s
T = v² (m/l)
T = (260)²(0.005/2)
T = 169 N
Tension of the wire(T) = 169 N
how does the statement " silence is golden " relate to ethics in communicating at the workplace.?
Answer:
Being silent most of the time is a good virtue under certain circumstances and environment. It is always advisable to remain quite silent and not be too quick to respond to situations or issues so as to avoid making and saying wrong words.
The ethics in a workplace involves communicating with others with less amount of talking as possible and more of body languages and signs. This is because the workplace is meant to be a serene place.
g 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 N • m. What is the angle (assumed to be less than 90°) between the wrench handle and the direction of the applied force?
Answer:
Angle = 18.41°
Explanation:
Torque = F•r•sin θ
where;
F = force
r = distance from the rotation point
θ = the angle between the force and the radius vector.
We are given;
Torque = 15 N.m
F = 95 N
r = 0.5 m
Thus, plugging in the relevant values ;
15 = 95 × 0.5 × sin θ
sin θ = 15/(95 × 0.5)
sin θ = 0.3158
θ = sin^(-1)0.3158
θ = 18.41°
Two plates with area 7.00×10−3 m27.00×10−3 m2 are separated by a distance of 4.80×10−4 m4.80×10−4 m . If a charge of 5.40×10−8 C5.40×10−8 C is moved from one plate to the other, calculate the potential difference (voltage) between the two plates. Assume that the separation distance is small in comparison to the diameter of the plates.
Answer:
The voltage is [tex]V = 418.60 \ Volts[/tex]
Explanation:
From the question we are told that
The area of the both plate is [tex]A = 7.00 *10^{-3} \ m^2[/tex]
The distance between the plate is [tex]d = 4.80*10^{-4}\ m[/tex]
The magnitude of the charge is [tex]q = 5.40 *10^{-8} \ C[/tex]
The capacitance of the capacitor that consist of the two plates is mathematically represented as
[tex]C = \frac{\epsilon _o A}{d}[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with a value [tex]e = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
So
[tex]C = \frac{8.85*10^{-12} * (7* 10^{-3})}{ 4.8*10^{-4}}[/tex]
[tex]C = 1.29 *10^{-10} \ F[/tex]
The potential difference between the plate is mathematically represented as
[tex]V = \frac{ Q}{C }[/tex]
[tex]V = \frac{ 5.4*10^{-8}}{1.29 *10^{-10}}[/tex]
[tex]V = 418.60 \ Volts[/tex]
