How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 20.0 cm in diameter to produce an electric field of 1450 N/C just outside the surface of the sphere
Answer:
1.007 × 10^(10) electron
Explanation:
We are given;
Electric Field;E = 1450 N/C
Diameter;d = 20 cm = 0.2 m
So, Radius: r = 0.2/2 = 0.1 m
Formula for Electric field just outside the sphere is given by the formula;
E = kq/r²
Where;
E is the magnitude of the electric field. q is the magnitude of the point charge r is distance from the point charge
k is a constant with a value of 9 x 10^(9) N.m²/C²
Making q the subject, we have;
q = Er²/k
Thus,
q = 1450 × 0.1²/(9 × 10^(9))
q = 1.61 × 10^(-9) C
Now, total charge q is also given by the formula;
q = Ne
Where;
e is charge on electron which is 1.6 × 10^(-19)
N is number of excess electrons
Making N the formula, we have;
N = q/e
N = (1.61 × 10^(-9))/(1.6 × 10^(-19))
N = 1.007 × 10^(10) electron
At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room temperature is 20° C.)
Answer:
The temperature of silver at this given resistivity is 2971.1 ⁰C
Explanation:
The resistivity of silver is calculated as follows;
[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\[/tex]
where;
Rt is the resistivity of silver at the given temperature
Ro is the resistivity of silver at room temperature
α is the temperature coefficient of resistance
To is the room temperature
T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature
[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)][/tex]
Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m
When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;
[tex]R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C[/tex]
Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C
A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential of the surface of this sphere and find it to be 14 kV relative to infinity. Find the excess charge on this sphere.
Answer:
The excess charge is [tex]Q = 3.5 *10^{-7} \ C[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 45 \ cm = 0.45 \ m[/tex]
The potential of the surface is [tex]V = 14 \ kV = 14 *10^{3} \ V[/tex]
The radius of the sphere is
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.45}{2}[/tex]
[tex]r = 0.225 \ m[/tex]
The potential on the surface is mathematically represented as
[tex]V = \frac{k * Q }{r }[/tex]
Where k is coulomb's constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
given from the question that there is no other charge the Q is the excess charge
Thus
[tex]Q = \frac{V* r}{ k}[/tex]
substituting values
[tex]Q = \frac{14 *10^{3} 0.225}{ 9*10^9}[/tex]
[tex]Q = 3.5 *10^{-7} \ C[/tex]
Refer the attached photo
Answer:
A
Explanation:
since the wooden bat is an opaque object placed after a translucent object, light will come through the plastic sheet but will be unable to go through the bat. hence the dark shadow of the bat on a lit sheet
If an object is determined to have a negative charge of 1.6 micro Coulomb, you can conclude that the object has an excess of
Answer:
The object has an excess of [tex]10^{13}[/tex] electrons.
Explanation:
When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:
[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]
The object has an excess of [tex]10^{13}[/tex] electrons.
If you're swimming underwater and knock two rocks together, you will hear a very loud noise. But if your friend above the water knocks two rocks together, you'll barely hear the sound.
Match the words.
The air-water interface is an example of boundary. The( )portion of the initial wave energy is way smaller than the( )portion. This makes the( ) wave hard to hear.
When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can( ) .
1. reflect more efficiently
2. transmitted
3. travel directly to your ears
4. boundary
5. reflected
6. discontinuity
Answer:
The air-water interface is an example of boundary. The transmitted portion of the initial wave energy is way smaller than the reflected portion. This makes the boundary wave hard to hear.
When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can travel directly to your ear.
Explanation:
The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.
Answer: The air-water interface is an example of boundary. The (transmitted) portion of the initial wave is way smaller than the (reflected) portion. This makes the (transmitted) wave hard to hear.
When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can (travel directly to your ears.)
Explanation:
The part of the sound wave that is transmitted across the boundary between air and water is much smaller than the part of the wave that is reflected. This is what makes it hard to hear your friend knocking two rocks together above the surface.
When you and the rocks are underwater, the sound that comes from knocking the rocks together can travel directly to your ears rather than having to be transmitted across mediums.
symbol of science hhshsjsiwtwwisjzhJava
Answer:
is this a company name.? java is a computer software right..
An 100 V/m electric field is directed along the x axis. If the potential at the origin is 300 V, what is potential at the point ( -2m, 0) point
Answer:
200volts
Explanation:
Pls see attached file
Answer:
100 V
Explanation:
Electric field E = 100 V/m
Potential at the origin = 300 V
Potential at point (-2m, 0) i.e 2 m behind the origin = ?
From the equation ΔV = EΔd,
ΔV = [tex]V_{0} - V_{x}[/tex]
where [tex]V_{0}[/tex] is the potential at origin,
and [tex]V_{x}[/tex] is the potential at point (-2, 0)
E = electric field
Δd = 0 - (-2) = 2 m
[tex]V_{0} - V_{x}[/tex] = 300 - [tex]x[/tex]
equating, we have
300 - [tex]x[/tex] = 100 x 2
300 - [tex]x[/tex] = 200
[tex]x[/tex] = 100 V
Two long, parallel wires carry currents in the same direction. If I1 = 10 A, and I2 = 20 A, and they are d = 1.0 m apart, what is the magnetic field at a point P midway between them?
Answer:
The magnetic field at mid point between two parallel wires is 1.2 x 10⁻⁵ T
Explanation:
Given;
current in the first wire, I₁ = 10 A
current in the second wire, I₂ = 20 A
distance between the two wires, d = 1.0 m
Magnetic field at mid point between two parallel wires is calculated as;
[tex]B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r}(I_1 +I_2)[/tex]
where;
r is the midpoint between the wires, = 0.5 m
μ₀ is the permeability of free space, = 4π x 10⁻⁷
[tex]B = \frac{\mu_o }{2\pi r}(I_1 +I_2)\\\\B = \frac{4\pi*10^{-7} }{2\pi *0.5}(10 +20)\\\\B = \frac{4\pi*10^{-7} *30}{2\pi *0.5}\\\\B = 1.2 *10^{-5} \ T[/tex]
Therefore, the magnetic field at mid point between two parallel wires is 1.2 x 10⁻⁵ T
Wind gusts create ripples on the ocean that have a wavelength of 3.03 cm and propagate at 3.37 m/s. What is their frequency (in Hz)?
Answer:
Their frequency is 111.22 Hz
Explanation:
Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration and is expressed in units of length (m).
Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
The propagation speed of a wave is the quantity that measures the speed at which the wave's disturbance propagates throughout its displacement. The speed at which the wave propagates depends on both the type of wave and the medium through which it propagates. Relate wavelength (λ) and frequency (f) inversely proportional using the following equation:
v = f * λ.
Then the frequency can be calculated as: f=v÷λ
In this case:
λ=3.03 cm=0.0303 m (1m=100 cm)v= 3.37 m/sReplacing:
[tex]f=\frac{3.37 \frac{m}{s} }{0.0303 m}[/tex]
Solving:
f=111.22 Hz
Their frequency is 111.22 Hz
Besides the gravitational force, a 2.80-kg object is subjected to one other constant force. The objectstarts from rest and in 1.20 s experiences a displacement of (4.20 i - 3.30 j) m, where the direction of jis the upward vertical direction. Determine the other force.
Answer:
the other force= (16.3i + 14.6j)N
EXPLANATION:
Given:
Mass=2.80-kg
t= 1.2s
Since the object started from rest, the origin is (0,0) which symbolize the the object's initial position.
We will need to calculate the magnitude of the displacement using the below formula;
d = (1/2)at2 + v0t + d0
But note that
d0 = 0,( initial position)
v0 = 0( initial position)
a is the net acceleration
d = √[4.202 + (-3.30)2] m = 5.34 m
Hence, the magnitude of the displacement is 5.34 m, then we can make 'a' the subject of formula in the above expression in order to calculate the value for acceleration, note that d0 = 0,( initial position) and v0 = 0( initial position)
d = (1/2)at2
a = 2d/t2 = 2(5.34)/(1.20)2 m/s2 = 7.42 m/s2
the net acceleration is 7.42 m/s2
Acceleration in terms of the vector can be calculated as
a=2(ri - r0)/t^2
Where t =1.2s which is the time
a= 2(4.2i - 3.30j)/ 1.2^2
a=( 5.83i - 4.58j)m/s
now the net force can now be calculated since we have known the value of acceleration, using the formula below;
F(x) = ma - mg
Where a = 5.83i - 4.58j)m/s and g= 9.8m/s
2.8(5.83i - 4.58j)m/s - (2.80 × 9.8)m/s^2
Therefore, the other force= (16.3i + 14.6j)N
A standard 1 kilogram weight is a cylinder 48.5 mm in height and 49.0 mm in diameter. What is the density of the material? kg/m3
Answer:
Density = 10,933.93 kg/m^3
the density of the material is 10,933.93 kg/m^3
Explanation:
Density is the mass per unit volume
Density = mass/volume = m/V
Volume of a cylinder V = πr^2 h
Given;
Height h = 48.5mm = 0.0485 m
Radius r = diameter/2 = 49mm÷2 = 24.5mm = 0.0245m
Substituting the values;
Volume V = π×(0.0245^2)×0.0485
V = 0.000091458438030 m^3
V = 0.000091458 m^3
The mass is given as;
Mass = 1 kg
So, the density can be calculated as;
Density = 1/0.000091458
Density = 10933.92825785 kg/m^3
Density = 10,933.93 kg/m^3
the density of the material is 10,933.93 kg/m^3
A friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back toward your friend. What should you do in order to maximize your speed right after your interaction with the ball?
A. You should catch the ball.
B. You should let the ball go past you without touching it.
C. You should deflect the ball back toward your friend.
D. More information is required to determine how to maximize your speed.
E. It doesn't matter. Your speed is the same regardless of what you do.
Answer:
C You should deflect the ball back toward your friend.
Explanation:
This is because it would result in a completely inelastic collision, and the final velocity of me would be found using,
with m= mass, V=velocity, i=initial, f=final:
mV(me,i) +mV(ball,i) = [m(me)+m(b)]V(f)
So V(f) would be just the momentum of the ball divided by just MV mass of the ball and it will be higher resulting in inelastic collision
Answer:
A. You should catch the ball.
Explanation:
Catching the ball maximizes your speed by converting most of the momentum of the flying ball into the momentum of you and the ball. Since the ice is smooth, the friction between your feet and the ice is almost negligible, meaning less energy is needed to set your body in motion. Catching the ball means that you and the ball undergoes an inelastic collision, and part of the kinetic energy of the ball is transferred to you, setting you in motion. Deflecting the ball will only give you a relatively small speed compared to catching the ball.
A cylindrical shell of radius 7.00 cm and length 2.21 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 15.2 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. (a) Find the net charge on the shell.
Answer:
The net charge on the shell is 30x10^-9C
Explanation:
Pls see attached file
Transverse waves are sent along a 4.50 m long string with a speed of 85.00 m/s. The string is under a tension of 20.00 N. What is the mass of the string (in kg)?
Answer:
m = 0.0125 kg
Explanation:
Let us apply the formula for the speed of a wave on a string that is under tension:
[tex]v = \sqrt{\frac{F}{\mu} }[/tex]
where F = tension force
μ = mass per unit length
Mass per unit length is given as:
μ = m / l
where m = mass of the string
l = length of the string
This implies that:
[tex]v = \sqrt{\frac{F}{m/l} }\\ \\v = \sqrt{\frac{F * l}{m} }[/tex]
Let us make mass, m, the subject of the formula:
[tex]v^2 = \frac{F * l}{m}\\\\m = \frac{F * l}{v^2}[/tex]
From the question:
F = 20 N
l = 4.50 m
v = 85 m/s
Therefore:
[tex]m = \frac{20 * 4.5}{85^2}\\\\m = \frac{90}{7225}\\ \\m = 0.0125 kg[/tex]
A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It strikes a target 1.7 seconds later. What is the vertical distance from where the projectile was launched to where it hit the target.
Answer:
30.67m
Explanation:
Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;
s = ut ± [tex]\frac{1}{2} at^2[/tex] ------------(i)
Where;
s = horizontal/vertical displacement
u = initial horizontal/vertical component of the velocity
a = acceleration of the projectile
t = time taken for the projectile to reach a certain horizontal or vertical position.
Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;
h = vt ± [tex]\frac{1}{2} at^2[/tex] ------------(ii)
Where;
h = vertical displacement
v = initial vertical component of the velocity = usinθ
a = acceleration due to gravity (since vertical motion is considered)
t = time taken for the projectile to hit the target
From the question;
u = 47m/s, θ = 0.6rads
=> usinθ = 47 sin 0.6
=> usinθ = 47 x 0.5646 = 26.54m/s
t = 1.7s
Take a = -g = -10.0m/s (since motion is upwards against gravity)
Substitute these values into equation (ii) as follows;
h = vt - [tex]\frac{1}{2} at^2[/tex]
h = 26.54(1.7) - [tex]\frac{1}{2} (10)(1.7)^2[/tex]
h = 45.118 - 14.45
h = 30.67m
Therefore, the vertical distance is 30.67m
(5 pt) You tie a cord to a pail of water, and your swing the pail in a vertical circular 0.700 m. What is the minimum speed must you give the pail at the highest point of the circle if no water is to spill from it
Answer:
The minimum speed required is 2.62m/s
Explanation:
The value of gravitational acceleration = g = 9.81 m/s^2
Radius of the vertical circle = R = 0.7 m
Given the mass of the pail of water = m
The speed at the highest point of the circle = V
The centripetal force will be needed must be more than the weight of the pail of water in order to not spill water.
Below is the calculation:
[tex]\frac{mV^{2}}{R} = mg[/tex]
[tex]V = \sqrt{gR}[/tex]
[tex]V = \sqrt{9.81 \times 0.7}[/tex]
[tex]V = 2.62 m/s[/tex]
Which one of the following is the shortest length?
A)
100 meters
C)
104 millimeters
E)
10 nanometers
B)
10² centimeters
D)
105 micrometers
Answer:
Option E (10 nanometers) is the shortest lengthExplanation:
From,
1cm = [tex]10^{-2}m[/tex]
1mm = [tex]10^{-3}m[/tex]
1nanometer = [tex]10^{-9[/tex]
1micrometer = [tex]10^{-6[/tex]
Therefore,
A) [tex]10^0[/tex] meters = 1meter
B) [tex]10^2[/tex] cm = [tex]10^2 * 10^{-2} = 1meter[/tex]
C) [tex]10^4[/tex] mm = [tex]10^4 * 10^{-3} = 10meter[/tex]
D) [tex]10^5[/tex] micrometer = [tex]10^5 * 10^{-6} = 0.1meter[/tex]
E) [tex]10[/tex] nanometer = [tex]10 * 10^-9 = 1*10^{-8}[/tex]
Therefore 10nanometers is the shortest length
For more information on length conversions, visit
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Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The longer one has length 8L, diameter 4D, and resistance R2. How do the resistances of these two resistors compare
Answer:
the resistance of the longer one is twice as big as the resistance of the shorter one.
Explanation:
Given that :
For the shorter cylindrical resistor
Length = L
Diameter = D
Resistance = R1
For the longer cylindrical resistor
Length = 8L
Diameter = 4D
Resistance = R2
So;
We all know that the resistance of a given material can be determined by using the formula :
[tex]R = \dfrac{\rho L }{A}[/tex]
where;
A = πr²
[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]
For the shorter cylindrical resistor ; we have:
[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]
since 2 r = D
[tex]R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}[/tex]
[tex]R = \dfrac{ 4 \rho L }{\pi \ D ^2}[/tex]
For the longer cylindrical resistor ; we have:
[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]
since 2 r = D
[tex]R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}[/tex]
[tex]R = \dfrac{32\rho L }{\pi \ (4 D) ^2}[/tex]
[tex]R = \dfrac{2\rho L }{\pi \ (D) ^2}[/tex]
Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :
[tex]\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D ^2}}{ \dfrac{2\rho L }{\pi \ (D) ^2}}[/tex]
[tex]\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D ^2}}* { \dfrac {\pi \ (D) ^2} {2\rho L}}[/tex]
[tex]\dfrac{R_s}{R_L} =2[/tex]
[tex]{R_s}=2{R_L}[/tex]
Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.
An electron, moving west, enters a magnetic field of a certain strength. Because of this field the electron curves upward. What is the direction of the magnetic field?
Answer:
Towards the west.
Explanation:
The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.
Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.
According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward
Four identical charges particles of charge 1Uc, 2Uc,
3Uc and 4Uc
are placed at x = lm, x=2m,
x=3m and
x=5m. The electric field intensity
at origin is?
Answer:
17.94 kN/C is the electric field intensity at the origin due to the charges.
Explanation:
From the question, we are told that
The distance of 1 μC from origin = 1 m
The distance of 2 μC from origin = 2 m
The distance of 3 μC from origin = 3 m
The distance of 4 μC from origin = 5 m
Therefore, for us to find the electric field intensity, we'll solve below:
The formula for Electric field intensity = ( k * q ) / ( r * r )
where , r is distance ,
k = 9 * 10^9 ,
and , q is charge .
now ,
electric field intensity at the origin = [ k * 10^(-6) / 1 * 1 ] +[ k * 2 * 10^(-6) / 2 * 2 ] + [ k * 3 * 10^(-6) / 3 * 3 ] + [ k * 4 * 10^(-6) / 5 * 5 ]
=> electric field intensity at the origin = k * 10^(-6) [ 1 + 1/2 + 1/3 + 4/25 ] N/C
=> electric field intensity at the origin = 9 * 10^9 * 10^(-6) * 1.99 N/C
=> electric field intensity at the origin = 17.94 kN/C
You have a circuit of three resistors in series connected to a battery. You add a fourth resistor, also in series, to the combination. As a result:_______.
A. Power delivered from the battery to combination increases.
B. It is impossible to give the answer without knowing the actual resistances and voltage of the battery.
C. Power delivered from the battery to combination is unchanged.
D. Power delivered from the battery to combination decreases.
Answer:
D
Explanation:
The power equation is P= V^2/R
Please let me know if this helped! Please rate it the brainlist if possible!
As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.
What is a resistors?A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.
Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.
A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.
The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.
Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.
Thus, the correct option is D.
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The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?
Answer:
The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.
Explanation:
The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.
The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.
Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.
I hope it helps you!
A rubber ball is attached to a string and whirled around in a circle. If the string is 1.0 m long (measured from the center of the baseball to the far end of the string) and the ball’s speed is 10 m/s, what is the ball’s centripetal acceleration?
Centripetal acceleration = (speed squared) / (radius)
Centripetal acceleration = (10 m/s)² / (1.0 m)
Centripetal acceleration = (100 m²/s²) / (1.0 m)
Centripetal acceleration = 100 m/s²
a point charge q is located at the center of a cube with edge length d. whatis the value of the flux over one face of the cube
Answer:
q/6Eo
Explanation:
See attached file pls
Determine the maximum height and range of a projectile fired at a height of 6 feet above the ground with an initial velocity of 100 feet per second at an angle of 40 degrees above the horizontal.Maximum heightRange Question 20 options:a) 70.56 feet183.38 feet b) 92.75 feet310.59 feet c) 92.75 feet183.38 feet d) 70.56 feet314.74 feet e)
Answer:
C is the correct answer
Explanation:
supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistance is 207.4 What is the temerature (in degree Celsius of this solution
Answer:
T₂ = 95.56°C
Explanation:
The final resistance of a material after being heated is given by the relation:
R' = R(1 + αΔT)
where,
R' = Final Resistance = 207.4 Ω
R = Initial Resistance = 154.9 Ω
α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹
ΔT = Change in Temperature = ?
Therefore,
207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]
207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT
1.34 - 1 = (0.0045°C⁻¹)ΔT
ΔT = 0.34/0.0045°C⁻¹
ΔT = 75.56°C
but,
ΔT = Final Temperature - Initial Temperature
ΔT = T₂ - T₁ = T₂ - 20°C
T₂ - 20°C = 75.56°C
T₂ = 75.56°C + 20°C
T₂ = 95.56°C
In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.
Answer:
The angular speed is [tex]w = 5.89 \ rad/s[/tex]
Explanation:
From the question we are told that
The time taken is [tex]t = 1.6 s[/tex]
The number of somersaults is n = 1.5
The total angular displacement during the somersault is mathematically represented as
[tex]\theta = n * 2 * \pi[/tex]
substituting values
[tex]\theta = 1.5 * 2 * 3.142[/tex]
[tex]\theta = 9.426 \ rad[/tex]
The angular speed is mathematically represented as
[tex]w = \frac{\theta }{t}[/tex]
substituting values
[tex]w = \frac{9.426}{1.6}[/tex]
[tex]w = 5.89 \ rad/s[/tex]
A rod has length 0.900 mm and mass 0.500 kgkg and is pivoted at one end. The rod is not uniform; the center of mass of the rod is not at its center but is 0.500 mm from the pivot. The period of the rod's motion as a pendulum is 1.49 ss. What is the moment of inertia of the rod around the pivot
Answer:
The moment of inertia is [tex]I =0.14 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The length of the rod is [tex]l = 0.900 \ m[/tex]
The mass of the rod is [tex]m = 0.500 \ kg[/tex]
The distance of the center of mass from the pivot is [tex]d = 0.500 \ m[/tex]
The period of the rod's motion is [tex]T = 1.49 \ s[/tex]
Generally the period of the motion is mathematically represented as
[tex]T = 2 \pi * \sqrt{\frac{I}{m* g * d} }[/tex]
Where [tex]I[/tex] is the moment of inertia about the pivot so making [tex]I[/tex] the subject of formula
[tex]I = [\frac{T}{2\pi } ]^2 * m * g * d[/tex]
substituting values
[tex]I = [\frac{1.49}{2* 3.142 } ]^2 * 0.5 * 9.8 * 0.5[/tex]
[tex]I =0.14 \ kg \cdot m^2[/tex]
Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing through the aorta (in grams of blood per second)?
Answer:
94.248 g/sec
Explanation:
For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:
[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]
And, the velocity of blood pumping is 30 cm^2
Now apply the following formula to solve the total current
[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]
Q = 94.248 g/sec
Basically we applied the above formula So, that the total current could come