To find the magnitude and direction of the electric field, let us find the horizontal and vertical components of the field separately, then we will use those values to calculate the total magnitude and direction.
The tension in the thread is 6.57×10⁻²N and the thread is aligned horizontally, so the tension force is directed entirely horizontally. The sphere is in static equilibrium, therefore the horizontal component of the electrostatic force acting on the sphere, Fx, must act in the opposite direction of the tension and have a magnitude of 6.57×10⁻²N. We know this equation relating a charge, an electric field, and the force that the field exerts on the charge:
F = Eq
F is the electric force, E is the electric field, and q is the charge
Let us adjust the equation for only the horizontal components of the above quantities:
Fx = (Ex)(q)
Fx is the horizontal component of the electric force and Ex is the horizontal component of the electric field.
Given values:
F = 6.57×10⁻²N
q = 6.80×10³C
Plug in these values and solve for Ex:
6.57x10⁻² = Ex(6.80×10³)
Ex = 9.66×10⁻⁶N/C
Since the sphere is in static equilibrium, the vertical component of the electrostatic force acting on the sphere, Fy, must have the same magnitude and act in the opposite direction of the sphere's weight. If we assume the weight to act downwards, then Fy must act upward.
We know the weight of the sphere is given by:
W = mg
W is the weight, m is the mass, and g is the acceleration of objects due to earth's gravity field near its surface.
We also know this equation:
F = Eq
Let us adjust for the vertical components:
Fy = (Ey)(q)
Set Fy equal to W and we get:
(Ey)(q) = mg
Given values:
q = 6.80×10³C
m = 0.018kg
g = 9.81m/s²
Plug in the values and solve for Ey:
(Ey)(6.80×10³) = 0.018(9.81)
Ey = 2.60×10⁻⁵N/C
Let's now use the Pythagorean theorem to find the total magnitude of the electric field:
E = [tex]\sqrt{Ex^{2}+Ey^{2}}[/tex]
E = 2.77×10⁻⁵N/C
The direction of the electric field is given by:
θ = tan⁻¹(Ey/Ex)
θ = 20.4° off the horizontal
A uniform electric field, with a magnitude of 370 N/C, is directed parallel to the positive x-axis. If the electric potential at x = 2.00 m is 1 000 V, what is the change in potential energy of a particle with a charge of + 2.80 x 10-3 C as it moves from x = 1.9 m to x = 2.1 m?
Answer:
[tex]\Delta U = 0.2072 J[/tex]
Explanation:
Potential difference between two points in constant electric field is given by the formula
[tex]\Delta V = E.\Delta x[/tex]
here we know that
[tex]E = 370 N/C[/tex]
also we know that
[tex]\Delta x = 2.1 - 1.9 = 0.2 m[/tex]
now we have
[tex]\Delta V = 370 (0.2) = 74 V[/tex]
now change in potential energy is given as
[tex]\Delta U = Q\Delta V[/tex]
[tex]\Delta U = (2.80 \times 10^{-3})(74)[/tex]
[tex]\Delta U = 0.2072 J[/tex]
A vertical spring (ignore its mass), whose spring constant is 594-N/m, is attached to a table and is compressed down by 0.196-m. What upward speed (in m/s) can it give to a 0.477-kg ball when released?
Answer:
Speed, v = 6.91 m/s
Explanation:
Given that,
Spring constant, k = 594 N/m
It is attached to a table and is compressed down by 0.196 m, x = 0.196 m
We need to find the speed of the spring when it is released. Here, the elastic potential energy is balanced by the kinetic energy of the spring such that,
[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\dfrac{kx^2}{m}}[/tex]
[tex]v=\sqrt{\dfrac{594\ N/m\times (0.196\ m)^2}{0.477\ kg}}[/tex]
v = 6.91 m/s
So, the speed of the ball is 6.91 m/s. Hence, this is the required solution.
The sides of a square increase in length at a rate of 3 m/sec. a. At what rate is the area of the square changing when the sides are 14 m long? b. At what rate is the area of the square changing when the sides are 25 m long?
The area of a square is given by:
A = s²
A is the square's area
s is the length of one of the square's sides
Let us take the derivative of both sides of the equation with respect to time t in order to determine a formula for finding the rate of change of the square's area over time:
d[A]/dt = d[s²]/dt
The chain rule says to take the derivative of s² with respect to s then multiply the result by ds/dt
dA/dt = 2s(ds/dt)
A) Given values:
s = 14m
ds/dt = 3m/s
Plug in these values and solve for dA/dt:
dA/dt = 2(14)(3)
dA/dt = 84m²/s
B) Given values:
s = 25m
ds/dt = 3m/s
Plug in these values and solve for dA/dt:
dA/dt = 2(25)(3)
dA/dt = 150m²/s
When the side of the square is 14 m, the rate at which the area is changing is 84 m²/s.
When the side of the square is 25 m, the rate at which the area is changing is 150 m²/s.
The given parameters;
rate at which the side of the square is increasing, [tex]\frac{dl}{dt}[/tex] = 3 m/sThe area of the square is calculated as;
A = L²
The change in the area is calculated as;
[tex]\frac{dA}{dt} = 2l\frac{dl}{dt}[/tex]
When the side of the square is 14 m, the rate at which the area is changing is calculated as;
[tex]\frac{dA}{dt} = 2l \frac{dl}{dt} \\\\\frac{dA}{dt} = 2 \times l \times \frac{dl}{dt}\\\\\frac{dA}{dt} = 2 \times 14 \times 3\\\\\frac{dA}{dt} = 84 \ m^2/s[/tex]
When the side of the square is 25 m, the rate at which the area is changing is calculated as;
[tex]\frac{dA}{dt} = 2l \frac{dl}{dt} \\\\\frac{dA}{dt} = 2 \times l \times \frac{dl}{dt}\\\\\frac{dA}{dt} = 2 \times 25 \times 3\\\\\frac{dA}{dt} = 150 \ m^2/s[/tex]
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At a carnival, you can try to ring a bell by striking a target with a 8.91-kg hammer. In response, a 0.411-kg metal piece is sent upward toward the bell, which is 3.88 m above. Suppose that 21.9 percent of the hammer's kinetic energy is used to do the work of sending the metal piece upward. How fast must the hammer be moving when it strikes the target so that the bell just barely rings?
Answer:
4 m/s
Explanation:
M = mass of the hammer = 8.91 kg
m = mass of the metal piece = 0.411 kg
h = height gained by the metal piece = 3.88 m
Potential energy gained by the metal piece is given as
PE = mgh
PE = (0.411) (9.8) (3.88)
PE = 15.6 J
KE = Kinetic energy of the hammer
Given that :
Potential energy of metal piece = (0.219) Kinetic energy of the hammer
PE = (0.219) KE
15.6 = (0.219) KE
KE = 71.2 J
v = speed of hammer
Kinetic energy of hammer is given as
KE = (0.5) M v²
71.2 = (0.5) (8.91) v²
v = 4 m/s
A car driving at an initial speed of 10.0 m/s accelerates on a straight road at 3.00 m/s^2. a) what is the speed of the car after one quarter of a mile? (1 mile =1.609km). B) the driver in part A slams on the brakes after reaching the quarter mile. If the car can decelerate at a rate of 4.50 m/s^2, what is the stopping distance of the car?
Answer:
The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.
Explanation:
Given that,
Initial speed = 10.0 m/s
Acceleration = 3.00 m/s^2
Distance [tex]d = \dfrac{1}{4}\times1.609[/tex]
[tex]d = 0.40225\ km=0.40225\times10^{3}\ m[/tex]
We need to calculate the speed of the car,
Using equation of motion
[tex]v^2-u^2=2as[/tex]
Where, u = initial velocity
v = final velocity
a = acceleration
Put the value in the equation
[tex]v^2=(10.0)^2+2\times3.00\times0.40225\times10^{3}[/tex]
[tex]v^2=2513.5[/tex]
[tex]v=50.13\ m/s[/tex]
(B). We need to calculate the stopping distance of the car,
Using equation of motion again
[tex]v^2=u^2+2as[/tex]
Here,initial velocity = 50.13 m/s
Final velocity = 0
Acceleration = -4.50 m/s²
Put the value in the equation
[tex]0=(50.13)^2-2\times4.50\times s[/tex]
[tex]s=\dfrac{(50.13)^2}{2\times4.50}[/tex]
[tex]s=279.22\ m[/tex]
Hence, The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.
A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the paper. Directed out of the plane of the paper is a magnetic field of 0.06 T. The field out of the paper decreases to 0.02 T in 12 milliseconds. What is the direction of the current induced?
Explanation:
It is given that,
Number of turns, N = 200
Area of cross section, A = 8.5 cm²
Magnetic field is directed out of the paper and is, B = 0.06 T
The magnetic field is out of the paper decreases to 0.02 T in 12 milliseconds. We need to find the direction of current induced. The induced emf is given by :
[tex]\epsilon=-N\dfrac{d\phi}{dt}[/tex]
Since, [tex]\epsilon=IR[/tex]
I is the induced current
[tex]I=-\dfrac{N}{R}\dfrac{d\phi}{dt}[/tex]
According to Lenz's law, the direction of induced current is such that it always opposes the change in current that causes it.
Here, the field is directed out of the plane of the paper, this gives the induced current in counterclockwise direction.
Two parallel plate capacitors are identical, except that one of them is empty and the other contains a material with a dielectric constant of 4.2 in the space between the plates. The empty capacitor is connected between the terminals of an ac generator that has a fixed frequency and rms voltage. The generator delivers a current of 0.29 A. What current does the generator deliver after the other capacitor is connected in parallel with the first one?
Answer:
current = 1.51 A
Explanation:
Initially the capacitor without any dielectric is connected across AC source
so the capacitive reactance of that capacitor is given as
[tex]x_c = \frac{1}{\omega c}[/tex]
now we have
[tex]i = \frac{V_{rms}}{x_c}[/tex]
here we know that
[tex]i = 0.29 A[/tex]
now other capacitor with dielectric of 4.2 is connected in parallel with the first capacitor
so here net capacitance is given as
[tex]c_{eq} = 4.2c + c = 5.2c[/tex]
now the equivalent capacitive reactance is given as
[tex]x_c' = \frac{1}{\omega(5.2c)}[/tex]
[tex]x_c' = \frac{x_c}{5.2}[/tex]
so here we have new current in that circuit is given as
[tex]i' = \frac{V_{rms}}{x_c'}[/tex]
[tex]i' = 5.2 (i) = 5.2(0.29)[/tex]
[tex]i' = 1.51 A[/tex]
A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point on the circle, the speed of the mass is 12 m/s. What is the magnitude of the force of the string on the mass at this position?
Answer:
31.1 N
Explanation:
m = mass attached to string = 0.50 kg
r = radius of the vertical circle = 2.0 m
v = speed of the mass at the highest point = 12 m/s
T = force of the string on the mass attached.
At the highest point, force equation is given as
[tex]T + mg =\frac{mv^{2}}{r}[/tex]
Inserting the values
[tex]T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}[/tex]
T = 31.1 N
Answer: 41N
Explanation :
T= mv^2/R + mgcos θ
At the highest point on the circle θ=0
Cos 0 = 1
T= mv^2/R + mg
m = 0.5kg
Velocity at the highest point (amplitude)= 12m/s
T = 0.5× 12^2/2 + 0.5×10
0.5×144/2 +5
T = 0.5×72 + 5
T = 36+5
T = 41N
Be sure to answer all parts. Convert: (a) 29.8°C (the melting point of gallium) to degrees Fahrenheit °F (b) 172.9°F (the boiling point of ethanol) to degrees Celsius °C (c) 4.2 K ,the boiling point of helium, to degrees Celsius °C
Answer:
(a) 85.64°F
(b) 78.28°C
(c) - 268.8°C
Explanation:
(a) 29.8°C into °F
To convert °C into °F, we use the formula given below
C / 100 = (F - 32) / 180
C / 5 = (F - 32) / 9
9 x 29.8 / 5 = F - 32
F = 85.64°F
(b) 172.9°F into °C
C / 5 = (F - 32) / 9
C = (172.9 - 32) x 5 / 9
C = 78.28°C
(c) 4.2 K into °C
C = K - 273
C = 4.2 - 273 = - 268.8°C
g 'A fire hose ejects a stream of water at an angle of 31.6o above the horizontal. The water leaves the nozzle with a speed of 29.3 m/s. Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?
Answer:
39.08 m
Explanation:
θ = 31.6 degree
u = 29.3 m/s
Let it hits at the maximum height h. It takes half of the time of flight to cover the distance.
T = 2 u Sinθ / g
t = T / 2 = u Sinθ / g = (29.3 x Sin 31.6) / 9.8 = 1.566 s
The horizontal distance covered in this time is
d = u Cosθ x t = 29.3 x Cos 31.6 x 1.566 = 39.08 m
Thus, the fire hose be located at a distance of 39.08 m from the building.
What is the beat frequency heard when two organ pipes, each open at both ends, are sounded together at their fundamental frequencies if one pipe is 60 cm long and the other is 68 cm long?(The speed of sound is 340 m/s)
Answer:
The beat frequency is 33.33 Hz.
Explanation:
Given that,
Length of first pipe =60 cm
Length of other pipe = 68 cm
Speed of sound = 340 m/s
We need to calculate the frequency
We know that,
When they operate at fundamental frequency then the length is given by,
[tex]L=\dfrac{\lambda}{2}[/tex]
The wavelength is given by
[tex]\lambda=2L[/tex]
For first organ pipe,
Using formula of frequency
[tex]f=\dfrac{v}{\lambda}[/tex]
[tex]f_{1}=\dfrac{v}{2L_{1}}[/tex]...(I)
Put the value into the formula
[tex]f_{1}=\dfrac{340}{2\times60\times10^{-2}}[/tex]
[tex]f_{1}=283.33\ Hz[/tex]
For second organ pipe,
[tex]f_{2}=\dfrac{v}{2L_{2}}[/tex]...(II)
Put the value in the equation (II)
[tex]f_{2}=\dfrac{340}{2\times68\times10^{-2}}[/tex]
[tex]f_{2}=250\ Hz[/tex]
Therefore the beat frequency
[tex]\Delta f=f_{1}-f_{2}[/tex]
[tex]\Delta f=283.33-250[/tex]
[tex]\Delta f=33.33\ Hz[/tex]
Hence, The beat frequency is 33.33 Hz.
The beat frequency heard when two organ pipes are sounded together at their fundamental frequencies, where one pipe is 60 cm long and the other is 68 cm long, is 33.33 Hz.
Explanation:To find the beat frequency heard when two organ pipes are sounded together at their fundamental frequencies, we need to calculate the frequencies of each pipe. The formula for frequency is f = v/λ, where v is the speed of sound and λ is the wavelength. Since the pipes are open at both ends, the wavelength is twice the length of the pipe. So, the frequency of the 60 cm pipe would be f₁ = v/(2L₁), and the frequency of the 68 cm pipe would be f₂ = v/(2L₂). Now, to calculate the beat frequency, we subtract the frequencies: beat frequency = |f₁ - f₂|.
Using the given speed of sound (340 m/s), we can substitute the lengths of the pipes into the frequency formula to find the frequencies:
f₁ = 340/(2 x 0.6) Hz = 283.33 Hz
f₂ = 340/(2 x 0.68) Hz = 250 Hz
Now, we can calculate the beat frequency:
beat frequency = |283.33 - 250| Hz = 33.33 Hz
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Two point charges each experience a 1-N electrostatic force when they are 2 cm apart. If they are moved to a new separation of 8 cm, what is the magnitude of the electric force on each of them?
Explanation:
Force between two point changes, F₁ = 1 N
Distance between them, r₁ = 2 cm = 0.02 m
We know that the electrostatic force is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
i.e.
[tex]F\propto \dfrac{1}{r^2}[/tex]
i.e
[tex]\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2[/tex]
Let F₂ is the force when the distance between the charges is 8 cm, r₂ = 0.08 m
[tex]F_2=\dfrac{F_1\times r_1^2}{r_2^2}[/tex]
[tex]F_2=\dfrac{1\ N\times (0.02\ m)^2}{(0.08\ m)^2}[/tex]
F₂ = 0.0625 N
So, the distance between the sphere is 8 N, the new force is equal to 0.0625 N. Hence, this is the required solution.
The heater element of a particular 120-V toaster is a 8.9-m length of nichrome wire, whose diameter is 0.86 mm. The resistivity of nichrome at the operating temperature of the toaster is 1.3 × 10-6 Ω ∙ m. If the toaster is operated at a voltage of 120 V, how much power does it draw
Answer:
Power, P = 722.96 watts
Explanation:
It is given that,
Voltage, V = 120 V
Length of nichrome wire, l = 8.9 m
Diameter of wire, d = 0.86 mm
Radius of wire, r = 0.43 mm = 0.00043 m
Resistivity of wire, [tex]\rho=1.3\times 10^{-6}\ \Omega-m[/tex]
We need to find the power drawn by this heater. Power is given by :
[tex]P=\dfrac{V^2}{R}[/tex]
And, [tex]R=\rho\dfrac{l}{A}[/tex]
[tex]P=\dfrac{V^2\times A}{\rho\times l}[/tex]
[tex]P=\dfrac{120^2\times \pi (0.00043)^2}{1.3\times 10^{-6}\times 8.9}[/tex]
P = 722.96 watts
So, the power drawn by this heater element is 722.96 watts. Hence, this is the required solution.
What If? Physicists often measure the momentum of subatomic particles moving near the speed of light in units of MeV/c, where c is the speed of light, and 1 MeV = 1.6 ✕ 10−13 kg · m2/s2. Based on this, what are the units of momentum for a high-speed subatomic particle in terms of fundamental SI units?
Answer:
(kg⋅m/s)
Explanation:
The unit of momentum is the product of the units of mass and velocity. In SI units, if the mass is in kilograms and the velocity is in meters per second then the momentum is in kilogram meters per second (kg⋅m/s)
This question is dealing with fundamental SI units. Thus, let's list the seven basic SI Units upon which other units are expressed;
Mass - kilogram (kg)
Length - meter (m)
Time - second (s)
Amount of substance - mole (mol)
Electric current - ampere (A)
Thermodynamic temperature - kelvin (K)
Luminous intensity - candela (cd)
Fundamental SI unit of momentum is Kg.m/s
Now, we want to write the SI Unit of momentum.
From the question, we are told that Physicists often measure the momentum of subatomic particles using the formula;
MeV/c, where c is the speed of light, and 1 MeV = 1.6 ✕ 10−13 kg.m²/s²
Now, we know that unit of speed is in m/s.
Thus, in units, momentum = MeV/c = (kg.m²/s²)/(m/s)
Simplifying this gives; Kg.m/s
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A garden hose with an inside diameter of 0.75 inches is used to fill a round swimming pool 3.0 m in diameter. How many hours will it take to fill the pool to a depth of 1.0 m if water flows from the hose at a speed of 0.30 m/s? Enter your answer without units.
Answer:
23
Explanation:
First, we need to convert the hose diameter from inches to meters.
0.75 in × (2.54 cm / in) × (1 m / 100 cm) = 0.0191 m
Calculate the flow rate given the velocity and hose diameter:
Q = vA
Q = v (¼ π d²)
Q = (0.30 m/s) (¼ π (0.0191 m)²)
Q = 8.55×10⁻⁵ m³/s
Find the volume of the pool:
V = π r² h
V = π (1.5 m)² (1.0 m)
V = 7.07 m³
Find the time:
t = V / Q
t = (7.07 m³) / (8.55×10⁻⁵ m³/s)
t = 82700 s
t = 23 hr
Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the other? Treat the ships as particles and assume each has a mass of 39,000 metric tons. (Give the magnitude of your answer in m/s2.)
Answer:
Acceleration of the ship, [tex]a=2.14\times 10^{-7}\ m/s^2[/tex]
Explanation:
It is given that,
Mass of both ships, [tex]m=39000\ metric\ tons=39\times 10^6\ kg[/tex]
Distance between two ships, d = 110 m
The gravitational force between two ships is given by :
[tex]F=G\dfrac{m^2}{d^2}[/tex]
[tex]F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}[/tex]
F = 8.38 N
Let a is the acceleration. Now, using second law of motion as :
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{8.38\ N}{39\times 10^6\ kg}[/tex]
[tex]a=2.14\times 10^{-7}\ m/s^2[/tex]
So, the acceleration of either ship due to the gravitational attraction of the other is [tex]2.14\times 10^{-7}\ m/s^2[/tex]. Hence, this is the required solution.
The gravitational attraction between the two ships can be determined by using Newton's law of gravitation. Afterwards, the acceleration of one ship due to this force is found by applying Newton's second law of motion.
Explanation:In this scenario, we're dealing with the concept of gravitational attraction between two massive bodies. We can find the gravitational force between the ships using Newton's law of gravitation:
F = G * (m1 * m2) / r²,
where F is the gravitational force, G is the gravitational constant (6.674 x 10^-11 N(m²/kg²)), m1 and m2 are the two masses, and r is the distance between them.
So, let's plug in the values. Each ship mass (m1 = m2) is 39,000 metric tons, equivalent to 3.9 x 10^7 kg. The distance between them (r) is 110 m, which we need to square. Calculating for F, we get an incredibly small value, which evidences the weak nature of gravitational force.
To find the acceleration of either ship (let's say m1), due to the gravitational force of the other, we will apply Newton's second law (F = m*a). Here, F is the gravitational force we found, m is the mass of the ship, and a is the acceleration we're looking for:
a = F / m,
Plugging the values found into this equation will result in the desired acceleration in m/s².
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The rate constant for this first‑order reaction is 0.150 s−1 at 400 ∘C. A⟶products How long, in seconds, would it take for the concentration of A to decrease from 0.860 M to 0.260 M?
Answer : The time taken for the concentration will be, 7.98 seconds
Explanation :
First order reaction : A reaction is said to be of first order if the rate is depend on the concentration of the reactants, that means the rate depends linearly on one reactant concentration.
Expression for rate law for first order kinetics is given by :
[tex]k=\frac{2.303}{t}\log\frac{[A]_o}{[A]}[/tex]
where,
k = rate constant = [tex]0.150s^{-1}[/tex]
t = time taken for the process = ?
[tex][A]_o[/tex] = initial concentration = 0.860 M
[tex][A][/tex] = concentration after time 't' = 0.260 M
Now put all the given values in above equation, we get:
[tex]0.150s^{-1}=\frac{2.303}{t}\log\frac{0.860}{0.260}[/tex]
[tex]t=7.98s[/tex]
Therefore, the time taken for the concentration will be, 7.98 seconds
To determine the time for the concentration of substance A to decrease in a first-order reaction, the formula t = (1/k) * ln([A]0/[A]t) is used with the known rate constant and initial and final concentrations.
Explanation:The question asks how long it would take for the concentration of substance A to decrease from 0.860 M to 0.260 M given a first-order reaction with a rate constant of 0.150 s−¹ at 400 °C.
For a first-order reaction, the time (t) it takes for the concentration of a reactant to change can be found using the formula:
t = (1/k) * ln([A]0/[A]t)
Where:
t = Timek = Rate constant[A]0 = Initial concentration of A[A]t = Final concentration of AIn this case, we can substitute the given values into the equation:
t = (1/0.150 s−¹) * ln(0.860 M / 0.260 M) = (1/0.150 s−¹) * ln(3.3077)
The time can be calculated by completing the computation for ln(3.3077) and dividing by the rate constant.
A silicon wafer is used to attenuate the intensity from a laser emitting at a wavelength of 0.8 µm. If the laser output power is 100 mW what is the wafer thickness required to attenuate the power to 1 mW?
Answer:
[tex]t =4.605 *10^{-3}[/tex]
Explanation:
given data:
wavelength of emission =[tex]0.8 \mu m[/tex]
output power = 100 mW
We can deduce value of obsorption coefficient from the graph obsorption coefficient vs wavelength
for wavelength [tex]0.8 \mu m[/tex] the obsorption coefficient value is 10^{3}
intensity can be expressed as a function of thickness as following:
[tex]I(t) = I_{O} *e^{-\lambda *t}[/tex]
putting all value to get thickness
[tex]1*10^{-3} =100*10^{-3}e^{-10^{3}*t}[/tex]
[tex]0.01 = e^{10^{3}t}[/tex]
[tex]t =4.605 *10^{-3}[/tex]
Assume that you can heat water with perfect insulation (all the heat from combustion of ethanol is transferred to water). What is the volume of ethanol required to heat 200 mL of water by 10 °C? The density of ethanol is 0.78 g/mL, the specific heat capacity of water is 4.184 J∙g−1∙°C−1and the heat of combustion of ethanol is -1368 kJ/mol. Show your work.
Answer:
volumme =0.36 ml
Explanation:
total heat required can be obtained by using following formula
[tex]q= mC \Delta T[/tex].......(1)
where,
m - mass of water,
C - specific heat capacity of water and = 4.184 j g^{-1} °C
[tex]\Delta T[/tex] - total change in temperature. = 10°C
The density of water is 1 g/cc. hence, 200 mL of water is equal to 200 g
putting all value in the above equation (1)
q = 200*4.184* 10 ° = 8368 J.
Therefore total number of moles of ethanol required to supply 8368 J of heat is
[tex]\frac {8368}{1368000} = 0.006117 moles.[/tex]
The molar mass of ethanol is 46 g/mol.
The mass of ethanol required is 46* 0.006117 = 0.28138 g
The density of ethanol is 0.78 g/ml.
The volume of ethanol required is
[tex]\frac {0.28138}{0.78} = 0.36 ml[/tex]
To heat 200 mL of water by 10 °C, you would need approximately 0.00785 mL of ethanol. This is calculated by converting the volume of water to its mass, calculating the heat energy using the specific heat capacity of water, and then converting that energy to the amount of ethanol required using the heat of combustion of ethanol.
Explanation:To calculate the amount of ethanol required to heat 200 mL of water by 10 °C, we need to use the formula: Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, we need to convert the volume of water to its mass. Since the density of water is 1 g/mL, the mass of 200 mL of water is 200 g.
Next, we can calculate the heat energy using the formula: Q = mcΔT. The specific heat capacity of water is 4.184 J/g °C, the mass of water is 200 g, and the change in temperature is 10 °C. Plugging in these values, we get: Q = (200 g)(4.184 J/g °C)(10 °C) = 8376 J.
Finally, we can convert this heat energy to the amount of ethanol required using the heat of combustion of ethanol. The heat of combustion of ethanol is -1368 kJ/mol. To convert from J to kJ, we divide the heat energy by 1000 to get: 8376 J / 1000 = 8.376 kJ.
Now we can calculate the volume of ethanol. Since the density of ethanol is 0.78 g/mL, we can use the formula: V = m/ρ, where V is the volume, m is the mass, and ρ is the density. The mass of ethanol can be calculated by rearranging the formula: m = Q/ΔH, where Q is the heat energy and ΔH is the heat of combustion. Plugging in the values, we get: m = 8.376 kJ / -1368 kJ/mol = -0.00611 mol.
Now we can calculate the volume of ethanol using the formula: V = m/ρ. The mass of ethanol is -0.00611 mol and the density of ethanol is 0.78 g/mL. Plugging in these values, we get: V = -0.00611 mol / (0.78 g/mL) = -0.00785 mL.
Since volume cannot be negative, the volume of ethanol required to heat 200 mL of water by 10 °C is 0.00785 mL.
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A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratio of their speeds, vproton/valpha is:
Explanation:
Charge on proton, q₁ = e
Charge on alpha particles, q₂ = 2e
The magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
Here, [tex]\theta=90=sin(90) = 1[/tex]
For proton, [tex]F_p=ev_pB[/tex]..........(1)
For alpha particle, [tex]F_a=2ev_aB[/tex]..........(2)
Since, a proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle. So,
[tex]ev_pB=2ev_aB[/tex]
[tex]\dfrac{v_p}{v_a}=\dfrac{2}{1}[/tex]
So, the ratio of the speed of proton to the alpha particle is 2 : 1 .Hence, this is the required solution.
If a proton and an alpha particle experience the same force in a magnetic field, the proton must be traveling at twice the speed of the alpha particle. This is because the force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field.
Explanation:The force exerted by a magnetic field on a moving charge depends on the charge of the particle, the speed of the particle, and the strength of the magnetic field. Given that a proton (charge e) and alpha particle (charge 2e) experience the same force in the same magnetic field, we can create an equation to solve for their speed ratio.
The force on a particle due to a magnetic field is given by F = qvB where q is the charge, v is the speed, and B is the magnetic field. Since the force on the proton and alpha particle are the same, we can set their force equations equal to each other.
This means that e * v_proton * B = 2e * v_alpha * B. Simplifying, the ratio v_proton/v_alpha = 2.
Therefore, the proton is moving twice as fast as the alpha particle.
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A police car on the side of the road (at rest) uses a radar gun to catch speeders as they approach it. The frequency the radar gun emits is 8 x 109 Hz and the speed limit is 65 mi/hr. What is the difference in frequency of the emitted and returned radar wave
Final answer:
Using the Doppler Effect formulae for electromagnetic waves, the difference in frequency (Dopler shift) experienced by radar signals upon hitting and returning from a moving vehicle allows a radar gun to calculate the vehicle's speed. The principle involves the use of shift in frequency and the speed of light to measure the speed at which the vehicle is moving.
Explanation:
The question involves the Doppler Effect in physics, specifically its application in a radar speed trap. To find the frequency shift, we must use the Doppler Effect formulae for electromagnetic waves. However, it seems there might have been a mix-up with the frequencies provided in several example problems. Since those seem to be examples rather than the actual frequencies we are working with, let's focus on finding the principle behind calculating the speed of the vehicle based on a known frequency shift of a radar emission and its return signal.
The frequency shift (Δf) in the Doppler Effect for electromagnetic waves such as radar can be calculated by the formula: Δf = (2 * f * v) / c, where f is the original frequency emitted by the radar gun, v is the speed of the vehicle, and c is the speed of light. The factor of 2 is because the radar signal experiences a frequency shift once when it hits the moving vehicle and another shift when the echo returns. The radar gun's internal processors calculate the difference in frequency (the Dopler shift) to find the speed of the vehicle. For accurate measurement, the radar unit must be able to discern even small frequency shifts to effectively differentiate speeds with fine resolution.
A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other movements. If the linear velocity of the ball relative to the elbow joint is 17.117.1 m/s at a distance of 0.4700.470 m from the joint and the moment of inertia of the forearm is 0.5500.550 kg·m2, what is the rotational kinetic energy of the forearm?
Answer:
364.4 J
Explanation:
I = Moment of inertia of the forearm = 0.550 kgm²
v = linear velocity of the ball relative to elbow joint = 17.1 m/s
r = distance from the joint = 0.470 m
w = angular velocity
Using the equation
v = r w
17.1 = (0.470) w
w = 36.4 rad/s
Rotational kinetic energy of the forearm is given as
RKE = (0.5) I w²
RKE = (0.5) (0.550) (36.4)²
RKE = 364.4 J
Answer:
363.96 J
Explanation:
v = 17.1 m/s
r = 0.47 m
I = 0.55 kgm^2
Let ω be the angular velocity
ω = v / r = 17.1 / 0.47 = 36.38 rad/s
The kinetic energy of rotation is
K = 1/2 I ω^2 = 0.5 x 0.55 x 36.38 x 36.38 = 363.96 J
9.96 kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the container. The container is now heated until the pressure is 600 kPa. Determine the temperature and total enthalpy when the heating is completed. Use data from the steam tables.
By using steam tables or similar thermodynamic property charts, you can determine the initial and final temperature and enthalpy of R-134a under specified pressure conditions in a rigid container. As the pressure changes with heating, you can find the corresponding changes in temperature and enthalpy.
Explanation:The subject matter of this question falls under the domain of thermodynamics in physics. It involves an element of heating, changes in pressure and the resultant alterations in temperature. Given the specifics, we can't directly provide a numerical solution in this setting as it requires the usage of refprop or other similar thermodynamic property database or the proper specific enthalpy and temperature charts for R-134a. However, conceptually, it can be handled via using steam tables (or more specifically refrigerant tables) for the substance R-134a. Initially, you lookup the properties using the given pressure and known mass and volume to calculate the initial temperature and enthalpy. Similarly, when the container is heated and the pressure raises to 600 kPa, you can again refer to the steam tables with this new pressure (and the volume which remains unchanged as the container is rigid) to determine the new temperature and enthalpy.
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A satellite orbits at a distance from the Earth's center of about 2.60 Earth radii and takes 5.89 hours to go around once. What distance (in meters) does the satellite travel in one day?
Answer:
424088766.068 m
Explanation:
Radius of the circular orbit that the satellite is 2.6 Earth radii (r) = 2.6 R
R = Radius of earth = 6371000 m (mean radius)
In order to find the distance that the satellite travels in 5.89 hours to complete one complete revolution is the circumference of the circular orbit
Circumference of a circle = 2×π×r
⇒Distance travelled in 5.89 hours = 2×π×2.6 R
⇒Distance travelled in 5.89 hours = 2×π×2.6×6371000
⇒Distance travelled in 5.89 hours = 104078451.3393m
Distance travelled in 1 hour = 104078451.3393/5.89 = 17670365.252 m
∴ Distance travelled in 24 hours = 17670365.252×24 = 424088766.068 m
Final answer:
The satellite orbits Earth at approximately 2.60 Earth radii and travels a total distance of about 4.23×10⁸ meters in one day.
Explanation:
The question pertains to calculating the total distance traveled by a satellite orbiting Earth at a given distance over the duration of one day. To solve this, we need to determine the circumference of the satellite's circular orbit and then calculate how many orbits it completes in a day.
To determine the satellite's orbit circumference, we use the formula for the circumference of a circle, C = 2πr, where r represents the orbit radius. Given that the satellite orbits at a distance of about 2.60 Earth radii, we can express this distance in meters by multiplying the Earth's radius, 6.37×10⁶ meters, by 2.60, yielding a radius of approximately 1.6562×10⁷ meters. Therefore, the satellite's orbit circumference is approximately 1.04×10⁸ meters.
Since the satellite completes one orbit in 5.89 hours, we can now calculate how many orbits it completes in a day (24 hours) by dividing 24 by 5.89, which is approximately 4.07 orbits per day. The total distance traveled in one day is the circumference multiplied by the number of orbits, resulting in a distance of approximately 4.23×10⁸ meters per day.
The magnetic flux that passes through one turn of a 18-turn coil of wire changes to 2.67 from 8.19 Wb in a time of 0.0386 s. The average induced current in the coil is 275 A. What is the resistance of the wire?
A motorcycle rider executes a loop-to-loop stunt in a loop with a diameter of 8.4 m At the top of the loop, the rider has a speed of 7.5 m/s. The combined mass of the rider and motorcycle is 175 kg. Determine the normal force between the motorcycle and track at the top of the loop
Answer:
The normal force between the motorcycle and track at the top of the loop = 2343.75 N
Explanation:
At the top of the loop the track is directed tangential to the motion, the normal to the tangent is along radius of circle. that is here we need to find centripetal force.
We have expression for centripetal force
[tex]F=\frac{mv^2}{r}[/tex]
Here mass, m = 175 kg
Velocity, v = 7.5 m/s
Diameter, d = 8.4 m
Radius, r = 0.5d = 0.5 x 8.4 = 4.2 m
Substituting
[tex]F=\frac{mv^2}{r}=\frac{175\times 7.5^2}{4.2}=2343.75N[/tex]
The normal force between the motorcycle and track at the top of the loop = 2343.75 N
The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)
Answer:
Charge, [tex]Q=1.56\times 10^{-8}\ C[/tex]
Explanation:
It is given that,
Electric field strength, E = 180000 N/C
Distance from a small object, r = 2.8 cm = 0.028 m
Electric field at a point is given by :
[tex]E=\dfrac{kQ}{r^2}[/tex]
Q is the charge on an object
[tex]Q=\dfrac{Er^2}{k}[/tex]
[tex]Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}[/tex]
[tex]Q=1.56\times 10^{-8}\ C[/tex]
So, the charge on the object is [tex]1.56\times 10^{-8}\ C[/tex]. Hence, this is the required solution.
The object's charge q is calculated to be approximately 1.57 × 10⁻⁵ C.
To find the object's charge q, we use the formula for the electric field due to a point charge:
E = kQ/r²
Here, E is the electric field, k is the Coulomb's constant (8.99 × 10⁹ N·m²/C²), Q is the charge, and r is the distance from the charge.
We are provided with:
Electric field, E = 180,000 N/C
Distance, r = 2.8 cm = 0.028 m
Rearranging the formula to solve for Q gives:
Q = Er²/k
Substituting in the given values:
Q = (180,000 N/C) × (0.028 m)² / (8.99 × 10⁹ N·m²/C²)
Q = (180,000) × (0.000784) / (8.99 × 10⁹)
Q ≈ 1.57 × 10⁻⁵ C
The object's charge q is approximately 1.57 × 10⁻⁵ C.
An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the air-water interface at a 45∘ angle relative to the vertical. You see a bright spot where the beam hits the bottom of the aquarium. The index of refraction of water is 1.33. How much water (in terms of height) should you add to the tank so the bright spot on the bottom moves 5.0 cm?
Final answer:
To find out how much water should be added to move the bright spot by 5.0 cm, apply Snell's Law to find the angle of refraction and use trigonometry to determine the change in horizontal distance of the spot due to the change in water depth.
Explanation:
The student is asking about the behavior of light as it passes through different media, specifically how the depth of the water in an aquarium would need to change to move the bright spot created by a laser. The scenario of shining a laser at an angle into water involves Snell's Law, which describes how light bends at the interface between two media with different indices of refraction. To calculate the necessary increase in water height to move the bright spot by 5.0 cm, we need to apply the principles of refraction and trigonometry.
First, we determine the initial angle of refraction using Snell's Law:
nairsin(θair) = nwatersin(θwater)1.00sin(45°) = 1.33sin(θwater)
Upon finding the angle of refraction θwater, we can use trigonometric functions to calculate the initial horizontal distance (Xinitial) from the point of incidence to the bright spot. To find the new distance (Xfinal) after adding water, we repeat the process with the new depth. The difference in the horizontal distances gives the amount the bright spot moved, 5.0 cm. Adjusting the depth to match this movement provides the answer to the student's question.
Density is a physical property that relates the mass of a substance to its volume. A. Calculate the density, in g/mL , of a liquid that has a mass of 0.155 g and a volume of 0.000235 L. B. Calculate the volume in milliliters of a 4.71-g sample of a solid with a density of 3.63 g/mL. C. Calculate the mass of a 0.293-mL sample of a liquid with a density of 0.930 g/mL.
Answer:
A) 0.660 g/ml
B) 1.297 ml
C) 0.272 g
Explanation:
Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes. This is where density [tex]D[/tex] appears as a physical characteristic property of matter that establishes a relationship between the mass [tex]m[/tex] of a body or substance and the volume [tex]V[/tex] it occupies:
[tex]D=\frac{m}{V}[/tex] (1)
Knowing this, let's begin with the answers:
Answer A:Here the mass is [tex]m=0.155g[/tex] and th volume [tex]V=0.000235L=0.235mL[/tex]
Solving (1) with these values:
[tex]D=\frac{0.155g}{0.235mL}[/tex] (2)
[tex]D=0.660g/mL[/tex] (3)
Answer B:In this case the mass of a sample is [tex]m=4.71g[/tex] and its density is [tex]D=3.63g/mL[/tex].
Isolating [tex]V[/tex] from (1):
[tex]V=\frac{m}{D}[/tex] (4)
[tex]V=\frac{4.71g}{3.63g/mL}[/tex] (5)
[tex]V=1.297mL[/tex] (5)
Answer C:In this case the volume of a sample is [tex]V=0.293mL[/tex] and its density is [tex]D=0.930g/mL[/tex].
Isolating [tex]m[/tex] from (1):
[tex]m=D.V[/tex] (6)
[tex]m=(0.930g/mL)(0.293mL)[/tex] (7)
[tex]m=0.272g[/tex] (8)
To compute density, volume or mass, utilize the density formula (Density = Mass / Volume). For problem A, the liquid's density is 0.660 g/mL. For B, the volume of the sample is 1.30 mL, and for C, the mass of the sample is 0.272 g.
Explanation:The formula to calculate density is Density = Mass / Volume. A. To calculate the density in g/mL of a liquid with a mass of 0.155 g and a volume of 0.000235 L, we first need to convert the volume from liters to milliliters: 0.000235 L is equal to 0.235 mL. Then, divide the mass by the volume to get the density: Density = 0.155 g / 0.235 mL = 0.660 g/mL.
B. To calculate the volume in milliliters of a 4.71-g sample of a solid with a density of 3.63 g/mL, divide the mass by the density: Volume = 4.71 g / 3.63 g/mL = 1.30 mL.
C. To calculate the mass of a 0.293-mL sample of a liquid with a density of 0.930 g/mL, multiply the volume by the density: Mass = 0.293 mL * 0.930 g/mL = 0.272 g.
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If a 0.9 kg ball is dropped from a height of 7 m, what is its kinetic energy when it hits the ground?
Answer:
Kinetic energy, [tex]E_k=61.74\ J[/tex]
Explanation:
It is given that,
Mass of ball, m = 0.9 kg
It is dropped form a height of 7 m, h = 7 m
When the ball is at certain height it will have only potential energy. As it hits the ground, its potential energy gets converted to kinetic energy as per the law of conservation of energy.
So, [tex]E_k=PE=mgh[/tex]
[tex]E_k=0.9\ kg\times 9.8\ m/s^2\times 7\ m[/tex]
[tex]E_k=61.74\ J[/tex]
So, the kinetic energy when it hits the ground is 61.74 Joules.
The kinetic energy of a 0.9 kg ball dropped from a height of 7 m is calculated using the conservation of mechanical energy, resulting in 61.74 Joules when it hits the ground.
The kinetic energy of a 0.9 kg ball when it hits the ground after being dropped from a height of 7 m. The principle used to determine this is the conservation of mechanical energy, which states that the sum of the potential and kinetic energies in a system remains constant if only conservative forces are doing work.
When the ball is at the height of 7 m, it posses gravitational potential energy (GPE) which can be calculated using the formula GPE = mgh, where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s²), and h is the height from which the ball is dropped.
Since the ball is dropped from rest, its initial kinetic energy is 0. When the ball reaches the ground, its potential energy is converted into kinetic energy.
Therefore, the kinetic energy (KE) of the ball when it hits the ground can be calculated using the potential energy formula, assuming no energy is lost:
KE = mgh = 0.9 kg x 9.8 m/s² x 7 m = 61.74 Joules.
This means that the ball will have a kinetic energy of 61.74 Joules when it impacts the ground.