Answer:
5.95 nm to 33.6 nm
Explanation:
Energy of a single photon can be written as:
[tex]E = \frac{hc}{\lambda}[/tex]
where, h is the Planck's constant, c is the speed of light and λ is the wavelength of light.
Consider the lowest energy of an electron that can break a DNA = 3.45 eV
1 eV = 1.6 ×10⁻¹⁹ J
⇒3.45 eV = 5.52×10⁻¹⁹ J
[tex]E = \frac{hc}{\lambda}\\ \Rightarrow \lambda = \frac{hc}{E}= \frac {6.63\times 10^{-34} m^2kg/s \times 3\times 10^8 m/s}{5.52 \times 10^{-19} J} = 3.60\times 10^{-7} m = 360 nm[/tex]
Consider the highest energy of an electron that can break a DNA = 20.9 eV
1 eV = 1.6 ×10⁻¹⁹ J
⇒20.9 eV = 33.4×10⁻¹⁹ J
[tex]E = \frac{hc}{\lambda}\\ \Rightarrow \lambda = \frac{hc}{E}= \frac {6.63\times 10^{-34} m^2kg/s \times 3\times 10^8 m/s}{33.4 \times 10^{-19} J} = 0.595\times 10^{-7} m = 59.5 nm[/tex]
The wavelength range of light that can cause DNA breaks is approximately 59.4 nm to 360 nm. This corresponds to ultraviolet light and part of the visible spectrum.
To find the range of light wavelengths that can cause DNA breaks, we need to convert the given energy range of 3.45 eV to 20.9 eV into wavelengths.
The energy of a photon (E) is related to its wavelength (λ) by the equation:
E = hc/λ
where h is Planck's constant (6.626 × 10⁻³⁴ J·s), c is the speed of light .(3.00 × 10⁸m/s), and λ is the wavelength in meters.
First, convert the energy from electron volts (eV) to Joules (J):
1 eV = 1.602 × 10⁻¹⁹ JSo, the energy range in Joules is:
3.45 eV = 3.45 × 1.602 × 10⁻¹⁹J = 5.53 × 10⁻¹⁹ J20.9 eV = 20.9 × 1.602 × 10⁻¹⁹J = 3.35 × 10⁻¹⁸ JNext, use the energy-wavelength relation to find the wavelengths:
For 3.45 eV: λ = hc/E = (6.626 × 10⁻³⁴J·s × 3.00 × 10⁸ m/s) / 5.53 × 10⁻¹⁹J ≈ 3.60 × 10⁻⁷ m ≈ 360 nmFor 20.9 eV: λ = hc/E = (6.626 × 10⁻³⁴ J·s × 3.00 × 10⁸ m/s) / 3.35 × 10⁻¹⁸ J ≈ 5.94 × 10⁻⁸ m ≈ 59.4 nmThus, the range of wavelengths that can cause DNA breaks is approximately 59.4 nm to 360 nm, corresponding to the ultraviolet (UV) and part of the visible spectrum.
Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of heat? A) 50.0 g Al, CAl = 0.903 J/g°C B) 50.0 g Cu, CCu = 0.385 J/g°C C) 25.0 g granite, Cgranite = 0.79 J/g°C D) 25.0 g Au, CAu = 0.128 J/g°C E) 25.0 g Ag, CAg = 0.235 J/g°C
Answer:
A) 50.0 g Al
Explanation:
We can calculate the temperature change of each substance by using the equation:
[tex]\Delta T=\frac{Q}{mC_s}[/tex]
where
Q = 200.0 J is the heat provided to the substance
m is the mass of the substance
[tex]C_s[/tex] is the specific heat of the substance
Let's apply the formula for each substance:
A) m = 50.0 g, Cs = 0.903 J/g°C
[tex]\Delta T=\frac{200}{(50)(0.903)}=4.4^{\circ}C[/tex]
B) m = 50.0 g, Cs = 0.385 J/g°C
[tex]\Delta T=\frac{200}{(50)(0.385)}=10.4^{\circ}C[/tex]
C) m = 25.0 g, Cs = 0.79 J/g°C
[tex]\Delta T=\frac{200}{(25)(0.79)}=10.1^{\circ}C[/tex]
D) m = 25.0 g, Cs = 0.128 J/g°C
[tex]\Delta T=\frac{200}{(25)(0.128)}=62.5^{\circ}C[/tex]
E) m = 25.0 g, Cs = 0.235 J/g°C
[tex]\Delta T=\frac{200}{(25)(0.235)}=34.0^{\circ}C[/tex]
As we can see, substance A) (Aluminium) is the one that undergoes the smallest temperature change.
The substance that would show the smallest temperature change upon gaining 200.0 J of heat is Au (Gold), as calculated using the formula for calculating heat (Q = mcΔT) and rearranging for ΔT, then substituting the given values.
Explanation:The substance that would show the smallest temperature change upon gaining 200.0 J of heat can be determined using the formula used to calculate heat (Q), which is Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the temperature change. We want to find the smallest temperature change, so we rearrange the equation to solve for ΔT, which gives us ΔT = Q/(mc). By substituting the given values for each substance into this equation, we find that the smallest temperature change is for Au (Gold).
For Au: ΔT = 200.0J / (25.0g x 0.128 J/g°C) = 62.5°C. All other substances have a smaller temperature change when they absorb 200.0J of heat, due to their higher specific heat capacity.
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Calculate the buoyant force (in N) on a 1.0 m^3 chunk of brass submerged in a bath of mercury.
Answer:
133280 N
Explanation:
Volume, V = 1 m^3
density of mercury, d = 13.6 x 10^3 kg/m^3
Buoyant force, F = Volume immersed x density of mercury x g
F = 1 x 13.6 x 1000 x 9.8
F = 133280 N
The newest CREE led has a life expectancy of mu = 50000 hours and its life probability density function is given by: f(t) = [e^(-t/mu)]/[mu] if t greater or = 0 and f(t) = 0 if t < 0. Calculate the chance that a led will last at least tau = 100000.
Answer:
change that a lead is 0.13533
Explanation:
µ = 50000
f(t) = [e^(-t/µ )]/[µ if t ≥ 0
f(t) = 0 if t < 0
τ = 100000
to find out
the chance that a led will last
solution
we know function is f(t) = [e^(-τ/µ)]/[µ]
τ = 100000
so we can say that probability (τ ≥ 100000 ) that is
= 1 - Probability ( τ ≤ 100000 )
that is function of F so
= 1 - f ( 100000 )
that will be
= 1 - ( 1 - [e^(-τ/µ)]/[µ] )
put all value here τ = 100000 and µ = 50000
= 1 - ( 1 - [e^(-100000/50000)] )
= 1 - 1 - [e^(-100000/50000)]
= 0.13533
so that change that a lead is 0.13533
Two equally charged spheres of mass 1.00 g are placed 2.00 cm apart. When released, the initial acceleration of each sphere is 256 m/s2 . What is the magnitude of the charge on each sphere? You may assume that the only force acting on each sphere is the electric force due to the other sphere.
Answer:
[tex]q = 0.107 \mu C[/tex]
Explanation:
As we know that net force is given by
[tex]F = ma[/tex]
here we have
m = 1.00 g = 0.001 kg
also we know that acceleration is given as
[tex]a = 256 m/s^2[/tex]
now force is given as
[tex]F = 0.001(256) = 0.256 N[/tex]
now by the formula of force we know that
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
[tex]0.256 = \frac{(9\times 10^9)q^2}{(0.02)^2}[/tex]
now for solving charge we have
[tex]q = 0.107 \mu C[/tex]
A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is removed from the tank. What is the pressure of the remaining gas in the tank?
Answer:
The pressure of the remaining gas in the tank is 6.4 atm.
Explanation:
Given that,
Temperature T = 13+273=286 K
Pressure = 10.0 atm
We need to calculate the pressure of the remaining gas
Using equation of ideal gas
[tex]PV=nRT[/tex]
For a gas
[tex]P_{1}V_{1}=nRT_{1}[/tex]
Where, P = pressure
V = volume
T = temperature
Put the value in the equation
[tex]10\times V=nR\times286[/tex]....(I)
When the temperature of the gas is increased
Then,
[tex]P_{2}V_{2}=\dfrac{n}{2}RT_{2}[/tex]....(II)
Divided equation (I) by equation (II)
[tex]\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}[/tex]
[tex]\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}[/tex]
[tex]P_{2}=\dfrac{10\times368}{2\times286}[/tex]
[tex]P_{2}= 6.433\ atm[/tex]
[tex]P_{2}=6.4\ atm[/tex]
Hence, The pressure of the remaining gas in the tank is 6.4 atm.
Describe one elastic collisions and one inelastic collisions that you observed in your daily activities. In the inelastic collision explain where did the lost energy (with mathematical expressions) go, in the elastic collision explain what energy (include mathematical expressions) is conserved.
Answer:
A typical example of an elastic collision that can be observed is the collision of billiard balls, while an inelastic collision is presented in cars collisions.
Explanation:
In an inelastic collision, the energy system is lost in making the permanent deformation over car's structures due to the impact. As can be stated below, the final and initial kinetic energy are expressed:
[tex] Ei =0.5*m1.v1_{i}^2+m2.v2_{i}^2[/text]
[tex] Ef =0.5*m1.v1_{f}^2+m2.v2_{f}^2[/text]
Where the subscripts 1 and 2 relate to each car. In the final energy equation Ef, the car's final velocity will be lower than the respective initial velocities.
[tex] v1_{f}<v1_{i}[/text]
[tex] v2_{f}<v2_{i}[/text]
Take into account that car's masses still being the same after the collision, therefore the energy losses are always because of cars velocities changes:
[tex] Ef<Ei[/text]
In the elastic collision, there will be little or negligible deformations and that won't make energy losses. But this statement doesn't affirm that billiard balls velocities will be the same. In fact, could happen that one ball increases its velocities if the other ball decreases its velocity, but taking into account that the energy will always conserve.
[tex] v2_{f}>v2_{i}[/text] if [tex] v1_{f}<v1_{i}[/text]
or
[tex] v1_{f}>v1_{i}[/text] if [tex] v2_{f}<v2_{i}[/text]
Under the assumption that balls masses still being the same:
[tex] Ef=Ei[/text]
A 500-g metal wire has a length of 50 cm and is under tension of 80 N. (a) What is the speed of a transverse wave in the wire? (b) If the wire is cut in half, what will be the speed of the wave?
Explanation:
It is given that,
Mass of the metal wire, m = 500 g = 0.5 kg
Tension in the wire, T = 80 N
Length of wire, l = 50 cm = 0.5 m
(a) The speed of the transverse wave is given by :
[tex]v=\sqrt{\dfrac{T}{M}}[/tex]
M is the mass per unit length or M = m/l
[tex]v=\sqrt{\dfrac{T.l}{m}}[/tex]
[tex]v=\sqrt{\dfrac{80\ N\times 0.5\ m}{0.5\ kg}}[/tex]
v = 8.94 m/s
(b) If the wire is cut in half, so l = l/2
[tex]v=\sqrt{\dfrac{T.l}{2m}}[/tex]
[tex]v=\sqrt{\dfrac{80\ N\times 0.5\ m}{2\times 0.5\ kg}}[/tex]
v = 6.32 m/s
Hence, this is the required solution.
A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms (milliseconds)?
Answer:
The average force exerted on the bullet are of F=9000 Newtons.
Explanation:
t= 2*10⁻³ s
m= 0.03 kg
V= 600 m/s
F*t= m*V
F= (m*V)/t
F= 9000 N
The average force exerted on the bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms is 9000 Newtons (N). when a bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gunpowder.
Given:
Mass of the bullet (m) = 0.0300 kg
The final velocity of the bullet (v) = 600 m/s
Time taken to reach the final velocity (t) = 2.00 ms = 2.00 × 10⁻³ s
acceleration (a) = (change in velocity) / (time taken)
a = (v - u) / t
a = (600 - 0 ) / (2.00 × 10⁻³)
Now, we can calculate the average force using Newton's second law:
force (F) = mass (m) × acceleration (a)
F = 0.0300 × [(600 ) / (2.00 × 10⁻³)]
F = 0.0300× (3.00 × 10⁵)
F = 9000 N
Therefore, the average force exerted on the bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms is 9000 Newtons (N).
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A uniform disk turns at 5.00 rev/s around a frictionless spindle. A non-rotating rod, of the same mass as the disk and length equal to the disk’s diameter, is dropped onto the freely spinning disk. They then turn together around the spindle with their centers superposed. What is the angular frequency in of the rev/scombination?
Answer:
Final angular speed equals 3 revolutions per second
Explanation:
We shall use conservation of angular momentum principle to solve this problem since the angular momentum of the system is conserved
[tex]L_{disk}=I_{disk}\omega \\\\L_{disk}=\frac{1}{2}mr^{2}\\\therefore L_{disk}=\frac{1}{2}mr^{2}\times10rad/sec[/tex]
After the disc and the dropped rod form a single assembly we have the final angular momentum of the system as follows
[tex]L_{final}=I_{disk+rod}\times \omega_{f} \\\\I_{disk+rod}=\frac{1}{2}mr_{disc}^{2}+\frac{1}{12}mL_{rod}^{2}\\I_{disk+rod}=\frac{1}{2}mr_{disc}^{2}+\frac{1}{12}m\times (2r_{disc})^{2}\\\\I_{disk+rod}=\frac{1}{2}mr_{disc}^{2}+\frac{1}{3}mr_{disc}^{2}\\\\L_{final}=\frac{5mr_{disc}^{2}}{6}\times \omega _{f}\\\\[/tex]
Equating initial and final angular momentum we have
[tex]\frac{5mr_{disc}^{2}}{6}\times \omega _{f}=\frac{1}{2}m_{disc}\times r_{disc}^{2}\times 10\pi rad/sec[/tex]
Solving for [tex]\omega_{f}[/tex] we get
[tex]\omega_{f}=6\pi rad/sec[/tex]
Thus no of revolutions in 1 second are 6π/2π
No of revolutions are 3 revolutions per second
A duck flying horizontally due north at 12.3 m/s passes over East Lansing, where the vertical component of the Earth's magnetic field is 4.78×10-5 T (pointing down, towards the Earth). The duck has a positive charge of 7.64×10-8 C. What is the magnitude of the magnetic force acting on the duck?
Answer:
4.49 x 10^-11 newton
Explanation:
v = 12.3 m/s along north = 12.3 j m/s
B = 4.78 x 10^-5 T downwards = 4.78 x 10^-5 k T
q = 7.64 x 10^-8 C
force on a charged particle when it is moving in a uniform magnetic field is given by
F = q (v x B )
F = 7.64 x 10^-8 {(12.3 i) x (4.78 x 10^-5 k)}
F = 4.49 x 10^-11 (- k) newton
magnitude of force = 4.49 x 10^-11 newton
Calculate the electric force an electron exerts upon a proton inside a He atom if they are d=2.7⋅10^-10m apart.
Hint: Fe=k⋅qp⋅qp/d^2 where k=9⋅10^9.
A) 6.8E-8N;
B) -212.7E-9N;
C) -6.31E-9N;
D) -57.6E-10N;
Explanation:
Charge of electron in He, [tex]q_e=1.6\times 10^{-19}\ kg[/tex]
Charge of proton in He, [tex]q_p=1.6\times 10^{-19}\ kg[/tex]
Distance between them, [tex]d=2.7\times 10^{-10}\ m[/tex]
We need to find the electric force between them. It is given by :
[tex]F=k\dfrac{q_eq_p}{d^2}[/tex]
[tex]F=-9\times 10^9\times \dfrac{(1.6\times 10^{-19}\ C)^2}{(2.7\times 10^{-10}\ m)^2}[/tex]
[tex]F=-3.16\times 10^{-9}\ N[/tex]
Since, there are two protons so, the force become double i.e.
[tex]F=2\times 3.16\times 10^{-9}\ N[/tex]
[tex]F=6.32\times 10^{-9}\ N[/tex]
So, the correct option is (c). Hence, this is the required solution.
51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00Ω by placing a 1.00-kΩ voltmeter across its terminals. (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio.
Explanation:
Given that,
Terminal voltage = 3.200 V
Internal resistance [tex]r= 5.00\ \Omega[/tex]
(a). We need to calculate the current
Using rule of loop
[tex]E-IR-Ir=0[/tex]
[tex]I=\dfrac{E}{R+r}[/tex]
Where, E = emf
R = resistance
r = internal resistance
Put the value into the formula
[tex]I=\dfrac{3.200}{1.00\times10^{3}+5.00}[/tex]
[tex]I=3.184\times10^{-3}\ A[/tex]
(b). We need to calculate the terminal voltage
Using formula of terminal voltage
[tex]V=E-Ir[/tex]
Where, V = terminal voltage
I = current
r = internal resistance
Put the value into the formula
[tex]V=3.200-3.184\times10^{-3}\times5.00[/tex]
[tex]V=3.18\ V[/tex]
(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf
[tex]\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }[/tex]
[tex]\dfrac{Terminal\ voltage}{emf}= \dfrac{159}{160}[/tex]
Hence, This is the required solution.
The current flowing in the circuit is 3.195 milliamps. The terminal voltage is calculated to be 2.984V. The ratio of the terminal voltage to the emf is 0.9325 which shows that the terminal voltage is 93.25% of the emf due to the voltage drop resulted from the internal resistance of the battery.
Explanation:The question is about the terminal voltage and internal resistance of a battery. To answer this, first we need to understand that terminal voltage is the potential difference (voltage) between two terminals of a battery, and it's slightly less than the emf due to internal resistance of the battery.
(a) The current (I) in the circuit is found using Ohm’s law: I = emf / (R_load + r) = 3.200V / (1.00kΩ + 5.00Ω) = 3.195 milliamps.
(b) The terminal voltage (V) is calculated by: V = emf - Ir = 3.200V - (3.195mA * 5.00Ω) = 2.984V.
(c) The ratio of the measured terminal voltage to the emf is V / emf = 2.984V / 3.200V = 0.9325. This shows that the terminal voltage is 93.25% of the emf, which accounts for the voltage drop due to the internal resistance of the battery.
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A pendulum clock with a brass suspension system is calibrated so that its period is 1 s at 20 degree C. If the temperature increases to 43 degree C, by how much does the period change? Answer in units of s.
Answer:
0.207 ms
Explanation:
First of all we need to find the length of the pendulum at 20 degrees. We know that the period is 1 s, and the formula for the period is
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
where L is the length of the pendulum and g is the gravitational acceleration. Solving the equation for L and using T = 1 s and g = 9.8 m/s^2, we find
[tex]L=g(\frac{T}{2\pi})^2=(9.8) (\frac{1}{2\pi})^2=0.248237 m[/tex]
Now we can find the new length of the pendulum at 43 degrees; the coefficient of thermal expansion of brass is
[tex]\alpha =18\cdot 10^{-6} 1/^{\circ}C[/tex]
And the new length of the pendulum is given by
[tex]L' = L (1+\alpha \Delta T)[/tex]
where in this case
[tex]\Delta T = 43-20 = 23^{\circ}[/tex] is the change in temperature
Substituting,
[tex]L'=(0.248237)(1+(18\cdot 10^{-6})(23))=0.248340 m[/tex]
So we can now calculate the new period of the pendulum:
[tex]T'=2\pi \sqrt{\frac{L'}{g}}=2\pi \sqrt{\frac{0.248340}{9.8}}=1.000208 s[/tex]
So the change in the period is
[tex]T'-T=1.000208 - 1.000000 = 0.000207 s = 0.207 ms[/tex]
Final answer:
The period of a pendulum clock with a brass suspension system will change by approximately 0.000414 seconds when the temperature increases from 20°C to 43°C.
Explanation:
A pendulum clock with a brass suspension system is calibrated to have a period of 1 second at 20 degrees Celsius. When the temperature increases to 43 degrees Celsius, the period of the pendulum will change. To calculate the change in period, you can use the formula T2 = T1 * (1 + α * (T2 - T1)), where T2 is the final temperature, T1 is the initial temperature, and α is the coefficient of linear expansion for the brass material. In this case, α is 18 × 10^-6 °C^-1.
Using the formula, we can plug in the values: T1 = 20°C, T2 = 43°C, and α = 18 × 10^-6 °C^-1. Subtracting T1 from T2 gives us 23, and multiplying this by α gives us 0.000414. Finally, multiplying this by the initial period of 1 second gives us a change in period of approximately 0.000414 seconds.
This exercise involves the formula for the area of a circular sector. The area of a sector of a circle with a central angle of 2π/11 rad is 25 m2. Find the radius of the circle. (Round your answer to one decimal place.)
Final answer:
The radius of the circle is approximately 1.392 m (rounded to one decimal place).
Explanation:
To find the radius of the circle, we need to use the formula for the area of a sector. The area of a sector is given by the formula A = (θ/2π) × πr², where θ is the central angle in radians and r is the radius. In this case, we are given that the central angle is 2π/11 radians and the area is 25 m². We can set up the equation as 25 = (2π/11) × πr² and solve for r.
Solution:
25 = (2π/11) × πr²
25 = (2π²/11) × r²
r² = 11/2π
r ≈ √(11/2π)
r ≈ 1.392 m (rounded to one decimal place)
A skier moving at 5.23 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping?
im not that smart but maybe 2 x the mass of the wieght will give u the answer
The allowed energies of a quantum system are 0.0 eV, 5.0 eV , and 8.5 eV .
What wavelengths appear in the system's emission spectrum?
Express your answers in nanometers in ascending order separated by commas.
The wavelengths in the system's emission spectrum, in ascending order, are [tex]\(146 \, \text{nm}\) and \(249 \, \text{nm}\).[/tex]
To find the wavelengths associated with the allowed energies of the quantum system, we can use the formula for the energy of a photon:
[tex]\[ E = \frac{hc}{\lambda} \][/tex]
where:
-[tex]\( E \)[/tex] is the energy of the photon,
- [tex]\( h \)[/tex] is Planck's constant[tex](\( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)),[/tex]
- [tex]\( c \)[/tex] is the speed of light [tex](\( 3.00 \times 10^8 \, \text{m/s} \)),[/tex]
- [tex]\( \lambda \)[/tex] is the wavelength of the photon.
Given the energies [tex]\(0.0 \, \text{eV}\), \(5.0 \, \text{eV}\), and \(8.5 \, \text{eV}\)[/tex], we need to convert these energies to joules, since the units in the formula for energy are in joules.
1.[tex]\(0.0 \, \text{eV}\) corresponds to \(0.0 \, \text{J}\),[/tex]
2. [tex]\(5.0 \, \text{eV}\) corresponds to \(5.0 \times 1.602 \times 10^{-19} \, \text{J}\),[/tex]
3. [tex]\(8.5 \, \text{eV}\) corresponds to \(8.5 \times 1.602 \times 10^{-19} \, \text{J}\).[/tex]
Now, we can use these energies to calculate the wavelengths of the photons:
1. For [tex]\(0.0 \, \text{J}\):[/tex]
[tex]\[ \lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3.00 \times 10^8 \, \text{m/s})}{0.0 \, \text{J}}} \][/tex]
2. For [tex]\(5.0 \times 1.602 \times 10^{-19} \, \text{J}\)[/tex]:
[tex]\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3.00 \times 10^8 \, \text{m/s})}{5.0 \times 1.602 \times 10^{-19} \, \text{J}} \]\[ \lambda \approx \frac{1.995 \times 10^{-25}}{5.0 \times 1.602} \, \text{m} \]\[ \lambda \approx 2.49 \times 10^{-8} \, \text{m} \][/tex]
3. For [tex]\(8.5 \times 1.602 \times 10^{-19} \, \text{J}\):[/tex]
[tex]\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3.00 \times 10^8 \, \text{m/s})}{8.5 \times 1.602 \times 10^{-19} \, \text{J}} \]\[ \lambda \approx \frac{1.995 \times 10^{-25}}{8.5 \times 1.602} \, \text{m} \]\[ \lambda \approx 1.46 \times 10^{-8} \, \text{m} \][/tex]
Now, let's convert these wavelengths to nanometers:
[tex]\( 2.49 \times 10^{-8} \, \text{m} = 249 \, \text{nm} \),[/tex]
[tex]. \( 1.46 \times 10^{-8} \, \text{m} = 146 \, \text{nm} \)[/tex]
So, the wavelengths in the system's emission spectrum, in ascending order, are [tex]\(146 \, \text{nm}\) and \(249 \, \text{nm}\).[/tex]
A circular coil of wire of 200 turns and diameter 6 cm carries a current of 7 A. It is placed in a magnetic field of 0.90 T with the plane of the coil making an angle of 30° with the magnetic field. What is the torque on the coil?
Answer:
3.08 Nm
Explanation:
N = 200, diameter = 6 cm, radius = 3 cm, I = 7 A, B = 0.90 T, Angle = 30 degree
The angle made with the normal of the coil, theta = 90 - 30 = 60 degree
Torque = N I A B Sin Theta
Torque = 200 x 7 x 3.14 x 0.03 x 0.03 x 0.90 x Sin 60
Torque = 3.08 Nm
jason hits a baseball off a tee toward right field. the ball has a horizontal velocity of 10 m/s and lands 5 meters from the tee. what is the height of the tee? show your work, including formula(s) and units.
Answer:
The height is 1,225 meters
Explanation:
DistanceX= speedX × time ⇒ time= (5 meters) ÷ (10 meters/second) = 0,5 seconds
DistanceY= high= (1/2) × g × (time^2) = (1/2) × 9,8 (meters/(second^2)) × 0,25 (second^2) = 1,225 meters
If we measure the temperature of a blackbody to be about 300 K (a typical air temperature in Florida), at what wavelength would this blackbody's intensity have its maximum, and where in the electromagnetic spectrum is this wavelength?
Answer:
9.66 x 10^-6 m
Explanation:
Use the Wein's displacement law
[tex]\lambda _{m}\times T = b[/tex]
Where, b is the Wein's constant
b = 2.898 x 10^-3 meter-kelvin
So, λm x 300 = 2.898 x 10^-3
λm = 9.66 x 10^-6 m
From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s. Determine the uncertainty for the slope of the line.
Solution:
Let the slope of the best fit line be represented by '[tex]m_{best}[/tex]'
and the slope of the worst fit line be represented by '[tex]m_{worst}[/tex]'
Given that:
[tex]m_{best}[/tex] = 1.35 m/s
[tex]m_{worst}[/tex] = 1.29 m/s
Then the uncertainity in the slope of the line is given by the formula:
[tex]\Delta m = \frac{m_{best}-m_{worst}}{2}[/tex] (1)
Substituting values in eqn (1), we get
[tex]\Delta m = \frac{1.35 - 1.29}{2}[/tex] = 0.03 m/s
A 14-Ω coffee maker and a 14-Ω frying pan are connected in series across a 120-V source of voltage. A 20-Ω bread maker is also connected across the 120-V source and is in parallel with the series combination. Find the total current supplied by the source of voltage.
Answer:
The total current supplied by the source of voltage = 10.29 A
Explanation:
We have a 14-Ω coffee maker and a 14-Ω frying pan are connected in series.
Effective resistance = 14 + 14 = 28Ω
Now we have 28Ω and 20Ω in parallel
Effective resistance
[tex]R=\frac{28\times 20}{28+20}=11.67\Omega[/tex]
So we have resistor with 11.67Ω in a 120 V source of voltage.
We have equation V = IR
Substituting
120 = I x 11.67
I = 10.29 A
The total current supplied by the source of voltage = 10.29 A
g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the incline of angle " (where 0 ≤ " ≤ 90°). Evaluate your results for " = 0, 30°, and 45°
The launch angle that results in the maximum range of a projectile up an incline depends on the initial speed and the angle of the incline. For conditions neglecting air resistance, the maximum range is obtained at 45 degrees. If air resistance is considered, the maximum angle is around 38 degrees.
Explanation:The range of a projectile launched up an incline depends on the launch angle. To determine the launch angle that results in the maximum range, we need to consider the initial speed and the angle of the incline. Figure 3.38(b) shows that for a fixed initial speed, the maximum range is obtained at 45 degrees. However, this is only true for conditions neglecting air resistance. If air resistance is considered, the maximum angle is around 38 degrees. It is also interesting to note that for every initial angle except 45 degrees, there are two angles that give the same range, and the sum of those angles is 90 degrees.
A car is travelling at a constant speed of 26.5 m/s. Its tires have a radius of 72 cm. If the car slows down at a constant rate to 11.7 m/s over 5.2 s, what is the magnitude of the angular acceleration of the tires during that time? (in rad/s^2)
Answer:
Magnitude of angular acceleration = -3.95 rad/s²
Explanation:
Angular acceleration is the ratio of linear acceleration and radius.
That is
[tex]\texttt{Angular acceleration}=\frac{\texttt{Linear acceleration}}{\texttt{Radius}}\\\\\alpha =\frac{a}{r}[/tex]
Radius = 72 cm = 0.72 m
Linear acceleration is rate of change of velocity.
[tex]a=\frac{11.7-26.5}{5.2}=-2.85m/s^2[/tex]
Angular acceleration
[tex]\alpha =\frac{a}{r}=\frac{-2.85}{0.72}=-3.95rad/s^2[/tex]
Angular acceleration = -3.95 rad/s²
Magnitude = 3.95 rad/s²
A 18-g paper clip is attached to the rim of a phonograph record with a diameter of 48 cm, spinning at 3.2 rad/s. What is the magnitude of its angular momentum (in kg m2/s)? Round your answer to the nearest ten-thousandth.
The magnitude of the angular momentum of the paperclip attached to the spinning vinyl record is approximately 0.0033 kg m²/s. This is calculated using the formulas for moment of inertia and angular momentum.
Explanation:To calculate the angular momentum of the paperclip, we first need to know the moment of inertia (I) of the paperclip. The moment of inertia can be calculated using the formula I = mR² where 'm' is the mass of the paperclip (converted into kg - 0.018 kg) and 'R' is the radius of the record player (converted into m - 0.24 m).
So, I = 0.018 kg * (0.24 m)² = 0.0010368 kg m².
Next, we use the formula for angular momentum (L), which is L = Iω, where ω is the angular velocity. Given ω = 3.2 rad/s, we plug these values into our formula:
L = 0.0010368 kg m² * 3.2 rad/s = t0.00331776 kg m²/s.
Thus, rounding to the nearest ten-thousandth, the magnitude of the angular momentum of the paperclip is 0.0033 kg m²/s.
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The magnitude of the angular momentum of the paper clip is approximately [tex]\( 0.0033 \text{ kg m}^2/\text{s} \)[/tex].
The magnitude of the angular momentum of the paper clip is given by the formula [tex]\( L = I\omega \)[/tex], where I is the moment of inertia of the paper clip and [tex]\( \omega \)[/tex] is the angular velocity of the record.
Given:
- Mass of the paper clip, [tex]\( m = 18 \) g \( = 0.018 \)[/tex] kg (after converting grams to kilograms)
- Diameter of the record, [tex]\( d = 48 \) cm \( = 0.48 \)[/tex] m (after converting centimeters to meters)
- Radius of the record, [tex]\( r = \frac{d}{2} = \frac{0.48}{2} = 0.24 \)[/tex]m
- Angular velocity, [tex]\( \omega = 3.2 \)[/tex] rad/s
Now, we calculate the moment of inertia I:
[tex]\[ I = mr^2 = 0.018 \times (0.24)^2 \] \[ I = 0.018 \times 0.0576 \] \[ I = 0.0010368 \text{ kg m}^2 \][/tex]
Next, we calculate the angular momentum L:
[tex]\[ L = I\omega \] \[ L = 0.0010368 \times 3.2 \] \[ L = 0.00331776 \text{ kg m}^2/\text{s} \][/tex]
Rounding to the nearest ten-thousandth, we get:
[tex]\[ L \approx 0.0033 \text{ kg m}^2/\text{s} \][/tex]
Imagine that two charged balls placed some distance apart strongly attract each other. Now imagine placing a pane of glass halfway between the two balls. Will insertion of this glass increase or decrease the magnitude of force that each ball feels, or will it have strictly no effect? Use diagrams to explain your reasoning carefully.
Answer:
decrease
Explanation:
If the two charged ball attracts each other, it means the charge on both the balls are opposite in nature.
As, we insert a glass slab, it means a dielectric is inserted in between the charges. The force between them is reduced.
A circular loop of radius 0.7cm has 520 turns of wire and carries a current of 3.9A. The axis of the loop makes an angle of 57 degrees with a magnetic field of 0.982T. Find the magnitude of the torque on the loop.
Answer:
Torque, [tex]\tau=0.1669\ N-m[/tex]
Explanation:
It is given that,
Radius of the circular loop, r = 0.7 cm = 0.007 m
Number of turns, N = 520
Current in the loop, I = 3.9 A
The axis of the loop makes an angle of 57 degrees with a magnetic field.
Magnetic field, B = 0.982 T
We need to find the magnitude of torque on the loop. It is given by :
[tex]\tau=\mu\times B[/tex]
[tex]\tau=NIABsin(90-57)[/tex]
[tex]\tau=520\times 3.9\ A\times \pi (0.007\ m)^2\times 0.982\ T\ cos(57)[/tex]
[tex]\tau=0.1669\ N-m[/tex]
[tex]\tau=0.167\ N-m[/tex]
So, the magnitude of torque is 0.1669 N-m. Hence, this is the required solution.
A small bag of sand is released from an ascending hot-air balloon whose upward constant velocity is vo = 1.55 m/s. Knowing that at the time of the release the balloon was 85.8 m above the ground, determine the time, T, it takes the bag to reach the ground from the moment of its release.
Answer:
t = 4.35 s
Explanation:
Since the balloon is moving upwards while the sand bag is dropped from it
so here the velocity of sand bag is same as the velocity of balloon
so here we can use kinematics to find the time it will take to reach the ground
[tex]\Delta y = v_y t + \frac{1}{2} gt^2[/tex]
here we know that since sand bag is dropped down so we have
[tex]\Delta y = -85.8 m[/tex]
initial upward speed is
[tex]v_y = 1.55 m/s[/tex]
also we know that gravity is downwards so we have
[tex]a = - 9.8 m/s^2[/tex]
so here we have
[tex]-85.8 = 1.55 t - \frac{1}{2}(9.8) t^2[/tex]
[tex]4.9 t^2 - 1.55 t - 85.8 = 0[/tex]
[tex]t = 4.35 s[/tex]
Astronauts on a distant planet set up a simple pendulum of length 1.20 m. The pendulum executes simple harmonic motion and makes 100 complete oscillations in 450 s. What is the magnitude of the acceleration due to gravity on this planet?
Answer:
Magnitude of the acceleration due to gravity on the planet = 2.34 m/s²
Explanation:
Time period of simple pendulum is given by
[tex]T=2\pi\sqrt{\frac{l}{g}}[/tex], l is the length of pendulum, g is acceleration due to gravity value.
We can solve acceleration due to gravity as
[tex]g=\frac{4\pi^2l}{T^2}[/tex]
Here
Length of pendulum = 1.20 m
Pendulum executes simple harmonic motion and makes 100 complete oscillations in 450 s.
Period, [tex]T=\frac{450}{100}=4.5s[/tex]
Substituting
[tex]g=\frac{4\pi^2\times 1.2}{4.5^2}=2.34m/s^2[/tex]
Magnitude of the acceleration due to gravity on the planet = 2.34 m/s²
The liquid nitrogen temperature is 63 K. Convert to ºC.
Answer:
[tex]T_c=-210.15^{\circ}C[/tex]
Explanation:
In this question we need to convert the temperature in kelvin to degree Celsius. The conversion from kelvin scale to Celsius scale is as follows :
[tex]T_k=T_c+273.15[/tex]
Here,
[tex]T_k=63\ K[/tex]
[tex]T_k-273.15=T_c[/tex]
[tex]63-273.15=T_c[/tex]
[tex]T_c=-210.15^{\circ}C[/tex]
Here, negative sign shows that the heat is released. So, the temperature at 63 K is equivalent to 210.15 °C. Hence, this is the required solution.
When you raise the temperature of air, the molecules move farther apart from each other. This lowers the density of the warm air. What will happen to this warm air? (In an ideal gas, increasing the temperature of the gas also increases its volume.)