Answer:
The speed of the water shoot out of the hole is 20 m/s.
(d) is correct option.
Explanation:
Given that,
Height = 20 m
We need to calculate the velocity
Using formula Bernoulli equation
[tex]\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]
Where,
v₁= initial velocity
v₂=final velocity
h₁=total height
h₂=height of the hole from the base
Put the value into the formula
[tex]v_{1}^2=2g(h_{2}-h_{1})[/tex]
[tex]v_{1}=\sqrt{2g(h_{2}-h_{1})}[/tex]
[tex]v_{1}=\sqrt{2\times9.8\times(20-0.005)}[/tex]
[tex]v_{1}=19.7\ m/s= approximate\ 20\ m/s[/tex]
Hence, The speed of the water shoot out of the hole is 20 m/s.
To practice Problem-Solving Strategy 11.1 for conservation of momentum problems.An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?
Answer:
0.081 m /s
Explanation:
According to the conservation of momentum, the momentum of a system is conserved when no external force is applied on a body.
momentum of the system before throwing = momentum of the system after throwing
Let v be the velocity of quarterback after throwing the football.
80 x 0 + 0.43 x 0 = 80 x v + 0.43 x 15
0 = 80 v + 6.45
v = - 0.081 m /s
The negative sign shows that he is moving in backward direction.
The quarterback will be moving backward at approximately 0.080625 m/s just after releasing the football due to the conservation of momentum.
Explanation:The student's question involves practicing Problem-Solving Strategy 11.1 for conservation of momentum problems. Specifically, the student is asked to calculate how fast an 80-kg quarterback will be moving backward just after releasing a football that has a mass of 0.43 kg and is thrown horizontally at a speed of 15 m/s. To solve this, we use the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it.
Before the throw, the total momentum of the system (quarterback and football) is zero because both are stationary with respect to horizontal motion. After the throw, the combined momentum must still be zero. We can set up the equation mQB × vQB + mball × vball = 0, where mQB is the mass of the quarterback, vQB is the quarterback's velocity after the throw, mball is the mass of the football, and vball is the velocity of the football. This simplifies to vQB = -(mball × vball) / mQB.
Plugging in the numbers gives us vQB = -(0.43 kg × 15 m/s) / 80 kg, which yields a velocity of vQB ≈ -0.080625 m/s. The negative sign indicates that the quarterback is moving in the opposite direction of the throw, which is backward.
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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.
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
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]
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
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.)
A Raman line is observed at 4768.5 À, when acetylene was irradiated by 4358.3 A radiation, Calculate the equilibrium vibrational frequency that causes the shift.
Answer:
The equilibrium vibrational frequency that causes the shift is [tex]0.56\times10^{14}\ Hz[/tex]
Explanation:
Given that,
Wavelength of Raman line [tex]\lambda'=4768.5\ A[/tex]
Wavelength [tex]\lambda=4358.3\ A[/tex]
We need to calculate the frequency
Using formula of frequency
[tex]f =\dfrac{c}{\lambda}[/tex]
For 4748.5 A
The frequency is
[tex]f'=\dfrac{3\times10^{8}}{4748.5\times10^{-10}}[/tex]
[tex]f' =6.32\times10^{14}\ Hz[/tex]
For 4358.3 A
The frequency is
[tex]f=\dfrac{3\times10^{8}}{4358.3\times10^{-10}}[/tex]
[tex]f=6.88\times10^{14}\ Hz[/tex]
We need to calculate the shift
[tex]\Delta f=f-f'[/tex]
[tex]\Delta f=(6.88-6.32)\times10^{14}\ Hz[/tex]
[tex]\Delta f=0.56\times10^{14}\ Hz[/tex]
Hence, The equilibrium vibrational frequency that causes the shift is [tex]0.56\times10^{14}\ Hz[/tex]
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.
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
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
A 100-kg box is a rest on the floor. The coefficient of static friction between the box and the floor is 0.40, while the coefficient of kinetic friction is 0.20. You apply a rightward horizontal force of P = 300 N. Part B What is the normal force acting on the block?
Answer:
The normal force acting on the block is 980 N.
Explanation:
Mass of box = 100 kg
Coefficient of static friction = 0.40
Coefficient of kinetic friction = 0.20
We need to calculate the normal force
We know that,
When the object is placed on the flat surface then the normal force is equal to the weight of the object.
So. The normal force = mass x acceleration due to gravity
[tex]N=mg[/tex]
Where, N = Normal force
m = mass of object
g = acceleration due to gravity
[tex]N=100\times9.8[/tex]
[tex]N=980\ N[/tex]
Hence, The normal force acting on the block is 980 N.
Final answer:
The normal force acts perpendicular to the floor supporting the box and is equal to 980 N.
Explanation:
The question asks about the normal force acting on a 100-kg box at rest on the floor. To find the normal force on the box, we can use the equation that relates the normal force to the weight of the object, which is W = mg, where m is the mass and g is the acceleration due to gravity.
Given that the mass of the box is 100 kg and the acceleration due to gravity is 9.80 m/s², the normal force can be calculated as:
Normal Force (N) = Mass (m) × Acceleration due to gravity (g)
Normal Force = 100 kg × 9.80 m/s² = 980 N
The normal force acts perpendicular to the floor supporting the box and is equal to 980 N. It is often confused with the applied force P, but the normal force is related to the box's weight, not the horizontal force applied.
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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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.
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
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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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
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.
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
17. (a) What is the terminal voltage of a large 1.54-V carbon-zinc dry cell used in a physics lab to supply 2.00 A to a circuit, if the cell’s internal resistance is 0.100 Ω? (b) How much electrical power does the cell produce? (c) What power goes to its load?
Answer:
a) 1.34 Volts
b) 3.08 W
c) 2.68 W
Explanation:
Given:
Emf of the cell, E = 1.54 V
current, i = 2.0 A
internal resistance, r = 0.100Ω
(a) Terminal voltage (V) = E - v
where,
v is the potential difference across the resistance 'r'
now,
according to the Ohm's Law, we have
v = i × r
substituting the values in the above equation we get
v = 2.0 × 0.100 = 0.2 Volts
thus,
Terminal voltage (V) = (1.54 - 0.2) = 1.34 V
(b) Now, the Total power (P) is given as
P = E × i = (1.54 × 2.0) = 3.08 W
(c) Power into its load = [terminal voltage, v] * i
= (1.34 × 2.0) = 2.68 W
Final answer:
Terminal voltage of a carbon-zinc cell is 1.34 V, producing 2.68 W of electrical power, with 2.68 W going to the load.
Explanation:
Terminal voltage: The formula to calculate terminal voltage is V = EMF - I * r, where V is the terminal voltage, EMF is the electromotive force, I is the current, and r is the internal resistance of the cell. So, V = 1.54 V - 2.00 A * 0.100 Ω = 1.34 V.
Electrical power: The electrical power produced by the cell is given by P = V * I, where P is power, V is voltage, and I is current. Substituting the values, P = 1.34 V * 2.00 A = 2.68 W.
Power to load: The power delivered to the load is equal to the voltage supplied to the load times the current flowing through it. Hence, the power to the load is P = V_load * I = 1.34 V * 2.00 A = 2.68 W.
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]
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 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]
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
Suppose a particle moves along a straight line with velocity v(t)=t2e−2tv(t)=t2e−2t meters per second after t seconds. How many meters has it traveled during the first t seconds?
The distance traveled by a particle with velocity function
[tex]v(t)=t^2e^-2t[/tex]over the first t seconds can be obtained by integrating that function from 0 to t using integration by parts.
Explanation:The distance traveled by a particle can essentially be obtained by integrating the velocity function over a given timeframe. In this case, the velocity function is given by
[tex]v(t)=t^2e^-2t.[/tex]
To calculate the distance traveled over the first t seconds, we need to integrate this function from 0 to t. Therefore, the integral ∫v(t)dt from 0 to t will provide the required solution. Use the method of integration by parts, choosing u=t^2 and
[tex]dv=e^-2t dt,[/tex]then complete the integration process following the standard procedure.
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A charge Q is located inside a rectangular box The electric flux through each of the six surfaces of the box is 1 2060 Nm2 C2 1590 Nm2 C 3 1690 Nm2 C 4 3430 Nm2 C 5 1870 Nm2 C and 6 5760 Nm2 C What is Q
Answer:
[tex]Q = 1.45 \times 10^{-7} C[/tex]
Explanation:
Here flux passing through each surface is given
so total flux through whole cube is given by
[tex]\phi = 2060 + 1590 + 1690 + 3430 + 1870 + 5760[/tex]
[tex]\phi = 16400 Nm^2 C[/tex]
now we also know that total flux through a closed surface depends on the total charge enclosed in the surface
So we will have
[tex]\frac{Q}{\epsilon_0} = 16400[/tex]
[tex]Q = (8.85 \times 10^{-12})(16400)[/tex]
[tex]Q = 1.45 \times 10^{-7} C[/tex]
The total charge [tex](\( Q \))[/tex] enclosed by the rectangular box is approximately [tex]\( 2.544 \times 10^{-4} \, \text{C} \)[/tex].
The electric flux [tex](\( \Phi \))[/tex] through a closed surface is given by Gauss's Law:
[tex]\[ \Phi = \frac{Q}{\varepsilon_0} \][/tex]
where:
- [tex]\( \Phi \)[/tex] is the electric flux,
- [tex]\( Q \)[/tex] is the total charge enclosed by the closed surface,
- [tex]\( \varepsilon_0 \)[/tex] is the permittivity of free space [tex](\( \varepsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \))[/tex].
For a closed box, the total electric flux [tex](\( \Phi_{\text{total}} \))[/tex] is the sum of the electric flux through each of its six surfaces. Let's denote the electric flux through each surface as [tex]\( \Phi_i \) where \( i = 1, 2, \ldots, 6 \)[/tex].
[tex]\[ \Phi_{\text{total}} = \sum_{i=1}^{6} \Phi_i \][/tex]
Now, we can set up an equation using the given values:
[tex]\[ \Phi_{\text{total}} = 12060 \, \text{Nm}^2/\text{C} + 1590 \, \text{Nm}^2/\text{C} + 1690 \, \text{Nm}^2/\text{C} + 3430 \, \text{Nm}^2/\text{C} + 1870 \, \text{Nm}^2/\text{C} + 5760 \, \text{Nm}^2/\text{C} \][/tex]
[tex]\[ \Phi_{\text{total}} = 28800 \, \text{Nm}^2/\text{C} \][/tex]
Now, use Gauss's Law to find the total charge [tex](\( Q \))[/tex] enclosed by the closed surface:
[tex]\[ Q = \Phi_{\text{total}} \cdot \varepsilon_0 \][/tex]
[tex]\[ Q = 28800 \, \text{Nm}^2/\text{C} \cdot 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \][/tex]
[tex]\[ Q \approx 2.544 \times 10^{-4} \, \text{C} \][/tex]
So, the total charge [tex](\( Q \))[/tex] enclosed by the rectangular box is approximately [tex]\( 2.544 \times 10^{-4} \, \text{C} \)[/tex].
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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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 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.
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 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.