Answer:
[tex]\Delta s\ =\ 21.33\ J/K[/tex]
Explanation:
Given,
Mass of the ice = [tex]m_i\ =\ 13.8\ kg\ =\ 0.0138\ kg[/tex]Temperature of the ice = [tex]T_i\ =\ 0^o\ C[/tex]Mass of the original water = [tex]m_w\ =\ 134\ g\ =\ 0.134\ kg[/tex]Temperature of the original water = [tex]T_w\ =\ 70.0^o\ C[/tex]Specific heat of water = [tex]S_w\ =\ 4190\ J/kg K[/tex]Latent heat of fusion of ice = [tex]L_f\ =\ 333\ kJ/kg[/tex]Let T be the final temperature of the mixture,
Therefore From the law of mixing, heat loss by the water is equal to the heat gained by the ice,
[tex]m_iL_f\ +\ m_is_w(T_f\ -\ 0)\ =\ m_ws_w(T_w\ -\ T_f)\\\Rightarrow 333000\times 0.0138\ +\ 0.0138\times 4190T_f\ =\ 0.134\times 4190\times(70.7\ -\ T_f)\\\Rightarrow 4595.4\ +\ 57.96T_f\ =\ 39789.96\ -\ 562.8T_f\\\Rightarrow 620.76T_f\ =\ 35194.56\\\Rightarrow T_f\ =\ \dfrac{35194.56}{620.76}\\\Rightarrow T_f\ =\ 56.69^o\ C[/tex]
Now, We know that,
Change in the entropy,
[tex]\Delta s\ =\ s_f\ -\ s_i\ =\ \dfrac{Q}{T}\\\Rightarrow \displaystyle\int_{s_i}^{s_f} ds\ =\ \displaystyle\int_{T_i}^{T_f}\dfrac{msdT}{T}\\\Rightarrow \Delta s =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)[/tex]
Now, change in entropy for the ice at 0^o\ C to convert into 0^o\ C water.
[tex]\Delta s_1\ =\ \dfrac{Q}{T}\\\Rightarrow \Delta s_1\ =\ \dfrac{m_iL_f}{T}\ =\ \dfrac{0.0138\times 333000}{273.15}\ =\ 16.82\ J/K.[/tex]
Change in entropy of the water converted from ice from [tex]273.15\ K[/tex] to water 330.11 K water.
From the equation (1),
[tex]\therefore \Delta s_2\ =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\\\Rightarrow \Delta s_2\ =\ 0.0138\times 4190\times \ln \left (\dfrac{273.15}{330.11}\ \right )\\\Rightarrow \Delta s_2\ =\ 6.88\ J/K[/tex]
Change in entropy of the original water from the temperature 342.85 K to 330.11 K
From the equation (1),
[tex]\therefore \Delta s_3\ =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\\\Rightarrow \Delta s_3\ =\ 0.134\times 4190\times \ln \left (\dfrac{330.11}{343.85} \right )\\\Rightarrow \Delta s_3\ =\ -2.36\ J/K[/tex]
Total entropy change = [tex]\Delta s\ =\ \Delta s_1\ +\ \Delta s_2\ +\ \Delta s_3\\\Rightarrow \Delta s\ =\ (16.82\ +\ 6.88\ -\ 2.36)\ J/K\\\Rightarrow \Delta s\ =\ 21.33\ J/K.[/tex]
Hence, the change in entropy of the system form then untill the system reaches the final temperature is 21.33 J/K
The Z0 boson, discovered in 1985, is themediator of
the weak nuclear force, and it typically decays veryquickly. Its
average rest energy is 91.19 GeV, but its shortlifetime shows up as
an intrinsic width of 2.5 GeV (rest energyuncertainty). What is the
lifetime of this particle?
Answer:
The life time of the particle is [tex]2.491\times 10^{- 25} s[/tex]
Solution:
As per the question:
Average rest energy of [tex]Z^{0}boson = 91.19 GeV[/tex]
Uncertainty in rest energy, [tex]\Delta E_{r} = 2.5 GeV = 2.5\times 10^{9}\times 1.6\times 140^{- 19} J = 4\times 10^{- 10} J[/tex]
Now,
From the Heisenberg's Uncertainty Principle, we can write:
[tex]\Delta E_{r}\times \Delta T \geq \frac{h}{2\pi}[/tex]
where
T = Life time of the particle
[tex]\Delta T \geq \frac{h}{2\pi\Delta E_{r}}[/tex]
[tex]\Delta T \geq \frac{6.262\times 10^{- 34}}{2\pi\times 4\times 10^{- 10}}[/tex]
[tex]\Delta T \simeq 2.491\times 10^{- 25} s[/tex]
The lifetime of the Z0 boson is approximately 8.95 x 10^-17 seconds.
Explanation:The Z0 boson is a particle that mediates the weak nuclear force. Its average rest energy is 91.19 GeV and it has an intrinsic width of 2.5 GeV. The lifetime of a particle can be calculated using the Heisenberg uncertainty principle, which relates the energy uncertainty to the time uncertainty. The relationship is given by the equation ΔE × Δt ≥ ℏ/2, where ℏ is the reduced Planck constant. By rearranging the equation and substituting the values, we can calculate the lifetime of the Z0 boson to be approximately 8.95 x 10^-17 seconds.
Learn more about Z0 boson here:https://brainly.com/question/35880581
#SPJ12
A 7.80-nc charge is located 1.81 m from a 4.30-nc point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other. N (b) Is the force attractive or repulsive? O attractive O repulsive
Explanation:
Given that,
Charge 1, [tex]q_1=7.8\ nC=7.8\times 10^{-9}\ C[/tex]
Charge 2, [tex]q_2=4.3\ nC=4.3\times 10^{-9}\ C[/tex]
Distance between charges, r = 1.81 m
(a) The electrostatic force that one charge exerts on the another is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{7.8\times 10^{-9}\times 4.3\times 10^{-9}}{(1.81)^2}[/tex]
[tex]F=9.21\times 10^{-8}\ N[/tex]
(b) As both charges are positively charged. So, the force of attraction between them is repulsive.
According to Archimedes' principle, the mass of a floating object equals the mass of the fluid displaced by the object. A 150-lbm swimmer is floating in a nearby pool; 95% of his or her body's volume is in the water while 5% of his or her body's volume is above water. Determine the density of the swimmer's body. The density of water is 0.036lbm/in^3.
Answer:
Density of the swimmer = [tex]0.0342\ lbm/in^3[/tex].
Explanation:
Assuming,
[tex]\rho[/tex] = density of the swimmer.[tex]\rho_w[/tex] = density of the water.[tex]m[/tex] = mass of the swimmer.[tex]m_w[/tex] = mass of the water displaced by the swimmer.[tex]V_w[/tex] = volume of the displaced water.[tex]V[/tex] = volume of the swimmer.Given:
[tex]m=150\ lbm.[/tex][tex]\rho_w = 0.036\ lbm/in^3.[/tex]The density of an object is defined as the mass of the object per unit volume.
Therefore,
[tex]\rho =\dfrac{m}{V}\ \Rightarrow m = \rho V\ \ .........\ (1).[/tex]
Since only 95% of the body of the swimmer is inside the water, therefore,
[tex]V_w = 95\%\ \text{of}\ V=\dfrac{95}{100}\times V = 0.95V.[/tex]
According to Archimedes' principle,
[tex]m=m_w\\[/tex]
Using (1),
[tex]\rho V=\rho_w V_w\\\rho V = 0.036\ lbm/in^3\times 0.95 V\\\rho=0.036\times 0.95\ lbm/in^3=0.0342\ lbm/in^3.[/tex]
A runner in a relay race runs 20 m north, turns around and runs south for 30 m, then turns north again and runs 40 m. The entire run took 30 seconds. What was the average speed of the runner? What was the average velocity of the runner?
Answer:
The average speed its 1 m/sThe average velocity its 1 m/s to the north.Explanation:
So, lets say the runner stars from the position [tex]x_0[/tex]. Lets make this point the origin of a coordinate system in which the vector i points north.
[tex]x_0 = (0,0)[/tex]
Now, in the first sections of the race, he runs 20 meters north, so, he finds himself at:
[tex]x_1 = x_0 + 20 m * i = (0,0) \ + (20 \ m,0)[/tex].
[tex]x_1 = (20 \ m,0)[/tex].
The, he runs 30 meters south
[tex]x_2 = x_1 - 30 \ m * i = (20 \ m,0)-(30 \ m,0)[/tex]
[tex]x_2 = (-10 \ m,0)[/tex]
Finally, he runs 40 meter north
[tex]x_3 = x_2 + 40 \ m * i = (-10 \ m,0)+(40 \ m,0)[/tex]
[tex]x_3 = (30 \ m,0)[/tex].
This is our displacement vector. Now, the average speed will be:
[tex]\frac{distance}{time}[/tex].
The distance its the length of the displacement vector,
[tex]d=\sqrt{x^2+y^2}[/tex]
[tex]d=\sqrt{(30 \ m)^2+0^2}[/tex]
[tex]d=30 \ m[/tex]
So, the average speed its:
[tex]\frac{30 \ m }{30 \ s} = 1\frac{m}{s}[/tex].
The average velocity, instead, its:
[tex]\vec{v} = \frac{displacement}{time}[/tex]
[tex]\vec{v} = \frac{(30 \ m ,\ 0)}{30 \ s}[/tex]
[tex]\vec{v} = (1 \ \frac{m}{s} ,\ 0)[/tex]
This is, 1 m/s north.
A 5.0-V battery is places in series with two 1.25-Ω resistors. Determine the current through each resistor.
Answer:
Current through each resistor is 2 A.
Explanation:
Given that,
Voltage of a battery, V = 5 volts
Resistance 1, R = 1.25 ohms
Resistance 2,R' = 1.25 ohms
Both resistors are connected in series. The equivalent resistance is given by :
R" = R + R'
R" = 1.25 + 1.25
R" = 2.5 ohms
The current flowing throughout all resistors is same in series combination of resistors. Current can be calculated using Ohm's law as :
[tex]I=\dfrac{V}{R"}[/tex]
[tex]I=\dfrac{5\ V}{2.5}[/tex]
I = 2 A
So, the current through each resistor is 2 A. Hence, this is the required solution.
When a car is on an inclined bank of angle θ and rounding a curve with no friction, what is the centripetal force equal to? a. The weight of the car b. N cos (θ) c. N sin (θ) d. Zero
Answer:
[tex]F_{net}=N\ sin\theta[/tex]
Explanation:
Let a car of m is on an incline bank of angle θ and it is rounding a curve with no friction. We need to find the centripetal force acting on it.
The attached free body diagram shows the car on the banked turn. It is clear that,
In vertical direction,
[tex]N\ cos\theta=mg[/tex]
In horizontal direction,
[tex]F_{net}=F_{centripetal}[/tex]
[tex]F_{net}=N\ sin\theta[/tex]
So, the centripetal force is equal to [tex]N\ sin\theta[/tex]. Hence, the correct option is (c).
The heat in an internal combustion engine raises the internal temperatures to 300 deg C. If the outside temperature is 25 deg C, what is the maximum efficiency of this engine? And is it possible to attin this efficiency?
Answer:48 %
Explanation:
Given
outside temperature [tex]=25^{\circ}\approx 298 K[/tex]
Internal temperature[tex]=300^{\circ}\approx 573 K[/tex]
Maximum efficiency is carnot efficiency which is given by
[tex]\eta =1-\frac{T_L}{T_H}[/tex]
[tex]\eta =1-\frac{298}{573}[/tex]
[tex]\eta =\frac{275}{573}=0.4799[/tex]
i.e. [tex]47.99 %\approx 48 %[/tex]
No it is not possible to achieve this efficiency because carnot engine is the ideal engine so practically it is not possible but theoretically it is possible
What is the magnitude of the electric field of a proton at a distance of 50 micrometers? _____________ (in units of N/C)
Answer:
Electric field at distance of 50 micrometer due to a proton is 0.576 N/C
Explanation:
We have given distance where we have to find the electric field [tex]r=50\mu m=50\times 10^{-6}m[/tex]
We know that charge on proton [tex]q=1.6\times 10^{-16}C[/tex]
Electric field due to point charge is given by [tex]E=\frac{Kq}{r^2}[/tex], here K is a constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]
So electric field [tex]E=\frac{9\times 10^9\times 1.6\times 10^{-16}}{(50\times 10^{-6})^2}=0.576N/C[/tex]
A basketball referee tosses the ball straight up for the starting tip-off. At what velocity (in m/s) must a basket ball player leave the grownd to rist 1.24 m above the floor in an attempt to get the ball?
Answer:4.93 m/s
Explanation:
Given
height to reach is (h )1.24 m
here Let initial velocity is u
using equation of motion
[tex]v^2-u^2=2ah[/tex]
here Final Velocity v=0
a=acceleration due to gravity
[tex]0-u^2=2\left ( -g\right )h[/tex]
[tex]u=\sqrt{2gh}[/tex]
[tex]u=\sqrt{2\times 9.81\times 1.24}[/tex]
[tex]u=\sqrt{24.328}[/tex]
u=4.93 m/s
Sitting in a second-story apartment, a physicist notices a ball moving straight upward just outside her window. The ball is visible for 0.20 s as it moves a distance of 1.19 m from the bottom to the top of the window. How long does it take before the ball reappears?
Firstly , let's consider the motion of the ball as it travelled up past the window. Given that the distance covered by the ball is 1.19 m in 0.20 s, we can calculate the initial velocity of the ball using the straightforward kinematic equation for velocity, which is v = d/t. Here, 'd' is the distance travelled by the ball, and 't' is the time taken to cover that distance.
Plugging the values into the equation:
vi = d/t
vi = 1.19 m / 0.20 s
vi = 5.95 m/s
Therefore, the initial velocity of the ball is approximately 5.95 m/s.
Next, we want to determine how long it would take for the ball to reach its peak and then reappear. The acceleration due to gravity (g) is 9.81 m/s^2 and acts downwards. Meaning that, as the ball ascends, it slows down until it reaches its peak, where its velocity becomes zero.
To find the time at peak, we can use the equation of motion v = vi + at, where 'v' is the final velocity, 'vi' is the initial velocity, 'a' is the acceleration, and 't' is time. At the peak, the final velocity 'v' becomes 0, thus:
t_peak = vi / g
t_peak = 5.95 m/s / 9.81 m/s²
t_peak = 0.61 s
Therefore, the time it takes to reach the peak is approximately 0.61 seconds.
As we're dealing with symmetrical motion (upwards and downwards motion are equal), the time for the ball to descend from its peak would be identical to the time it took to ascend, which means:
Time for the ball to reappear, t_reappear = 2 * t_peak
t_reappear = 2 * 0.61 s
t_reappear = 1.22 s
Hence, it should take approximately 1.22 seconds before the ball reappears.
https://brainly.com/question/29545516
#SPJ12
Air enters a heat exchanger at a rate of 5000 cubic feet per minute at a temperature of 55 °F and pressure of 14.7 psia. The air is heated by hot water flowing in the same exchanger at a rate of 11,200 pounds per hour with a decrease in temperature of 10 °F. At what temperature does the air leave the heat exchanger?
Answer:
75 °F
Explanation:
Air has a specific heat at constant pressure of:
Cpa = 0.24 BTU/(lbm*F)
The specific heat of water is:
Cpw = 1 BTU/(lbm*F)
The first law of thermodynamics:
Q = L + ΔU
The heat exchanger is running at a steady state, so ΔU = 0. Also does not perform or consume any work L = 0.
Then:
Q = 0.
We split the heat into the heat transferred by the air and the heat trnasferred by the water:
Qa + Qw = 0
Qa = -Qw
The heat exchanged by the air is
Qa = Ga * Cpa * (tfin - ti)
And the heat exchanged by the water is:
Qw = Gw * Cpw * Δt
Replacing:
Ga * Cpa * (tfin - ti) = -Gw * Cpw * Δt
tfin - ti = (-Gw * Cpw * Δt) / (Ga * Cpa)
tfin = (-Gw * Cpw * Δt) / (Ga * Cpa) + ti
The G terms are mass flows, however we have volume flow of air.
With the gas state equation we calculate the mass:
p * V = m * R * T
m = (p * V) / (R * T)
55 °F = 515 °R
The gas constant for air is R = 53.35 (ft*lb)/(lbm* °R)
14.7 psi = 2117 lb/ft^2
m = (2117 * 5000) / (53.35 * 515) = 385 lbm
The mass flow is that much amount per minute
The mass flow of water is
11200 lbm/h = 186.7 lbm/min
Then:
tfin = (-186.7 * 1 * (-10)) / (385 * 0.24) + 55 = 75 °F
(a) A 1.00-μF capacitor is connected to a 15.0-V battery. How much energy is stored in the capacitor? ________ μJ (b) Had the capacitor been connected to a 6.00-V battery, how much energy would have been stored?________ μJ
Answer:
(a) [tex]E_{ c} = 112.5 \mu J[/tex]
(b) [tex]E'_{ c} = 18 \mu J[/tex]
Solution:
According to the question:
Capacitance, C = [tex]1.00\mu F = 1.00\times 10^{- 6} F[/tex]
Voltage of the battery, [tex]V_{b} = 15.0 V[/tex]
(a)The Energy stored in the Capacitor is given by:
[tex]E_{c} = \frac{1}{2}CV_{b}^{2}[/tex]
[tex]E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}[/tex]
[tex]E_{ c} = 112.5 \mu J[/tex]
(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:
[tex]E'_{c} = \frac{1}{2}CV'_{b}^{2}[/tex]
[tex]E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}[/tex]
[tex]E'_{ c} = 18 \mu J[/tex]
(a) The energy stored in the capacitor is [tex]1.125 \times 10^{-4} \ J[/tex]
(b) The energy stored in the capacitor is [tex]1.8 \times 10^{-5} \ J[/tex]
The given parameters;
charge of the capacitor, q = 1 -μFvoltage across the capacitor, V = 15The energy stored in the capacitor is calculated as follows;
[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 15^2\\\\E = 1.125 \times 10^{-4} \ J[/tex]
When the battery voltage changes to 6 V, the energy stored in the capacitor is calculated as follows;
[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 6^2\\\\E = 1.8 \times 10^{-5} \ J\\\\[/tex]
Learn more here:https://brainly.com/question/24943210
The pilot of an airplane carrying a package of mail to a remote outpost wishes to release the package at the right moment to hit the recovery location A. What angle θ with the horizontal should the pilot’s line of sight to the target make at the instant of release? The airplane is flying horizontally at an altitude of 86 m with a velocity of 283 km/h.
Answer:
The angle is [tex]\theta\approx 14.61[/tex] degrees.
Explanation:
Se the attached drawing if you need a visual aid for the explanation. Let [tex]\theta[/tex] be the angle of elevation of the plante which in itself is the same drop angle that the pilot measures. Let [tex]d[/tex] be the horizontal distance from the target and [tex]h[/tex] the height of the plane. We know that the package is dropped without any initial vertical speed, that means that it has a y-position equation of the form:
[tex]y(t)=-\frac{1}{2}gt^2+h[/tex]
If we set [tex]y(t)=0[/tex] we are setting the condition that the package is in the ground. We can then solve for t and get the flight time of the package.
[tex]0=-\frac{1}{2}gt^2+h\implies t_f=\sqrt{\frac{2h}{g}}[/tex].
If the flight time is -[tex]t_f[/tex] then the distance b can be found in meters by taking into account that the horizontal speed of the plane is [tex]v=283\, Km/h=78.61 \, m/s[/tex].
[tex]d=v\cdot t_f=78.61\cdot \sqrt{\frac{2h}{g}}[/tex]
The angle is thus
[tex]\theta=\arctan{\frac{h}{v\cdot t_f}}=\arctan{\frac{h}{v\cdot \sqrt{\frac{2\cdot h}{g}}}\approx 14.61 [/tex] degrees.
A parallel-plate capacitor consists of two plates, each with an area of 27 cm^2 separated by 3.0 mm. The charge on the capacitor is 4.8 nC . A proton is released from rest next to the positive plate. How long does it take for the proton to reach the negative plate? Steps please with right answer.
Answer:
Explanation:
Capacity of a parallel plate capacitor C = ε₀ A/ d
ε₀ is permittivity whose value is 8.85 x 10⁻¹² , A is plate area and d is distance between plate.
C =( 8.85 X10⁻¹² X 27 X 10⁻⁴ ) / 3 X 10⁻³
= 79.65 X 10⁻¹³ F.
potential diff between plate = Charge / capacity
= 4.8 X 10⁻⁹ / 79.65 X 10⁻¹³
= 601 V
Electric field = V/d
= 601 / 3 x 10⁻³
= 2 x 10⁵ N/C
Force on proton
= charge x electric field
1.6 x 10⁻¹⁹ x 2 x 10⁵
= 3.2 x 10⁻¹⁴
Acceleration a = force / mass
= 3.2 x 10⁻¹⁴ / 1.67 x 10⁻²⁷
= 1.9 x 10¹³ m s⁻²
Distance travelled by proton = 3 x 10⁻³
3 x 10⁻³ = 1/2 a t²
t = [tex]\sqrt{\frac{3\times2\times10^{-3}}{1.9\times10^{13}} }[/tex]
t = 1.77 x 10⁻⁸ s
At a construction site a pipe wrench struck the ground with a speed of 23 m/s. (a) From what height was it inadvertently dropped? (b)How long was it falling?
Answer:26.96 m,2.34 s
Explanation:
Given
Wrench hit the ground with a speed of 23 m/s
Applying equation of motion
[tex]v^2-u^2=2as[/tex]
Here u=0 because it is dropped from a height of S m
[tex]23^2-0=2\times 9.81\times s[/tex]
[tex]s=\frac{529}{2\times 9.81}=26.96 m[/tex]
Time required by wrench to hit the ground
v=u+at
[tex]23=9.81\times t[/tex]
[tex]t=\frac{23}{9.81}=2.34 s[/tex]
The first accurate measurements of the properties of high-pressure gases were made by E. H. Amagat in France between 1869 and 1893. Before developing the dead-weight gauge, he worked in a mineshaft and used a mercury manometer for measurements of pressure to more than 400 bar. Estimate the height of the manometer required.
Answer:
Height, h = 300.27 meters
Explanation:
Given that,
Pressure of the gas, [tex]P=400\ bar=4\times 10^7\ Pa[/tex]
We need to find the height of the manometer required. The pressure at a height is given by :
[tex]P=\rho gh[/tex]
Where
[tex]\rho[/tex] is the density of mercury, [tex]\rho=13593\ kg/m^3[/tex]
h is the height of the manometer required.
[tex]h=\dfrac{P}{\rho g}[/tex]
[tex]h=\dfrac{4\times 10^7}{13593\times 9.8}[/tex]
h = 300.27 meters
So, the height of the manometer required is 300.27 meters. Hence, this is the required solution.
A piano string having a mass per unit length of 5.00 g/m is under a tension of 1350 N. Determine the speed of transverse waves in this string.
Answer:
The speed of transverse waves in this string is 519.61 m/s.
Explanation:
Given that,
Mass per unit length = 5.00 g/m
Tension = 1350 N
We need to calculate the speed of transverse waves in this string
Using formula of speed of the transverse waves
[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]
Where, [tex]\mu[/tex] = mass per unit length
T = tension
Put the value into the formula
[tex]v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}[/tex]
[tex]v =519.61\ m/s[/tex]
Hence, The speed of transverse waves in this string is 519.61 m/s.
A 60.0 kg astronaut is freely floating in space and pushes on a freely floating 120.0 kg spacecraft with a force of 30.0 N for 1.50 s. (a) Compare the forces exerted on the astronaut and the spacecraft, and (b) compare the acceleration of the astronaut to the acceleration of the spacecraft
Answer:
Explanation:
mass of astronaut, m = 60 kg
mass of space craft, M = 120 kg
t = 1.5 s
Force on space craft = 30 N
(a) According to Newton's third law
Force on spacecraft by the astronaut = Force on astronaut by the space craft
Force on astronaut by the space craft = 30 N
(b) According to Newton's second law
Force = mass x acceleration
Let a be the acceleration of the astronaut
30 = 60 x a
a = 0.5 m/s^2
Let A be the acceleration of the spacecraft
30 = 120 x A
a = 0.25 m/s^2
The total length of the cord is L = 7.00 m, the mass of the cord is m = 7.00 g, the mass of the hanging object is M = 2.50 kg, and the pulley is a fixed a distance d = 4.00 m from the wall. You pluck the cord between the wall and the pulley and it starts to vibrate. What is the fundamental frequency (in Hz) of its vibration?
Answer:
frequency = 19.56 Hz
Explanation:
given data
length L = 7 m
mass m = 7 g
mass M = 2.50 kg
distance d = 4 m
to find out
fundamental frequency
solution
we know here frequency formula is
frequency = [tex]\frac{v}{2d}[/tex] ...........1
so here d is given = 4
and v = [tex]\sqrt{\frac{T}{\mu} }[/tex] ..........2
tension T = Mg = 2.50 × 9.8 = 24.5 N
and μ = [tex]\frac{m}{l}[/tex] = [tex]\frac{7*10^{-3} }{7}[/tex] = [tex]10^{-3}[/tex] kg/m
so from equation 2
v = [tex]\sqrt{\frac{24.5}{10^{-3}} }[/tex]
v = 156.52
and from equation 1
frequency = [tex]\frac{v}{2d}[/tex]
frequency = [tex]\frac{156.52}{2(4)}[/tex]
frequency = 19.56 Hz
Final answer:
The fundamental frequency of the vibrating cord is approximately 14.18 Hz.
Explanation:
To determine the fundamental frequency of the vibrating cord, we can use the equation for the fundamental frequency of a vibrating string:
f1 = 1/2L * sqrt(T / μ)
Where f1 is the fundamental frequency, L is the total length of the cord, T is the tension in the cord, and μ is the linear density of the cord.
Plugging in the given values, we have: f1 = 1/2 * 7.00 * sqrt(90 / 0.007)
Simplifying this equation gives us the fundamental frequency of the cord: f1 ≈ 14.18 Hz
The masses of the earth and moon are 5.98 x 1024 and 7.35 x 1022 kg, respectively. Identical amounts of charge are placed on each body, such that the net force (gravitational plus electrical) on each is zero. What is the magnitude of the charge placed on each body?
Answer:
The magnitude of charge on each is [tex]5.707\times 10^{13} C[/tex]
Solution:
As per the question:
Mass of Earth, [tex]M_{E} = 5.98\times 10^{24} kg[/tex]
Mass of Moon, [tex]M_{M} = 7.35\times 10^{22} kg[/tex]
Now,
The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:
[tex]F_{G} = \frac{GM_{E}M_{M}}{d^{2}}[/tex] (1)
Now,
If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:
[tex]F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}[/tex] (2)
Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:
Equating eqn (1) and (2):
[tex]\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}[/tex]
[tex](6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}[/tex]
[tex]\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q[/tex]
Q = [tex]\pm 5.707\times 10^{13} C[/tex]
To find the velocity and acceleration vectors for uniform circular motion and to recognize that this acceleration is the centripetal acceleration. Suppose that a particle's position is given by the following expression: r(t)=R[cos(ωt)i^+sin(ωt)j^] =Rcos(ωt)i^+Rsin(ωt)j^.The particle's motion can be described by ____________.(A) an ellipse starting at time t=0 on the positive x axis(B) an ellipse starting at time t=0 on the positive y axis(C) a circle starting at time t=0 on the positive x axis(D) a circle starting at time t=0 on the positive y axis
Answer:
(C) a circle starting at time t=0 on the positive x axis
Explanation:
particle's position is
r(t)=R[cos(ωt)i^+sin(ωt)j^] =Rcos(ωt)i^+Rsin(ωt)j^
this is a parametric equation of a circle, because the axis at x and y are the same = R.
for t=0:
r=Ri^
so: circle starting at time t=0 on the positive x axis
On the other hand:
[tex]v=\frac{dx}{dt}= Rw[-sin(wt)i+cos(wt)j]\\a=\frac{dv}{dt}= Rw^{2}[-cos(wt)i-sin(wt)j][/tex]
The value of the magnitude of the acceleration is:
[tex]a=Rw^{2}(cos^{2}(wt)+sin^{2}(wt))=Rw^{2}[/tex]
we can recognise that this represent the centripetal acceleration.
During constant linear acceleration, what changes uniformly? A. acceleration B. distance C. displacement D. velocity E. all of these
Answer:D-velocity
Explanation:
Given
Constant linear acceleration i.e. acceleration is constant
a=constant
and we know [tex]\frac{\mathrm{d}V}{\mathrm{d} t}=a[/tex]
Thus velocity is changing uniformly to give constant acceleration
While displacement and distance are function of time thus they are not changing uniformly.
During constant linear acceleration, the following quantity changes uniformly: velocity. Therefore option D is correct.
Constant linear acceleration refers to a situation where an object's acceleration remains constant over a specific period of time. In this case, the rate of change of velocity is uniform, meaning the velocity increases or decreases by the same amount in equal time intervals.
Velocity is the rate of change of displacement and is directly affected by acceleration. During constant linear acceleration, the velocity changes uniformly. It either increases or decreases at a constant rate, depending on the direction of acceleration.
Therefore,
During constant linear acceleration, the following quantity changes uniformly: velocity.
Know more about linear acceleration:
https://brainly.com/question/13385172
#SPJ6
An arrow is shot straight up in the air with an initial speed of 250 ft/s. If on striking the ground, it embeds itself 4.00 in into the ground, find the magnitude of the acceleration (assumed constant) required to stop the arrow in units of ft/sec^2
Answer:
93750 ft/s²
Explanation:
t = Time taken
u = Initial velocity = 250 ft/s (It is assumed that it is speed of the arrow just when it enter the ground)
v = Final velocity = 0
s = Displacement = 4 in = [tex]\frac{4}{12}=\frac{1}{3}\ feet[/tex]
a = Acceleration
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-250^2}{2\times \frac{1}{3}}\\\Rightarrow a=-93750\ ft/s^2[/tex]
The magnitude of acceleration is 93750 ft/s²
Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately 4.0 cm/yr. Assuming this to be a constant rate, how many years will pass before the radius of the Moon's orbit increases by 3.84x 10^7 m (10%)?
Answer:
967500000 years
Explanation:
The Speed at which the radius of the orbit of the Moon is increasing is 4 cm/yr
Converting to m
1 m = 100 cm
[tex]1\ cm=\frac{1}{100}\ m[/tex]
[tex]4\ cm\y=\frac{4}{100}=0.04\ m/yr[/tex]
The distance by which the radius increases is 3.84×10⁷ m
Time = Distance / Speed
[tex]\text{Time}=\frac{3.87\times 10^7}{0.04}\\\Rightarrow \text{Time}=967500000\ yr[/tex]
967500000 years will pass before the radius of the orbit increases by 10%.
Final answer:
The radius of the Moon's orbit increases by approximately 4 cm/year. To calculate the number of years it will take for the radius to increase by 3.84 × 10^6 m, we can use the formula Time = Change in Distance / Rate of Increase. The answer is approximately 9.6 × 10^7 years.
Explanation:
The radius of the Moon's orbit is increasing at a rate of approximately 4 cm/year. To find out how many years will pass before the radius of the Moon's orbit increases by 3.84 × 10^6 m, we can use the formula:
Time = Change in Distance / Rate of Increase
Substituting the given values, we get:
Time = (3.84 × 10^6 m) / (4 cm/year)
Now, we need to convert the units so that they are consistent. 3.84 × 10^6 m is equivalent to 3.84 × 10^8 cm. Substituting this value into the equation, we get:
Time = (3.84 × 10^8 cm) / (4 cm/year)
Canceling out the units, we find that:
Time = 9.6 × 10^7 years.
Why is it easier to use a potentiometer in a circuit rather than two separte resistors in series with one another?
Answer:
Because by potentiometer we can drop voltage according to our requirement but in set of two series resistors we can not do this.
Explanation:
As we know that the potentiometer can act as a variable resistor but resistors have constant value.
In any circuit if we insert resistors they will drop a constant value if we insert a potentiometer we will drop the value of voltage by according to our will.
Therefore, by the above discussion it can be say that the potentiometer is better than two resesters in series.
A 35-mm single lens reflex (SLR) digital camera is using a lens of focal length 35.0 mm to photograph a person who is 1.80 m tall and located 3.60 m from the lens. (a) How far is the CCD sensor from the lens when the person is in focus?
(b) How tall is the person's image on the CCD sensor?
Answer:
a) 35.44 mm
b) 17.67 mm
Explanation:
u = Object distance = 3.6 m
v = Image distance
f = Focal length = 35 mm
[tex]h_u[/tex]= Object height = 1.8 m
a) Lens Equation
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{3600}\\\Rightarrow \frac{1}{v}=\frac{713}{25200} \\\Rightarrow v=\frac{25200}{713}=35.34\ mm[/tex]
The CCD sensor is 35.34 mm from the lens
b) Magnification
[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{35.34}{3600}[/tex]
[tex]m=\frac{h_v}{h_u}\\\Rightarrow -\frac{35.34}{3600}=\frac{h_v}{1800}\\\Rightarrow h_v=-\frac{35.34}{3600}\times 1800=-17.67\ mm[/tex]
The person appears 17.67 mm tall on the sensor
The position vector of an object of mass 0.5 kg is given by r 2.0t2i+ 3.0tj m where t represents time. Which ONE of the following statements is FALSE when a time of t 2.0 s has elapsed. a) The object has moved a distance of 10 m. b) The momentum of the object is 41+1.5/ kg.m.s c) The kinetic energy of the object is 18 J d) The force on the object acts perpendicular to the direction of the unit vectorj e) The magnitude of the force acting on the object is 4 N
Answer:
Statement E is false.
Explanation:
Statement a says that the object would have moved a distance of 10 m.
At t = 0
x0 = 0
y0 = 0
At t= 2
x2 = 2 * 2^2 = 8
y2 = 3 * 2 = 6
[tex]d2 = \sqrt{x2^2 + y2^2}[/tex]
[tex]d2 = \sqrt{8^2 + 6^2} = 10 m[/tex]
Statement A is true.
Statement B says that the momentum of the object would be 4*i+1.5*j
To test this we need to know the speed.
The speed is the first derivative by time:
v = 4*t*i + 3*j
At t = 2 the speed is
v2 = 8*i + 3*j
The momentum is
P = m * v
P2 = 0.5 * (8*i + 3*j) = (4*i+1.5*j)
This statement is true.
Statement C says that the kinetic energy of the object would be 18 J.
The magnitude of the speed is
[tex]v = \sqrt{8^2 + 3^2} = 8.54 m/s[/tex]
The kinetic energy is
Ek = 1/2 * m * v^2
Ek = 1/2 * 0.5 * 8.54^2 = 18 J
Statement C is true
Statement D says that the force acts perpenducular to the direction of the unit vector j.
The acceleration is the second derivative respect of time
a = 4*i + 0*j
Since there is no acceleration in the direction of j we can conclude that the force is perpendicular to the direction of j.
Statement D is true.
Statement E says that the force acting on the object is of 4 N
f = m * a
f = 0.5 * 4 = 2 N
Statement E is false.
We attach two blocks of masses m1 = 7 kg and m2 = 7 kg to either end of a spring of spring constant k = 1 N/m and set them into oscillation. Calculate the angular frequency ω of the oscillation.
Answer:
The angular frequency [tex]\omega[/tex] of the oscillation is [tex]0.58s^{-1}[/tex]
Explanation:
For this particular situation, the angular frequency of the system is given by
[tex]\omega=\sqrt{\frac{m_1+m_2}{m_1m_2}k}=\sqrt{\frac{7 kg+5 kg }{7kg *5 kg}1\frac{N}{m}}=\sqrt{\frac{3}{35s^2}}\approx 0.58s^{-1}[/tex]
Consider the scenario where a person jumps off from the edge of a 1 m high platform and lands on the ground Suppose his initial jumping speed was 3 m/s. For how long was this person in the air?
Answer:
For 0.24 sec the person was in the air.
Explanation:
Given that,
Height = 1 m
Initial velocity = 3 m/s
We need to calculate the time
Using equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Where, u = initial velocity
s = height
Put the value into the formula
[tex]1 =3\times t+\dfrac{1}{2}\times9.8\times t^2[/tex]
[tex]4.9t^2+3t-1=0[/tex]
[tex]t = 0.24\ sec[/tex]
On neglecting the negative value of time
Hence, For 0.24 sec the person was in the air.
Consider two metallic rods mounted on insulated supports. One is neutral, the other positively charged. You bring the two rods close to each, but without contact, and briefly ground the the neutral rod by touching it with your hand. What would be resulting charge (if any) on initially neutral rod?
Hi!
Letw call A to the initially neutral rod, and B to the positively charged. When they are close to each other, the positive charges in B attract the negative charges in A, and repelle the positive ones. If you ground A, negative charges from ground (your body, in this case), flow to A attracted by the positive charges in B, and positive charges in A flow to ground, so finally A results negatively charged
Answer:
Negative charges
Explanation:
The procedure described above is known in physics as charging by electrostatic induction. If we desire to impart negative charges to a hitherto neutral rod, we bring a positively charged rod near it without allowing the two insulated rods to touch each other. If the neutral rod is earthed, negative charges remain on the rod.