Answer:
a.V=8tj+k
b.a=8j
Explanation:
Given:
Position r= i+4t^2j +tk
Nb r is position in metre and time in seconds
a.velocity is change in position/ change in time
v= ∆r/∆t =dr/dt
V=d ( i+ 4t^2j+tk)/dr
Differenting with respect to (t)
V=8tj+K
b.acceleration = change in velocity/change in time
a= ∆v/∆r =dv/dt
a=d (8tj+k)/dt
a= 8j
Answer:
(a) velocity, v = 8t j + k
(b) acceleration, a = 8 j
Explanation:
The position of the particle as a function of time is given as;
r = i + 4t² j + t k --------------------(i)
(a) To get the expression of its velocity, v, find the derivative of its position with respect to time by differentiating equation (i) with respect to t as follows;
v = dr / dt = 0 + 8t j + k
v = dr / dt = 8t j + k
v = 8t j + k ----------------------(ii)
Therefore, the equation/expression for the particle's velocity (v) is
v = 8t j + k
(b) To get the expression of its acceleration, a, find the derivative of its velocity with respect to time by differentiating equation (ii) with respect to t as follows;
a = dv / dt = t j + 0
a = dv / dt = t j
a = 8 j
Therefore, the expression for the particle's acceleration, a, is a = 8 j
A toy helicopter is flying in a straight line at a constant speed of 4.5 m/s. If a projectile is launched vertically with an initial speed of v0 = 28 m/s, what horizontal distance d should the helicopter be from the launch site S if the projectile is to be traveling downward when it strikes the helicopter? Assume that the projectile travels only in the vertical direction.
Final answer:
To find the horizontal distance, we use the equation d = horizontal velocity * T, where T is the time of flight. Using the given values, the horizontal distance is 25.71 m.
Explanation:
To determine the horizontal distance the helicopter should be from the launch site, we need to find the time it takes for the projectile to reach the helicopter. Since the projectile only travels in the vertical direction, we can use the equation:
T = (2 * v0) / g
where T is the time of flight, v0 is the initial vertical velocity, and g is the acceleration due to gravity. Plugging in the values, we get:
T = (2 * 28 m/s) / 9.8 m/s^2 = 5.71 s
Now, we can find the horizontal distance using the equation:
d = horizontal velocity * T
The horizontal velocity of the helicopter is its constant speed, which is 4.5 m/s. Plugging in the values, we get:
d = 4.5 m/s * 5.71 s = 25.71 m
A point charge is placed at the center of a spherical Gaussian surface. The electricflux ΦEischangedif(a) a second point charge is placed outside the sphere(b) the point charge is moved outside the sphere(c) the point charge is moved off center, but still inside the original sphere(d) the sphere is replaced by a cube of one-tenth the volume (the original charge remains in thecenter)
Answer:
(b) the point charge is moved outside the sphere
Explanation:
Gauss' Law states that the electric flux of a closed surface is equal to the enclosed charge divided by permittivity of the medium.
[tex]\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}[/tex]
According to this law, any charge outside the surface has no effect at all. Therefore (a) is not correct.
If the point charge is moved off the center, the points on the surface close to the charge will have higher flux and the points further away from the charge will have lesser flux. But as a result, the total flux will not change, because the enclosed charge is the same.
Therefore, (c) and (d) is not correct, because the enclosed charge is unchanged.
How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, relative to infinity, is 1.50 kV?
Answer:
[tex]q=2.08*10^{-8}C\\ q=20.8nC[/tex]
Explanation:
Given data
Electric potential V=1.50 kV
Diameter d=25.0 cm
radius=diameter/2=25/2=12.5 cm=0.125 m
to find
We are asked to find excess charge
Solution
As inside the sphere the electric field is zero everywhere and potential is same at every point inside the sphere and on its surface
So we can use the value of the potential to get the charge q on the sphere
[tex]V=\frac{1}{4\pi E}\frac{q}{R}\\ q=4\pi ERV\\Where\\4\pi E=\frac{1}{9.0*10^{9}N.m^{2}/C^{2} }\\ q=(\frac{1}{9.0*10^{9}N.m^{2}/C^{2} })(0.125m)(1.50*10^{3}V )\\q=2.08*10^{-8}C\\ q=20.8nC[/tex]
To make the potential of a copper sphere's center 1.50 kV relative to infinity, an excess charge of approximately 2.09 x 10^-8 C must be placed on it.
Explanation:The potential V of a charged sphere is given by the equation V = kQ/R, where k is Coulomb's constant (8.99 x 109 Nm2/C2), Q is the charge on the sphere, and R is the radius. We can rearrange this to calculate the sphere's charge: Q = VR/k. The sphere's radius is half its diameter, hence 25.0 cm / 2 = 12.5 cm = 0.125 m. Substituting the given values in, we have Q = 1.50 x 103 V x 0.125 m / 8.99 x 109 Nm2/C2, recalling that 1 kV = 103 V. Doing the calculation, we find the excess charge Q to be approximately 2.09 x 10-8 C.
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Calculate the speed of an electron (in m/s) after it accelerates from rest through a potential difference of 160 V.
Answer:
v = 7.5*10⁶ m/s
Explanation:
While accelerating through a potential difference of 160 V, the electron undergoes a change in the electric potential energy, as follows:
ΔUe = q*ΔV = (-e)*ΔV = (-1.6*10⁻¹⁹ C) * 160 V = -2.56*10⁻¹⁷ J (1)
Due to the principle of conservation of energy, in absence of non-conservative forces, this change in potential energy must be equal to the change in kinetic energy, ΔK:
ΔK = Kf -K₀
As the electron accelerates from rest, K₀ =0.
⇒ΔK =Kf = [tex]\frac{1}{2}*me*vf^{2}[/tex] (2)
From (1) and (2):
ΔK = -ΔUe = 2.56*10⁻¹⁷ J = [tex]\frac{1}{2}*me*vf^{2}[/tex]
where me = mass of the electron = 9.1*10⁻³¹ kg.
Solving for vf:
[tex]vf =\sqrt{\frac{2*(2.56e-17J)}{9.1e-31kg} } =7.5e6 m/s[/tex]
⇒ vf = 7.5*10⁶ m/s
Final answer:
The speed of an electron after accelerating from rest through a potential difference of 160 V can be calculated using the conservation of energy principle, resulting in a velocity of approximately 5.93 × 10^6 m/s.
Explanation:
To calculate the speed of an electron after it accelerates from rest through a potential difference, we use the concept of conservation of energy where the electrical potential energy converted into kinetic energy is expressed as qAV = ½mv². Here, q is the charge of the electron (-1.60 × 10^-19 C), V is the potential difference (160 V), m is the mass of the electron (9.11 × 10^-13 kg), and v is the velocity of the electron we need to find. Rearranging the equation to solve for v and substituting the given values gives:
v = sqrt(2 × q × V / m) = √(2 × (-1.60 × 10^-19 C) × 160 V / (9.11 × 10^-13kg))
Performing the calculation yields a velocity of approximately 5.93 × 10^6 m/s.
A 1.75 µF capacitor and a 6.00 µF capacitor are connected in series across a 3.00 V battery. How much charge (in µC) is stored on each capacitor?
Answer:
Explanation:
Given
First capacitor magnitude [tex]C=1.75\ \mu F[/tex]
Second capacitor magnitude [tex]C=6.00\ \mu F[/tex]
Voltage of battery [tex]V=3.00\ V[/tex]
Both capacitor are connected in series so net capacitor is given by
[tex]\frac{1}{C_{net}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}[/tex]
[tex]\frac{1}{C_{net}}=\frac{1}{1.75}+\frac{1}{6}[/tex]
[tex]C_{net}=\frac{6\times 1.75}{1.75+6}[/tex]
[tex]C_{net}=1.35\ \mu F[/tex]
So charge Across each capacitor is given by
[tex]Q=C_{net}\times V[/tex]
[tex]Q=1.35\times 10^{-6}\times 3[/tex]
[tex]Q=4.064\ \mu C[/tex]
A stainless-steel-bottomed kettle, its bottom 22cm cm in diameter and 2.2 mmmm thick, sits on a burner. The kettle holds boiling water, and energy flows into the water from the kettle bottom at 800 WW .What is the temperature of the bottom surface of the kettle? Thermal conductivity of stainless steel is 14 W/(m?K).
Answer:
103.3°C
Explanation:
k = Thermal conductivity of stainless steel = 14 W/m² K
P = Power = 800 W
d = Diameter = 22 cm
t = Thickness of kettle = 2.2 mm
[tex]\Delta T[/tex] = Change in temperature = [tex](T_2-100)[/tex]
100°C boiling point of water
Power is given by
[tex]P=\dfrac{kA\Delta T}{t}\\\Rightarrow P=\dfrac{kA(T_2-T_1)}{t}\\\Rightarrow T_2=\dfrac{Pt}{kA}+T_1\\\Rightarrow T_2=\dfrac{800\times 2.2\times 10^{-3}}{14\times \dfrac{\pi}{4} 0.22^2}+100\\\Rightarrow T_2=103.3\ ^{\circ}C[/tex]
The temperature of the bottom of the surface 103.3°C
The temperature of the bottom surface of the kettle is approximately 1.593 Kelvin higher than the boiling water.
Explanation:To calculate the temperature of the bottom surface of the kettle, we can use the formula for heat conduction: q = (k * A * ΔT) / t. Here, q is the heat transfer rate, k is the thermal conductivity of stainless steel, A is the surface area of the bottom of the kettle, ΔT is the temperature difference between the bottom surface of the kettle and the boiling water, and t is the thickness of the bottom of the kettle. Rearranging the formula, we can solve for ΔT as:
ΔT = (q * t) / (k * A)
Plugging in the given values, we find:
ΔT = (800 W * 0.0022 m) / (14 W/(m·K) * π * (0.22 m/2)²)
Calculating this, we find ΔT ≈ 1.593 K. Therefore, the temperature of the bottom surface of the kettle would be approximately 1.593 Kelvin higher than the boiling water.
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a. How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter to produce an electric field of 1350 N/C just outside the surface of the sphere.b. What is the electric field at a point 10.0 cm outside the surface of the sphere?
Answer:
Explanation:
The electric field outside the sphere is given as,
E = k Q /r²
here Q = n x 1.6 x 10⁻¹⁹ C
where n is the number of electons
if the dimeter of sphere d= 25 cm= 0.25 m
then the radius r = 0.125 m
we get
n= E r²/ k x 1.6 x 10⁻¹⁹ C
n = 1350N/C x (0.125m)² / (8.99 x 10⁹ N m²/C² x 1.6 x 10⁻¹⁹ C)
n = 14664731646
Why do optical astronomers often put their telescopes at the tops of mountains, whereas radio astronomers sometimes put their telescopes in deep valleys?
Explanation:
The visible light coming from celestial bodies from space are affected most by the atmosphere. The visible light suffer blurring and absorption. So, to avoid such situation optical telescopes are kept in mountains where atmosphere is thin.
Whereas radio signal are not affected by the atmosphere so, they need not be placed in mountains.However, it is affected by various noises from the man made devices. So, to avoid these noises radio telescopes are kept in deep valleys.
Final answer:
Optical telescopes are placed on mountains to minimize atmospheric interference, reduce light pollution, and capture clearer images, while radio telescopes are located in valleys to shield them from man-made radio interference for clearer radio signal observations.
Explanation:
Optical astronomers often put their telescopes at the tops of mountains because these locations offer several advantages that are critical for observing the cosmos. Sir Isaac Newton mentioned that a serene and quiet air, often found on the tops of high mountains, is beneficial for reducing the confusion of rays caused by the atmosphere's tremors. This is echoed by modern practices where observatories are located in high, dry, and dark sites to minimize atmospheric interference, reduce light pollution, and avoid water vapor which absorbs infrared light.
The Andes Mountains in Chile, the desert peaks of Arizona, and the summit of Maunakea in Hawaii are examples of such ideal locations. On the other hand, radio astronomers sometimes place their telescopes in deep valleys to shield them from radio interference from human-made sources such as cell phones and electrical circuits. The natural geography of valleys can act as a barrier against these disturbances, providing a clearer signal for radio observations.
An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed of the plane (its speed in still air) is 320.0 km/h (about 200 mi/h), in which direction should the pilot head? (b) What is the speed of the plane over the ground? Draw a vector diagram.
Answer:
a) 75.5 degree relative to the North in north-west direction
b) 309.84 km/h
Explanation:
a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it
So the pilot should head to the West-North direction at an angle of
[tex]cos(\alpha) = 80/320 = 0.25[/tex]
[tex]\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0[/tex] relative to the North-bound.
b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is
[tex]320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h[/tex]
Answer:
a) Ф=14°, north of west
b) 310 km/h
Explanation:
we have,
[tex]v_{p/G}[/tex], velocity of plane relative to the ground(west)
[tex]v_{p/A}[/tex],velocity of plane relative to the air(320 km/h)
[tex]v_{A/G}[/tex],velocity of air relative to the ground(80 km/h, due south).
[tex]v_{p/G}[/tex]=[tex]v_{p/A}[/tex]+[tex]v_{A/G}[/tex].........(1)
a) sin(Ф)=[tex]\frac{v_{A/G}}{v_{p/A}}[/tex]
=[tex]\frac{80km/h}{320km/h}[/tex]
Ф=14°, north of west
b) using Pythagorean theorem
[tex]v_{p/G}=\sqrt{v^2_{p/A}+v^2_{A/G}}[/tex]
[tex]v_{p/G}[/tex]=310 km/h
note:
diagram is attached
A 2-ft3 tank contains a gas at 2 atm(g) and 60 oF. This tank is connected to a second tank containing the same gas at atmospheric pressure and 60 oF. The two tanks are connected and allowed to reach equilibrium. The final conditions are measured to be 1 atm(g) and 60oF. What is the volume of the second tank
Answer : The volume of the second tank is, [tex]4ft^3[/tex]
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex]
or,
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 2 atm
[tex]P_2[/tex] = final pressure of gas = 1 atm
[tex]V_1[/tex] = initial volume of gas = [tex]2ft^3[/tex]
[tex]V_2[/tex] = final volume of gas = ?
Now put all the given values in the above equation, we get:
[tex]2atm\times 2ft^3=1atm\times V_2[/tex]
[tex]V_2=4ft^3[/tex]
Therefore, the volume of the second tank is, [tex]4ft^3[/tex]
The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s. A seismograph records the arrival of the transverse waves 56.4 s after that of the longitudinal waves. How far away was the earthquake? Answer in units of km.
Answer:
[tex]d=691.71km[/tex]
Explanation:
The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:
[tex]t=\frac{d}{v_t}-\frac{d}{v_l}[/tex]
Here d is the distance at which the earthquake take place and [tex]v_t, v_l[/tex] is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:
[tex]t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km[/tex]
The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by various atmospheric processes, including lightning. What is the excess charge on the surface of the earth?
Answer:
The excess charge on earth's surface was calculated to be 4.56 × 10⁵ C
Explanation:
Using the formula for an electric field;
E = kQ/r²
k = 1/(4πε₀) = 8.99 × 10⁹ Nm²/C²
E = 100N/C
r = radius of the earth = 6400 km = 6400000m
Q = Er²/k = 100 × (6400000)²/(8.99 × 10⁹)
Q = 455617.4 C = 4.56 × 10⁵ C
Hope this helps!!!
The excess charge on the surface of the earth is 4.55×10⁵C
To calculate the charge on the surface of the earth, we apply the equation of electric field strength, that is:
[tex]E = k\frac{Q}{r^2}[/tex]
here, E is the electric field strength = 100N/C (given)
k = 9×10⁹ Nm²/C²
Q is the charge, and
r = distance = radius of earth = 6.4×10⁶ m
Now,
[tex]Q = \frac{r^2E}{k}\\ \\=\frac{(6.4*10^6)^2*100}{9*10^9}\\\\Q=4.55*10^5 C[/tex] is the charge acquired on the surface of the earth.
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A steel wire in a piano has a length of 0.600 m and a mass of 5.200 ✕ 10−3 kg. To what tension must this wire be stretched so that the fundamental vibration corresponds to middle C (fC = 261.6 Hz on the chromatic musical scale)?
Answer:
854.39 N
Explanation:
The formula for the fundamental frequency of a stretched string is given as,
f = 1/2L√(T/m)..................... Equation 1
Where f = fundamental frequency, L = Length of the wire, T = Tension, m = mass per unit length.
Given: f = 261.6 Hz, L = 0.6 m, m = (5.2×10⁻³/0.6) = 8.67×10⁻³ kg/m.
Substitute into equation 1
261.6 = 1/0.6√(T/8.67×10⁻³)
Making T the subject of the equation,
T = (261.6×0.6×2)²(8.67×10⁻³)
T =854.39 N
Hence the tension of the wire is 854.39 N.
We have that for the Question "To what tension must this wire be stretched so that the fundamental vibration corresponds to middle C "
Answer:
Tension = [tex]512.43N[/tex]
From the question we are told
a piano has a length of 0.6m and a mass of [tex]5.2 * 10^{−3} kg[/tex]
Generally the equation for frequency is mathematically given as
[tex]F = \frac{V}{2L}[/tex]
where,
[tex]V = \sqrt\frac{T}{U}[/tex]
Therefore,
[tex]261.6 = \frac{V}{2*0.76}\\\\V = 261.6*2*0.6\\\\V = 313.92m/s[/tex]
so,
[tex]v^2 = \frac{T}{U}\\\\T = V^2 * U\\\\T = 512.43N[/tex]
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A hydrogen atom has a single proton at its center and a single electron at a distance of approximately 0.0539 nm from the proton. a. What is the electric potential energy in joules?b. What is the significance of the sign of the answer?
To solve this problem we will apply the concepts related to electrostatic energy, defined as,
[tex]U_e = \frac{kq_1q_2}{d^2}[/tex]
Here,
k = Coulomb's constant
[tex]q_{1,2}[/tex] = Charge of each object (electron and proton, at this case the same)=
d = Distance
Replacing with our values we have that
[tex]U_e = \frac{(9*10^9Nm^2)(1.6*10^{-19}C)(-1.6*10^{-19}C)}{0.0539*10^{-9}m}[/tex]
[tex]U_e = -4.27*10^{-18}J[/tex]
Therefore for the Part A the answer is [tex]-4.27*10^{-18}J[/tex] and por the Part B the sign indicates that the force between the proton and electron is attractive
During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive force of 115 N to the tire's rim. The mass of the wheel is 1.70 kg and, for the purpose of this problem, assume that all of this mass is concentrated on the outside radius of the wheel. The diameter of the wheel is 50.0 cm. A chain passes over a sprocket that has a diameter of 9.50 cm. In order for the wheel to have an angular acceleration of 4.90 rad/s2, what force, in Newtons, must be applied to the chain? (Enter the magnitude only.) N
Answer:
616.223684211 N
Explanation:
[tex]F_r[/tex] = Resistive force on the wheel = 115 N
F = Force acting on sprocket
[tex]r_2[/tex] = Radius of sprocket = 4.75 cm
[tex]r_1[/tex] = Radius of wheel = 25 cm
Moment of inertia is given by
[tex]I=mr_1^2\\\Rightarrow I=1.7\times 0.25^2\\\Rightarrow I=0.10625\ kgm^2[/tex]
Torque
[tex]\tau=I\alpha\\\Rightarrow \tau=0.10625\times 4.9\\\Rightarrow \tau=0.520625\ Nm[/tex]
Torque is given by
[tex]\tau=Fr_2-F_rr_1\\\Rightarrow F=\dfrac{\tau+F_rr_1}{r_2}\\\Rightarrow F=\dfrac{0.520625+115\times 0.25}{0.0475}\\\Rightarrow F=616.223684211\ N[/tex]
The force on the chain is 616.223684211 N
Earth's atmosphere is constantly bombarded by cosmic ray protons that originate somewhere in space. If the protons all passed through the atmosphere, assume that each square meter of Earth's surface would intercept protons at the average rate of 1110 protons per second. What would be the electric current intercepted by the total surface area of the planet?
Answer:
Explanation:
Each square meter of earth surface intercepts 1110 protons per second.
Total surface of the earth will intercept no of proton equal to
= 1110 x area of surface
= 1110 x 5.1 x 10¹⁴ m²
= 5661 x 10¹⁴
Total charge
= 1.6 x 10⁻¹⁹ x 5661 x 10¹⁴ coulomb per second
= .09 A .
electric current intercepted by the total surface area of the planet = .09 A.
An electric motor exerts a constant torque of 10 Nm on a grindstone mounted on its shaft. The moment of inertia of the grindstone is I = 2.0 kgm2 . The system starts from rest. A. Determine the angular acceleration of the shaft when this torque is applied. B. Determine the kinetic energy of the shaft after 8 seconds of operation. C. Determine the work done by the motor in this time. D. Determine the average power delivered by the motor over this time interval
Answer:
Angular acceleration = 5 rad /s ^2
Kinetic energy = 0.391 J
Work done = 0.391 J
P =6.25 W
Explanation:
The torque is given as moment of inertia × angular acceleration
angular acceleration = torque/ moment of inertia
= 10/2= 5 rad/ s^2
The kinetic energy is = 1/2 Iw^2
w = angular acceleration/time
=5/8= 0.625 rad /s
1/2 × 2× 0.625^2
=0.391 J
The work done is equal to the kinetic energy of the motor at this time
W= 0.391 J
The average power is = torque × angular speed
= 10× 0.625
P = 6.25 W
A. The angular acceleration is 5 rad/s². B. The kinetic energy of the shaft after 8 seconds is 16000 J. C. The work done by the motor in this time is 1600 J. D. The average power delivered by the motor over this time interval is 200 W.
Explanation:A. Angular acceleration can be found using the rotational analog to Newton's second law a = net τ/I. The moment of inertia I is given and the torque τ can be found from the given torque value. So, τ = 10 Nm and I = 2.0 kgm2. Substituting these values, we get a = 10 Nm / 2.0 kgm2 = 5 rad/s².
B. Using the equation K = (1/2) I ω², where I is the moment of inertia and ω is the angular velocity, we can calculate the initial angular velocity ω₁ as 0 rad/s. Then, the final angular velocity ω can be found using the relationship ω = ω₁ + αt. So, ω = 0 rad/s + (5 rad/s²)(8 s) = 40 rad/s. Finally, substituting these values into the kinetic energy equation, we have K = (1/2)(2.0 kgm2)(40 rad/s)² = 16000 J.
C. The work done by the motor can be found using the equation W = τθ, where τ is the torque and θ is the angular displacement. In this case, θ = ω₁t + (1/2)αt² = 0 rad/s(8 s) + (1/2)(5 rad/s²)(8 s)² = 160 rad. Therefore, W = (10 Nm)(160 rad) = 1600 J.
D. The average power can be calculated using the equation P = W/t, where W is the work done and t is the time interval. Substituting the values we found in the previous calculations, we have P = 1600 J / 8 s = 200 W.
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Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At which points in time does a zero net force act on the ball? Ignore air resistance.
(A) When you hold the ball still in your hands after catching it
(B) Just after the ball first leaves your hands.
(C) At the instant the ball reaches its highest point.
(D) At the instant the falling ball hits your hands.
(E) When you hold the ball still in your hands before it is thrown.
Answer:
(A) When you hold the ball still in your hands after catching it
(E) When you hold the ball still in your hands before it is thrown.
Explanation:
According to Newton's 1st law, objects subjected to 0 net force would maintain a constant velocity or staying at rest. This is not the case for the ball when it leaves your hands. This ball would always be subjected to gravitational acceleration g so its velocity changes. So only (A) and (E) are correct when the ball stays still in your hands.
Final answer:
A zero net force acts on the basketball when it is held still after catching it (A) and before throwing it (E) because these are the points where there is no acceleration due to balanced forces. Points B, C, and D have non-zero net forces because the ball is accelerating due to the force of gravity.
Explanation:
When considering a basketball being thrown straight up and ignoring air resistance, a zero net force acts on the basketball:
(A) When holding the ball still after catching it because there is no acceleration and gravity is balanced by the upward force from your hands.
(E) When holding the ball still before throwing it for the same reason as in (A).
The points at which the net force is not zero are:
(B) Just after the ball leaves the hand as the only force acting on the ball is gravity, resulting in acceleration and therefore a net force downwards.
(C) At the highest point because gravity is still acting downwards, although the ball is temporarily at rest.
(D) As the ball hits the hands because the hands apply an upward force to decelerate the ball, which is different from the force of gravity, hence not zero net force.
Overall, the net force on an object is related to its acceleration due to Newton's second law of motion, and since acceleration occurs whenever the ball is in motion and not at rest, there is a net force acting on the ball during these periods.
A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across a 2.25 mm length of the wire? VV = nothing VV SubmitRequest Answer Part B What would the potential difference in part A be if the wire were silver instead of copper, but all else was the same?
Answer:
a) 2.063*10^-4
b) 1.75*10^-4
Explanation:
Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:
[tex]A=\frac{\pi }{4}d^{2} \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\[/tex]
a) The Potential difference across a 2.00 in length of a 14-gauge copper
wire:
L= 2.00 m
From Table Copper Resistivity [tex]p[/tex]= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:
[tex]R=\frac{pL}{A}[/tex]
=0.0165Ω
The Potential difference across the copper wire is:
V=IR
=2.063*10^-4
b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m
The Resistance of the Silver wire is:
[tex]R=\frac{pL}{A}[/tex]
=0.014Ω
The Potential difference across the Silver wire is:
V=IR
=1.75*10^-4
A 2-m telescope can collect a given amount of light in 1 hour. Under the same observing conditions, how much time would be required for a 6-m telescope to perform the same task? A 12-m telescope?
Answer
given,
diameter of telescope = 2 m
time to collect area = 1 hour
Diameter of the another telescope = 6 m
The light collected by the telescope is directly proportional to the area of its primary mirror.
Area of the mirror is directly proportional to the square of diameter.
so,
6 m diameter telescope will carry (6/2)² = 9 times more light than 2 m telescope.
Time taken to collect light = 60/9 = 6.67 minutes.
now, For 12 m telescope
12 m diameter telescope will carry (12/2)² = 36 times more time than 2 m telescope.
Time taken by 12 m telescope to collect light= 60/36 = 1.7 minutes.
To find the time required for a 6-m telescope and a 12-m telescope to perform the same task as a 2-m telescope, we can use a formula and rearrange it to solve for the time taken.
Explanation:To find out how much time would be required for a 6-m telescope and a 12-m telescope to perform the same task as a 2-m telescope, we can use the formula:
(light collected by telescope 1) × (time taken by telescope 1) = (light collected by telescope 2) × (time taken by telescope 2)
Let's solve for the time taken by the 6-m telescope and the 12-m telescope:
For the 6-m telescope: (light collected by 2-m telescope) × (1 hour) = (light collected by 6-m telescope) × (time taken by 6-m telescope)For the 12-m telescope: (light collected by 2-m telescope) × (1 hour) = (light collected by 12-m telescope) × (time taken by 12-m telescope)By rearranging the equation and solving for the time, we can find the answer.
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If the separation between the first and the second minima of a single-slit diffraction pattern is 6.0 mm, what is the distance between the screen and the slit?
Incomplete question.The complete question is here
If the separation between the first and the second minima of a single-slit diffraction pattern is 6.0 mm, what is the distance between the screen and the slit? The light wavelength is 500 nm and the slit width is 0.16 mm.
Answer:
The distance between first and second minimum is 1.92m
Explanation:
Given data
Wavelength of light λ=500nm
Slit width D=0.16 mm
The distance between first and second minima is 6.0 mm
To find
Distance L between the screen and the slit
Solution
As we know that
Yn=(nλL)/D
Where is L is distance between the slit and screen
D is slit width
n is order of minimum
Yn is distance of nth minimum from center maximum
λ is wavelength of light
The Distance between first and second minimum is given as:
Y₂-Y₁=6 mm
(2λL)/D-(1λL)/D=6 mm
(2λL-1λL)/D=6 mm
(λL/D)=6 mm
[tex]L=\frac{6mm(0.16mm)}{500nm}\\ L=1.92m[/tex]
The distance between first and second minimum is 1.92m
Final answer:
The screen distance is 0.19 meters, calculated using the separation between minima, slit width, and light wavelength in single-slit diffraction.
Explanation:
Solve for the distance between the screen and the slit (L).
1. Given Values:
Separation between minima (Δx) = 6.0 mm (convert to meters: 0.0060 m)
Wavelength of light (λ) = 500 nm (convert to meters: 500 x 10⁻⁹ m)
Slit width (a) = 0.16 mm (convert to meters: 0.16 x 10⁻³ m)
2. Relate Separation and Screen Distance:
We can use the relationship between the angular separation (Δθ) and the linear separation (Δx) on the screen:
Δθ ≈ Δx / L
3. Relate Separation and Minimum Position:
We know the separation (Δθ) is related to the difference between the positions of the first (m = 1) and second (m = 2) minima:
Δθ = θ₂ - θ₁ = (λ / a)
4. Combine Equations:
Substitute the equation for Δθ from step 3 into the equation from step 2:
(λ / a) ≈ Δx / L
5. Solve for Screen Distance (L):
Now we can arrange the equation to solve for L:
L ≈ Δx * a / λ
6. Plug in the Values:
L ≈ (0.0060 m) * (0.16 x 10⁻³ m) / (500 x 10⁻⁹ m)
7. Calculate and Round:
L ≈ 0.192 m (round to two significant figures)
Therefore, the distance between the screen and the slit is approximately 0.19 meters.
A point charge gives rise to an electric field with magnitude 2 N/C at a distance of 4 m. If the distance is increased to 20 m, then what will be the new magnitude of the electric field?
Answer:
0.08 N/C
Explanation:
Electric Field: This is defined as the force per unit charge exerted at a point. The expression for electric field is given as,
E = Kq/r².............................. Equation 1
Where E = Electric Field, q = Charge, k = proportionality constant, r = distance.
making q the subject of the equation,
q = Er²/k............................... Equation 2
Given: E = 2 N/C, r = 4 m,
Substitute into equation 2
q = 2(4)²/k
q = 32/k C.
When r is increased to 20 m,
E = k(32/k)/20²
E = 32/400
E = 0.08 N/C.
Hence the electric Field = 0.08 N/C
The initial electric field from a charge is 2 N/C at 4m distance. When the distance increases fivefold, the strength of the electric field decreases by the square of this factor, resulting in a new electric field magnitude of 0.08 N/C.
Explanation:This question is about the relationship between an electric field and its distance from a point charge. According to the formula for the magnitude of the electric field generated by a point charge, which is E = kQ / r^2, where E is the electric field strength, k is Coulomb's constant, Q is the charge, and r is the distance from the charge, the electric field is inversely proportional to the square of the distance.
At a distance of 4m, the electric field's magnitude is 2 N/C. If you increase the distance to 20m (which is 5 times more than the initial distance), the new electric field magnitude will be the initial magnitude divided by the square of this factor, i.e., 2 N/C divided by 5^2 = 2 N/C / 25, which equals to 0.08 N/C.
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A very large sheet of a conductor carries a uniform charge density of 4.00 pC/mm2 on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Answer:
451977.40113 N/C
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
[tex]\sigma[/tex] = Surface charge density = [tex]4\ pc/mm^2[/tex]
Electric field near the surface of a charged conductor is given by
[tex]E=\dfrac{\sigma}{\epsilon_0}\\\Rightarrow E=\dfrac{4\times 10^{-12}\times 10^{6}}{8.85\times 10^{-12}}\\\Rightarrow E=451977.40113\ N/C[/tex]
The electric field is 451977.40113 N/C
A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of the same mass on the same spring be on the Moon, where the acceleration due to gravity is one sixth that of Earth? Show your work.
To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.
The extension of the spring due to the weight of the object on Earth is 0.3m, then
[tex]F_k = F_{W,E}[/tex]
[tex]kx_1 = mg[/tex]
The extension of the spring due to the weight of the object on Moon is a value of [tex]x_2[/tex], then
[tex]kx_2 = mg_m[/tex]
Recall that gravity on the moon is a sixth of Earth's gravity.
[tex]kx_2 = m\frac{g}{6}[/tex]
[tex]kx_2 = \frac{1}{6} mg[/tex]
[tex]kx_2 = \frac{1}{6} kx_1[/tex]
[tex]x_2 = \frac{1}{6} x_1[/tex]
We have that the displacement at the earth was [tex]x_1 = 0.3m[/tex], then
[tex]x_2 = \frac{1}{6} 0.3[/tex]
[tex]x_2 = 0.05m[/tex]
Therefore the displacement of the mass on the spring on Moon is 0.05m
The same spring and mass system will have a displacement of 0.05 m on the Moon compared to 0.3 m on Earth due to the Moon's lower gravitational acceleration.
Explanation:The subject of the question relates to how a spring and mass system will behave on Earth versus the Moon. The solution requires understanding of Hooke's Law and gravitation. Hooke's Law states that the displacement of a spring is directly proportional to the force applied. On Earth, when a 0.500 kg mass is suspended and displaced by 0.3 m, this sets up a certain relationship of force (F = mg, where g is Earth's gravity, 9.8 m/s²): F = 0.500 kg * 9.8 m/s² = 4.9 Newtons. Displacement, d, is proportional to the force:F = kd, so d = F/k where k is the spring constant.
On the Moon, g is not equal to Earth's, it's 1.67 m/s². The same hanging mass on the Moon would exert a force of Fm=0.500 kg * 1.67 m/s² = 0.835 Newtons. Since the spring constant doesn't change, and F = kd still holds, the new displacement on the Moon will be greater because the force is smaller. Displacement on the Moon (dM) will be dM = Fm/k, which is Fm divided by the same k we had on Earth. Not knowing k, we do know dM = (Fm/F) * d, and (Fm/F) is equal to the ratio of the Moon's gravity to Earth's gravity, 1/6, so dM = (1/6)*0.3 m = 0.05 m.
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When a tractor moves with uniform velocity its heavier wheel rotates slowly than its lighter wheel why?
Answer:
Because of Moment of inertia.
Explanation:
Larger wheels have larger moment of inertia,when they rotate rotational energy is stored in spinnning of motion so they rotate slowly while lighter wheels move faster comparitively because they use smaller moment of inertia.The thing is to keeping tyre in contact with road,when a vehicla hits a
jolt it is difficult with heavy tyre to be in contact with road. When vehicle loses contact the driver will lose steering control which result in giving a way to sliding friction.
In the SI system of units, dynamic viscosity of water μ at temperature T (K) can be computed from μ=A10B/(T-C), where A=2.4×10-5, B=250 K and C=140 K. (6 points) (a) Determine the dimensions of A. (b) Determine the kinematic viscosity of water at 20°C. Express your results in both SI and BG units.
Answer:
0.00000103149529075 m²/s
0.0103149529075 stokes
Explanation:
A = [tex]2.4\times 10^-5[/tex]
B = 250 K
C = 140 K
T = 20°C
[tex]\rho[/tex] = Density of water = 998 kg/m³
Viscosity is given by
[tex]\mu=A\times 10^{\dfrac{B}{T-C}}\\\Rightarrow \mu=2.4\times 10^{-5}\times 10^{\dfrac{250}{273.15+20-140}}\\\Rightarrow \mu=0.00102943230017\ Pas[/tex]
Kinematic viscosity is given by
[tex]\nu=\dfrac{\mu}{\rho}\\\Rightarrow \nu=\dfrac{0.00102943230017}{998}\\\Rightarrow \nu=0.00000103149529075\ m^2/s[/tex]
The kinematic viscosity is 0.00000103149529075 m²/s
In BG units [tex]0.00000103149529075\times 10^4=0.0103149529075\ stokes[/tex]
In the SI system of units, the dynamic viscosity of water can be computed using the equation μ=A10B/(T-C). The dimensions of A are (m⋅s²)/kg. To find the kinematic viscosity of water at 20°C, we can use the formula ν=μ/ρ, where μ is the dynamic viscosity and ρ is the density of water. The kinematic viscosity of water at 20°C is approximately 0.0001 m²/s in SI units and 1 cm²/s in BG units.
Explanation:(a) Dimensions of A:
To determine the dimensions of A, we need to rearrange the equation and analyze the units on both sides. We know that viscosity has units of kilograms per meter per second (kg/m/s). Plugging in the given values for B and C, we have μ=A10B/(T-C). The dimensions of 10B/(T-C) are (dimensionless). Therefore, the dimensions of A must be kg/m/s multiplied by the inverse of the dimensions of 10B/(T-C), which is 1/(kg/m/s) or (m⋅s²)/kg.
(b) Kinematic viscosity at 20°C:
To find the kinematic viscosity of water at 20°C, we can use the formula ν=μ/ρ, where μ is the dynamic viscosity and ρ is the density of water. The density of water at 20°C is approximately 998 kg/m³. Plugging in the given values for A and C, we can calculate μ using the equation μ=A10B/(T-C). Substituting the values into the formula ν=μ/ρ, we get ν≈μ/ρ≈(A10B/(T-C))/ρ.
In the SI unit, ν≈(2.4×10-5×10250 K/(293 K-140 K))/998 kg/m³≈0.0001 m²/s.
In the BG unit, ν≈0.0001 m²/s×104 cm²/1 m²≈1 cm²/s.
A dock worker applies a constant horizontal force of 80.5 N to a block of ice on a smooth horizontal floor. The frictional force is negligible.The block starts from rest and moves a distance 13.0 m in a time 4.50 s.(a)What is the mass of the block of ice?----- I found this to be 56.9kg and got it right.(b)If the worker stops pushing at the end of 4.50 s, how far does the block move in the next 4.20s ? -
Answer:
(a). The mass of the block of ice is 62.8 kg.
(b). The distance is 24.192 m.
Explanation:
Given that,
Horizontal force = 80.5 N
Distance = 13.0 m
Time = 4.50 s
(a). We need to calculate the acceleration of the block of ice
Using equation of motion
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Put the value into the formula
[tex]13=0+\dfrac{1}{2}\times a\times(4.50)^2[/tex]
[tex]a=\dfrac{13\times2}{(4.50)^2}[/tex]
[tex]a=1.28\ m/s^2[/tex]
We need to calculate the mass of the block of ice
Using formula of force
[tex]F = ma[/tex]
[tex]m=\dfrac{F}{a}[/tex]
Put the value into the formula
[tex]m=\dfrac{80.5}{1.28}[/tex]
[tex]m=62.8\ kg[/tex]
(b). If the worker stops pushing at the end of 4.50 s,
We need to calculate the velocity
Using equation of motion
[tex]v =u+at[/tex]
Put the value into the formula
[tex]v=0+1.28\times4.50[/tex]
[tex]v=5.76\ m/s[/tex]
We need to calculate the distance
Using formula of distance
[tex]v = \dfrac{d}{t}[/tex]
[tex]d=v\times t[/tex]
Put the value into the formula
[tex]d=5.76\times4.20[/tex]
[tex]d=24.192\ m[/tex]
Hence, (a). The mass of the block of ice is 62.8 kg.
(b). The distance is 24.192 m.
Nuclear reactors use fuel rods to heat water and generate steam. Is this process endothermic or exothermic?
Explanation:
Exothermic reaction is defined as the reaction in which release of heat takes place. This also means that in an exothermic reaction, bond energies of reactants is less than the bond energies of products.
Hence, difference between the energies between the reactants and products releases as heat and therefore, enthalpy of the system will decrease.
Whereas in an endothermic reaction, heat is supplied from outside and absorbed by the reactant molecules. Hence, enthalpy of the system increases.
As water acts as a coolent and when fuel rods in a nuclear reactor are immersed in it then heat created by coolent is absorbed by water and then it changes into steam.
Since, absorption of heat occurs in the nuclear reactor. Therefore, it is an endothermic reaction.
Thus, we can conclude that nuclear reactors use fuel rods to heat water and generate steam. This process is endothermic.
Answer:
The heat produced by the nuclear reactor is an exothermic process while the heat absorbed by the water to convert into steam is an endothermic process.
Explanation:
Nuclear reactor being the heat of a nuclear power plant uses the radioactive uranium fuel to generate the heat by the process of nuclear fission in a controlled manner.
The processing of uranium is carried out into small ceramic pellets which are stacked together into sealed metal tubes known as fuel rods.
Usually more than 200 such rods are bunched together leading to the formation of a fuel assembly.
The core of the reactor is often made up of a couple hundred assemblies, according to its power level.
Inside the reactor vessel, these fuel rods are immersed into water which serve as both a coolant and moderator. The moderator helps slow down the neutrons produced by fission to sustain the chain reaction.
A sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz. What is the maximum speed of the needle?
Answer:
v = 0.2035 m/s
Explanation:
given,
Amplitude,A = 0.0127 m
Frequency, f = 2.55 Hz
maximum speed of the needle = ?
we know,
x = A cos ω t
velocity of the motion
v = - A ω sin ω t
for maximum speed
sin ω t = 1
v = - A ω
ω = 2 π f
now,
v = A 2 π f
v = 0.0127 x 2 π x 2.55
v = 0.2035 m/s
Hence, the maximum speed of the needle is equal to 0.2035 m/s
A light beam in glass (n = 1.5) reaches an air-glass interface, at an angle of 60 degrees from the surface. What is the angle of the refracted light beam from the normal in air?Note: sin(30) = 0.5, sin(45) = 0.71, sin(60) = 0.87 (all angles are in degrees)
Answer:
θ₂ = 35.26°
Explanation:
given,
refractive index of air, n₁ = 1
refractive index of glass, n₂ = 1.5
angle of incidence, θ₁ = 60°
angle of refracted light, θ₂ = ?
using Snell's Law
n₁ sin θ₁ = n₂ sin θ₂
1 x sin 60° = 1.5 sin θ₂
sin θ₂ = 0.577
θ₂ = sin⁻¹(0.577)
θ₂ = 35.26°
Hence, the refracted light is equal to θ₂ = 35.26°
When a light beam passes from one medium to another, its angle of incidence and angle of refraction are related by Snell's law. The angle of the refracted light beam from the normal in air is 90 degrees.
Explanation:When a light beam passes from one medium to another, its angle of incidence and angle of refraction are related by Snell's law. The formula for Snell's law is:
n1×sin(θ1) = n2×sin(θ2)
In this case, the light beam is passing from glass (with a refractive index of 1.5) to air. Given that the angle of incidence is 60 degrees, we can use Snell's law to find the angle of refraction from the normal in air.
Using Snell's law, we have:
1.5×sin(60) = 1×sin(θ2)
Simplifying the equation, we find:
sin(θ2) = 1.5×sin(60) = 1.5×0.87 = 1.31
However, the maximum value for the sine function is 1, so the angle of refraction cannot exceed 90 degrees (the maximum angle for a light beam in air). Therefore, the angle of the refracted light beam from the normal in air is 90 degrees.
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