Answer:
Speration between the slits [tex]d=2.86\ \mu m[/tex]
Explanation:
Given that,
Wavelength of light, [tex]\lambda=612\ nm=612\times 10^{-9}\ m[/tex]
It is incident on a double slit produces a second-order maximum at an angle of 25 degree.
The condition for maxima is given as :
[tex]dsin\theta=m\lambda[/tex]
d = seperation between the slits
m = order, m = 2
[tex]d=\dfrac{m\lambda}{sin\theta}[/tex]
[tex]d=\dfrac{2\times 612\times 10^{-9}\ m}{sin(25)}[/tex]
d = 0.00000289
or
[tex]d=2.86\mu m[/tex]
So, the seperation between the silts is [tex]2.86\ \mu m[/tex]. Hence, this is the required solution.
In a model of the hydrogen atom, the electron travels in circular orbits around the proton. What is the electric potential, in volts, due to the proton on an electron in an orbit with radius 2.08 x 10^-10 m?
Answer:
6.93 volts
Explanation:
q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C
r = radius of the orbit = 2.08 x 10⁻¹⁰ m
V = Electric potential due to proton on electron
Electric potential due to proton on electron is given as
[tex]V = \frac{kq}{r}[/tex]
Inserting the values
[tex]V = \frac{(9\times 10^{9})(1.6\times 10^{-19})}{2.08\times 10^{-10}}[/tex]
V = 6.93 volts
Final answer:
The electric potential due to the proton on an electron in an orbit with a radius of 2.08 x 10^-10 m in a hydrogen atom model is approximately -13.6 volts.
Explanation:
In the context of the model of a hydrogen atom, the electric potential due to the proton on an electron in an orbit with radius 2.08 x 10-10 m can be calculated using the formula for the electric potential V due to a point charge, which is V = k * Q / r. Here, k is the Coulomb's constant (approximately 8.99 x 109 N m2/C2), Q is the charge of the proton (1.602 x 10-19 C), and r is the distance from the proton to the electron, which is the radius of the orbit.
To find the electric potential, simply plug these values into the formula:
V = (8.99 x 109 N m2/C2) * (1.602 x 10-19 C) / (2.08 x 10-10 m)
After calculating, the electric potential is found to be approximately -13.6 volts, with the negative sign indicating that the potential energy associated with the electron is negative, which is common for bound states such as electrons in an atom.
At a football game, an air gun fires T-shirts into the crowd. The gun is fired at an angle of 46 degree from the horizontal with an initial speed of 27 m/s. A fan who is sitting 60 m horizontally from the gun, but high in the stands, catches a T-shirt. A) How long does it take for the T-shirt to reach the fan? B) At what height h is the fan from the ground?
Answer:
a) [tex]t=3.199 seconds[/tex]
b) [tex]h = 11.97 m[/tex]
Explanation:
Since this problem belongs to the concept of projectile motion
a) we know,
[tex]Vcos\theta=\frac{R}{t}[/tex]
Where,
V = initial speed
Θ = angle with the horizontal
R = horizontal range
t = Time taken to cover the range 'R'
Given:
V = 27m/s
R = 60m
Θ = 46°
thus,
the equation becomes
[tex]27\times cos46^o=\frac{60}{t}[/tex]
or
[tex]t=\frac{60}{27\times cos46^o}[/tex]
[tex]t=3.199 seconds[/tex]
b)The formula for height is given as:
[tex]h = Vsin\theta \times t-\frac{1}{2}\times gt^2\\[/tex]
where,
g = acceleration due to gravity = 9.8m/s²
substituting the values in the above equation we get
[tex]h = 27\times sin46^o\times 3.199-\frac{1}{2}\times 9.8\times 3.199^2\\[/tex]
or
[tex]h = 62.124-50.14[/tex]
or
[tex]h = 11.97 m[/tex]
It takes 2.22 seconds for the T-shirt to reach the fan. The fan is located at a height of 24.57 meters from the ground.
Explanation:To find the time it takes for the T-shirt to reach the fan, we need to solve for the time in the horizontal motion. The horizontal distance from the gun to the fan is given as 60 m. Since the horizontal motion is constant velocity, we can use the equation d = vt and solve for t. Plugging in the values, we get t = 60 m / 27 m/s = 2.22 s.
To find the height of the fan from the ground, we need to solve for the vertical motion. The equation for vertical motion is y = yt + (1/2)gt^2, where y is the vertical displacement, yt is the initial vertical velocity, and g is the acceleration due to gravity. In this case, the vertical displacement is unknown, the initial vertical velocity is 0 m/s, and the acceleration due to gravity is 9.8 m/s^2. Plugging in the values and solving for y, we get y = (1/2)(9.8 m/s^2)(2.22 s)^2 = 24.57 m.
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A 7.8 µF capacitor is charged by a 9.00 V battery through a resistance R. The capacitor reaches a potential difference of 4.20 V at a time 3.21 s after charging begins. Find R.
Answer:
655128 ohm
Explanation:
C = Capacitance of the capacitor = 7.8 x 10⁻⁶ F
V₀ = Voltage of the battery = 9 Volts
V = Potential difference across the battery after time "t" = 4.20 Volts
t = time interval = 3.21 sec
T = Time constant
R = resistance
Potential difference across the battery after time "t" is given as
[tex]V = V_{o} (1-e^{\frac{-t}{T}})[/tex]
[tex]4.20 = 9 (1-e^{\frac{-3.21}{T}})[/tex]
T = 5.11 sec
Time constant is given as
T = RC
5.11 = (7.8 x 10⁻⁶) R
R = 655128 ohm
A rope with a mass density of 1 kg/m has one end tied to a vertical support. You hold the other end so that the rope is horizontal and has a tension of 4 N. If you move the end of the rope back and forth, you produce a transverse wave in the rope with a wave speed of 2 m/s. If you double the amount of tension you exert on the rope, what is the wave speed?
Answer:
[tex]v' = 2.83 m/s[/tex]
Explanation:
Velocity of wave in stretched string is given by the formula
[tex]v = \sqrt{\frac{T}{\mu}}[/tex]
here we know that
T = 4 N
also we know that linear mass density is given as
[tex]\mu = 1 kg/m[/tex]
so we have
[tex]v = \sqrt{\frac{4}{1}} = 2 m/s[/tex]
now the tension in the string is double
so the velocity is given as
[tex]v' = \sqrt{\frac{8}{1}} = 2\sqrt2 m/s[/tex]
[tex]v' = 2.83 m/s[/tex]
The force exerted by the wind on the sails of a sailboat is 390 N north. The water exerts a force of 180 N east. If the boat (including its crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration? Serway, Raymond A.; Jewett, John W.. Physics for Scientists and Engineers (MindTap Course List) (Page 121). Cengage Learning. Kindle Edition.
Answer:
Acceleration= 1,59 (meters/(second^2))
Direction= NE; 65,22° above the east direction.
Explanation:
Resulting force= ( ((180N)^2) + ((390N)^2) ) ^ (1/2) = 429,53 N
Angle obove the east direction= ((cos) ^ (-1)) (180N / 429,53 N) = 65,22°
Acceleration= Resulting force / mass = (429,53 N) / (270 kg) =
= (429,53 kg × (meters/(second^2))) / (270kg) = 1,59 (meters/(second^2))
A wheel, originally rotating at 126 rad/s undergoes a constant angular deceleration of 5.00 rad/s2. What is its angular speed after it has turned through an angle of 628 radians
Answer:
97.96 [tex]\frac{rad}{s}[/tex]
Explanation:
The initial angular velocity is [tex]w_{0}[/tex] = 126 rad / s.
The constant angular deceleration is 5.00 rad / s2.
The constant angular deceleration is, by definition: dw / dt.
[tex]\frac{dw}{dt}=-5 \frac{rad}{s^{2} }[/tex]
Separating variables
[tex]dw=-5 dt[/tex]
Integration (limits for w: 0 to W0; limits for t: 0 to t)
[tex]w= w_{0}-5t[/tex]
W is by definition [tex]\frac{d\alpha }{dt}[/tex], where [tex]\alpha[/tex] is the angle.
[tex]\frac{d\alpha}{dt}=w_{0} -5t[/tex]
Separating variables
[tex]d\alpha=(w_{0} -5t )dt[/tex]
Integration (limits for [tex]\alpha[/tex]: 0 to 628; limits for t: 0 to t)
[tex]\alpha =w_{0}t-(\frac{5}{2})t^{2}[/tex]
[tex]128=126t-(\frac{5}{2})t^{2}[/tex]
Put in on the typical form of a quadratic equation:
[tex]\frac{5}{2}t^{2}-126t+628=0[/tex]
Solve by using the quadratic equation formula and discard the higher result because it lacks physical sense.
t=5.608 s
Evaluate at this time the angular velocity:
[tex]w(t=5.608)=126-5*5.608[/tex]
[tex]w(5.608)=97.96 \frac{rad}{s}[/tex]
To find the final angular speed of a rotating wheel given the initial speed, angular deceleration and the angle turned, we use an equation for angular motion. We substitute the given values and solve the equation to find the final angular speed.
Explanation:
To solve this question, we will be using the equation for angular motion,
ω² = ω0² + 2αθ
, where ω is the final angular speed we want to find, ω0 = 126 rad/s is the initial angular speed, α = -5.00 rad/s² is the angular deceleration, and θ = 628 rad is the angle turned through. This equation is analogous to the equation for linear motion, v² = u² + 2as, where v is final velocity, u initial velocity, a acceleration and s distance. We substitute the given values into the equation to find the final angular speed. Keep in mind, because we are dealing with deceleration, our α value is a negative.
Solve the equation: ω² - ω0² = 2αθ, which gives: ω² = (ω0² + 2αθ), ω = sqrt((ω0² + 2αθ)), ω = sqrt((126rad/s)² + 2*(-5.00 rad/s²)*628 rad). So, when you calculate the square root of the total, you will find the answer for the final angular speed.
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A wind turbine has 12,000 kg blades that are 37 m long. The blades spin at 24 rpm .If we model a blade as a point mass at the midpoint of the blade, what is the inward force necessary to provide each blade's centripetal acceleration
Answer:
[tex]F_c = 1.4 \times 10^6 N[/tex]
Explanation:
As we know
[tex]f = 24 rpm[/tex]
so we will have
[tex]f = 24 \frac{1}{60} hz = 0.4hz[/tex]
now angular frequency is given as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 0.8\pi[/tex]
Now the inwards is given as centripetal force
[tex]F_c = m\omega^2 r[/tex]
[tex]F_c = (12000)(0.8\pi)^2(\frac{37}{2})[/tex]
[tex]F_c = 1.4 \times 10^6 N[/tex]
Final answer:
The inward force necessary to provide each blade's centripetal acceleration is approximately 4,755,076 Newtons.
Explanation:
To calculate the inward force necessary to provide each blade's centripetal acceleration, we can use the formula for centripetal force:
[tex]Fc = m \times \omega^2 \times r[/tex]
where Fc is the centripetal force, m is the mass of the blade, ω is the angular velocity, and r is the radius of the blade.
In this case, the mass of the blade is 12,000 kg, the angular velocity is 24 rpm (which can be converted to radians per second by multiplying by 2π/60), and the radius of the blade is 37 m.
Plugging these values into the formula, we get:
Fc = 12000 kg * (24 * 2π/60)^2 * 37 m
Fc ≈ 4,755,076 N
So, the inward force necessary to provide each blade's centripetal acceleration is approximately 4,755,076 Newtons.
How much heat is needed to melt 13.74 kg of silver that is initially at 20°C?
Answer:
[tex]Q = 4.63 \times 10^6 J[/tex]
Explanation:
As we know that melting point of silver is
T = 961.8 degree C
Latent heat of fusion of silver is given as
L = 111 kJ/kg
specific heat capacity of silver is given as
[tex]s = 240 J/kg C[/tex]
now we will have
[tex]Q = ms\Delta T + mL[/tex]
[tex]\Delta T = 961.8 - 20 [/tex]
[tex]\Delta T = 941.8 degree C[/tex]
now from above equation
[tex]Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)[/tex]
[tex]Q = 3.1 \times 10^6 + 1.53 \times 10^6[/tex]
[tex]Q = 4.63 \times 10^6 J[/tex]
During a volleyball serve, the ball leaves the hand with an initial velocity of 10 m/s angled 41 degrees from the horizontal. What are the horizontal and vertical velocities of the ball?
Answer:
7.55 m/s, 6.56 m/s
Explanation:
v = 10 m/s, theta = 41 degree
Horizontal component of velocity = v x Cos theta = 10 x Cos 41 = 7.55 m/s
Vertical component of velocity = v x Sin theta = 10 x Sin 41 = 6.56 m/s
Final answer:
The ball's horizontal velocity is approximately 7.57 m/s, and its vertical velocity is approximately 6.46 m/s immediately after the serve.
Explanation:
The student's question about the initial horizontal and vertical velocities of a volleyball serve involves breaking down the initial velocity into its component parts using trigonometric functions. Given an initial velocity (v) of 10 m/s and an angle (θ) of 41 degrees from the horizontal, the horizontal component (vx) is calculated using cosine, and the vertical component (vy) is calculated using sine.
Using the formula:
vx = v ∙ cos(θ)
vy = v ∙ sin(θ)
For this problem:
vx = 10 m/s ∙ cos(41°)
vy = 10 m/s ∙ sin(41°)
Plugging in the values yields:
vx ≈ 7.57 m/s
vy ≈ 6.46 m/s
The ball's horizontal velocity is approximately 7.57 m/s, and its vertical velocity is approximately 6.46 m/s immediately after the serve.
A 4-kg hammer is lifted to a height of 10 m and dropped from rest. What was the velocity (in m/s) of the hammer when it was at a height of 4 m from the earth? O12 O5 O11 O109.5
Answer:
v = 10.84 m/s
Explanation:
using the equation of motion:
v^2 = (v0)^2 + 2×a(r - r0)
due to the hammer starting from rest, vo = 0 m/s and a = g , g is the gravitational acceleration.
v^2 = 2×g(r - r0)
v = \sqrt{2×(-9.8)×(4 - 10)}
= 10.84 m/s
therefore, the velocity at r = 4 meters is 10.84 m/s
A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial Kinetic energy. What is the mangnitude of the change in momentum of the stone?
Answer:
change in momentum of the stone is 1.635 kg.m/s
Explanation:
let m = 0.500kg ball and M be the mass of the stone, v be the velocity of the ball and V be the velocity of the stone
the initial kinetic energy of the ball is 1/2(0.500)(20^2) = 100 J
the kinetic energy of the ball after rebounding is 70/100(100) = 70 J
Kb = 1/2mv^2
v = \sqrt{2k/m} = \sqrt{2(70)/0.500} = 16.73 m/s
from the conservation of linear momentum, we know that:
mvi + MVi = mvf + MVf
MVf - MVi = mvi - mvf
MVf - MVi = (0.500)(20) - (0.500)(16.73)
= 1.635 kg.m/s
therefore, the change is momentum of the stone is 1.635 kg.m/s
A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2. Compute the magnitude of the resultant acceleration (in m/s2) of a point on its rim after it has turned through 60.0°.
Answer:
0.42 m/s²
Explanation:
r = radius of the flywheel = 0.300 m
w₀ = initial angular speed = 0 rad/s
w = final angular speed = ?
θ = angular displacement = 60 deg = 1.05 rad
α = angular acceleration = 0.6 rad/s²
Using the equation
w² = w₀² + 2 α θ
w² = 0² + 2 (0.6) (1.05)
w = 1.12 rad/s
Tangential acceleration is given as
[tex]a_{t}[/tex] = r α = (0.300) (0.6) = 0.18 m/s²
Radial acceleration is given as
[tex]a_{r}[/tex] = r w² = (0.300) (1.12)² = 0.38 m/s²
Magnitude of resultant acceleration is given as
[tex]a = \sqrt{a_{t}^{2} + a_{r}^{2}}[/tex]
[tex]a = \sqrt{0.18^{2} + 0.38^{2}}[/tex]
[tex]a[/tex] = 0.42 m/s²
The magnitude of the resultant acceleration of the point on the flywheel's rim after it has turned through 60.0° is approximately 0.424 m/s².
Explanation:To compute the magnitude of the resultant acceleration of a point on the flywheel's rim, we need to use the equation:
θ = ω0t + 0.5αt2
Where θ is the angle turned, ω0 is the initial angular velocity, and α is the angular acceleration.
Plugging in the given values, the equation becomes:
60.0° = 0 + 0.5 * 0.600 rad/s² * t2
Simplifying, we get:
t2 = (60.0° * 2) / 0.600 rad/s²
t2 = 0.200 rad/s²
Therefore, the magnitude of the resultant acceleration is:
a = ωt = 0.600 rad/s² * sqrt(0.200 rad/s²) ≈ 0.424 m/s²
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A constant force of 120 N pushes a 55 kg wagon across an 8 m level surface. If the wagon was initially at rest, what is the final kinetic energy of the wagon?
Answer:
The kinetic energy of the wagon is 967.0 J
Explanation:
Given that,
Force = 120 N
Mass = 55 kg
Height = 8 m
We need to calculate the kinetic energy of the wagon
Using newtons law
[tex]F = ma[/tex]
[tex]\dfrac{120}{55}=a[/tex]
[tex]a =2.2\ m/s^2[/tex]
Using equation of motion
[tex]v^2 =u^2+2as[/tex]
Where,
v = final velocity
u = initial velocity
s = height
Put the value in the equation
[tex]v^2=0+2\times2.2\times8[/tex]
[tex]v=5.93\ m/s[/tex]
Now, The kinetic energy is
[tex]K.E=\dfrac{1}{2}mv^2[/tex]
[tex]K.E=\dfrac{1}{2}\times55\times(5.93)^2[/tex]
[tex]K.E=967.0\ J[/tex]
Hence, The kinetic energy of the wagon is 967.0 J
In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 3.5 kg book is pushed from rest through a distance of 0.74 m by the horizontal 21 N force from the broom and then has a speed of 1.2 m/s, what is the coefficient of kinetic friction between the book and floor?
Answer:
0.51
Explanation:
m = mass of the book = 3.5 kg
F = force applied by the broom on the book = 21 N
a = acceleration of the book
v₀ = initial speed of the book = 0 m/s
v = final speed of the book = 1.2 m/s
d = distance traveled = 0.74 m
Using the equation
v² = v₀² + 2 a d
1.2² = 0² + 2 a (0.74)
a = 0.973 m/s²
f = kinetic frictional force
Force equation for the motion of the book is given as
F - f = ma
21 - f = (3.5) (0.973)
f = 17.6 N
μ = Coefficient of kinetic friction
Kinetic frictional force is given as
f = μ mg
17.6 = μ (3.5 x 9.8)
μ = 0.51
One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 24 m/s. The masses of the two objects are 3.8 and 8.4 kg. Determine the final speed of the two-object system after the collision for the case (a) when the large-mass object is the one moving initially and the case (b) when the small-mass object is the one moving initially.
Answer:
Part a)
v = 16.52 m/s
Part b)
v = 7.47 m/s
Explanation:
Part a)
(a) when the large-mass object is the one moving initially
So here we can use momentum conservation as the net force on the system of two masses will be zero
so here we can say
[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v[/tex]
since this is a perfect inelastic collision so after collision both balls will move together with same speed
so here we can say
[tex]v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}[/tex]
[tex]v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}[/tex]
[tex]v = 16.52 m/s[/tex]
Part b)
(b) when the small-mass object is the one moving initially
here also we can use momentum conservation as the net force on the system of two masses will be zero
so here we can say
[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v[/tex]
Again this is a perfect inelastic collision so after collision both balls will move together with same speed
so here we can say
[tex]v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}[/tex]
[tex]v = \frac{(8.4\times 0 + 3.8\times 24)}{3.8 + 8.4}[/tex]
[tex]v = 7.47 m/s[/tex]
A completely reversible refrigerator is driven by a 10-kW compressor and operates with thermal energy reservoirs at 250 K and 310 K. Calculate the rate of cooling provided by this refrigerator.
The rate of cooling provided by a reversible refrigerator, working with a 10-kW compressor and thermal energy reservoirs at 250 K and 310 K, is calculated using the refrigerator's Carnot coefficient and the power of the compressor. With these specific conditions, the cooling rate is calculated to be 50 kW.
Explanation:The question pertains to the functioning of a reversible refrigerator, specifically the cooling rate provided by the appliance which uses a 10-kW compressor and operates with thermal energy reservoirs at 250 K and 310 K. To determine the cooling rate, we must first understand some basics about the refrigerator's operation.
A reversible refrigerator absorbs heat Qc from a cold reservoir and discards it into a hot reservoir, while work W is done on it. This work is represented by the power exerted by the compressor. The refrigerator functions by moving a coolant through coils, which absorbs heat from the contents of the refrigerator and releases it outside.
The coefficient of performance or Carnot coefficient (KR) of the refrigerator can be computed using the formula KR = Tc / (Th - Tc) where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir, both measured in Kelvin. In this situation, KR = 250 / (310 - 250) = 5.
Since the work done W is the compressor power P, we can use the formula Qc = KR * P to find the cooling rate. Substituting the known values, Qc = 5 * 10 kW = 50 kW. Therefore, the rate of cooling provided by this refrigerator is 50 kW.
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A wire 73.6 cm long carries a 0.720 A current in the positive direction of an x axis through a magnetic field with an x component of zero, a y component of 0.000270 T, and a z component of 0.00770 T. Find the (a) x, (b) y, and (c) z components of the magnetic force on the wire.
The magnetic force on the wire has an x-component of 0 N, a y-component of 0.004056224 N in the negative y-direction, and a z-component of 0.000143712 N in the positive z-direction.
The question involves calculating the magnetic force on a current-carrying wire in a magnetic field using the right-hand rule and the formula F = I × (L × B), where F is the force, I is the current, L is the length of the wire, and B is the magnetic field. We know the current I is 0.720 A, the length L is 73.6 cm (which we will convert to meters), and the magnetic field components are By = 0.000270 T and Bz = 0.00770 T.
First, let's convert the length of the wire from centimeters to meters: L = 73.6 cm = 0.736 m.
The force on the wire in the x, y, and z directions (Fx, Fy, Fz) can be calculated using the cross product of the current direction (along the x-axis) and the magnetic field components. Since there is no x component for the magnetic field (Bx = 0), the force in the x-direction (Fx) will be zero.
Using the right-hand rule, the force in the y-direction (Fy) will be:
Fy = I × (L × Bz) = 0.720 A × (0.736 m × 0.00770 T) = 0.004056224 N, pointing in the negative y-direction (since the current is in the positive x-direction and Bz is positive).
Similarly, the force in the z-direction (Fz) is calculated as:
Fz = I × (L × By) = 0.720 A × (0.736 m × 0.000270 T) = 0.000143712 N, pointing in the positive z-direction.
A 100 kg individual consumes 1200 kcal of food energy a day. Calculate
(a) the altitude change, in m, if the food energy content was converted entirely into lifting the indi-
vidual under normal earth gravity.
(b) the velocity, in m/s, if the food energy content was converted entirely into accelerating the indi-
vidual from rest.
(c) the final temperature, in ◦C, of a 100 kg mass of liquid water initially at the normal human body
temperature and heated with the energy content of the food. You can use a liquid water specific
heat of 4.1 kJ/kg K.
Answer:
(a) 5142.86 m
(b) 317.5 m/s
(c) 49.3 degree C
Explanation:
m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J
(a) Let the altitude be h
Q = m x g x h
504 x 10^4 = 100 x 9.8 x h
h = 5142.86 m
(b) Let v be the speed
Q = 1/2 m v^2
504 x 10^4 = 1/2 x 100 x v^2
v = 317.5 m/s
(c) The temperature of normal human body, T1 = 37 degree C
Let the final temperature is T2.
Q = m x c x (T2 - T1)
504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)
T2 = 49.3 degree C
ly charged particles are held 24 x 103m apart and then released from rest. The initial acceleration of the first particle is observed to be 7.0 m/s2 and that of the second to be 9.0 m/s2. The mass of the first particle is 6.3 x 107 kg. (a) What is the mass of the second particle? kg (b) What is the magnitude of the charge of each particle?
Answer:
Part a)
[tex]m_2 = 4.9 \times 10^7 kg[/tex]
Part b)
[tex]q_1 = q_2 = 5312.6 C[/tex]
Explanation:
Part a)
As we know that both charge particles will exert equal and opposite force on each other
so here the force on both the charges will be equal in magnitude
so we will have
[tex]F = m_1a_1 = m_2a_2[/tex]
here we have
[tex]6.3 \times 10^7(7) = m_2(9)[/tex]
now we have
[tex]m_2 = 4.9 \times 10^7 kg[/tex]
Part b)
Now for the force between two charges we can say
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
now we have
[tex](6.3 \times 10^7)(7) = \frac{(9\times 10^9)q^2}{(24\times 10^3)^2}[/tex]
now we have
[tex]q_1 = q_2 = 5312.6 C[/tex]
700*.135An inventor claims to have developed a heat engine that receives 700 kJ of heat from a source at 500 K and produces 300 kJ of net work while rejecting the waste heat to a sink at 290 K. Is this a reasonable claim?
Answer:
output work is not possible to have more than 294 kJ value so this is not reasonable claim
Explanation:
As we know that the efficiency of heat engine is given as
[tex]\eta = 1 - \frac{T_2}{T_1}[/tex]
now we will have
[tex]T_2 = 290 K[/tex]
[tex]T_1 = 500 K [tex]
[tex]\eta = 1 - \frac{290}{500}[/tex]
[tex]\eta = 0.42[/tex]
now we know that efficiency is defined as
[tex]\eta = \frac{Work}{Heat}[/tex]
[tex]0.42 = \frac{W}{700}[/tex]
[tex]W = 294 kJ[/tex]
So output work is not possible to have more than 294 kJ value
To determine if the inventor's claim of developing a heat engine is reasonable, we can calculate the maximum theoretical efficiency of the engine and compare it to the claimed net work output and heat input.
Explanation:A heat engine operates between a hot reservoir and a cold reservoir, absorbing heat from the hot reservoir and converting some of it into work, while rejecting the remaining heat to the cold reservoir. The efficiency of a heat engine is determined by the temperature of the reservoirs. In this case, the inventor claims that the heat engine receives 700 kJ of heat from a source at 500 K and produces 300 kJ of net work while rejecting waste heat to a sink at 290 K. To determine if this claim is reasonable, we can calculate the theoretical maximum efficiency of the heat engine using the Carnot efficiency formula.
The Carnot efficiency formula is given by:
Efficiency = 1 - (Tc / Th)
where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir, both measured in Kelvin. In this case, Tc = 290 K and Th = 500 K.
Plugging in these values into the formula, we have:
Efficiency = 1 - (290 K / 500 K) = 1 - 0.58 = 0.42 or 42%
Therefore, the maximum theoretical efficiency for a heat engine operating between these temperatures is 42%. Since the claimed net work output of the heat engine is 300 kJ, we can calculate the maximum heat input by dividing the net work output by the efficiency:
Maximum heat input = Net work output / Efficiency = 300 kJ / 0.42 = 714.3 kJ
Since the claimed heat input is 700 kJ, which is slightly less than the maximum calculated heat input, it is reasonable to say that the inventor's claim is possible.
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Which of the following can penetrate the deepest (Please explain)
A) 3MeV electron
B) 10MeV alpha
C) 0.1 MeV auger
D) 400keV proton
Answer: 3MeV electron
Explanation:
m_e={9.1\times 10^{-31} m_α=4\times m_e m_a={9.1\times 10^{-31}
m_p=1.67\times 10^{-27}
(a) K.E. Energy of electron =[tex]\frac{1}{2}\times{m_e}\times{v_{e} ^{2}}[/tex]=3MeV
[tex]v_{e} ^{2}=\frac{2\times3\times1.6\times10^{-19}\times10^{6} }{9.1\times 10^{-31} }[/tex]=1.05\times10^{18}
[tex]v_e=\sqrt{1.05\times10^{18} } = 1.025\times10^{9}\frac{m}{s}[/tex]
(b) K.E. Energy of alpha particle =[tex]\frac{1}{2}\times{m_\alpha}\times{v_{\alpha} ^{2}}[/tex]=10MeV
[tex]v_{\alpha} ^{2}= \frac{2\times10\times1.6\times10^{-19}\times10^{6} }{4\times9.1\times 10^{-31} }[/tex]=0.88\times10^{18}
[tex]v_\alpha=\sqrt{0.88\times10^{18} } =.94\times10^{9}\frac{m}{s}[/tex]
(c) K.E. Energy of auger particle =[tex]\frac{1}{2}\times{m_a}\times{v_{a} ^{2}}[/tex]=0.1MeV
[tex]v_{a} ^{2}=\frac{2\times0.1\times1.6\times10^{-19}\times10^{6} }{9.1\times 10^{-31} }[/tex]=0.035\times10^{18}
[tex]v_a=\sqrt{0.035\times10^{18} } =.19\times10^{9}\frac{m}{s}[/tex]
(d) K.E. Energy of proton particle =[tex]\frac{1}{2}\times{m_p}\times{v_{p} ^{2}}[/tex]=400keV
[tex]v_{p} ^{2}=\frac{2\times400\times1.6\times10^{-19}\times10^{3} }{1.67\times 10^{-27} }[/tex]=0.766\times10^{14}
[tex]v_p=\sqrt{0.766\times10^{14} } =0.875\times10^{7}[tex]\frac{m}{s}[/tex]
from (a),(b),(c),and (d) we can clearly say that the velocity of the electron is more so the penetration of the electron will be deepest.
A right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 lb/ft3. How much work is required to pump all of the fuel to a point 13 ft above the top of the tank?
Given:
Height of tank = 8 ft
and we need to pump fuel weighing 52 lb/ [tex]ft^{3}[/tex] to a height of 13 ft above the tank top
Solution:
Total height = 8+13 =21 ft
pumping dist = 21 - y
Area of cross-section = [tex]\pi r^{2}[/tex] = [tex]\pi 4^{2}[/tex] =16[tex]\pi[/tex] [tex]ft^{2}[/tex]
Now,
Work done required = [tex]\int_{0}^{8} 52\times 16\pi (21 - y)dy[/tex]
= [tex]832\pi \int_{0}^{8} (21 - y)dy[/tex]
= 832[tex]\pi([/tex][tex][ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\[/tex])
= 113152[tex]\pi[/tex] = 355477 ft-lb
Therefore work required to pump the fuel is 355477 ft-lb
3.A robot is on the surface of Mars. The angle of depression from a camera in the robot to a rock on the surface of Mars is 13.78°. The camera is 184.0 cm above the surface. How far is the camera from the rock? (Round to the nearest tenth as needed.)
Answer:
750.25 cm
Explanation:
θ = 13.78°, h = 184 cm
Let the distance between rock and camera is d.
Tan θ = h / d
tan 13.78 = 184 / d
d = 750.25 cm
By utilizing trigonometry and the tangent of an angle in a right triangle, it is determined that the camera (at a height of 184.0 cm from the Mars surface with an angle of depression of 13.78° to a rock) is approximately 761.4 cm away from the rock.
Explanation:This question involves trigonometry, specifically the tangent of an angle in a right triangle. The robot captures an image of a rock with an angle of depression of 13.78°. The camera is 184.0 cm above the Mars surface. We can create a right triangle where the angle at the camera is 13.78°, the opposite side is 184.0 cm (the height of the camera), and the adjacent side is the distance between the camera and the rock, which we will call x.
By definition, tan(angle) = opposite/adjacent. Here, the angle is 13.78°, the opposite side is 184.0 cm and the adjacent side is x (unknown). To find x, we can use the following formula: x = opposite/tan(angle).
Therefore, x = 184.0 cm / tan(13.78°). Using a calculator, x is approximately 761.4 cm (to the nearest tenth). So the camera is approximately 761.4 cm away from the rock.
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(a) Write the energy equation for an elastic collision. (b) For an inelastic collision.
Answer:
Explanation:
There are two types of collision.
(a) Elastic collision: When there is no loss of energy during the collision, then the collision is said to be elastic collision.
In case of elastic collision, the momentum is conserved, the kinetic energy is conserved and all the forces are conservative in nature.
The momentum of the system before collision = the momentum of system after collision
The kinetic energy of the system before collision = the kinetic energy after the collision
(b) Inelastic collision: When there is some loss of energy during the collision, then the collision is said to be inelastic collision.
In case of inelastic collision, the momentum is conserved, the kinetic energy is not conserved, the total mechanical energy is conserved and all the forces or some of the forces are non conservative in nature.
The momentum of the system before collision = the momentum of system after collision
The total mechanical energy of the system before collision = total mechanical of the system after the collision
A 38.0 kg satellite has a circular orbit with a period of 1.30 h and a radius of 7.90 × 106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 30.0 m/s2, what is the radius of the planet
Answer:
5.44×10⁶ m
Explanation:
For a satellite with period t and orbital radius r, the velocity is:
v = 2πr/t
So the centripetal acceleration is:
a = v² / r
a = (2πr/t)² / r
a = (2π/t)² r
This is equal to the acceleration due to gravity at that elevation:
g = MG / r²
(2π/t)² r = MG / r²
M = (2π/t)² r³ / G
At the surface of the planet, the acceleration due to gravity is:
g = MG / R²
Substituting our expression for the mass of the planet M:
g = [(2π/t)² r³ / G] G / R²
g = (2π/t)² r³ / R²
R² = (2π/t)² r³ / g
R = (2π/t) √(r³ / g)
Given that t = 1.30 h = 4680 s, r = 7.90×10⁶ m, and g = 30.0 m/s²:
R = (2π / 4680 s) √((7.90×10⁶ m)³ / 30.0 m/s²)
R = 5.44×10⁶ m
Notice we didn't need to know the mass of the satellite.
An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electron is 4.52×105 m/s pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.395 m east of point A?
The speed of the electron when it reaches point B, which is 0.395 m east of point A, is approximately 3.86×10⁵ m/s, directed towards the east.
Explanation:The initial speed of the electron is provided as 4.52×10⁵ m/s. Given the electric field, we can calculate the force on the electron as F = qE, where q is the charge of the electron (-1.60 × 10⁻¹⁹ C) and E is the electric field (1.55 N/C). Hence, the force acting on the electron is F = -1.60 × 10⁻¹⁹ C * 1.55 N/C = -2.48 × 10⁻¹⁹ N.
Using F = ma, we can calculate the acceleration of the electron. Knowing the mass of the electron is 9.11 × 10⁻³¹ kg, the acceleration is a = F/m = -2.48 × 10⁻¹⁹ N / 9.11 × 10⁻³¹ kg = -2.72 × 10¹¹ m/s².
Given the distance of the movement is 0.395 m, we can use the equation v² = u² + 2as to solve for the final velocity 'v', where 'u' is the initial velocity, 'a' is the acceleration and 's' is the distance. Substituting the known values, we get v = sqrt((4.52×10⁵ m/s)² - 2 * 2.72 × 10¹¹ m/s² * 0.395 m) ≈ 3.86 x 10⁵ m/s (approximately).
So, the speed of the electron when it reaches point B is approximately 3.86 x 10⁵ m/s, directed towards the east.
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A charged isolated metal sphere of diameter 15 cm has a potential of 8500 V relative to V = 0 at infinity. Calculate the energy density in the electric field near the surface of the sphere.
Answer:
[tex]u = 0.057 J/m^3[/tex]
Explanation:
Energy density near the surface of the sphere is given by the formula
[tex]u = \frac{1}{2}\epsilon_0 E^2[/tex]
also for sphere surface we know that
[tex]E = \frac{V}{R}[/tex]
R = radius of sphere
V = potential of the surface
now we have
[tex]u = \frac{1}{2}\epsilon_0 (\frac{V^2}{R^2})[/tex]
now from the above formula we have
[tex]u = \frac{1}{2}(8.85 \times 10^{-12})(\frac{8500^2}{0.075^2})[/tex]
[tex]u = 0.057 J/m^3[/tex]
A clarinetist, setting out for a performance, grabs his 3.230 kg3.230 kg clarinet case (including the clarinet) from the top of the piano and carries it through the air with an upward force of 30.10 N.30.10 N. Find the case's vertical acceleration. Indicate an upward acceleration as positive and a downward one as negative.
Answer:
- 0.5 m/s²
Explanation:
m = mass of the clarinet case = 3.230 kg
W = weight of the clarinet case in downward direction
a = vertical acceleration of the case
Weight of the clarinet case is given as
W = mg
W = 3.230 x 9.8
W = 31.654 N
F = Upward force applied = 30.10 N
Force equation for the motion of the case is given as
F - W = ma
30.10 - 31.654 = 3.230 a
a = - 0.5 m/s²
Final answer:
By applying Newton's second law and accounting for the forces acting on the clarinet case, it is found to experience a downward vertical acceleration of 0.489 m/s², due to the net force acting on it being in the downward direction.
Explanation:
To find the vertical acceleration of the clarinet case, first identify the forces acting on it. The force of gravity (weight) pulls it downward, which can be calculated by multiplying the mass of the case by the acceleration due to gravity (9.81 m/s²). The equation for weight is W = mg, where m is mass and g is gravity.
Substituting the given values, W = 3.230 kg * 9.81 m/s² = 31.68 N. The net force [tex]F_{net}[/tex] acting on the case is the upward force by the clarinetist subtracting the weight of the case, [tex]F_{net}[/tex] = 30.10 N - 31.68 N = -1.58 N. Applying Newton's second law, F = ma, and solving for acceleration (a), a = [tex]F_{Net/m}[/tex], we find the case's vertical acceleration as a = -1.58 N / 3.230 kg = -0.489 m/s².
Therefore, the case experiences a downward acceleration of 0.489 m/s², indicating that it is slowing in its ascent or accelerating downward.
A second's pendulum is taken to the moon. What will be the time period of the pendulum at the moon? (Acceleration due to gravity at the surface of the moon is 1/6 on the surface of earth).
Answer:
4.89 seconds
Explanation:
The time period of a pendulum is given by
[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]
For a second pendulum on earth, T = 2 second
[tex]2 = 2\pi \sqrt{\frac{l}{g}}[/tex] ...... (1)
Now the time period is T when the pendulum is taken to moon and gravity at moon is 1/6 of gravity of earth
[tex]T = 2\pi \sqrt{\frac{l}{\frac{g}{6}}}[/tex] ...... (2)
Divide equation (2) by equation (1)
[tex]\frac{T}{2} = \sqrt{6}[/tex]
T = 4.89 seconds
Fluids at rest possess no flow energy. a)-True b)-False
I believe the answer is A: True