Answer:
The length of the simple pendulum is 2.4 meters.
Explanation:
Time period of simple pendulum is given by :
[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]
L is the length of pendulum
The time period of the rope is given by :
[tex]T=2\pi\sqrt{\dfrac{2L'}{3g}}[/tex]
L' is the length of the rod, L' = 3.6 m
It is given that, the rod have the same period as a simple pendulum and we need to find the length of simple pendulum i.e.
[tex]2\pi\sqrt{\dfrac{L}{g}}=2\pi\sqrt{\dfrac{2L'}{3g}}[/tex]
On solving the above equation as :
[tex]\dfrac{L}{g}=\dfrac{2L'}{3g}[/tex]
L = 2.4 m
So, the length of the thin rod that is hung vertically from one end and set into small amplitude oscillation 2.4 meters. Hence, this is the required solution.
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
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]
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.
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 person is standing a distance D = 5.8 m in front of a flat, vertical mirror. The distance from the ground to his eyes is H = 1.6 m. An object is placed on the ground a distance d = D/2 = 2.9 m in front of the mirror. At what height h should the bottom of the mirror be so that the person can see the bottom of the object?
Answer:
it is at height of y = 0.533 m from ground
Explanation:
As per law of reflection we know that angle of incidence = angle of reflection
so here we have
[tex]tan\theta_i = tan\theta_r[/tex]
here we know that
[tex]tan\theta_i = \frac{y}{d}[/tex]
also we have
[tex]tan\theta_r = \frac{H - y}{D}[/tex]
now we have
[tex]\frac{H - y}{D} = \frac{y}{d}[/tex]
here we have
[tex]\frac{1.6 - y}{5.8} = \frac{y}{2.9}[/tex]
[tex]3y = 1.6[/tex]
[tex]y = 0.533 m[/tex]
To see the bottom of the object in the mirror, the bottom of the mirror should be at a height of half the height of the person.
Explanation:To see the bottom of the object in the mirror, the person should be able to see an image of the bottom of the object reflected in the mirror.
Using the law of reflection, we can determine the height h at which the bottom of the mirror should be. The angle of incidence for the light from the bottom of the object is equal to the angle of reflection, so the height of the mirror should be equal to the height of the person's eyes, H, plus the height of the bottom of the object, h. We can calculate h using similar triangles:
h/H = d/D
where d is the distance from the object to the mirror and D is the distance from the person to the mirror. Substituting the given values, we have:
h = (d/D) * H
Substituting d = D/2, we get:
h = (D/2D) * H = H/2
Therefore, the bottom of the mirror should be at a height of half the height of the person.
On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision?
Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J
The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic friction between the block and the plane is μk = 0.24. Determine the distance the block will slide before it stops.
Answer:
The distance the block will slide before it stops is x= 3.4 m .
Explanation:
m1= 0,02 kg
V1= 400 m/s
m2= 2 kg
V2= ?
g= 9.8 m/s²
μ= 0.24
N = m2 * g = 19.6 N
Fr= μ * N
Fr= 4.704 N
due to the conservation of the amount of movement:
m1*V1 = m2*V2
V2= 4 m/s
Fr = m2*a
a= Fr/m2
a= -2.352 m/s²
matching momentum and amount of movement
Fr*t=m2*V2
t= 1.7 sec
x= V2*t - (a*t²)/2
x= 3.4 m
Answer:
The distance the block will slide before it stops is 3.3343 m
Explanation:
Given;
mass of bullet, m₁ = 20-g = 0.02 kg
speed of the bullet, u₁ = 400 m/s
mass of block, m₂ = 2-kg
coefficient of kinetic friction, μk = 0.24
Step 1:
Determine the speed of the bullet-block system:
From the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the bullet-block system after collision
(0.02 x 400) + (2 x 0) = v (0.02 + 2)
8 = v (2.02)
v = 8/2.02
v = 3.9604 m/s
Step 2:
Determine the time required for the bullet-block system to stop
Apply the principle of conservation momentum of the system
[tex]v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f[/tex]
when the system stops, vf = 0
3.9604 -2.352t = 0
2.352t = 3.9604
t = 3.9604/2.352
t = 1.684 s
Thus, time required for the system to stop is 1.684 s
Finally, determine the distance the block will slide before it stops
From kinematic, distance is the product of speed and time
[tex]S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2[/tex]
Now, recall that t = 1.684 s
S = 3.9604(1.684) - 1.176(1.684)²
S = 6.6693 - 3.3350
S = 3.3343 m
Thus, the distance the block will slide before it stops is 3.3343 m
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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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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Answer True or False to each case. Coulomb's law states that the force between static charges Q1 and Q2 is K Q1Q2/r2 where K is a universal constant. is a force of attraction. may be attractive or repulsive. is K Q1Q2/r where K is a universal constant. is proportional to 1/r2.
Explanation:
The Coulomb's law states that the force acting on two charges is directly proportional to the product of charges and inversely proportional to the square of distance between them . Mathematically, it is given by
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Where
k is the electrostatic constant
q₁ and q₂ are charges
r is the distance between them.
The SI unit of electric force is Newton. It can be attractive or repulsive. The attraction or repulsion depend on charges. If both charges are positive, the force is repulsive and if both are opposite charges, the force is attractive.
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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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
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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For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser pointer (λ = 685 nm) through the slit onto a screen 5.4 m away, a diffraction pattern appears. The bright band in the center of the pattern is 7.0 cm wide. What is the width of the slit?
Answer:
105.6857 nm
Explanation:
Wave length of laser pointer = λ = 685 nm = 685×10⁻⁹ m
Distance between screen and slit = L = 5.4 m
Width of bright band = 2x = 7 cm
⇒x = 3.5 cm = 3.5×10⁻² m
The first minimum occurs at
sin θ = λ/d
θ = λ/d (since the angle is very small, sin θ ≈ θ)
Width of slit
d = λL/x
⇒d = 685×10⁻⁹×5.4/3.5×10⁻²
⇒d = 1056.857×10⁻⁷ m
∴ The width of the slit is 105.6857 nm
The North American Plate is moving west at a rate of approximately 20 mm/yr. How long will it take for New York to move 10° longitude west of its current position? (Assume 110 km/°longitude at New York's latitude.)
Answer:
55000000 years
Explanation:
Rate of North American Plate moving (Velocity) = 20 mm/yr
Degrees New York has to move = 10° west
New York's longitude = 110 km/°
Distance of New York = 110×10 = 1100 km
= 1100×10⁶ mm
[tex]\text {Time taken by the continental plate}=\frac {\text {Distance of New York}}{\text {Rate of North American Plate moving (Velocity)}}\\\Rightarrow \text {Time taken by the continental plate}=\frac{1100\times 10^6}{20}\\\Rightarrow \text {Time taken by the continental plate}=55\times 10^6\ years[/tex]
∴ Time taken by New York to move 10° west is 55000000 years
For an electron moving in a direction opposite to the electric field A) both its potential energy and itelectric potential remain constant its potential energy decreases and itselectricpotential increases. its potential energy decreases and its electric potential decreases. its potential energy increases and itselectric potential decreases. its potential energy increases and its electric potential increases
Answer:
Its potential energy decreases and itselectricpotential increases.
Explanation:
The electric potential does not depend on the charge but only on the magnitude of the electric field. In particular:
- Electric potential decreases when moving in the same direction of the electric field lines
- Electric potential increases when moving in the opposite direction to the field lines
So in this case, since the electron is moving in a direction opposite to the field, the electric potential increases.
However, the electric potential energy of a charge is given by
[tex]U=qV[/tex]
where
q is the charge
V is the electric potential
Here we said that the electric potential is increasing: however, the charge q of an electron is negative. This means that the product (qV) is increasing in magnitude but it is negative, so the potential energy of the electron is decreasing.
For an electron moving in a direction opposite to the electric field its potential energy decreases and its electric potential increases.
What is electric potential?Electric potential is the amount of work needed to move a unit charge from a point to a specific point against an electric field.it is denoted by U.
However, the electric potential energy of a charge is given by
[tex]\rm{U=qv}[/tex]
q is denoted for the charge.
V is denoted for the electric potential.
The above relation shows that the electric potential is inversely proportional to the electric potential energy.
The electric potential is independent of the charge. but it depends on the direction in which the charge is moving. If the charge moves in the direction of the electric field electric potential decreases and vice versa.
Here the electron is moving in a direction opposite to the field, the electric potential increases while electric potential energy decreases.
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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.
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.
An ideal gas experiences a change of internal energy of 284 J/kg when its temperature is increased by 12oC (at a constant volume). What is its specific heat
Answer:
23.5 J/Kg °C
Explanation:
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree Celcius.
c = Q / m ΔT
Here, Q / m = 282 J/kg, ΔT = 12
So, the specific heat = 282 / 12 = 23.5 J/Kg °C
A flat sheet of ice has a thickness of 2.20 cm. It is on top of a flat sheet of crystalline quartz that has a thickness of 1.50 cm. Light strikes the ice perpendicularly and travels through it, and then through the quartz. In the time it takes the light to travel through the two sheets, how far would it have traveled in a vacuum?
Answer:
[tex]Distance_{vaccum}=5.19cm[/tex]
Explanation:
The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under
Total time = Time taken through ice + Time taken through quartz
Time taken through ice = Thickness of ice / (speed of light in ice)
[tex]T_{ice}=\frac{2.20\times 10^{-2} \times \mu _{ice}}{V_{vaccum}}[/tex]
[tex]T_{quartz}=\frac{1.50\times 10^{-2} \times \mu _{quartz}}{V_{vaccum}}[/tex]
Thus in the same time the it would had covered a distance of
[tex]Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}][/tex]
we have
[tex]\mu _{ice}=1.309\\\\\mu _{quartz}=1.542[/tex]
Applying values we have
[tex]Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542][/tex]
[tex]Distance_{vaccum}=5.19cm[/tex]
The light, after going through ice and quartz, would have traveled approximately 5.198 cm in vacuum. This is calculated considering indices of refraction for each material and their thicknesses.
Explanation:The subject of this question is in the domain of Physics, specifically optics. The problem wants us to compare the distances light travels in different media to its travel in vacuum. The key to solve this matter is knowing the speed of light changes depending on the medium it passes through: it's slower in media like ice or quartz than in vacuum, due to these materials' indices of refraction.
The index of refraction for ice is approximately 1.31 and for crystalline quartz is approximately 1.544. The index of refraction (n) is the ratio of the speed of light in vacuum (c), to the speed of light in the medium (v), or n = c/v. To find out the distances light travels in vacuum for the time it takes to travel through the ice and quartz, we need to multiply the thickness of each material by their respective indices of refraction.
For the ice, it's 2.20 cm * 1.31 = 2.882 cm and for the quartz, it's 1.50 cm * 1.544 = 2.316 cm. So the total distance light would have traveled in vacuum in the same time it passes through both sheets is approximately 2.882 cm + 2.316 cm = 5.198 cm.
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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.
The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 2.15 x 10^5 kg train moving at 0.710 m/s. (a) What is the force constant of the spring? (N/m)
(b) What speed would the train be going if it only compressed the spring 0.640 m?
(c) What force does the spring exert when compressed 0.640 m? (Enter the magnitude only.)
Answer:
(a) the constant of the spring is 27095.375 J/m2
(b) 0.052 m/s
(c) F=17341.04 N
Explanation:
Hello
The law of conservation of energy states that the total amount of energy in any isolated system (without interaction with any other system) remains unchanged over time.
if we assume that there is no friction, then, the kinetic energy of the train (due to movement) will be equal to the energy accumulated in the spring
Step 1
energy of the train (kinetic)
[tex]E_{k}=\frac{m*v^{2} }{2}\\ E_{k}=\frac{2.15*10^{5}kg*(0.71\frac{m}{s}) ^{2} }{2}\\E_{k}=54190.75 J\\ \\[/tex]
step 2
energy of the spring
[tex]E_{s}=\frac{Kx^{2} }{2}\\[/tex]
where K is the constant of the spring and x the length compressed.
[tex]E_{s}=\frac{Kx^{2} }{2}=54190.75 J\\\frac{Kx^{2} }{2}=54190.75 J\\\\k=\frac{2*54190.75 J}{x^{2}}\\k=\frac{2*54190.75 J}{(2.0 m)^{2} }\\ k=27095.375\frac{J}{m^{2} } \\\\[/tex]
(a) the constant of the spring is 27095.375 J/m2
(b)
[tex]x=0.640 m\\\\E_{s}=\frac{27095.375 (\frac{j}{m^{2} })*(0.640m)^{2} }{2}\\ E_{s}=5549.1328 J\\[/tex]
equal to train energy
[tex]E_{k}=\frac{2.15*10^{5}kg*(v) ^{2} }{2}\\\frac{2.15*10^{5}kg*(v) ^{2} }{2}=5549.1328 J\\v^{2}=\frac{2*5549.1328 J}{2.15*10^{5}kg}\\v=0.052 \frac{m}{s} \\\\[/tex]
(b) 0.052 m/s
(c)
[tex]F= kx\\\\F=27095.375 \frac{J}{m^{2} }*(0.640 m)\\ F=17341.04 N\\F=17341.04[/tex]
(c) F=17341.04 N
I hope it helps
The force constant of the spring is 38.03 N/m. If the train compressed the spring 0.640 m, the train would be going at a speed of 0.418 m/s. The force the spring exerts when compressed 0.640 m is 24.34 N.
Explanation:To solve these questions we use conservation of energy principles and the spring equation, F = k*x. The initial kinetic energy of the train is converted into potential energy in the spring at the point of maximum compression.
(a) What is the force constant of the spring?
Here we equate kinetic energy with potential energy by using the equations KE = 0.5 * m * v^2 and PE = 0.5 * k * x^2. We can solve for k to get k = m * v^2 / x^2, so it's (2.15 x 10^5 kg * (0.710 m/s)^2)/(2.00 m)^2 = 38.03 N/m.
(b) What speed would the train be going if it only compressed the spring 0.640 m?
Here we rearrange the previous equation for velocity, v = sqrt(k * x^2 / m) which we then place values into to calculate v = sqrt(38.03 N/m * (0.640 m)^2)/(2.15 x 10^5 kg) = 0.418 m/s.
(c) What force does the spring exert when compressed 0.640 m?
Here we refer back to Hooke's Law (F = k*x), which states that the force required to compress or extend a spring by some distance x is proportional to that distance. The spring constant (k) is the proportionality constant in this relationship. So, the force is F = (38.03 N/m * 0.640 m) = 24.34 N.
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A coil is wrapped with 300 turns of wire on the perimeter of a square frame (side length = 20 cm). Each turn has the same area as the frame, and the total resistance of the coil is 1.5 Ω. A uniform magnetic field perpendicular to the plane of the coil changes in magnitude at a constant rate from 0.50 T to 0.90 T in 2.0 s. What is the magnitude of the induced emf in the coil while the field is changing?
Answer:
EMF = 2.4 V
Explanation:
As we know by Lenz law that induced EMF is given by rate of change in magnetic flux in the coil
As we know that magnetic flux is given by
[tex]\phi = NBA[/tex]
here we know that
[tex]N = 300[/tex]
[tex]Area = 0.20 \times 0.20 = 0.04 m^2[/tex]
now we for induced EMF we have
[tex]EMF = NA\frac{dB}{dt}[/tex]
here we have
[tex]EMF = (300)(0.04)(\frac{0.90 - 0.50}{2})[/tex]
[tex]EMF = 2.4 V[/tex]
The magnitude of the induced emf in a coil, while the magnetic field is changing, can be calculated using Faraday's law. The law states that the induced emf is equal to the negative of the rate of change of the magnetic flux linked through the circuit. We use this formula to calculate the change in magnetic field, take into accounts of coil's area and number of turns then divide by the change in time.
Explanation:The magnitude of the induced emf in a coil, while the magnetic field is changing, can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf is equal to the negative rate of change of the magnetic fluxlinked through the circuit.
In this case, the magnetic flux, Φ, through the coil can be calculated using the formula Φ = BAN, where B is the magnetic field, A is the area of each turn (which can be calculated using the given side length) and N is the number of turns (300 in this case).
To calculate the rate of change of the magnetic flux use the formula ΔΦ/Δt, where ΔΦ is the change in magnetic flux and Δt is the change in time. In this case, the magnetic field changes from 0.50 T to 0.90 T in 2.0 s, so the change in magnetic flux is (B_final - B_initial)*A*N, and Δt is 2.0 s.
The magnitude of the induced emf (which is also the negative rate of change of the magnetic flux) is given by Faraday's law, |ε| = |ΔΦ/Δt|.?
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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 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 Boeing 777 aircraft has a mass of 300,000 kg. At a certain instant during its landing, its speed is 27.0 m/s. If the braking force is 445,000 N, what is the speed of the airplane 10.0 s later?
Answer:
Speed of the airplane 10.0 s later = 12.2 m/s
Explanation:
Mass of Boeing 777 aircraft = 300,000 kg
Braking force = 445,000 N
Deceleration
[tex]a=\frac{445000}{300000}=1.48m/s^2[/tex]
Initial velocity, u = 27 m/s
Time , t = 10 s
We have equation of motion, v =u +at
v = 27 + (-1.48) x 10 = 27 - 14.8 = 12.2 m/s
Speed of the airplane 10.0 s later = 12.2 m/s
After applying the deceleration due to the braking force for 10 seconds, the speed of the Boeing 777 airplane reduces to 12.17 m/s.
Explanation:Calculating the Final Speed of a Boeing 777 After Braking
To determine the speed of a Boeing 777 aircraft after a 10-second interval of braking, we will use the relationship between force, mass, initial velocity, and acceleration provided by Newton’s second law of motion. The braking force applied on the aircraft is 445,000 N acting opposite to the direction of motion. We can calculate the deceleration using the formula:
F = ma
Where F is the force, m is the mass, and a is the acceleration (deceleration in this case since the force is applied opposite to the direction of motion). The mass (m) of the airplane is 300,000 kg, so we can solve for 'a' as follows:
a = F / m = 445,000 N / 300,000 kg = 1.483 m/s² (deceleration)
Next, we use the kinematic equation to find the final velocity (vf) after 10 seconds:
vf = vi + at
Where vi is the initial velocity, and t is the time. We know that the initial velocity (vi) is 27.0 m/s and the time (t) is 10.0 s.
Plugging the numbers in:
vf = 27.0 m/s - (1.483 m/s² × 10.0 s) = 27.0 m/s - 14.83 m/s = 12.17 m/s
Therefore, the speed of the Boeing 777 airplane 10 seconds later is 12.17 m/s.
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
(a) A uniform disk of mass 21 kg, thickness 0.5 m, and radius 0.6 m is located at the origin, oriented with its axis along the axis. It rotates clockwise around its axis when viewed from above (that is, you stand at a point on the axis and look toward the origin at the disk). The disk makes one complete rotation every 0.7 s. What is the rotational angular momentum of the disk? What is the rotational kinetic energy of the disk? (Express your answer for rotational angular momentum in vector form.)
Answer:L=33.93
R.E.=152.30
Explanation:
Given
mass of disk[tex]\left ( m\right )[/tex]=21kg
thickness[tex]\left ( t\right )[/tex]=0.5m
radius[tex]\left ( r\right )[/tex]=0.6m
t=0.7 sec for every complete rotation
therefore angular velocity[tex]\left ( \omega \right ) =\frac{\Delta \theta }{\Delta t}[/tex]
[tex]\left ( \omega \right )=\frac{ 2\pi }{0.7}=8.977[/tex] rad/s
Rotational angular momentum is given by
[tex]L=I\omega [/tex]
[tex]I[/tex]=[tex]\frac{mr^2}{2}[/tex]
I=3.78 [tex]kg-m^2[/tex]
[tex]L=3.78\times 8.977[/tex]
L=33.93 [tex]\hat{j}[/tex]
considering disk is rotating in z-x plane
Rotational kinetic energy=[tex]\frac{I\omega ^{2}}{2}[/tex]=152.30 J
The rotational angular momentum of the disk is 33.89 kg.m²/s directed into the page, and the rotational kinetic energy of the disk is 151.5 J.
Explanation:To calculate the rotational angular momentum of the disk, we will first need to compute the moment of inertia and angular velocity. The moment of inertia is given by the formula for a solid disk, which is I = 1/2mR². Here, m is the mass of the disk (in this case, 21 kg), and R is the radius of the disk (which is 0.6 m). So, I = 0.5*21*(0.6)² = 3.78 kg.m².
Furthermore, the angular velocity (ω) can be calculated using the formula 2π/T, where T is the period, which is the time it takes for one complete rotation (0.7 s in this situation). Therefore, ω = 2π/0.7 = 8.96 rad/s.
Finally, the rotational angular momentum (L), is given by the product of the moment of inertia and angular velocity, L = I*ω. Therefore, L = 3.78 kg.m² * 8.96 rad/s = 33.89 kg.m²/s.
The direction of the rotational angular momentum is along the axis of rotation, which depends on the right-hand rule. Since it is rotating clockwise when viewed from above, the vector is directed into the page (negative z-direction).
To compute the rotational kinetic energy (KE), we use the formula KE = 1/2 I ω². Substituting the values, we get KE = 0.5*3.78 kg.m²*(8.96 rad/s)² = 151.5 J.
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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]