The velocity of the fish relative to water is approximately 10.44 m/s when it hits the water. The fish inherits the eagle's horizontal velocity at the start and then accelerates downward due to gravity.
Explanation:The eagle's speed is given as 3.00 m/s horizontally. Therefore, when the fish falls, it also initially has a horizontal velocity of 3.00 m/s because it was brought to that speed by the eagle. The vertical component of the fish's velocity can be calculated using the second equation of motion: vf = vi + a*t, which simplifies to vf = g*t in the absence of initial vertical velocity. The time 't' it takes for the fish to hit the water can be calculated using the equation d = 0.5*a*t2, which gives t = √(2d/g). Substituting 5.00 m for 'd' and 9.81 m/s2 (approx gravitational acceleration) for 'g', we get t ≈ 1.017 seconds. Replacing 't' in the velocity equation with 1.017 seconds, which results in a vertical velocity of 9.965 m/s. The resultant velocity of the fish relative to the water when it hits would then be calculated using Pythagoras theorem since the horizontal and vertical movements are independent and at right angles to each other, giving a result of √((3.00 m/s)2 + (9.965 m/s)2) = 10.44 m/s.
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The velocity of the fish relative to the water when it hits the water is approximately 10.36 m/s at an angle of about 73.74° below the horizontal.
To calculate the velocity of the fish relative to the water when it hits the water, we need to analyze both the vertical and horizontal components of its motion separately.
Given Data:
Horizontal speed of the eagle (and the fish when dropped): [tex]v_{horizontal} = 3.00 \, \text{m/s}[/tex]
Height from which the fish falls: [tex]h = 5.00 \, \text{m}[/tex]
Step 1: Calculate the time it takes for the fish to fall 5m.
To find the time of fall, we can use the kinematic equation for vertical motion:
[tex]h = \frac{1}{2} g t^2[/tex]
where:
[tex]h[/tex] is the height (5.00 m),
[tex]g[/tex] is the acceleration due to gravity (approximately [tex]9.81 \, \text{m/s}^2[/tex]),
[tex]t[/tex] is the time in seconds.
Rearranging the equation to solve for [tex]t[/tex], we have:
[tex]t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5.00}{9.81}} \approx \sqrt{1.02} \approx 1.01 \, \text{s}[/tex]
Step 2: Calculate the vertical velocity of the fish just before hitting the water.
The vertical velocity can be calculated using:
[tex]v_{vertical} = g t[/tex]
Substituting the time we found:
[tex]v_{vertical} = 9.81 \, \text{m/s}^2 \times 1.01 \, \text{s} \approx 9.91 \, \text{m/s}[/tex]
Step 3: Combine the horizontal and vertical velocities to find the resultant velocity.
The horizontal velocity is constant at 3.00 m/s, and the vertical velocity just before impact is approximately 9.91 m/s. We can use the Pythagorean theorem to find the magnitude of the resultant velocity:
[tex]v_{resultant} = \sqrt{v_{horizontal}^2 + v_{vertical}^2}[/tex]
Substituting the values:
[tex]v_{resultant} = \sqrt{(3.00)^2 + (9.91)^2} = \sqrt{9 + 98.20} = \sqrt{107.20} \approx 10.36 \, \text{m/s}[/tex]
Step 4: Find the direction of the velocity relative to the horizontal using tangent.
The angle [tex]\theta[/tex] can be calculated using:
[tex]\theta = \tan^{-1}\left(\frac{v_{vertical}}{v_{horizontal}}\right)[/tex]
Substituting our values:
[tex]\theta = \tan^{-1}\left(\frac{9.91}{3.00}\right) \approx \tan^{-1}(3.30) \approx 73.74^\circ[/tex]
A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. Calculate the magnitude of the tension in the wire if the angle between the cable and the horizontal is θ = 47°.
Answer:
T = 2010 N
Explanation:
m = mass of the uniform beam = 150 kg
Force of gravity acting on the beam at its center is given as
W = mg
W = 150 x 9.8
W = 1470 N
T = Tension force in the wire
θ = angle made by the wire with the horizontal = 47° deg
L = length of the beam
From the figure,
AC = L
BC = L/2
From the figure, using equilibrium of torque about point C
T (AC) Sin47 = W (BC)
T L Sin47 = W (L/2)
T Sin47 = W/2
T Sin47 = 1470
T = 2010 N
The tension in the cable supporting a 150 kg beam can be calculated by equating the weight of the beam to the vertical component of the tension in the cable. Solve 150 kg * 9.8 m/s² = T * sin(47°) for the tension T to find the force exerted by the cable.
Explanation:The question concerns the calculation of the tension in the cable supporting a uniform beam. We begin by recognizing that this is a static situation with a beam of mass 150 kg in equilibrium. Thus, the total of the forces in the vertical direction must equal zero.
The forces acting on the beam are its weight (downwards) and the vertical component of the tension in the cable (upwards). The weight of the beam can be calculated by multiplying its mass by gravity g (approximately 9.81 m/s²), resulting in a downward force of 1471.5 N. The vertical component of the tension can be calculated using the sine function: T_vertical = T * sin(θ), where T is the total tension in the cable.
By setting the weight equal to the vertical component of the tension, we can solve for T: 150 kg * 9.8 m/s² = T * sin(47°). Solve this equation for the tension T to find the magnitude of the force exerted by the cable. This tension, along with its vertical and horizontal components, help maintain the beam in its horizontal position.
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A 64.0-kg ice skater is moving at 4.04 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.890 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. kN (b) Compare this force with her weight. Frope W =
We have that for the Question it can be said that the magnitude of the force exerted by the horizontal rope on her arms and the ratio of the Force to the weight is
F=1150.561NF/W=1.8325
From the question we are told
A 64.0-kg ice skater is moving at 4.04 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.890 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. kN (b) Compare this force with her weight. F-rope W =
Generally the equation for the force applied is mathematically given as
[tex]F=\frac{( mv^2)}{R}\\\\Therefore\\\\F=\frac{( mv^2)}{R}\\\\F=\frac{( (64)(4.0)^2)}{0.890}\\\\[/tex]
F=1150.561N
b)
Generally the equation for the Weight is mathematically given as
W=mg
Therefore
W=64*9.81
W=627.84N
Therefore
The Force to weight ratio is
[tex]F/W=1150.561N/627.84N[/tex]
F/W=1.8325
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The force exerted by the rope on the skater's arms as she moves in a circular path is 1.167 kN. This force is about 1.860 times her weight, which is 627.2 N.
Explanation:The skater is experiencing centripetal force exerted by the rope, which causes her to move in a circular path. The magnitude F of this force can be calculated using the formula F = mv²/r, where m is the skater's mass (64.0 kg), v is her velocity (4.04 m/s), and r is the radius of her circular path (0.890 m).
By substituting the given numbers into this formula, we get: F = (64.0 kg)(4.04 m/s)² / 0.890 m = 1166.67 N. In kilonewtons, this force is 1.167 kN.
To compare this force with her weight, we can calculate the weight (W) using the formula W = mg, where g is the acceleration due to gravity (around 9.8 m/s²). Substituting the given mass into this formula gives us: W = (64.0 kg)(9.8 m/s²) = 627.2 N.
Comparing these two forces shows that the force exerted by the rope on her arms is about 1.860 times her weight.
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A semi with a mass of 9,565 kg and going at a velocity of 55 m/s and hits a parked car(992 kg) at rest. If all the momentum is transfered to the car (the semi is now at rest), at what speed does the car move forward (assume an elastic collison, no decimals in the answer and leave no spaces between units and answer).
Answer:
Speed of car, v₁ = 55 m/s
Explanation:
It is given that,
Mass of Semi, m₁ = 9565 kg
Initial velocity of semi, u₁ = 55 m/s
Mass of car, m₂ = 992 kg
Initial velocity of car, u₂ = 0 (at rest)
Since, the collision between two objects is elastic and all the momentum is transferred to the car i.e final speed of semi, v₂ = 0
Let the speed of the car is v₁. Using conservation of linear momentum as :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
[tex]9565\ kg\times 55\ m/s+992\ kg\times 0=9565\ kg\times v_1+0[/tex]
v₁ = 55 m/s
Hence, the car move forward with a speed of 55 m/s.
What area must the plates of a capacitor be if they have a charge of 5.7uC and an electric field of 3.1 kV/mm between them? O 0.151 m O 0.147m2 0.169 ? O 0.208 m e
Answer:
Area of the plates of a capacitor, A = 0.208 m²
Explanation:
It is given that,
Charge on the parallel plate capacitor, [tex]q = 5.7\ \mu C=5.7\times 10^{-6}\ C[/tex]
Electric field, E = 3.1 kV/mm = 3100000 V/m
The electric field of a parallel plates capacitor is given by :
[tex]E=\dfrac{q}{A\epsilon_o}[/tex]
[tex]A=\dfrac{q}{E\epsilon_o}[/tex]
[tex]A=\dfrac{5.7\times 10^{-6}\ C}{3100000\ V/m\times 8.85\times 10^{-12}\ F/m}[/tex]
A = 0.208 m²
So, the area of the plates of a capacitor is 0.208 m². Hence, this is the required solution.
Four point charges of magnitude 3.6 mu or micro CC are at the corners of a square of side 4 m. (a) Find the electrostatic potential energy if all of the charges are negative.
The electrostatic potential energy if all of the charges are negative is
U = -1.026 x [tex]10^{-5}[/tex] J.
We have,
The electrostatic potential energy U of a system of point charges can be calculated using the formula:
U = 1/4π∈ [tex]\sum_{i = 1}^n \sum_{j > i}^n q_iq_j/r_{ij}[/tex]
Now,
n is the number of charges.
q are the magnitudes of charges.
r is the distance between the charges.
∈ is the vacuum permittivity
∈ = 8.85 x [tex]10^{-12}[/tex] C² / N - m²
Now,
Given that all charges are negative.
[tex]q_i[/tex] = - 3.6 mu
And they are at the corners of a square with a side of 4m, the distances between charges are all diagonals of the square is [tex]r_{ij} = 4\sqrt{2}m[/tex]
Substituting the values.
U = 1/4π∈ [tex]\sum_{i = 1}^n \sum_{j > i}^n q_iq_j/r_{ij}[/tex]
U = -1.026 x [tex]10^{-5}[/tex] J
Thus,
The electrostatic potential energy if all of the charges are negative is
U = -1.026 x [tex]10^{-5}[/tex] J.
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A 0.30-m-radius automobile tire accelerates from rest at a constant 2.0 rad/s2. What is the centripetal acceleration of a point on the outer edge of the tire after 5.0 s
Answer:
The centripetal acceleration is 30 m/s²
Explanation:
Given that,
Radius = 0.30 m
Acceleration = 2.0 rad/s²
Time = 5.0
We need to calculate the centripetal acceleration
Using formula of angular velocity
[tex]\omega=a\times t[/tex]
Where,
a = acceleration
t = time
Put the value into the formula
[tex]\omega=2.0\times5.0[/tex]
[tex]\omega=10\ rad/s[/tex]
Using formula of centripetal acceleration
[tex]a_{c}=r\omega^2[/tex]
[tex]a_{c}=0.30\times10^2[/tex]
[tex]a_{c}=30\ m/s^2[/tex]
Hence, The centripetal acceleration is 30 m/s²
The centripetal acceleration of a point on the outer edge of the tire after 5.0 seconds is [tex]\( 30 \, \text{m/s}^2 \)[/tex].
The centripetal acceleration of a point on the outer edge of the tire after 5.0 seconds is given by the formula [tex]\( a_c = \omega^2 r \)[/tex], where [tex]\( \omega \)[/tex] is the angular velocity and [tex]\( r \)[/tex] is the radius of the tire.
First, we need to find the angular velocity at 5.0 seconds. Since the tire starts from rest and accelerates at a constant rate.
We can use the formula [tex]\( \omega = \omega_0 + \alpha t \)[/tex], where [tex]\( \omega_0 \)[/tex] is the initial angular velocity (0 rad/s in this case, since the tire starts from rest), [tex]\( \alpha \)[/tex] is the angular acceleration ([tex]2.0 rad/s^2[/tex]), and [tex]\( t \)[/tex] is the time (5.0 s).
Plugging in the values, we get:
[tex]\( \omega = 0 + (2.0 \, \text{rad/s}^2)(5.0 \, \text{s}) = 10.0 \, \text{rad/s} \)[/tex]
Now that we have the angular velocity, we can calculate the centripetal acceleration:
[tex]\( a_c = \omega^2 r = (10.0 \, \text{rad/s})^2(0.30 \, \text{m}) \)[/tex]
[tex]\( a_c = 100 \, \text{s}^{-2} \times 0.30 \, \text{m} \)[/tex]
[tex]\( a_c = 30 \, \text{m/s}^2 \)[/tex]
A particle's position is given by x = 7.00 - 9.00t + 3t2, in which x is in meters and t is in seconds. (a) What is its velocity at t = 1 s? (b) Is it moving in the positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time t; if not, answer "0". (f) Is there a time after t = 3 s when the particle is moving in the negative direction of x? If so, give the time t; if not, answer "0".
[tex]x(t)=7.00\,\mathrm m-\left(9.00\dfrac{\rm m}{\rm s}\right)t+\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]
a. The particle has velocity at time [tex]t[/tex],
[tex]\dfrac{\mathrm dx(t)}{\mathrm dt}=-9.00\dfrac{\rm m}{\rm s}+\left(6\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]
so that after [tex]t=1\,\mathrm s[/tex] it will have velocity [tex]\boxed{-3.00\dfrac{\rm m}{\rm s}}[/tex].
b. The sign of the velocity is negative, so it's moving in the negative [tex]x[/tex] direction.
c. Its speed is 3.00 m/s.
d. The particle's velocity changes according to
[tex]\dfrac{\mathrm d^2x(t)}{\mathrm dt^2}=6\dfrac{\rm m}{\mathrm s^2}[/tex]
which is positive and indicates the velocity/speed of the particle is increasing.
e. Yes. The velocity is increasing at a constant rate. Solving for [tex]\dfrac{\mathrm dx(t)}{\mathrm dt}=0[/tex] is trivial; this happens when [tex]\boxed{t=1.50\,\mathrm s}[/tex].
f. No, the velocity is positive for all [tex]t[/tex] beyond 1.50 s.
As a car drives with its tires rolling freely without any slippage, the type of friction acting between the tires and the road is 100) A) static friction. B) kinetic friction. C) a combination of static and kinetic friction. D) neither static nor kinetic friction, but some other type of friction. E) It is impossible to tell what type of friction acts in this situation.
Final answer:
The type of friction acting between the tires and the road when a car's tires roll freely without any slippage is static friction, as the tire's contact point with the road is momentarily at rest.
Explanation:
When a car's tires roll freely without any slippage, the type of friction acting between the tires and the road is static friction. This is because the bottom of the tire is at rest with respect to the ground for a moment in time, ensuring there's no relative movement between the contact surfaces. Static friction is what allows the car to move forward as it prevents the tires from slipping on the surface of the road. Once slipping occurs, for instance, if the tires are spinning without moving the car forward, it becomes kinetic friction. However, in a scenario where rolling without slipping occurs, it's due to the presence of static friction, which is necessary for proper motion and control of the vehicle.
A floating ice block is pushed through a displacement d = (23 m) i - (9 m) j along a straight embankment by rushing water, which exerts a force F = (200 N) i - (149 N) j on the block. How much work does the force do on the block during the displacement?
Answer:
Work done, W = 5941 joules
Explanation:
It is given that,
Force exerted on the block, [tex]F=(200i-149j)\ N[/tex]
Displacement, [tex]x=(23i-9j)\ m[/tex]
Let W is the work done by the force do on the block during the displacement. Its formula is given by :
[tex]W=F.d[/tex]
[tex]W=(200i-149j){\cdot} (23i-9j)[/tex]
Since, i.i = j.j = k.k = 1
[tex]W=4600+1341[/tex]
W = 5941 joules
So, the work done by the force do on the block during the displacement is 5941 joules. Hence, this is the required solution.
Work is the energy transferred to an object by the application of force along a displacement. The work done by the force on the block during the displacement is 5941 J.
Work is the energy transferred to an object by the application of force along a displacement.
Given Here,
Displacement d = (23 m) i - (9 m) j
Force exerted on the block F = (200 N) i - (149 N) j
Work formula,
W = F.d
W = (200 N) i - (149 N) j . (23 m) i - (9 m) j
Since i.i = j.j = k.k = 1
Hence,
W = 4600 + 1341
W = 5941 J
Hence we can conclude that the work done by the force on the block during the displacement is 5941 J.
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A boy and a girl are riding on a merry-go-round that is turning. The boy is twice as far as the girl from the merry-go-round's center. If the boy and girl are of equal mass, which statement is true about the boy's moment of inertia with respect to the axis of rotation? g
Answer:
The answer would be "His moment of inertia is 4 times the girl's"
The statement true about the boy's moment of inertia with respect to the axis of rotation is "His moment of inertia is 4 times the girl's".
What is moment of inertia?The moment of inertia is the amount of rotation obtained by an object when it is in state of motion or rest.
A boy and a girl are riding on a merry-go-round that is turning. The boy is twice as far as the girl from the merry-go-round's center. The boy and girl are of equal mass.
Moment of inertia is given by I = mr²where m is the mass and r is the distance from the axis of rotation.
For girl, I = mr²
For boy with twice the distance from axis
I = m(2r)²I = 4mr²
On comparison, we have The boy's moment of inertia is 4 times less than the girl's.
Thus, the statement true about the boy's moment of inertia with respect to the axis of rotation is "His moment of inertia is 4 times the girl's".
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Calculate the kinetic energy of a 1158-kg car moving at 55 km/h.
Answer:
The kinetic energy is 135183.99 J
Explanation:
Given that,
Mass = 1158 kg
Velocity = 55 km/h = 15.28 m/s
We need to calculate the kinetic energy
The kinetic energy is equal to the half of the product of the mass and square of velocity.
Using formula of kinetic energy
[tex]K.E=\dfrac{1}{2}mv^2[/tex]
[tex]K.E=\dfrac{1}{2}\times1158\times15.28^2[/tex]
[tex]K.E=135183.99\ J[/tex]
Hence, The kinetic energy is 135183.99 J
A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5m/s2 for 4s. After the wind stops, the sailboat is traveling to the left with a velocity of 3.0m/s.Assuming the acceleration from the wind is constant, what was the initial velocity of the sailboat before the gust of wind?Answer using a coordinate system where rightward is positive.
Answer:
+7.0 m/s
Explanation:
Let's take rightward as positive direction.
So in this problem we have:
a = -2.5 m/s^2 acceleration due to the wind (negative because it is leftward)
t = 4 s time interval
v = -3.0 m/s is the final velocity (negative because it is leftward)
We can use the following equation:
v = u + at
Where u is the initial velocity
We want to find u, so if we rearrange the equation we find:
[tex]u = v - at = (-3.0 m/s) - (-2.5 m/s^2)(4 s)=+7.0 m/s[/tex]
and the positive sign means the initial direction was rightward.
A wheelhas a radius of 4.8 m. Howfar (path length) does a point on the circumference travel if thewheel is rotated through angles of 30°, 30 rad, and 30 rev,respectively?
Answer:
(a) 2.512 m
(b) 144 m
(c) 904.32 m
Explanation:
radius, r = 4.8 m
(a) for 30 degree
As we know that in 360 degree it rotates a complete round that means circumference.
In 360 degree, it rotates = 2 x π x r
in 30 degree, it rotates = 2 x π x r x 30 / 360
= 2 x 3.14 x 4.8 x 30 / 360
= 2.512 m
(b) for 30 rad
As we know that in one complete rotation, it rotates by 2π radian.
so,
for 2π radian it rotates = 2 x π x r
for 30 radian, it rotates = 2 x π x r x 30 / 2 π = 144 m
(c) For 30 rev
In one complete revolution, it travels = 2 x π x r
in 30 rev, it travels = 2 x π x r x 30 = 2 x 3.14 x 4.8 x 30 = 904.32 m
A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 82.0 N is applied to the rim of the wheel. The wheel has radius 0.150 m . Starting from rest, the wheel has an angular speed of 12.8 rev/s after 3.88 s. What is the moment of inertia of the wheel?
Answer:
The moment of inertia of the wheel is 0.593 kg-m².
Explanation:
Given that,
Force = 82.0 N
Radius r = 0.150 m
Angular speed = 12.8 rev/s
Time = 3.88 s
We need to calculate the torque
Using formula of torque
[tex]\tau=F\times r[/tex]
[tex]\tau=82.0\times0.150[/tex]
[tex]\tau=12.3\ N-m[/tex]
Now, The angular acceleration
[tex]\dfrac{d\omega}{dt}=\dfrac{12.8\times2\pi}{3.88}[/tex]
[tex]\dfrac{d\omega}{dt}=20.73\ rad/s^2[/tex]
We need to calculate the moment of inertia
Using relation between torque and moment of inertia
[tex]\tau=I\times\dfrac{d\omega}{dt}[/tex]
[tex]I=\dfrac{I}{\dfrac{d\omega}{dt}}[/tex]
[tex]I=\dfrac{12.3}{20.73}[/tex]
[tex]I= 0.593\ kg-m^2[/tex]
Hence, The moment of inertia of the wheel is 0.593 kg-m².
Answer:
0.593 kg-m²
Explanation:
edg.
A hypothetical atom has three energy levels: the ground-state level and levels 1.50 eV and 5.00 eV above the ground state. What is the longest wavelength in the line spectrum for this atom? Let Planck's constant h = 4.136 x 10^-15 ev s, and the speed of light c = 3.00 x 10^8 m/s. (a)355 nm (b)780 nm (c)882 nm (d) 827 nm
Answer:
option (d)
Explanation:
E1 = 1.5 eV = 1.5 x 1.6 x 10^-19 J, E2 = 5 eV = 5 x 1.6 x 10^-19 J, c = 3 x 10^8 m/s, h = 6.62 x 10^-34 Js
Wavelength associated with 1.5 eV is λ1.
E1 = h c / λ1
λ1 = h c / E1
λ1 = (6.62 x 10^-34 x 3 x 10^8) / (1.5 x 1.6 x 10^-19)
λ1 = 8.275 x 10^-7 m = 827 nm
Wavelength associated with 5 eV is λ2.
E2 = h c / λ2
λ2 = h c / E2
λ2 = (6.62 x 10^-34 x 3 x 10^8) / (5 x 1.6 x 10^-19)
λ2 = 2.4825 x 10^-7 m = 248 nm
So, the longest wavelength is 827 nm
A janitor opens a 1.10 m wide door by pushing on it with a force of 47.5 N directed perpendicular to its surface. HINT (a) What magnitude torque (in N · m) does he apply about an axis through the hinges if the force is applied at the center of the door? (b) What magnitude torque (in N · m) does he apply at the edge farthest from the hinges?
(a) 26.1 Nm
The magnitude of the torque exerted by a force acting perpendicularly to a surface is given by:
[tex]\tau = Fr[/tex]
where
F is the magnitude of the force
r is the distance from the pivot
In this situation,
F = 47.5 N is the force applied
[tex]r=\frac{1.10 m}{2}=0.55 m[/tex] is the distance from the hinges (the force is applied at the center of the door)
So, the magnitude of the torque is
[tex]\tau = (47.5 N)(0.55 m)=26.1 Nm[/tex]
(b) 52.3 Nm
In this case, the force is applied at the edge of the door farthest from the hinges. This means that the distance from the hinges is
r = 1.10 m
So, the magnitude of the torque is
[tex]\tau =(47.5 N)(1.10 m)=52.3 Nm[/tex]
The magnitude of torque applied about the axis of the door when the force is applied at the center is 26.125 N.m
The magnitude of the torque applied at the edge farthest from the center is 52.25 N.m.
The given parameters;
width of the door, w = 1.10 mapplied force, F = 47.5 NThe torque applied by the janitor is the product of applied force and perpendicular distance.
τ = F.r
The torque applied about the axis of the door when the force is applied at the center.
[tex]\tau = F \times \frac{r}{2} \\\\\tau = 47.5 \times \frac{1.1}{2} \\\\\tau = 26.125 \ N.m[/tex]
The magnitude of the torque applied at the edge farthest from the center ;
[tex]\tau = F\times r\\\\\tau = 47.5 \times 1.1\\\\\tau = 52.25 \ N.m[/tex]
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Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.87 Earth years in its circular orbit around its sun, which is a star with a measured mass of 3.59 x 1030 kg. Find the radius of the exoplanet's orbit.
Answer: [tex]4.487(10)^{11}m[/tex]
Explanation:
This problem can be solved using the Third Kepler’s Law of Planetary motion:
“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
This law states a relation between the orbital period [tex]T[/tex] of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size [tex]a[/tex] of its orbit:
[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex] (1)
Where:
[tex]T=3.87Earth-years=122044320s[/tex] is the period of the orbit of the exoplanet (considering [tex]1Earth-year=365days[/tex])
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]M=3.59(10)^{30}kg[/tex] is the mass of the star
[tex]a[/tex] is orbital radius of the orbit the exoplanet describes around its star.
Now, if we want to find the radius, we have to rewrite (1) as:
[tex]a=\sqrt[3]{\frac{T^{2}GM}{4\pi^{2}}}[/tex] (2)
[tex]a=\sqrt[3]{\frac{(122044320s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(3.59(10)^{30}kg)}{4\pi^{2}}}[/tex] (3)
Finally:
[tex]a=4.487(10)^{11}m[/tex] This is the radius of the exoplanet's orbit
Given the orbital period and mass of the exoplanet, its radius of orbit is 4.488 × 10¹¹m.
Given the data in the question;
Orbital period;[tex]T = 3.87 \ Earth\ years = [ 3.87yrs*365days*24hrs*60min*60sec = 122044320s[/tex] Mass of the Planet; [tex]M = 3.59*10^{30}kg[/tex]Radius of the exoplanet's orbit; [tex]r= \ ?[/tex]To determine the radius of the exoplanet's orbit, we use the equation from Kepler's Third Law:
[tex]T^2 = \frac{4\pi^2 }{GM}r^3\\[/tex]
Where, T is the period of the orbit of the exoplanet, G is the Gravitational Constant, M is the mass of the star and r is orbital radius.
We make "r", the subject of the formula
[tex]r = \sqrt[3]{\frac{T^2GM}{4\pi ^2} }[/tex]
We substitute our given values into the equation
[tex]r = \sqrt[3]{\frac{(122044320s)^2*(6.67430 * 10^{-11} m^3/kg s^2)*(3.59*10^{30}kg)}{4*\pi ^2} } \\\\r = \sqrt[3]{\frac{(1.48948*10^{16}s^2)*(6.67430 * 10^{-11} m^3/kg s^2)*(3.59*10^{30}kg)}{4*\pi ^2} }\\\\r = \sqrt[3]{\frac{3.5689*10^{36}m^3}{4*\pi ^2} }\\\\r = \sqrt[3]{9.04*10^{34}m^3}\\\\r = 4.488*10^{11}m[/tex]
Therefore, given the orbital period and mass of the exoplanet, its radius of orbit is 4.488 × 10¹¹m.
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Two parallel-plate capacitors have the same plate area, but the plate gap in capacitor 1 is twice as big as capacitor 2. If capacitance of the first capacitor is C, then capacitance of the second one is:
Answer:
Capacitance of the second capacitor = 2C
Explanation:
[tex]\texttt{Capacitance, C}=\frac{\varepsilon_0A}{d}[/tex]
Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.
Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.
We have
[tex]\texttt{Capacitance, C}_1=\frac{\varepsilon_0A_1}{d_1}=C[/tex]
Similarly for capacitor 2
[tex]\texttt{Capacitance, C}_2=\frac{\varepsilon_0A_2}{d_2}=\frac{\varepsilon_0A_1}{\frac{d_1}{2}}=2\times \frac{\varepsilon_0A_1}{d_1}=2C[/tex]
Capacitance of the second capacitor = 2C
A UHF antenna is oriented at an angle of 47o to a magnetic field that changes at a rate of 0.23 T/s. What is the induced emf of the antenna if it has a diameter of 13.4 cm? O 5.4 mV ? 2.2 ? 0027 ?
Answer:
Induced emf, [tex]\epsilon=2.2\ mV[/tex]
Explanation:
It is given that,
Rate of change of magnetic field, [tex]\dfrac{dB}{dt}=0.23\ T/s[/tex]
A UHF antenna is oriented at an angle of 47° to a magnetic field, θ = 47°
Diameter of the antenna, d = 13.4 cm
Radius, r = 6.7 m = 0.067 m
We need to find the induced emf of the antenna. It is given by :
[tex]\epsilon=-\dfrac{d\phi}{dt}[/tex]
Where
[tex]\phi[/tex] = magnetic flux, [tex]\phi=BA\ cos\theta[/tex]
So, [tex]\epsilon=\dfrac{d(BA\ cos\theta)}{dt}[/tex]
B = magnetic field
[tex]\epsilon=A\dfrac{d(B)}{dt}\ cos\theta[/tex]
[tex]\epsilon=\pi r^2\times \dfrac{dB}{dt}\times cos(47)[/tex]
[tex]\epsilon=\pi (0.067\ m)^2\times 0.23\ T-s\times cos(47)[/tex]
[tex]\epsilon=0.0022\ V[/tex]
[tex]\epsilon=2.2\ mV[/tex]
So, the induced emf of the antenna is 2.2 mV. Hence, this is the required solution.
An isotope of Uranium, Z = 92 and A = 235, decays by emitting an alpha particle. Calculate the number of neutrons in the nucleus left behind after the radioactive decay.
Answer:
141
Explanation:
The atomic number (Z) corresponds to the number of protons:
Z = p
while the mass number (A) corresponds to the number of protons+neutrons:
A = p + n
So the number of neutrons in a nucleus is equal to the difference between mass number and atomic number:
n = A - Z
For the initial nucleus of Uranium, Z = 92 and A = 235, so the initial number of neutrons is
n = 235 - 92 = 143
An alpha particle carries 2 protons and 2 neutrons: so, when the isotope of Uranium emits an alpha particle, it loses 2 neutrons. Therefore, the number of neutrons after the decay will be
n = 143 - 2 = 141
A graduated cylinder contains 63.0 mL of water. A piece of gold, which has a density of 19.3 g/ cm3, is added to the water and the volume goes up to 64.5 mL. Calculate the mass in grams of the gold that was added to the water. Explain how you got your answer.
Answer:
29.0 g
Explanation:
The mass of the piece of gold is given by:
m = dV
where
m is the mass
d is the density
V is the volume of the piece of gold
The density of gold is
d = 19.3 g/cm^3
while the volume of the sample is equal to the volume of displaced water, so
V = 64.5 mL - 63.0 mL = 1.5 mL
And since
1 mL = 1 cm^3
the volume is
V = 1.5 cm^3
So the mass of the piece of gold is:
m = (19.3 g/cm^3)(1.5 cm^3)=29.0 g
28. A stone is projected at a cliff of height h with an initial speed of 42.0 mls directed at angle θ0 = 60.0° above the horizontal. The stone strikes at A, 5.50 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the grou
Answer:
a) 51.8 m, b) 27.4 m/s, c) 142 m
Explanation:
Given:
v₀ = 42.0 m/s
θ = 60.0°
t = 5.50 s
Find:
h, v, and H
a) y = y₀ + v₀ᵧ t + ½ gt²
0 = h + (42.0 sin 60.0) (5.50) + ½ (-9.8) (5.50)²
h = 51.8 m
b) vᵧ = gt + v₀ᵧ
vᵧ = (-9.8)(5.5) + (42.0 sin 60.0)
vᵧ = -17.5 m/s
vₓ = 42.0 cos 60.0
vₓ = 21.0 m/s
v² = vₓ² + vᵧ²
v = 27.4 m/s
c) vᵧ² = v₀ᵧ² + 2g(y - y₀)
0² = (42.0)² + 2(-9.8)(H - 51.8)
H = 142 m
(a) The height of the cliff is 51.82 m
(b) The final velocity of the stone before it strikes A is 27.36 m/s
(c) The maximum height reached by the stone is 67.50 m
The given parameters;
initial velocity of the object, u = 42 m/sangle of projection, = 60° time of motion of the stone, t = 5.5 s(a) The height of the cliff is calculated as;
[tex]h_y = v_0_yt - \frac{1}{2} gt^2\\\\h_y = (v_0\times sin(\theta)) t \ - \ \frac{1}{2} gt^2\\\\h_y = (42\times sin(60))\times 5.5 \ - \ (0.5\times 9.8\times 5.5^2)\\\\h_y = 200.05 - 148.23\\\\h_y = 51.82 \ m[/tex]
(b) The final velocity of the stone before it strikes A is calculated as;
The vertical component of the final velocity
[tex]v_f_y = v_0_y - gt\\\\v_f_y = (v_0 \times sine (\theta)) - gt\\\\v_f_y = (42\times sin(60) - (9.8\times 5.5)\\\\v_f_y = -17.53 \ m/s[/tex]
The horizontal component of the final velocity
[tex]v_x_f = v_0\times cos(\theta)\\\\v_x_f = 42 \times cos(60)\\\\v_x_f = 21 \ m/s[/tex]
The resultant of the velocity at point A;
[tex]v_f = \sqrt{v_x_f^2 + v_y_f^2} \\\\v_f = \sqrt{(21)^2 + (-17.53)^2} \\\\v_f = 27.36 \ m/s[/tex]
(c) The maximum height reached by the projectile;
[tex]H = \frac{v_o^2 sin^2(\theta)}{2g} \\\\H = \frac{[42 \times sin(60)]^2}{2\times 9.8} \\\\H = 67.50 \ m[/tex]
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For a rigid body in rotational motion, what can be stated about the angular velocity of all of the particles? The linear velocity?
Answer:
Explanation:
In case of rotational motion, every particle is rotated with a same angular velocity .
The relation between linear velocity and angular velocity is
V = r w
Where v is linear velocity, r is the radius of circular path and w be the angular velocity.
Here w is same for all the particles but every particle has different radius of circular path I which they are rotating, so linear velocity is differnet for all.
A metal sphere of radius 10 cm carries a charge of +2.0 μC uniformly distributed over its surface. What is the magnitude of the electric field due to this sphere at a point 5.0 cm outside the sphere's surface? (k=1/4πϵ0=8.99×109 N · m2/C2) A metal sphere of radius cm carries a charge of μC uniformly distributed over its surface. What is the magnitude of the electric field due to this sphere at a point cm outside the sphere's surface? ( N · m2/C2) 4.0×109 N/C 4.0×107 N/C 8.0×107 N/C 4.2×106 N/C 8.0×109 N/C
Answer:
[tex]8.0\cdot 10^5 N/C[/tex]
Explanation:
Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.
Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:
[tex]E=k\frac{q}{(R+r)^2}[/tex]
where
[tex]k=8.99\cdot 10^9 N m^2C^{-2}[/tex] is the Coulomb's constant
[tex]q=2.0 \mu C=2.0 \cdot 10^{-6}C[/tex] is the charge on the sphere
[tex]R=10 cm = 0.10 m[/tex] is the radius of the sphere
[tex]r=5.0 cm = 0.05 m[/tex] is the distance from the surface of the sphere
Substituting, we find
[tex]E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C[/tex]
The magnitude of the electric field due to this sphere at the given distance is 8 x 10⁵ N/C.
The given parameters;
Radius of the sphere, r = 10 cm = 0.1 mCharge of the sphere, Q = 2 μC = 2 x 10⁻⁶ CDistance outside the sphere, x = 5 cm = 0.05 mThe magnitude of the electric field due to this sphere at the given distance is calculated using Coulomb's law;
[tex]E = \frac{kQ}{R^2}[/tex]
where;
R is the total distance from the center to the external distance of the sphere;R = (0.1 + 0.05) m = 0.15 m
[tex]E = \frac{8.99\times 10^{9} \times 2\times 10^{-6}}{(0.15)^2} \\\\ E= 8 \times 10^{5} \ N/C[/tex]
Thus, the magnitude of the electric field due to this sphere at the given distance is 8 x 10⁵ N/C.
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The equation that describes a transverse wave on a string is y = (0.0120 m)sin[(927 rad/s)t - (3.00 rad/m)x] where y is the displacement of a string particle and x is the position of the particle on the string. The wave is traveling in the +x direction. What is the speed v of the wave?
The speed of the wave is 309 m/s.
Explanation:The equation given is y = (0.0120 m)sin[(927 rad/s)t - (3.00 rad/m)x], where y represents the displacement of a string particle and x represents the position of the particle on the string. The wave is traveling in the +x direction. To find the speed v of the wave, we need to determine the wave velocity. The wave velocity is given by the formula v = ω/k, where ω is the angular frequency and k is the wave number.
In the equation y = (0.0120 m)sin[(927 rad/s)t - (3.00 rad/m)x], the angular frequency is 927 rad/s and the wave number is 3.00 rad/m. Therefore, the wave velocity is v = 927 rad/s / 3.00 rad/m = 309 m/s.
The tension in the string of a simple pendulum is:
a. constant
b. Maximum in the extreme position
c. Zero in the mean position
d. None of the above
Answer:
D. None of the above
Explanation:
There are only two forces acting on a pendulum:
- The force of gravity (downward)
- The tension in the string
We can consider the axis along the direction of the string: here we have the tension T, acting towards the pivot, and the component of the weight along this direction, acting away from the pivot. Their resultant must be equal to the centripetal force, so we can write:
[tex]T-mg cos \theta = m\frac{v^2}{r}\\T=m\frac{v^2}{r}+mg cos \theta[/tex]
where
T is the tension in the string
[tex]\theta[/tex] is the angle between the tension and the vertical
m is the mass
g is the acceleration of gravity
v is the speed of the pendulum
r is the length of the string
From the formula we see that the value of the tension, T, depends only on the value of v (the speed) and [tex]\theta[/tex], the angle. We notice that:
- Since [tex]\theta[/tex] and v constantly change, T must change as well
- At [tex]\theta=0^{\circ}[/tex] (equilibrium position), [tex]cos \theta=1[/tex] (maximum value), and also the speed v is maximum, so the tension has the maximum value at the equilibrium position
- For [tex]\theta[/tex] increasing, the [tex]cos \theta[/tex] decreases and the speed v decreases as well, so the tension T decreases: this means that the value of the tension will be minimum in the extreme positions.
So the correct answer is D. None of the above
A rifle with a weight of 35 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle in m/s. (b) If a 650-N man holds the rifle firmly against his shoulder, find the recoil speed of the man and rifle in m/s.
(a) 0.30 m/s
The total momentum of the rifle+bullet system before the shot is zero:
[tex]p_i = 0[/tex]
The total momentum of the system after the shot is the sum of the momenta of the rifle and of the bullet:
[tex]p_f = m_r v_r + m_b v_b[/tex]
where we have
[tex]m_r = \frac{W}{g}=\frac{35 N}{9.8 m/s^2}=3.57 kg[/tex] is the mass of the rifle
[tex]v_r[/tex] is the final velocity of the rifle
[tex]m_b = 4.5 g = 0.0045 kg[/tex] is the mass of the bullet
[tex]v_b = 240 m/s[/tex] is the final velocity of the bullet
Since the total momentum must be conserved, we have
[tex]p_i = p_f[/tex]
So
[tex]m_r v_r + m_b v_b=0[/tex]
and so we can find the recoil velocity of the rifle:
[tex]v_r = - \frac{m_b v_b}{m_r}=-\frac{(0.0045 kg)(240 m/s)}{3.57 kg}=-0.30 m/s[/tex]
And the negative sign means it travels in the opposite direction to the bullet: so, the recoil speed is 0.30 m/s.
(b) 0.016 m/s
The mass of the man is equal to its weight divided by the acceleration of gravity:
[tex]m=\frac{W}{g}=\frac{650 N}{9.8 m/s^2}=66.3 kg[/tex]
This time, we have to consider the system (man+rifle) - bullet. Again, the total momentum of the system before the shot is zero:
[tex]p_i = 0[/tex]
while the total momentum after the shot is
[tex]p_f = m_r v_r + m_b v_b[/tex]
where this time we have
[tex]m_r = 66.3 kg+3.57 kg=69.9 kg[/tex] is the mass of the rifle+person
[tex]v_r[/tex] is the final velocity of the man+rifle
[tex]m_b = 4.5 g = 0.0045 kg[/tex] is the mass of the bullet
[tex]v_b = 240 m/s[/tex] is the final velocity of the bullet
Since the total momentum must be conserved, we have
[tex]m_r v_r + m_b v_b=0[/tex]
and so we can find the recoil velocity of the man+rifle:
[tex]v_r = - \frac{m_b v_b}{m_r}=-\frac{(0.0045 kg)(240 m/s)}{66.9 kg}=-0.016 m/s[/tex]
So the recoil speed is 0.016 m/s.
The recoil speed of the rifle is 0.0031 m/s when held loosely away from the shoulder. When a 650 N man holds the rifle firmly against his shoulder, the effective mass of the rifle-man system is 28.0 kg, resulting in a recoil speed of 0 m/s.
Explanation:To calculate the recoil speed of the rifle in m/s, we use the principle of conservation of momentum. The momentum of the rifle before firing is equal to the momentum of the bullet after firing. The momentum of an object is calculated by multiplying its mass by its velocity. Given that the mass of the bullet is 4.5 g (0.0045 kg) and the velocity is 240 m/s, we can find the momentum of the bullet. Then, using the principle of conservation of momentum, we can calculate the recoil speed of the rifle.
(a) The momentum of the bullet is calculated as:
Momentum = mass x velocity = 0.0045 kg x 240 m/s = 0.108 kg·m/s
Since the momentum of the bullet before firing is equal to the momentum of the rifle after firing, we can write:
0.108 kg·m/s = mass of the rifle x recoil speed of the rifle
Rearranging the equation, we can solve for the recoil speed of the rifle:
Recoil speed of the rifle = 0.108 kg·m/s ÷ mass of the rifle = 0.108 kg·m/s ÷ 35 N = 0.0030857 m/s
(b) When a 650 N man holds the rifle firmly against his shoulder, the effective mass of the rifle-man system is 28.0 kg. To find the recoil speed of the man and rifle together, we can again use the principle of conservation of momentum. The initial momentum of the rifle-man system is zero, as they are at rest. Therefore, the final momentum of the system after firing must also be zero. We can write:
0 = (mass of the rifle + mass of the man) x recoil speed of the system
Rearranging the equation, we can solve for the recoil speed of the system:
Recoil speed of the system = 0 ÷ (mass of the rifle + mass of the man) = 0 ÷ (28 kg + 650 N ÷ 9.8 m/s²) = 0 m/s
A water main is to be constructed with a 12.512.5% grade in the north direction and a 2020% grade in the east direction. Determine the angle thetaθ required in the water main for the turn from north to east.
Final answer:
The angle theta required for the water main to turn from north to east can be determined by treating the grades as the rise over run and using vector addition with trigonometry to find the resultant direction and its angle east of north.
Explanation:
To determine the angle theta (θ) required for the water main to make the turn from north to east with a 12.5% grade in the north direction and a 20% grade in the east direction, we need to use vector addition and trigonometry. The water main's path can be viewed as containing two components: one in the north direction and one in the east direction. We will represent these as vectors and find their resultant to determine the required angle between them.
The grades can be understood as the rise over run, which means the north vector has a rise of 12.5 units for every 100 units of run, and the east vector has a rise of 20 units for every 100 units of run. To find the resultant vector, we can use the arctangent function:
θ = arctan(opposite/adjacent) = arctan(east grade / north grade) = arctan(0.20 / 0.125) = arctan(1.6)
Once we calculate θ, this will give us the angle between the north direction and the resultant direction of the water main. Keep in mind that this angle should be measured east of north to reflect the correct orientation of the water main's turn.
From Center Station, a train departs every 30 minutes on the Fast Line and a train departs every 50 minutes on the State Line. If two trains depart from Center Station at 8:00 A.M., one on each of the two lines, what is the next time that two trains, one on each line, will depart at the same time?
Explanation:
You need to find the least common multiple (LCM) of 30 and 50. First, write the prime factorization of each:
30 = 2×3×5
50 = 2×5²
The LCM must contain all the factors of both, so:
LCM = 2×3×5²
LCM = 150
It will take 150 minutes (or 2 hours and 30 minutes) before two trains depart at the same time again.
The next time that two trains, one on each line, will depart from Center Station at the same time is at 10:30 A.M. This is calculated by finding the least common multiple of the two train schedules.
Explanation:This question requires finding the least common multiple (LCM) of the two train schedules, which represents the time duration when both trains will depart again at the same time. The LCM of 30 and 50 is 150 minutes. This means, from 8:00 A.M., it will be the next 150 minutes or 2 hours and 30 minutes when both trains will depart from Center Station at the same time. Therefore, the answer is 10:30 A.M.
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Suppose the earth is shaped as a sphere with radius 4,0004,000 miles and suppose it rotates once every 24 hours. How many miles along the equator does it rotate each hour? (approximation is acceptable)
Answer:
1047 miles
Explanation:
The radius of the Earth is
[tex]r = 4000[/tex] (miles)
So its circumference, which is the total length of the equator, is given by
[tex]L=2\pi r= 2\pi(4000)=25133 mi[/tex]
Now we know that the Earth rotates once every 24 hours. So the distance through which the equator moves in one hour is equal to its total length divided by the number of hours, 24:
[tex]L' = \frac{25133 mi}{24h}=1047 mi[/tex]