Final answer:
The magnetic field at a point on the +x-axis near a wire carrying current in the +y-direction will be directed into the page, as determined by the right-hand rule. To cancel the magnetic field at a specific point on the +x-axis, a second wire carrying equal but opposite current must be arranged parallel to the first.
Explanation:
The question concerns the determination of the magnetic field direction around a long, straight wire carrying a current. According to the right-hand rule, if you point your right thumb in the direction of the current, then the direction in which your fingers curl will show the direction of the magnetic field. In this case, the current is running in the +y direction, so if one were to stand at a point on the +x-axis near the wire and apply the right-hand rule, the fingers would curl into the page, indicating that the magnetic field at that point is directed into the page.
An experimenter wanting to make the total magnetic field at the coordinate x = 1.0 m zero must arrange a second wire parallel to the first with current in the opposite direction. This second wire would also need to be carrying the same magnitude of current but in the -y direction to cancel out the magnetic field from the first wire at the point of interest.
A 2.7-kg cart is rolling along a frictionless, horizontal track towards a 1.1-kg cart that is held initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is +3.7 m/s, and the second cart's velocity is -1.6 m/s. (a) What is the total momentum of the system of the two carts at this instant? (b) What was the velocity of the first cart when the second cart was still at rest?
Answer:
Part a)
P = 8.23 kg m/s
Part b)
v = 3.05 m/s
Explanation:
Part a)
momentum of cart 1 is given as
[tex]P_1 = m_1v_1[/tex]
[tex]P_1 = (2.7)(3.7) = 9.99 kg m/s[/tex]
Momentum of cart 2 is given as
[tex]P_2 = m_2v_2[/tex]
[tex]P_2 = (1.1)(-1.6) = -1.76 kg m/s[/tex]
Now total momentum of both carts is given as
[tex]P = P_1 + P_2[/tex]
[tex]P = 8.23 kg m/s[/tex]
Part b)
Since two carts are moving towards each other due to mutual attraction force and there is no external force on two carts so here momentum is always conserved
so here we will have
[tex]P_i = P_f[/tex]
[tex](2.7 kg)v = 8.23[/tex]
[tex]v = 3.05 m/s[/tex]
Red light from three separate sources passes through a diffraction grating with 6.60×105 slits/m. The wavelengths of the three lines are 6.56 ×10−7m (hydrogen), 6.50 ×10−7m (neon), and 6.97 ×10−7m (argon). Part A Calculate the angle for the first-order diffraction line of first source (hydrogen). Express your answer using three significant figures.
Answer:· Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed on a screen that is 2.74 m from the grating. In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 3.16 m. What is the difference between these wavelengths? . I know to apply the equation din(theta) = m*wavelength but I'm not sure how to find all the missing variables or to get the difference in wavelengths
A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 7.2 m from this surface, the potential is150 V What is the radius of the sphere?
Answer:
The radius of the sphere is 3.6 m.
Explanation:
Given that,
Potential of first sphere = 450 V
Radial distance = 7.2 m
If the potential of sphere =150 V
We need to calculate the radius
Using formula for potential
For 450 V
[tex]V=\dfrac{kQ}{r}[/tex]
[tex]450=\dfrac{kQ}{r}[/tex]....(I)
For 150 V
[tex]150=\dfrac{kQ}{r+7.2}[/tex]....(II)
Divided equation (I) by equation (II)
[tex]\dfrac{450}{150}=\dfrac{\dfrac{kQ}{r}}{\dfrac{kQ}{r+7.2}}[/tex]
[tex]3=\dfrac{(r+7.2)}{r}[/tex]
[tex]3r=r+7.2[/tex]
[tex]r=\dfrac{7.2}{2}[/tex]
[tex]r=3.6\ m[/tex]
Hence, The radius of the sphere is 3.6 m.
The radius of the sphere whose surface has a potential difference of 450 V is 3.6 m.
What is the radius of the sphere?We know that the potential difference can be written as,
[tex]V = k\dfrac{Q}{R}[/tex]
We know that at R= R, Potential difference= 450 V,
[tex]450 = k\dfrac{Q}{R}[/tex]
Also, at R = (R+7.2), Potential difference = 150 V,
[tex]150 = k\dfrac{Q}{(R+7.2)}[/tex]
Taking the ratio of the two,
[tex]\dfrac{450}{150} = \dfrac{kQ}{R} \times \dfrac{(R+7.2)}{kQ}\\\\\dfrac{450}{150} = \dfrac{(R+7.2)}{R}\\\\R = 3.6\ m[/tex]
Hence, the radius of the sphere whose surface has a potential difference of 450 V is 3.6 m.
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The shortest air column inside a resonator vibrates with a frequency of 250 Hz, if the next harmonic is 750 Hz, and the speed of sound is 343 m/s.
a. Is this resonator closed at one end or open at both ends? Explain.
b. Find the length of the resonator.
Answer:
Part a)
the two frequencies are in ratio of odd numbers so it must be closed at one end
Part b)
L = 34.3 cm
Explanation:
Part a)
Since the shortest frequency is known as fundamental frequency
It is given as
[tex]f_o = 250 Hz[/tex]
next higher frequency is given as
[tex]f_1 = 750 Hz[/tex]
since the two frequencies here are in ratio of
[tex]\frac{f_1}{f_o} = \frac{750}{250} = 3 : 1[/tex]
since the two frequencies are in ratio of odd numbers so it must be closed at one end
Part b)
For the length of the pipe we can say that fundamental frequency is given as
[tex]f_o = \frac{v}{4L}[/tex]
here we have
[tex]250 = \frac{343}{4(L)}[/tex]
now we will have
[tex]L = \frac{343}{4\times 250}[/tex]
[tex]L = 34.3 cm[/tex]
A typical adult ear has a surface area of 2.90 × 10-3 m2. The sound intensity during a normal conversation is about 2.19 × 10-6 W/m2 at the listener's ear. Assume that the sound strikes the surface of the ear perpendicularly. How much power is intercepted by the ear?
Answer:
[tex]6.35\cdot 10^{-9} W[/tex]
Explanation:
The relationship between power and intensity of a sound is given by:
[tex]I=\frac{P}{A}[/tex]
where
I is the intensity
P is the power
A is the area considered
In this problem, we know
[tex]A=2.90\cdot 10^{-3}m^2[/tex] is the surface area of the ear
[tex]I = 2.19\cdot 10^{-6} W/m^2[/tex] is the intensity of the sound
Re-arranging the equation, we can find the power intercepted by the ear:
[tex]P=IA=(2.19\cdot 10^{-6} W/m^2)(2.90\cdot 10^{-3} m^2)=6.35\cdot 10^{-9} W[/tex]
A golf club (mass 0.5kg) hits a golf ball (mass 0.03kg) with a constant force of 25N over a time of 0.02 seconds. What is the magnitude of the impulse delivered to the ball? Select one: o a. 0.05 Ns b. 1250 Ns C.1.67 x102 Ns d.12.5Ns o e.8.00 x 104 Ns
Answer:
0.5 Ns
Explanation:
When a large force acting on a body for a very small time it is called impulsive force.
Impulse = force × small time
Impulse = 25 × 0.02 = 0.5 Ns
It is a vector quantity
Which is not a simple harmonic motion (S.H.M.) (a) Simple Pendulum (b) Projectile motion (c) None (d) Spring motion
Answer:
b) Projectile MOTION
Explanation:
SHM is periodic motion or to and fro motion of a particle about its mean position in a straight line
In this type of motion particle must be in straight line motion
So here we can say
a) Simple Pendulum : it is a straight line to and fro motion about mean position so it is a SHM
b) Projectile motion : it is a parabolic path in which object do not move to and fro about its mean position So it is not SHM
d) Spring Motion : it is a straight line to and fro motion so it is also a SHM
So correct answer will be
b) Projectile MOTION
Final answer:
Projectile motion is not a simple harmonic motion because it does not meet the conditions for SHM.
Explanation:
Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force is proportional to the displacement. The three conditions that must be met to produce SHM are: a linear restoring force, a constant force constant, and no external damping forces. Based on these conditions, the answer to the question is (b) Projectile motion, as it does not meet the conditions for SHM. A projectile follows a parabolic path and does not have a linear restoring force.
ml(d^2θ/dt^2) =-mgθ
1. From the linearized equation, justify Galileo’s observation that the period of a pendulum depends only on its length and not on the mass or on the initial displacement.
The equation of motion of a pendulum is:
[tex]\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,[/tex]
where [tex]\ell[/tex] it its length and [tex]g[/tex] is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for small angles ([tex]\theta \ll 1[/tex]), we can use:
[tex]\sin\theta \simeq \theta.[/tex]
Additionally, let us define:
[tex]\omega^2\equiv\dfrac{g}{\ell}.[/tex]
We can now write:
[tex]\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.[/tex]
The solution to this differential equation is:
[tex]\theta(t) = A\sin(\omega t + \phi),[/tex]
where [tex]A[/tex] and [tex]\phi[/tex] are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:
[tex]T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.[/tex]
This justifies that the period depends only on the pendulum's length.
A 300 g bird is flying along at 6.0 m/s and sees a 10 g insect heading straight towards it with a speed of 30 m/s. The bird opens its mouth wide and swallows the insect. a. What is the birds speed immediately after swallowing the insect? b. What is the impulse on the bird? c. If the impact lasts 0.015 s, what is the force between the bird and the insect?
Answer:
(a): The bird speed after swallowing the insect is V= 4.83 m/s
(b): The impulse on the bird is I= 0.3 kg m/s
(c): The force between the bird and the insect is F= 20 N
Explanation:
ma= 0.3 kg
va= 6 m/s
mb= 0.01kg
vb= 30 m/s
(ma*va - mb*vb) / (ma+mb) = V
V= 4.83 m/s (a)
I= mb * vb
I= 0.3 kg m/s (b)
F*t= I
F= I/t
F= 20 N (c)
This physics problem uses the principle of conservation of momentum to calculate the bird's speed after swallowing the insect, the impulse experienced by the bird, and the force between the bird and the insect.
Explanation:This is a physics problem relating to the conservation of momentum. Let's start by defining some facts, where m bird = 0.3 kg and v bird = 6.0 m/s are the mass and speed of the bird before the incident and m insect = 0.01 kg and v insect = 30 m/s are the mass and speed of the insect.
a. To find the bird's speed immediately after swallowing the insect, we need to apply the conservation of momentum principle: initial total momentum = final total momentum, which can be written as m bird * v bird + m insect * v insect = (m bird + m insect) * v final.
b. The impulse on the bird equals the change in momentum of the bird, thus equals to the final momentum of the bird - initial momentum of the bird.
c. The force between the bird and the insect is obtained from the definition of impulse: Force * time = impulse, or Force = Impulse/time.
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Consider a torque ~τ that is constant in both magnitude and direction, and acts on a rigid body of mass 10 kg at a point 1 m from the pivot. How much work does the torque do on the rigid body, if it turns through an angle of 180◦ while the torque is acting? Assume the acceleration due to gravity is 10 m/s2 .
Answer:
314 Joule
Explanation:
m = 10 kg, g = 10 m/s^2, d = 1 m, angle turn = 180 degree = π radian
work = torque x angle turn
torque = force x perpendicular distance
torque = m x g x d = 10 x 10 x 1 = 100 Nm
work = 100 x π
work = 100 x 3.14 = 314 Joule
Final answer:
The work done by a constant torque on a rigid body rotating through an angle of 180° can be found by multiplying the torque by the angle. The torque can be calculated using the equation τ = Iα, where I is the moment of inertia and α is the angular acceleration. Substituting the given values, we find that the work done by the torque on the rigid body is 18000 kg·m^2·rad/s^2.
Explanation:
The work done by a torque on a rigid body is given by the formula W = τθ, where τ is the torque and θ is the angle through which the body rotates. In this case, the torque is constant in both magnitude and direction, so we can use W = τθ. Given that the torque is constant and the body turns through an angle of 180°, we can calculate the work done as follows:
Since τ is constant, we can write W = τθ = τ (180° - 0°). The work done by the torque is equal to the torque multiplied by the change in angle. Substitute the given values into the formula: W = (τ) (180° - 0°) = (τ) (180°). The work done by the torque is equal to the torque multiplied by 180°.
To find the value of the torque, we need to use the equation τ = Iα, where I is the moment of inertia and α is the angular acceleration. In this case, the rigid body has a mass of 10 kg and a distance of 1 m from the pivot. The moment of inertia for a point mass rotating about a fixed axis is given by I = m(r^2), where m is the mass and r is the perpendicular distance from the axis. Substitute the given values into the formula: I = (10 kg)((1 m)^2) = 10 kg·m^2. Since α = a/r and the acceleration due to gravity is 10 m/s^2, we have α = (10 m/s^2)/(1 m) = 10 rad/s^2.
Substitute the values of τ and α into the equation τ = Iα: τ = (10 kg·m^2)(10 rad/s^2) = 100 kg·m^2·rad/s^2. Therefore, the torque is 100 kg·m^2·rad/s^2.
Finally, substitute the values of τ and θ into the equation W = τθ: W = (100 kg·m^2·rad/s^2 )(180°) = 18000 kg·m^2·rad/s^2.
The maximum magnitude of the magnetic field of an electromagnetic wave is 13.5 μΤ. (3396) Problem 3: 笄What is the average total energy density (in μ1m3) of this electromagnetic wave? Assume the wave is propagating in vacuum.
Answer:
The average total energy density of this electromagnetic wave is [tex]72.5\ \mu\ J/m^3[/tex].
Explanation:
Given that,
Magnetic field [tex]B = 13.5\mu T[/tex]
We need to calculate the average total energy density
Using formula of energy density
[tex]Energy\ density =\dfrac{S}{c}[/tex]....(I)
Where, S = intensity
c = speed of light
We know that,
The intensity is given by
[tex]S = \dfrac{B^2c}{2\mu_{0}}[/tex]
Put the value of S in equation (I)
[tex]Energy\ density =\dfrac{\dfrac{B^2c}{2\mu_{0}}}{c}[/tex]
[tex]Energy\ density = \dfrac{(13.5\times10^{-6})^2}{2\times4\pi\times10^{-7}}[/tex]
[tex]Energy\ density = 0.0000725\ J/m^3[/tex]
[tex]Energy\ density = 72.5\times10^{-6}\ J/m^3[/tex]
[tex]Energy\ density = 72.5\ \mu\ J/m^3[/tex]
Hence, The average total energy density of this electromagnetic wave is [tex]72.5\ \mu\ J/m^3[/tex].
A girl is standing on a trampoline. Her mass is 65 kg and she is able to jump 3 m. What is the spring constant for the trampoline? (logger pro?)
Answer:
k = 212.55 newton per meter
Explanation:
A girl is standing on a trampoline. Her mass is 65 kg and she is able to jump 3 meters.
We have to find the spring constant.
Since by Hooke's law,
F = -kx
Where F = force applied by the spring
k = spring constant
x = displacement
And we know force applied by the spring will be equal to the weight of the girl.
So, F = mg
Therefore, (-mg) = -kx
65×(9.81) = k×(3)
k = [tex]\frac{(65)(9.81)}{3}[/tex]
k = 212.55 N per meter
Therefore, spring constant of the spring is 212.55 Newton per meter.
Consider two charges, q1=3C and q2=2C 2m apart from each other. Calculate the electric force between them. Is the force attractive or repulsive?
Answer:
Electric force between the charges, [tex]F=1.35\times 10^{10}\ N[/tex]
Explanation:
It is given that,
Charge 1, q₁ = 3 C
Charge 2, q₂ = 2 C
Distance between them, r = 2 m
We need to find the electric force between them. The formula for electric force is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
k is the electrostatic constant
[tex]F=9\times 10^9\times \dfrac{3\ C\times 2\ C}{(2\ m)^2}[/tex]
[tex]F=1.35\times 10^{10}\ N[/tex]
So, the force between the charges is [tex]1.35\times 10^{10}\ N[/tex]. Hence, this is the required solution.
A coil is made of 150 turns of copper wire wound on a cylindrical core. If the mean radius of the turns is 6.5 mm and the diameter of the wire is 0.4 mm, calculate the resistance of the coil!
Answer:
0.84 Ω
Explanation:
r = mean radius of the turn = 6.5 mm
n = number of turns of copper wire = 150
Total length of wire containing all the turns is given as
L = 2πnr
L = 2 (3.14)(150) (6.5)
L = 6123 mm
L = 6.123 m
d = diameter of the wire = 0.4 mm = 0.4 x 10⁻³ m
Area of cross-section of the wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (0.4 x 10⁻³)²
A = 1.256 x 10⁻⁷ m²
ρ = resistivity of copper = 1.72 x 10⁻⁸ Ω-m
Resistance of the coil is given as
[tex]R = \frac{\rho L}{A}[/tex]
[tex]R = \frac{(1.72\times 10^{-8}) (6.123))}{(1.256\times 10^{-7}))}[/tex]
R = 0.84 Ω
Two forces are applied to a 5.0-kg crate; one is 3.0 N to the north and the other is 4.0 N to the east. The magnitude of the acceleration of the crate is: a. 1.0 m/s^2 b. 2.8 m/s^2 c.7.5 m/s^2 d. 10.0 m/s^2
Answer:
The acceleration of the crate is 1 m/s²
Explanation:
It is given that,
Mass of the crate, m = 5 kg
Two forces applied on the crate i.e. one is 3.0 N to the north and the other is 4.0 N to the east. So, there resultant force is :
[tex]F_{net}=\sqrt{3^2+4^2} =5\ N[/tex]
We need to find the acceleration of the crate. It is given by using the second law of motion as :
[tex]a=\dfrac{F_{net}}{m}[/tex]
[tex]a=\dfrac{5\ N}{5\ kg}[/tex]
a = 1 m/s²
So, the acceleration of the crate is 1 m/s². Hence, this is the required solution.
The magnitude of the acceleration of a crate with forces of 3.0 N north and 4.0 N east applied to it is 1.0 m/s². This is found using the Pythagorean theorem to calculate the resultant force and Newton's Second Law to calculate acceleration.
The forces are 3.0 N to the north and 4.0 N to the east on a 5.0-kg crate. Since the forces are perpendicular, we can use the Pythagorean theorem to find the resultant force. The resultant force (Fr) is √(3.02 + 4.02) N, which is 5.0 N. According to Newton's Second Law, F = ma, hence acceleration (a) is Fr divided by the mass (m). Calculating acceleration: a = 5.0 N / 5.0 kg = 1.0 m/s2. Therefore, the correct answer is a. 1.0 m/s2.
A system gains 757 kJ757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ.+176 kJ. How much work is done? ????=w= kJkJ Choose the correct statement. Work was done on the system. Work was done by the system.
•If a system gains 757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ. How much work is done is - 581 kJ
•The correct statement is: Work was done by the system
Let Change in internal energy ΔU = 176 kJ
Let Heat gained by the system (q) = 757 kJ
Using the First law of thermodynamics
ΔU = q + w
Where:
ΔU represent change in internal energy
q represent heat added to system and w is work done.
Let plug in the formula
176 kJ = 757 kJ + w
w = 176 kJ - 757 kJ
w= - 581 kJ
Based on the above calculation the negative sign means that work is done by the system
Inconclusion:
•If a system gains 757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ. How much work is done is - 581 kJ
•The correct statement is: Work was done by the system
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The amount of work done by the system, based on given heat gain and change in internal energy is 581 kJ, meaning the work was done by the system.
Explanation:The question asks about the amount of work done by or on a system in the field of thermodynamics. According to the first law of thermodynamics, the change in internal energy of a system (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W), or written as ΔU = Q - W. In this case, the heat added to your system was 757 kJ and the change in internal energy of the system was +176 kJ.
So we have: 176 kJ = 757 kJ - W. Subtracting 757 kJ from both sides of the equation would give us W = 757 kJ - 176 kJ. This results in the value of W = 581 kJ. Conclusively, since W is positive, we say that work was done by the system.
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What is the force on your eardrum if its area is 1.00 cm^2, and you are swimming 3.0 m below water level?
Answer:
Force on eardrum
[tex]F=29400\times 1\times 10^{-4}=2.94N[/tex]
Explanation:
Force = Pressure x Area
Pressure = hρg
Height, h = 3 m
ρ = 1000 kg/m³
g = 9.8 m/s²
Pressure = hρg = 3 x 1000 x 9.8 = 29400 N/m²
Area = 1 cm²
Force on eardrum
[tex]F=29400\times 1\times 10^{-4}=2.94N[/tex]
A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Friction is absent. The spring constant of the spring is 3600 N/m. The piston has a negligible mass and a radius of 0.028 m. (a) When the air beneath the piston is completely pumped out, how much does the atmospheric pressure cause the spring to compress? (b) How much work does the atmospheric pressure do in compressing the spring?
Answer:
a) 0.0693 m
b) Work done = 8.644 J
Explanation:
Given:
Spring constant, k = 3600 N/m
Radius of the piston, r = 0.028 m
Now, we know that the atmospheric pressure at STP = 1.01325 × 10⁵ Pa = 101325 Pa
Now,
The force ([tex]F_P[/tex]) due to the atmospheric pressure on the piston will be:
[tex]F_P[/tex] = Pressure × Area of the piston
on substituting the values we get,
[tex]F_P[/tex] = 101325 × πr²
F = 101325 × π × (0.028)² = 249.56 N
also,
Force on spring is given as:
F = kx
where,
x is the displacement in the spring
on substituting the values we get,
249.56 N = 3600N/m × x
or
x = 0.0693 m
thus, the compression in the spring will be = 0.0693 m
b) Applying the concept of conservation of energy
we have,
Work done by the atmospheric pressure in compressing the spring = Potential energy gained by the spring
mathematically,
[tex]W = \frac{1}{2}kx^2[/tex]
on substituting the values we get,
[tex]W = \frac{1}{2}\times 3600\times (0.0693)^2[/tex]
W = 8.644 J
a) x = 0.0693 m
b) W = 8.644 J
Given :
Spring constant, K = 3600 N/m
Radius of the piston, r = 0.028 m
Solution :
Now the atmospheric pressure at STP = 1.01325 × 10⁵ Pa = 101325 Pa
Force due to the atmospheric pressure on the piston is,
Force = Pressure × Area of the piston
on substituting the values we get,
[tex]\rm F_P = 101325\times \pi r^2[/tex]
[tex]\rm F_P = 249.56\;N[/tex]
a) We know that the force on spring is given by,
F = Kx
where, k is spring constant and x is the displacement in the spring.
[tex]249.56 = 3600\times x[/tex]
[tex]\rm x = 0.0693\;m[/tex]
b) We know that the Work Done is given by,
[tex]\rm W= \dfrac{1}{2} k x^2[/tex]
[tex]\rm W = 0.5\times 3600\times (0.0693)^2[/tex]
W = 8.644 J
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A 0.15 kg baseball is pushed with 100 N force. what will its acceleration be?
Answer:
The acceleration of the ball is 666.67 m/s²
Explanation:
It is given that,
Mass of the baseball, m = 0.15 kg
Applied force to it, F = 100 N
We need to find the acceleration of the ball. It can be calculated using Newton's second law of motion as :
F = ma
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{100\ N}{0.15\ kg}[/tex]
[tex]a=666.67\ m/s^2[/tex]
So, the acceleration of the ball is 666.67 m/s². Hence, this is the required solution.
Calculate the mass of a 0.9 m^3 block of a material having a density of 12500 kg/m^3.
Answer: The mass of the object will be 11250 kg.
Explanation:
Density is defined as the mass contained per unit volume.
[tex]Density=\frac{mass}{Volume}[/tex]
Given :
Density of the object= [tex]12500kg/m^3[/tex]
Mass of object = ?
Volume of the object = [tex]0.9m^3[/tex]
Putting in the values we get:
[tex]12500kg/m^3=\frac{mass}{0.9m^3}[/tex]
[tex]12500kg/m^3=\frac{mass}{0.9m^3}[/tex]
[tex]mass=11250kg[/tex]
Thus the mass of the object is 11250 kg.
If there were no air resistance, a penny dropped from the top of a skyscraper would reach the ground 9.3 s later. To the nearest integer, what would the penny's speed in m/s be right as it reaches the ground if it was dropped from rest?
Answer:
To the nearest integer, the penny's speed in m/s be right as it reaches the ground if it was dropped from rest = 91 m/s
Explanation:
We have equation of motion
S = ut + 0.5at²
Here u = 0, a = g and t = 9.3 s
We have equation of motion v = u +at
Substituting
v = u +at
v = 0 + 9.8 x 9.3 = 91.14 m/s
To the nearest integer, the penny's speed in m/s be right as it reaches the ground if it was dropped from rest = 91 m/s
A sign is held in equilbrium by 7 vertically hanging ropes attached to the ceiling. If each rope has an equal tension of 53 Newtons, what is the mass of the sign in kg?
Answer:
37.86 kg
Explanation:
The weight of sign board is equally divided on each rope. It means the tension in all the ropes is equal to the weight of the sign board in equilibrium condition.
Tension in each rope = 53 N
Tension in 7 ropes = 7 x 53 N = 371 N
Thus, The weight of sign = 371 N
Now, weight = m g
where m is the mass of sign.
m = 371 / 9.8 = 37.86 kg
The steady-state diffusion flux through a metal plate is 7.8 × 10-8 kg/m2-s at a temperature of 1220˚C ( 1493 K) and when the concentration gradient is -500 kg/m4. Calculate the diffusion flux at 1000˚C ( 1273 K) for the same concentration gradient and assuming an activation energy for diffusion of 145,000 J/mol.
To calculate the diffusion flux at 1000˚C for the same concentration gradient, use the Arrhenius equation.
J ≈ 2.4 × 10-12 kg/m2-s
Explanation:To calculate the diffusion flux at 1000˚C (1273 K) for the same concentration gradient, we can use the Arrhenius equation:
J = J0 * exp(-Q/RT)
Where J is the diffusion flux, J0 is the pre-exponential factor, Q is the activation energy for diffusion, R is the gas constant, and T is the absolute temperature.
Given that the diffusion flux at 1220˚C (1493 K) is 7.8 × 10-8 kg/m2-s and the activation energy for diffusion is 145,000 J/mol, we can calculate the diffusion flux at 1000˚C as:
J = (7.8 × 10-8) * exp(-145000/(8.314*1273))
J ≈ 2.4 × 10-12 kg/m2-s
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To calculate the steady-state diffusion flux at a different temperature, use the Arrhenius equation to find the diffusion coefficients at the two temperatures, and find the ratio based on the fact that the diffusion flux is proportional to the diffusion coefficient when the concentration gradient is constant. Input the known values into the equation to solve for the unknown diffusion flux.
Explanation:The steady-state diffusion through a metal plate can be calculated using the Arrhenius equation, which relates the diffusion coefficient (D) to temperature. The equation is D = D0e^-(Q/RT), where D0 is the pre-exponential factor, Q is the activation energy for diffusion, R is the gas constant and T is the temperature in K.
Given that the diffusion flux (J) is defined as J = -D×(dc/dx), where dc/dx is the concentration gradient. We can find that when the concentration gradient remains the same, the ratio of the two diffusion fluxes at different temperatures can be represented as J1/J2 = D1/D2.
Substitute the Arrhenius equation into the ratio, we get J1/J2 = e^(Q/R)×(1/T1-1/T2). Then you can use the given values, namely Q = 145,000 J/mol, R = 8.314 J/(mol×K), and temperatures T1 = 1493K , T2 = 1273K, as well as the known J1, to calculate J2.
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If the axes of the two cylinders are parallel, but displaced from each other by a distance d, determine the resulting electric field in the region R>R3, where the radial distance R is measured from the metal cylinder's axis. Assume d<(R2−R1). Express your answer in terms of the variables ρE, R1, R2, R3, d, R, and appropriate constants.
Answer:
E = ρ ( R1²) / 2 ∈o R
Explanation:
Given data
two cylinders are parallel
distance = d
radial distance = R
d < (R2−R1)
to find out
Express answer in terms of the variables ρE, R1, R2, R3, d, R, and constants
solution
we have two parallel cylinders
so area is 2 [tex]\pi[/tex] R × l
and we apply here gauss law that is
EA = Q(enclosed) / ∈o ......1
so first we find Q(enclosed) = ρ Volume
Q(enclosed) = ρ ( [tex]\pi[/tex] R1² × l )
so put all value in equation 1
we get
EA = Q(enclosed) / ∈o
E(2 [tex]\pi[/tex] R × l) = ρ ( [tex]\pi[/tex] R1² × l ) / ∈o
so
E = ρ ( R1²) / 2 ∈o R
Final answer:
The resulting electric field in the specified region can be calculated using Gauss' Law. The equation for the electric field in that region is [tex]E = 2\pi R_1^2ho_E[/tex].
Explanation:
The resulting electric field in the region R>R3 is:
[tex]E = 2\pi R_1^2ho_E[/tex]
where R_1 is the radius of the inner cylinder, and ρ_E is the charge density. This expression is obtained by applying Gauss' Law for the region where R1 < r < R2.
On takeoff, the combined action of the engines and the wings of an airplane exert a force of 8.00 × 103 N on the plane upward at an angle of 65.0" above the horizontal. The plane rises with constant velocity in the vertical direction while continuing to accelerate in the horizontal direction. (3 marks) a. What is the weight of the plane? b. What is the horizontal acceleration of the plane? .
Answer:
a) 7250.5 N
b) 4.6 m/s²
Explanation:
a)
F = applied force = 8000 N
θ = angle with the horizontal = 65 deg
Consider the motion along the vertical direction :
[tex]F_{y}[/tex] = Applied force in vertical direction in upward direction = F Sinθ = 8000 Sin65 = 7250.5 N
[tex]F_{g}[/tex] = weight of the plane in vertical direction in downward direction = ?
[tex]a_{y}[/tex] = Acceleration in vertical direction = 0 m/s²
Taking the force in upward direction as positive and in downward direction as negative, the force equation along the vertical direction can be written as
[tex]F_{y}-F_{g} = m a_{y}[/tex]
[tex]7250.5 -F_{g} = m (0)[/tex]
[tex]F_{g}[/tex] = 7250.5 N
b)
m = mass of the plane
force of gravity is given as
[tex]F_{g} = mg [/tex]
[tex]7250.5 = m(9.8) [/tex]
m = 739.85 kg
Consider the motion along the horizontal direction
[tex]F_{x}[/tex] = Applied force in horizontal direction = F Cosθ = 8000 Cos65 = 3381 N
[tex]a_{x}[/tex] = Acceleration in horizontal direction
Acceleration in horizontal direction is given as
[tex]a_{x}=\frac{F_{x}}{m}[/tex]
[tex]a_{x}=\frac{3381}{739.85}[/tex]
[tex]a_{x}[/tex] = 4.6 m/s²
Water is boiled in a pan on a stove at sea level. During 10 min of boiling, it is observed that 200 g of water has evaporated. Then the rate of heat transfer to the water is 225.7 kJ/min 45.1 kJ/min 53.5 kJ/min 0.84 kJ/min 41.8 kJ/min
Answer:
Rate of heat is 45.1 kJ/min
Explanation:
Heat required to evaporate the water is given by
Q = mL
here we know that
[tex]L = 2.25 \times 10^6 J/kg[/tex]
now we have
[tex]Q = (0.200)(2.25 \times 10^6 J/kg)[/tex]
[tex]Q = 452.1 kJ[/tex]
now the power is defined as rate of energy
[tex]P = \frac{Q}{t}[/tex]
[tex]P = \frac{452.1 kJ}{10}[/tex]
[tex]P = 45.1 kJ/min[/tex]
The same 1710 kg artificial satellite is placed into circular orbit at the same altitude of 2.6x10° m around an exoplanet with the same radius as the Earth, but twice the mass. a. What is the orbital speed of the satellite? b. What is the period of the satellite? C. What is the kinetic energy of the satellite? d. What is the total energy of the satellite?
Given:
mass of satellite, m = 1710 kg
altitude, h = [tex]2.6\times 10^{6} m[/tex]
G = [tex]6.67\times 10^{-11} [/tex]
we know
mass of earth, [tex]M_{E}[/tex] = [tex]5.972\times 10^{24} kg[/tex]
Here, according to question we will consider
[tex]2M_{E}[/tex] = [tex]11.944\times 10^{24} kg[/tex]
radius of earth, [tex]R_{E}[/tex] = [tex]6.371\times 10^{6} m[/tex]
Formulae Used and replacing [tex]M_{E}[/tex] by [tex]2M_{E}[/tex] :
1). [tex]v = \sqrt{\frac{2GM_{E}}{R_{E} + h}}[/tex]
2). [tex]T = \sqrt{\frac{4\pi ^{2}(R_{E} + h)^{3}}{2GM_{E}}}[/tex]
3). [tex]KE = \frac{1}{2}mv^{2}[/tex]
4). [tex]Total Energy, E = -\frac{2GM_E\times m}{2(R_{E} + h)}[/tex]
where,
v = orbital velocity of satellite
T = time period
KE = kinetic energy
Solution:
Now, Using Formula (1), for orbital velocity:
[tex]v = \sqrt{\frac{6.67 \times 10^{-11} \times 11.944 \times 10^{24}}{6.371 \times 10^{6} + 2.6 \times 10^{6}}[/tex]
v = [tex]9.423 \times 10^{3}[/tex] m/s
Using Formula (2) for time period:
[tex]T = \sqrt{\frac{4\pi ^{2}(6.371\times 10^{6} + 2\times 10^{6})^{3}}{6.67\times 10^{-11}\times 9.44\times 10^{24}}}[/tex]
[tex]T = 6.728\times 10^{3} s[/tex]
Now, Using Formula(3) for kinetic energy:
[tex]KE = \frac{1}{2}(9.44\times 10^{24})(9.42\times 10^{3})^{2}[/tex]
[tex]KE = \frac{1}{2}(1710)(9.42\times 10^{3})^{2} = 7.586\times 10^{10} J[/tex]
Now, Using Formula(4) for Total energy:
[tex]E = -\frac{6.67\times 10^{-11}\times 9.44\times 10^{24}\times 1710}{2( 6.371\times 10^{6} + 2.6\times 10^{6})}[/tex]
[tex]E = - 7.59\times 10^{10} J[/tex]
A 16.0 kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 25.0 N. Starting from rest, the sled attains a speed of 1.00 m/s in 8.00 m. Find the coefficient of kinetic friction between the runners of the sled and the snow. You Answered
Answer:
[tex]\mu_k = 0.15[/tex]
Explanation:
according to the kinematic equation
[tex]v^{2} - u^{2} = 2aS[/tex]
Where
u is initial velocity = 0 m/s
a = acceleration
S is distance = 8.00 m
final velocity = 1.0 m/s
[tex]a = \frac {v^{2}}{2S}[/tex]
[tex]a = \frac {1{2}}{2*8.6}[/tex]
a = 0.058 m/s^2
from newton second law
Net force = ma
[tex]f_{net} = ma[/tex]
F - f = ma
2[tex]5 - \mu_kN = ma[/tex]
[tex]25 - \mu_kmg = ma[/tex]
[tex]\frac {25 - ma}{mg} =\mu_k[/tex]
[tex]\frac {25 - 16*0.058}{16*9.81} = 0.15[/tex]
[tex]\mu_k = 0.15[/tex]
A proton experiences a force of 3.5x 10^-9 N when separated from a second charge by a distance of 1.6 mm. a) What is the size of the second charge? b) How many fundamental charges make up this charge in part a)?
Answer:
(a) 6.22 x 10^-6 C
(b) 3.8 x 10^13
Explanation:
Let the second charge is q2 = q
q1 = 1.6 x 10^-19 C
F = 3.5 x 10^9 N
d = 1.6 mm = 1.6 x 10^-3 m
(a) Use the formula of Coulomb's law
F = K q1 x q2 / d^2
3.5 x 10^-9 = 9 x 10^9 x 1.6 x 10^-19 x q / (1.6 x 10^-3)^2
q = 6.22 x 10^-6 C
(b)
Let the number of electrons be n
n = total charge / charge of one electron
n = 6.22 x 10^-6 / (1.6 x 10^-19) = 3.8 x 10^13
Absolute pressure in tank is P1 = 260 kPa and local ambient absolute pressure is P2 =100 kPa. If liquid density in pipe is 13600 kg/m3 , compute liquid height, h=..?.. m ? Use g =10 m/s2
Answer:
1.176m
Explanation:
Local ambient pressure(P1) = 100 kPa
Absolute pressure(P2)=260kPa
Net pressure=absolute pressure-local ambient absolute pressure
Net pressure=P1(absolute pressure)-P2(local ambient absolute pressure)
Net pressure=260-100=160kPa
Pressure= ρgh
160kPa=13600*10*h
h=[tex]\frac{160000}{136000}[/tex]
h=1.176m