Answer:
(A.) Resonance Frequency
(B.) Totally resistance
Explanation:
A. The resonance frequency is the frequency at which the net reactance of the inductor and capacitor is zero, this is the frequency at which the capacitive reactance of the capacitor and inductive reactance of the inductor cancels each other out. Resistance is low here and the frequency is high
B. Since the net reactance of the inductor and capacitor is zero as explained in (A) above, the total reactance of the circuit will be entirely from the resistor (resistance)
Explain why atomic radius decreases as you move to the right across a period for main-group elements but not for transition elements.Match the words in the left column to the appropriate blanks in the sentences on the right.
Answer:
Explained.
Explanation:
Only the first question has been answered
In a period from left to right the nuclear charge increases and hence nucleus size is compressed. Thus, atomic radius decreases.
In transition elements, electrons in ns^2 orbital remain same which is the outer most orbital having 2 electrons and the electrons are added to (n-1) d orbital. So, outer orbital electron experience almost same nuclear attraction and thus size remains constant.
The atomic radius of main-group elements decreases as you move to the right across a period due to increased positive charge, while the atomic radius of transition elements remains relatively constant.
Explanation:The atomic radius of main-group elements decreases as you move to the right across a period because the number of protons in the nucleus increases. This increased positive charge pulls the electrons closer to the nucleus, reducing the size of the atom. In contrast, the atomic radius of transition elements remains relatively constant as you move across a period because their outermost electrons are in different energy levels or subshells. The addition of protons does not significantly affect the size of the atom.
Moving across a period, the number of protons in the nucleus increases, leading to a greater positive charge. This increased positive charge exerts a stronger pull on the electrons, pulling them closer to the nucleus and resulting in a smaller atomic radius.
However, electron shielding, or the repulsion between electrons in different energy levels, also plays a role. As you move across a period, the number of electrons in the same energy level (shell) remains constant, providing consistent shielding effects. This partial counteraction to the increased positive charge contributes to the overall trend of decreasing atomic radius.
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A motor does 30 kJ of work and gains 4 kJ as heatfrom the surroundings. What is the change in the internal energy of the motor?
Answer:
ΔU= *-26 KJ
Explanation:
Given that
Work done by motor W= 30 KJ
Heat gains by motor Q= 4 KJ
Sign convention:
If heat is added to the system then it is taken as positive and if heat is rejected from the system then it is taken as negative.
If work done by the system then it is taken as positive and if work is done on the system then it is taken as negative.
From first law of thermodynamics
Q = W + ΔU
ΔU=Change in internal energy
Q=Heat transfer
W=Work
Now by putting the values
4 = 30 + ΔU
ΔU= -26 KJ
Answer:
Internal energy ∆U=-26KJ
Explanation:
Given that:
Work done by the motor=+30KJ
Heat gained by the motor=+4KJ
In solving thermodynamical questions it is reasonable to use the sign convention this
Heat is positive if it is added to a system,but becomes negative if the system rejects heats.
Work is positive if the system does work,but becomes negative if work is done on the system.
Using the thermodynamics first law
∆U=Q-W
∆U= 4-30=-26KJ
For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.100kg{\rm kg} of turkey. The slices of turkey are weighed on a plate of mass 0.400kg{\rm kg} placed atop a vertical spring of negligible mass and force constant of 200N/m{\rm N/m} . The slices of turkey are dropped on the plate all at the same time from a height of 0.250m{\rm m} . They make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (SHM). You may assume that the collision time is extremely small.What is the amplitude of oscillations A of the scale after the slices of turkey land on the plate?
Answer:
0.02268 m
Explanation:
[tex]m_1[/tex] = Mass of turkey slices = 0.1 kg
[tex]m_2[/tex] = Mass of plate = 0.4 kg
[tex]u_1[/tex] = Initial Velocity of turkey slices = 0 m/s
[tex]u_2[/tex] = Initial Velocity of plate = 0 m/s
[tex]v_1[/tex] = Final Velocity of turkey slices
[tex]v_2[/tex] = Final Velocity of plate
k = Spring constant = 200 N/m
x = Compression of spring
g = Acceleration due to gravity = 9.81 m/s²
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.25+0^2}\\\Rightarrow v=2.21472\ m/s[/tex]
The final velocity of the turkey slice is 2.21472 m/s = v₁
For the spring
[tex]x=\frac{m_1g}{k}\\\Rightarrow x=\frac{0.1\times 9.81}{200}\\\Rightarrow x=0.004905\ m[/tex]
As the linear momentum is conserved
[tex]m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_2=\frac{m_1v_1}{m_1+m_2}\\\Rightarrow v_2=\frac{0.1\times 2.21472}{0.1+0.4}\\\Rightarrow v_2=0.442944\ m/s[/tex]
Here the kinetic and potential energy of the system is conserved
[tex]\frac{1}{2}(m_1+m_2)v_2^2+\frac{1}{2}kx^2=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{(m_1+m_2)v_2^2+kx^2}{k}}\\\Rightarrow A=\sqrt{\frac{(0.1+0.4)0.442944^2+200\times 0.004905^2}{200}}\\\Rightarrow A=0.02268\ m[/tex]
The amplitude of oscillations is 0.02268 m
Doug’s average driving speed is 1 kilometers per hour faster than Thor’s. In the same length of time it takes Doug to drive 390 kilometers, Thor drives only 384 kilometers. What is Doug’s average speed?
Answer:
Doug speed will be 65 km/hr
Explanation:
Let the Thor's speed is x km/hr
So Doug's speed = x+1 km/hr
We have given that Doug and Thor take same time to cover 390 km and 384 km respectively
We know that time is given by
[tex]time=\frac{distance}{speed}[/tex]
So time taken by Doug to cover the distance
[tex]time=\frac{390}{x+1}[/tex]
And time taken by Thor to cover the distance
[tex]time=\frac{384}{x}[/tex]
As both times are equal
So [tex]\frac{390}{x+1}=\frac{384}{x}[/tex]
[tex]6x=384[/tex]
[tex]x=64km/hr[/tex]
So Doug speed will be 64+1 = 65 km/hr
A man cleaning his apartment pushes a vacuum cleaner with a force of magnitude 84.5 N. The force makes an angle of 33.9 ◦ with the horizontal floor. The vacuum cleaner is pushed 2.62 m to the right along the floor. Calculate the work done by the 84.5 N force.
Answer:
183.75641 Joules
Explanation:
F = Force of the vacuum cleaner = 84.5 N
s = Displacement of the vacuum cleaner = 2.62 m
[tex]\theta[/tex] = Angle the force makes with the horizontal = 33.9°
Work done is given by
[tex]W=F\times scos\theta\\\Rightarrow W=84.5\times 2.62\times cos33.9\\\Rightarrow W=183.75641\ J[/tex]
The work done by the force of the vacuum cleaner is 183.75641 Joules
The work done by the man pushing the vacuum cleaner with a force of 84.5 N at an angle of 33.9° over a distance of 2.62 m is approximately 184.8 joules.
Explanation:To calculate the work done by a force, you can use the formula W = F × d × cos(θ), where W is the work done, F is the magnitude of the force, d is the distance the object moves, and θ is the angle the force makes with the horizontal direction of movement. In this case, the man pushes a vacuum cleaner with a force of 84.5 N at an angle of 33.9° over a distance of 2.62 m. We multiply the force by the distance and the cosine of the angle to find:
W = 84.5 N × 2.62 m × cos(33.9°)
Calculating cosine of 33.9 degrees and multiplying with the force and distance, we get:
W = 84.5 × 2.62 × 0.8326 ≈(joules)
The man does approximately 184.8 joules of work pushing the vacuum cleaner.
A thermosensory neuron in the skin converts heat energy to nerve impulses via a conversion called
Answer:
Sensory transduction
Explanation:
The term sensory transduction refers to the conversion process where the sensory energy is converted in order to change the potential of a membrane.
In other words, it can defined as the process of energy conversion such that stimulus can be transmitted or received by the sensory receptors and the nervous system may initiate with the sensory receptors.
Transduction takes in all of the five receptors of the body. Thus skin is also one of the receptors and hence conversion of heat energy into impulses takes place with the help of thermo-sensory neuron.
How bad is the heavy traffic? You can walkwalk 1212 miles in the same time that it takes to travel 3232 miles by car. If the car's rate is 55 milesmiles per hour faster than your walkingwalking rate, find the average rate of each.
Answer:
Speed by walking is 33 miles per hour
And speed of by car is 88 miles per hour
Explanation:
We have given that it takes same time to walk 1212 miles as 3232 miles by car
Now let the speed by walk is x
As speed by car is 55 miles per hour faster than by walk = x+55
As time is same and we know that time is given as [tex]time=\frac{distance}{speed}[/tex]
So [tex]\frac{1212}{x}=\frac{3232}{x+55}[/tex]
[tex]1212(x+55)=3232x[/tex]
[tex]1212x+66660=3232x[/tex]
x = 33 miles per hour
So speed by walking is 33 miles per hour
And so speed of car = 33+55 =88 miles per hour
A car moves horizontally with a constant acceleration of 3 m/s2. A ball is suspended by astring from the ceiling of the car. The ball does not swing, being at rest with respect to thecar. What angle does the string make with the vertical?
Answer:
β = 16.7°
Explanation:
The sum of forces on the x-axis are:
[tex]T*sin\beta=m*a[/tex]
The sum of forces on the y-axis are:
[tex]T*cos\beta=m*g[/tex]
By dividing x-axis by the y-axis equation:
[tex]tan\beta=a/g[/tex]
Solving for β:
[tex]\beta=atan(a/g)[/tex]
β = 16.7°
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 3.0 N/cm. The block becomes attached to the spring and compresses the spring 11 cm before momentarily stopping.
(a) While the spring is being compressed, what work is done on the block by the gravitational force on it?
(b) What work is done on the block by the spring force while the spring is being compressed?
(c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)
(d) If the speed at impact is doubled, what is the maximum compression of the spring?
Final answer:
The work done by gravity on a block dropped onto a spring is 0.22638 J, the spring does 1.815 J of work compressing, and if the speed at impact is doubled, the maximum spring compression becomes 22 cm. The speed of the block before impact cannot be determined without the drop height.
Explanation:
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 3.0 N/cm. The block becomes attached to the spring and compresses it 11 cm before momentarily stopping.
(a) To calculate work done by gravity, use Work = mgh, where m is the mass of the block, g is the acceleration due to gravity (9.8 m/s²), and h is the height (11 cm = 0.11 m, since we need consistent units). The work done by gravity is (0.21 kg)(9.8 m/s²)(0.11 m) = 0.22638 J (since potential energy lost by the block is equal to the work done by gravity).
(b) The work done by the spring is Work = 1/2 kx², converting k to N/m gives us 300 N/m. The compression x is 0.11 m, so the work done is (1/2)(300 N/m)(0.11 m)² = 1.815 J.
(c) Since we're assuming friction is negligible and using energy conservation, the potential energy (mgh) at the beginning will be equal to the kinetic energy (1/2 mv²) just before impact. Solving for v gives v = √(2gh), where h is the drop height. However, without the drop height, we cannot calculate the exact velocity.
(d) Doubling the speed will increase the kinetic energy by a factor of four (since KE = 1/2 mv²). To find the new compression distance, we set the new kinetic energy equal to the spring potential energy (1/2 kx²) and solve for x. The maximum compression x will be twice the original compression, or 22 cm.
A box slides down a 31° ramp with an acceleration of 0.99 m/s2. Determine the coefficient of kinetic friction between the box and the ramp.μk=______.
Answer:[tex]\mu [/tex]=0.48
Explanation:
Given
inclination [tex]\theta =31^{\circ}[/tex]
Acceleration of object[tex]=0.99 m/s^2[/tex]
Now using FBD
[tex]mg\sin \theta -f_r=ma[/tex]
[tex]mg\sin \theta -\mu mg\cos \theta =ma[/tex]
[tex]a=g\sin \theta -\mu g\cos \theta [/tex]
[tex]0.99=5.04-\mu 8.4[/tex]
[tex]\mu 8.4=4.057[/tex]
[tex]\mu =0.48[/tex]
Before railroad were invented, goods often traveled along canals, with mules pulling barges from the bank. If a mule is exerting a 12,000N force for 10km, and the rope connecting the mule to the barge is at a 20 degree angle from the direction of travel, how much work did the mole do on the barge?
A. 12MJ
B. 11MJ
C. 4.1MJ
D. 6MJ
Answer:
W = 112.76MJ
Explanation:
the work is:
[tex]W = F_xD[/tex]
where [tex]F_x[/tex] is the force executed in the direction of the displacement and the d the displacement.
so:
W = 12000Ncos(20)(10000)
we use the cos of the angule because it give us the proyection in the axis x of the force, that means the force in the direction of the displacement.
W = 112.76MJ
A rod of length 35.50 cm has linear density (mass per length) given by λ = 50.0 + 23.0x where x is the distance from one end, and λ is measured in grams/meter. (a) What is its mass? g (b) How far from the x = 0 end is its center of mass? m
Answer:
(a)20.65g
(b)0.19m
Explanation:
(a) The total mass would be it's mass per length multiplied by the total lenght
0.355(50 + 23*0.355) = 20.65 g
(b) The center of mass would be at point c where the mass on the left and on the right of c is the same
Hence the mass on the left side would be half of its total mass which is 20.65/2 = 10.32 g
[tex]c(50 + 23c) = 10.32[/tex]
[tex]23c^2 + 50c - 10.32 = 0 [/tex]
[tex]c \approx 0.19m[/tex]
A potter's wheel has the shape of a solid uniform disk of mass 7 kg and radius 0.65 m. It spins about an axis perpendicular to the disk at its center. A small 2.1 kg lump of very dense clay is dropped onto the wheel at a distance 0.41 m from the axis.
What is the moment of inertia of the system about the axis of spin?
Answer:
1.832 kgm^2
Explanation:
mass of potter's wheel, M = 7 kg
radius of wheel, R = 0.65 m
mass of clay, m = 2.1 kg
distance of clay from centre, r = 0.41 m
Moment of inertia = Moment of inertia of disc + moment f inertia of the clay
I = 1/2 MR^2 + mr^2
I = 0.5 x 7 x 0.65 x 0.65 + 2.1 x 0.41 x 0.41
I = 1.47875 + 0.353
I = 1.832 kgm^2
Thus, the moment of inertia is 1.832 kgm^2.
The moment of inertia of the system about the axis of spin is mathematically given as
I = 1.832 kgm^2
What is the moment of inertia of the system about the axis of spin?Question Parameter(s):
A potter's wheel has the shape of a solid uniform disk of mass of 7 kg
and a radius of 0.65 m
A small 2.1 kg lump of very dense clay
the wheel at a distance of 0.41 m from the axis.
Generally, the equation for the moment of inertia is mathematically given as
I = 1/2 MR^2 + mr^2
I = 0.5 x 7 (0.65)^2 + 2.1 (0.41)^2
I = 1.47875 + 0.353
I = 1.832 kgm^2
In conclusion moment of inertia is
I = 1.832 kgm^2
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A 3,000-kg truck traveling 8 m/s collides with a 500-kg car that is at rest. After the collision, the car is traveling at 10 m/s. How fast will the truck be moving?
The final velocity of the truck is 6.33 m/s
Explanation:
We can solve this problem by using the law of conservation of momentum: the total momentum of the truck-car system must be conserved before and after the collision (if there are no external forces), so we can write
[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]
where:
[tex]m_1 = 3000 kg[/tex] is the mass of the truck
[tex]u_1 = 8 m/s[/tex] is the initial velocity of the truck
[tex]v_1[/tex] is the final velocity of the truck
[tex]m_2 = 500 kg[/tex] is the mass of the car
[tex]u_2 = 0[/tex] is the initial velocity of the car
[tex]v_2 = 10 m/s[/tex] is the final velocity of the car
And by solving the equation for [tex]v_1[/tex], we find the velocity of the truck after the collision:
[tex]v_1 = \frac{m_1 u_1-m_2 v_2}{m_1}=\frac{(3000)(8)-(500)(10)}{3000}=6.33 m/s[/tex]
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a person throws a rock at 3 M/s down over the edge of a very tall cliff on Earth how far will the rock have fallen in 4 seconds if the rock never hit the bottom?
The rock will be at 90.4 m from the top of the cliff.
Explanation:
The rock is thrown with the “initial velocity” 3 m/s. We need to find how much distance does the rock traveled in 4 seconds (t).
From the “kinematic equations” take
[tex]s=u t+\frac{1}{2} a t^{2}[/tex]
Where, “s” is distance traveled, “u” initial velocity of the object, “t” time the object traveled and “a” acceleration due to gravity is [tex]9.8 \mathrm{m} / \mathrm{s}^{2}.[/tex]
Substitute the given values in the above formula,
[tex]s=3 \times 4+\frac{1}{2} \times 9.8 \times 4^{2}[/tex]
[tex]s=12+\frac{1}{2} \times 9.8 \times 16[/tex]
[tex]s=12+\frac{1}{2} \times 156.8[/tex]
[tex]s=12+78.4[/tex]
[tex]s=90.4[/tex]
The rock is at height of 90.4 m from the top of the cliff.
A 531.7-W space heater is designed for operation in Germany, where household electrical outlets supply 230 V (rms) service. What is the power output of the heater when plugged into a 120-V (rms) electrical outlet in a house in the United States? Ignore the effects of temperature on the heater's resistance.
Answer:
P=144.74W
Explanation:
We can model the power output of a resistance by using the following formula:
[tex]P=\frac{V^{2}}{R}[/tex]
Wehre P is the power output, V is the rms voltage and R is the resistance. The resistance of the space heater will remain the same, so we can calculate it from the power output in Germany and its rms voltage. So when solving for R, we get:
[tex]R=\frac{V^{2}}{P}[/tex]
and we can now use the provided data:
[tex]R=\frac{(230V)^{2}}{531.7W}[/tex]
which yields:
R= 99.49 Ω
Once we know what the heater's resistance is, we can now go ahead and calculate the power outpor of the heater in the U.S.
[tex]P=\frac{V^{2}}{R}[/tex]
so
[tex]P=\frac{(120V)^{2}}{99.49\Omega}[/tex]
P=144.74W
A rectangular wire loop is pulled out of a region of uniform magnetic field B at a constant speed v. What is true about the induced emf in the loop while the loop is pulled out of the region of uniform magnetic field
Answer:
There is a constant emf induced in the loop.
Explanation:
In the uniform magnetic field suppose the rectangular wire loop of length L and width b is moved out with a uniform velocity v. suppose any instance x length of the loop is out of the magnetic field and L-x length is inside the loop.
Area of loop outside the field = b(L-x)
we know that flux φ= BA
B= magnitude of magnetic field , A= area
and emf [tex]\epsilon= \frac{d\phi}{dt}[/tex]
[tex]\epsilon=B\frac{dA}{dt}[/tex]
[tex]\epsilon=B\frac{db(L-x)}{dt}[/tex]
[tex]\epsilon=Bb\frac{d(L-x)}{dt}[/tex]
B,b and L are constant and dx/dt = v
⇒ε = -Bbv
which is a constant hence There is a constant emf induced in the loop.
The power needed to accelerate a projectile from rest to its launch speed v in a time t is 42.0 W. How much power is needed to accelerate the same projectile from rest to a launch speed of 2v in a time of t?
Answer:168 W
Explanation:
Given
Power needed [tex]P=42 W[/tex]
initial Launch velocity is v
Energy of projectile when it is launched [tex]E=\frac{1}{2}mv^2[/tex]
[tex]Power=\frac{Energy}{time}[/tex]
[tex]Power=\frac{E}{t}[/tex]
[tex]42=\frac{\frac{1}{2}mv^2}{t}--------1[/tex]
Power when it is launched with 2 v
[tex]E_2=\frac{1}{2}m(2v)^2=\frac{4}{2}mv^2[/tex]
[tex]P=\frac{2mv^2}{t}---------2[/tex]
Divide 1 & 2 we get
[tex]\frac{42}{P}=\frac{1}{2\times 2}[/tex]
[tex]P=42\times 4=168 W[/tex]
To accelerate the projectile to twice its launch speed, four times the power is needed.
Explanation:To find the power needed to accelerate the projectile from rest to a launch speed of 2v in a time of t, we need to recognize that power is directly proportional to the change in kinetic energy. The change in kinetic energy from rest to launch speed v is given by KE = (1/2)mv^2, and the change in kinetic energy from rest to launch speed 2v is given by KE' = (1/2)m(2v)^2 = 4(1/2)mv^2 = 4KE.
Since power is directly proportional to the change in kinetic energy, the power needed to accelerate the projectile to a launch speed of 2v is four times the power needed to accelerate it to a launch speed of v. Therefore, the power needed is 4(42.0 W) = 168.0 W.
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A 0.140-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. The glider has a head-on collision with a 0.299-kg glider that is moving to the left with a speed of 2.28 m/s. Find the final velocity (magnitude and direction) of each glider if the collision is elastic.
Answer:
v1 = 2.76 m/s and v2 = - 0.32 m/s
Explanation:
m1 = 0.140 kg
m2 = 0.299 kg
u1 = 0.80 m/s
u2 = - 2.28 m/s
Let the speed after collision is v1 and v2.
Use conservation of momentum
m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2
0.140 x 0.80 - 0.299 x 2.28 = 0.140 x v1 + 0.299 x v2
0.112 - 0.68 = 0.14 v1 + 0.299 v2
0.14 v1 + 0.299 v2 = - 0.568 ..... (1)
By the use of coefficient of restitution, the value of e = 1 for elastic collision
[tex]e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}[/tex]
u2 - u1 = v1 - v2
- 2.28 - 0.8 = v1 - v2
v1 - v2 = 3.08
v1 = 3.08 + v2
Put in equation (1)
0.14 (3.08 + v2) + 0.299 v2 = - 0.568
0.43 + 0.44 v2 = - 0.568
v2 = - 0.32 m/s
and
v1 = 3.08 - 0.32 = 2.76 m/s
Thus, v1 = 2.76 m/s and v2 = - 0.32 m/s
It is thought that bonding of adhesives occurs at the molecular level. What is the technical name of the force that holds glue to its bonding materials?
Answer:
Van der waals forces.
Explanation:
When we spread glue to stick any two substances as A and B with adhesives C. then there are adhesive force between substance A and C and adhesive force between substance B and C and cohesive force between C itself will act. In all adhesive and cohesive forces van der waals forces will apply at molecular level because there is no chemical bonding between adhesive and surface but lots of small attractive forces.
Answer:
Van der Waals force
Explanation:
The technical name given to the force that holds glue to its bonding materials is called Van der Waals force.
The forces of Van der Waals is defined by attraction and repulsion between atoms, molecules, and surfaces and other intermolecular forces. They differ from covalent and ionic bond in that they are caused by correlations in the varying polarizations of the nearby particles (as a result of quantum dynamics).
person throws a ball horizontally from the top of a building that is 24.0 m above the ground level. The ball lands 100 m down range from the base of the building. What was the initial velocity of the ball? Neglect air resistance and use g = 9.81 m/s2.
Answer:45.24 m/s
Explanation:
Given
Height of Building h=24 m
Range of ball R=100 m
Considering Vertical motion of ball
using [tex]y=u_yt+\frac{a_yt^2}{2} [/tex]
initial vertical velocity is zero therefore [tex]u_y=0[/tex]
[tex]24=0+\frac{9.8\times t^2}{2}[/tex]
[tex]t=\sqrt{\frac{48}{9.8}}[/tex]
[tex]t=2.21 s[/tex]
Now considering Horizontal Motion
[tex]R=u_xt+\frac{a_xt^2}{2}[/tex]
[tex]100=u_x\times 2.21+0[/tex] , as there is no horizontal acceleration
[tex]u_x=45.24 m/s[/tex]
The physics problem can be addressed by using the principles of projectile motion. The time of flight determined by the vertical motion is used to calculate the horizontal initial velocity. The initial velocity of the ball is approximately 45.24m/s.
Explanation:This is a problem in Physics based on the principles of Projectile Motion. We need to determine the initial velocity of the ball. The key point in this problem is that the horizontal motion of the projectile (in this case, the ball) is determined purely by the initial horizontal velocity, and is unaffected by the vertical motion. This is called the independence of the horizontal and vertical motions.
The time the ball is in the air is governed entirely by its vertical motion. Thus, we can find the time of flight by using the equation for vertical motion: y = 1/2gt², where y is the vertical displacement (24m in this case), g is the acceleration due to gravity (9.81 m/s²), and t is the time. So, t = sqrt(2y/g) = sqrt(2*24/9.81) = 2.21s.
Using this time, we can find the initial horizontal velocity using the equation for horizontal motion: x = vxt where x is the horizontal displacement (100m in this case), vx is the horizontal velocity, and t is the time. Rearranging the equation we get: vx = x/t which is approximately 45.24m/s . So, the initial velocity of the ball is around 45.24m/s
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Two speakers emit the same sound wave, identical frequency, wavelength, and amplitude. What other quantity would be necessary to determine if constructive or destructive interference occurs at a particular point some distance from the speakers?
Answer:
Phase Difference
Explanation:
When the sound waves have same wavelength, frequency and amplitude we just need the phase difference between them at a particular location to determine whether the waves are in constructive interference or destructive interference.
Interference is a phenomenon in which there is superposition of two coherent waves at a particular location in the medium of propagation.
When the waves are in constructive interference then we get a resultant wave of maximum amplitude and vice-versa in case of destructive interference.
For constructive interference the waves must have either no phase difference or a phase difference of nλ, where n is any natural number.For destructive interference the waves must have a phase difference of n×0.5λ, where n is any odd number.You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you pull the plates apart to a larger separation, do the following quantities increase, decrease, or stay the same?
a. C
b. Q
c. E between the plates
d. delta-V
Answer:
a. C will decrease
b. Q will remain the same
c. E will decrease
d. Delta-V will increase
Explanation:
Justification for C:
As we know that for parallel plate capacitors, capacitance is calculated using:
C = (ϵ_r * ϵ_o * A) / d - Say it Equation 1
Where:
ϵ_r - is the permittivity of the dielectric material between two plates
ϵ_o - Electric Constant
A - Area of capacitor's plates
d - distance between capacitor plates
From equation 1 it is clear that capacitance will decrease if distance between the plates will be increased.
Justification of Q
As charge will not be able to travel across the plates, therefore it will remain the same
Justification of E
As we know that E = Delta-V / Delta-d, thus considering Delta-V is increasing on increasing Delta-d (As justified below) as both of these are directly proportional to each other, therefore Electric field (E) will remain constant as capacitors' plates are being separated.
Moreover, as the E depends on charge density which remains same while plates of capacitor are being separated therefore E will remain the same.
Justification of Delta-V
As we know that Q = C * V, therefore considering charge remains the same on increasing distance between plates, voltage must increase to satisfy the equation.
A 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 Nm. What is the angel (assumed to be less than 90 degrees) between the wrench handle and the direction the force is applied?
Answer:
9.1°
Explanation:
Torque = distance from pivot * perpendicular force
15 Nm = 0.5 m * perpendicular force
perpendicular force = 30 N
So the bertical component of the applied force that caused a turning effect (torque) was 30 N. Now we use this information to find the angle that would produce a vertical component of 30N from an applied force of 95 N.
Fy = FSinθ
15 = 95 Sinθ
θ = 9.1°
A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800.0 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating backward at 1.20m/s2.
a. What is the force of friction between the losing player’s feet and the grass?
b. What force does the winning player exert on the ground to move forward if his mass plus equipment is 110.0 kg?
Answer:
Part a)
[tex]F_f = 692 N[/tex]
Part b)
[tex]F_f = 932 N[/tex]
Explanation:
Part A)
As we know by Force equation on the losing player
[tex]F - F_f = ma[/tex]
so we will have
[tex]800 - F_f = 90\times 1.20[/tex]
[tex]800 - F_f = 108[/tex]
[tex]F_f = 692 N[/tex]
Part b)
As we know that the winning player is also moving with same acceleration
so we will have
[tex]F_f - F = ma[/tex]
[tex]F_f - 800 = 110\times 1.20[/tex]
[tex]F_f = 932 N[/tex]
Abnormal protrusion of the eye out of the orbit is known as
Answer:
Exophthalmos
Explanation:
Exophthalmos is a disorder which can be either bilateral or unilateral. Sometimes it is also known by other names like Exophthalmus, Excophthamia, Exobitism.
It is basically the bulging of eye anterior out of orbit which if left unattended may result in eye openings even while sleeping consequently resulting in comeal dryness and damage which ultimately may lead to blindness.
It is commonly caused by trauma or swelling of eye surrounding tissues resulting from trauma.
You are standing 2.5 m directly in front of one of the two loudspeakers shown in the figure. They are 3.0 m apart and both are playing a 686 Hz tone in phase. Part A As you begin to walk directly away from the speaker, at what distances from the speaker do you hear a minimum sound intensity? The room temperature is 20 degrees C. Express your answer numerically using two significant figures. If there is more than one answer, enter your answers in ascending order separated by commas
Answer:
L = 3.8 m
Explanation:
As we know that the frequency of sound is given as
[tex]f = 686 Hz[/tex]
speed of the sound is given as
[tex]v = 332 + 0.6 t[/tex]
[tex]v = 332 + (0.6 \times 20)[/tex]
[tex]v = 344 m/s[/tex]
now we have wavelength of sound is given as
[tex]\lambda = \frac{v}{f}[/tex]
[tex]\lambda = \frac{344}{686}[/tex]
[tex]\lambda = 0.50 m[/tex]
now we have path difference at initial position given as
[tex]\Delta L = \sqrt{L^2 + d^2} - L[/tex]
[tex]\Delta L = \sqrt{3^2 + 2.5^2} - 2.5[/tex]
[tex]\Delta L = 3.9 - 2.5 = 1.4 m[/tex]
now we know that for minimum sound intensity we have
[tex]\Delta L = \frac{2N + 1}{2}\lambda[/tex]
[tex]\Delta L = \frac{2N + 1}{2}(0.50)[/tex]
so we have
N = 2
[tex]\Delta L = 1.25 m[/tex]
so we have
[tex]\sqrt{2.5^2 + L^2} - L = 1.25[/tex]
[tex]2.5^2 + L^2 = L^2 + 1.25^2 + 2.5L[/tex]
[tex]L = 1.875 m[/tex]
Now for N = 1
[tex]\Delta L = 0.75 m[/tex]
so we have
[tex]\sqrt{2.5^2 + L^2} - L = 0.75[/tex]
[tex]2.5^2 + L^2 = L^2 + 0.75^2 + 1.5L[/tex]
[tex]L = 3.8 m[/tex]
so the next minimum intensity will be at L = 3.8 m
A uniform ladder of length L and mass m leans against a frictionless vertical wall, making an angle of 54° with the horizontal. The coefficient of static friction between the ladder and the ground is 0.32. If your mass is four times that of the ladder, what percentage of the way up the ladder can you climb before the ladder begins to slip?
Answer:
h=0.425 L
Explanation:
Given that
θ = 54°
Coefficient of friction μ = 0.32
Mass of rod = m
Lets take mass of man = M = 4 m
C is the center of mass of the rod.
By balancing force in y and x direction
R= Fr
R = Fr= μ N
N = mg + Mg = mg + 4 m g ( M =4m)
N = 5 m g
Lets take distance cover by man is h along rod before sliding
Now taking moment about the lower end
M g h cosθ + m g cosθ L/2 = R L sinθ
2 M g h cosθ + m g cosθ L = 2 R L sinθ
Now by putting the value of R and M
8 m g h cosθ + m g cosθ L = 2 μ N L sinθ
8 m g h cosθ + m g cosθ L = 10 m g μ L sinθ
8 h cosθ + cosθ L = 10 μ L sinθ
8 h + L = 10 μ L tanθ
Now putting the value of θ and μ
8 h + L = 10 x 0.32 x tan54° x L
8 h + L = 4.4 L
8 h = 3.4 L
h=0.425 L
The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alternately stretches, as you bring your weight down onto your forward foot, and contracts to push you off the ground. A 70 kg runner has an Achilles tendon that is 15 cm long and has a cross-section area of 110 mm² typical values for a person of this size. 1. By how much will the runner's Achilles tendon stretch if the force on it is 8.0 times his weight? Young's modulus for tendor is 0.15 x 10¹⁰N/m². Express your answer to two significant figures and include the appropriate units. 2. What fraction of the tendon's length does this correspond.
Answer:
A) 0.5cm B) 1/30
Explanation:
The weight of the man = mass * acceleration due to gravity where the mass is 78kg and acceleration due to gravity is 9.81m/s^2
W = m * g = 78 * 9.81= 686.7 N
The force acting on the tendon is 8 times of the weight
Force = 8 * weight of the body = 8 * 686.7 = 5493.6 N
Young modulus of the tendon(e) = (F/A)/ (DL/L) where A is the cross sectional area in square meters, DL is the change in length of the tendon in meters and L is the original length of the tendon
e = (FL)/(ADL) cross multiply and make DL subject of the formula
DL = (FL) / (AL)
Convert the cross sectional area A into square meters and the length also
A = 110 / 1000000 since 1/1000 m = 1mm, 1/1000000 m^2 = 1 mm^2 and 1/100m = 1 cm
A = 0.00011 m ^2 and L = 0.15m
Substitute the values in the derived equation
DL = (5493.6 * 0.15)/ (1.5 * 10^ 9 * 1.1* 10^-4)
DL = 824.04 / 1.65 * 10^ 5
DL = 499.42 * 10^-5 = 499.42 *10^ -5 / 100 to convert it to meters
DL = 0.49942cm approx 0.5cm
B) fraction of the DL to L = 0.5 / 15 = 1/30
A box has a weight of 150 N and is being pulled across a horizontal floor by a force that has a magnitude of 110 N. The pulling force can point horizontally, or it can point above the horizontal at an angle θ. When the pulling force points horizontally, the kinetic frictional force acting on the box is twice as large as when the pulling force points at the angle θ. Find θ.
When the pulling force points at an angle θ above the horizontal, the frictional force acting on the box is -20N. By using the equation for frictional force and the weight of the box, we can determine that the coefficient of friction is 0.133. To find the angle θ, we use trigonometric ratios and find that it is 150°.
Explanation:Given that the weight of the box is 150N and the pulling force has a magnitude of 110N, we can determine the angle θ at which the pulling force is directed. Let's assume the angle θ is above the horizontal. The weight of the box, 150N, is equal to the normal force acting on the box. The frictional force between the box and the floor can be calculated as the difference between the force of the pulling and the weight of the box, which is 110N - 150N = -40N. Since the kinetic frictional force acting on the box is twice as large when the pulling force points horizontally, the frictional force when the pulling force points at an angle θ is -20N.
We can use the equation for frictional force, which is F_friction = μN, where F_friction is the frictional force, μ is the coefficient of friction, and N is the normal force. As the frictional force is -20N, we can substitute this value into the equation and solve for the coefficient of friction. Therefore, -20N = μ(150N), which gives us μ = -20N/150N = -0.133. Since the coefficient of friction is always positive, the actual value of μ is 0.133.
Now, let's use trigonometric ratios to find the angle θ. Since the weight of the box acts vertically downward and the pulling force has a horizontal component of 110N and a vertical component of -150N × sin(θ), the vertical components of the weight and the pulling force must cancel each other. Therefore, -150N × sin(θ) = 150N, which simplifies to sin(θ) = -1/2. Taking the inverse sine of -1/2, we get θ = -30° or 150°. However, since the pulling force is directed above the horizontal, the angle must be 150°.