Answer:
The reflected resistance in the primary winding is 6250 Ω
Explanation:
Given;
number of turns in the primary winding, [tex]N_P[/tex] = 50 turns
number of turns in the secondary winding, [tex]N_S[/tex] = 10 turns
the secondary load resistance, [tex]R_S[/tex] = 250 Ω
Determine the turns ratio;
[tex]K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5[/tex]
Now, determine the reflected resistance in the primary winding;
[tex]\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms[/tex]
Therefore, the reflected resistance in the primary winding is 6250 Ω
A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is the period if the amplitude of the motion is doubled
Answer:
The period of the motion will still be equal to T.
Explanation:
for a system with mass = M
attached to a massless spring.
If the system is set in motion with an amplitude (distance from equilibrium position) A
and has period T
The equation for the period T is given as
[tex]T = 2\pi \sqrt{\frac{M}{k} }[/tex]
where k is the spring constant
If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.
Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.
Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the concentric spheres. When compared to the number of field lines N1 going through the sphere of radius R, the number of electric field lines N2 going through the sphere of radius 2R is
Answer:
N2 = ¼N1
Explanation:
First of all, let's define the terms;
N1 = number of electric field lines going through the sphere of radius R
N2 = number of electric field lines going through the sphere of radius 2R
Q = the charge enclosed at the centre of concentric spheres
ε_o = a constant known as "permittivity of the free space"
E1 = Electric field in the sphere of radius R.
E2 = Electric field in the sphere of radius 2R.
A1 = Area of sphere of radius R.
A2 = Area of sphere of radius 2R
Now, from Gauss's law, the electric flux through the sphere of radius R is given by;
Φ = Q/ε_o
We also know that;
Φ = EA
Thus;
E1 × A1 = Q/ε_o
E1 = Q/(ε_o × A1)
Where A1 = 4πR²
E1 = Q/(ε_o × 4πR²)
Similarly, for the sphere of radius 2R,we have;
E2 = Q/(ε_o × 4π(2R)²)
Factorizing out to get;
E2 = ¼Q/(ε_o × 4πR²)
Comparing E2 with E1, we arrive at;
E2 = ¼E1
Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;
N2 = ¼N1
A speeding car has a velocity of 80 mph; suddenly it passes a cop car but does not stop. When the speeding car passes the cop car, the cop immediately accelerates his vehicle from 0 to 90 mph in 4.5 seconds. The cop car has a maximum velocity of 90 mph. At what time does the cop car meet the speeding car and at what distance?
Answer:
Distance= 4 miles
Time = 36.3 seconds
Explanation:
80 mph = 178.95 m/s
90 mph = 201.32 m/s
V = u +at
201.32= 0+a(4.5)
201.32/4.5= a
44.738 m/s² = a
Acceleration of the cop car
= 44.738 m/s²
Distance traveled at 4.5seconds
For the cop car
S= ut + ½at²
S= 0(4.5) + ½*44.738*4.5
S= 100.66 meters
Distance traveled at 4.5seconds
For the speeding car
4.5*178.95=805.275
The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal
X/178.95= (704.675+x)/201.32
X-0.89x= 626.37
0.11x= 626.37
X= 5694.3 meters
The time = 5694.3/178.95
Time =31.8 seconds
So the distance they meet
= 5694.3+805.275
= 6499.575 meters
= 4.0 miles
The Time = 4.5+31.8
Time = 36.3 seconds
A brick weighs 50.0 N, and measures 30.0 cm × 10.0 cm × 4.00 cm. What is the maximum pressure it can exert on a horizontal surface due to its weight?
Answer:
Pressure, P = 1250 Pa
Explanation:
Given that,
Weight of a brick, F = 50 N
Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm
We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.
A = 40 cm × 10 cm = 400 cm² = 0.04 m²
So,
Pressure,
[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]
So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.
The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.
What is pressure?The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.
The given data in the problem is;
W is the weight of a brick = 50 N
The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm
A is the area,
The area is found as;
A=40 cm × 10 cm = 400 cm² = 0.04 m²
The pressure is the ratio of the force and area
[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]
Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.
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A square loop 16.0 cm on a side has a resistance of 6.35 Ω . It is initially in a 0.510 T magnetic field, with its plane perpendicular to B , but is removed from the field in 40.5 ms.
Required:
Calculate the electric energy dissipated in this process.
Answer:
Explanation:
change in magnetic flux = .16 x .16 x .510 - 0
= .013056 weber .
rate of change of flux = change in flux / time
= .013056 / 40.5 x 10⁻³
= .32237
voltage induced = .32237 V
electrical energy dissipated = v² / R where v is voltage , R is resistance
= .32237² / 6.35
= 16.36 x 10⁻³ J .
A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86 times 10^6 m/s.
a. Find the magnitude of the maximum magnetic force that could be exerted on the proton.
b. What is the magnitude of the maximum acceleration of the proton?
c. Would the field exert the same magnetic force on an electron moving through the field with the same speed? (Assume that the electron is moving in the direction as the proton.)
1. Yes
2. No
.Answer;
Using Fmax=qVB
F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)
ANS=1.29*10^-12 N
2. Using Amax=Fmax/ m
Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)
ANS=1.93*10^15 m/s^2*
3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton
A fan rotating with an initial angular velocity of 1500 rev/min is switched off. In 2.5 seconds, the angular velocity decreases to 400 rev/min. Assuming the angular acceleration is constant, answer the following questions.
How many revolutions does the blade undergo during this time?
A) 10
B) 20
C) 100
D) 125
E) 1200
Answer:
The blade undergoes 40 revolutions, so neither of the given options is correct!
Explanation:
The revolutions can be found using the following equation:
[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
Where:
α is the angular acceleration
t is the time = 2.5 s
[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min
First, we need to find the angular acceleration:
[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]
Now, the revolutions that the blade undergo are:
[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]
Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!
I hope it helps you!
Did the kinetic frictional coefficient (for the wood/aluminum and felt/aluminum cases) vary with area of contact
Answer:
Explanation:
Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.
While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.
Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''
The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.
A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 cm carries a current of 0.25 A. The relative permeability of the core is 80.
(a) What is the magnetic field in the core?
(b) What part of the magnetic field is due to atomic currents?
Answer:
A) 0.0267 T
B) 0.0263 T
Explanation:
Given that
The number of turns, N = 400
Radius of the wire, r = 6 cm = 0.06 m
Current in the wire, I = 0.25 A
Relative permeability, K(m) = 80
See the attached picture for the calculation
A stereo speaker produces a pure "G" tone, with a frequency of 392 Hz. What is the period T of the sound wave produced by the speaker?
Answer:
The period is [tex]T = 0.00255 \ s[/tex]
Explanation:
From the question we are told that
The frequency is [tex]f = 392 \ Hz[/tex]
Generally the period is mathematically represented as
[tex]T = \frac{1}{f}[/tex]
=> [tex]T = \frac{1}{ 392}[/tex]
=> [tex]T = 0.00255 \ s[/tex]
The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various atmosphericprocesses, including lightning.
What is the excess charge on the surface of the earth? inC
Answer:
[tex]q = -461532.5 \ C[/tex]
Explanation:
From the question we are told that
The electric filed is [tex]E = 102 \ N/C[/tex]
Generally according to Gauss law
=> [tex]E A = \frac{q}{\epsilon_o }[/tex]
Given that the electric field is pointing downward , the equation become
[tex]- E A = \frac{q}{\epsilon_o }[/tex]
Here [tex]q[/tex] is the excess charge on the surface of the earth
[tex]A[/tex] is the surface area of the of the earth which is mathematically represented as
[tex]A = 4\pi r^2[/tex]
Where r is the radius of the earth which has a value [tex]r = 6.3781*10^6 m[/tex]
substituting values
[tex]A = 4 * 3.142 * (6.3781*10^6 \ m)^2[/tex]
[tex]A =5.1128 *10^{14} \ m^2[/tex]
So
[tex]q = -E * A * \epsilon _o[/tex]
Here [tex]\epsilon_o[/tex] s the permitivity of free space with value
[tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}[/tex]
[tex]q = -461532.5 \ C[/tex]
The molecules in Tyler are composed of carbon and other atoms that share one or more electrons between two atoms, forming what is known as a(n) _____ bond.
Answer:
covalent
Explanation:
covalent bonds share electrons
A car accelerates uniformly from rest and reaches a speed of 22.7 m/s in 9.02 s. Assume the diameter of a tire is 58.5 cm. (a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. rev (b) What is the final angular speed of a tire in revolutions per second? rev/s
(a) The car is undergoing an acceleration of
[tex]a=\dfrac{22.7\frac{\rm m}{\rm s}-0}{9.02\,\mathrm s}\approx2.52\dfrac{\rm m}{\mathrm s^2}[/tex]
so that in 9.02 s, it will have covered a distance of
[tex]x=\dfrac a2(9.02\,\mathrm s)^2\approx102\,\mathrm m[/tex]
The car has tires with diameter d = 58.5 cm = 0.585 m, and hence circumference π d ≈ 1.84 m. Divide the distance traveled by the tire circumference to determine how many revolutions it makes:
[tex]\dfrac{102\,\mathrm m}{1.84\,\mathrm m}\approx55.7\,\mathrm{rev}[/tex]
(b) The wheels have average angular velocity
[tex]\omega=\dfrac{\omega_f+\omega_i}2=\dfrac{\theta_f-\theta_i}{\Delta t}[/tex]
where [tex]\omega[/tex] is the average angular velocity, [tex]\omega_i[/tex] and [tex]\omega_f[/tex] are the initial and final angular velocities (rev/s), [tex]\theta_i[/tex] and [tex]\theta_f[/tex] are the initial and final angular displacements (rev), respectively, and [tex]\Delta t[/tex] is the duration of the time between initial and final measurements. The second equality holds because acceleration is constant.
The wheels start at rest, so
[tex]\dfrac{\omega_f}2=\dfrac{55.7\,\rm rev}{9.02\,\rm s}\implies\omega_f\approx12.4\dfrac{\rm rev}{\rm s}[/tex]
Convert 7,348 grams to kilograms
Which statement about friction is true? (1 point)
o
Static friction and kinetic friction in a system always act in opposite directions of each other and in the same direction as the
applied force
Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the
applied force
Static friction and kinetic friction in a system always act in opposite directions of each other and in the opposite direction of the
applied force
O
Static friction and kinetic friction in a system always act in the same direction as each other and in the same direction as the
applied force.
Answer:static friction and kinetic friction in a system always act in the same direction as each other and n the opposite direction of the applie force . Is the correct answer
Explanation:
Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the applied force. The correct option is B.
What is friction?Friction is the force that prevents one hard material from scooting or rolling over the other.
Frictional forces, such as the locomotion required to walk without dropping, are advantageous, but they also create a significant amount of resistance to motion.
We can control cars because of friction between the tires and the road: more precisely, because there are three types of friction: rolling friction, starting friction, and sliding friction.
Friction reduces the speed of moving objects and can even stop them from moving. The friction between the objects generates heat. As a result, energy is wasted in the machines. Friction will cause wear and tear on the machine parts.
In a system, static and kinetic friction always act in the same direction and in the opposite direction of the applied force.
Thus, the correct option is B.
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If a disk rolls on a rough surface without slipping, the acceleration of the center of gravity (G) will _ and the friction force will b
Answer:
Will be equal to alpha x r; less than UsN
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.50 s apart. The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.
Required:
How far away did the impact occur?
Answer:
The distance is [tex]d = 193.6 \ m[/tex]
Explanation:
From the question we are told that
The time interval between the sounds is k[tex]t_1 = k + t_2[/tex] = 0.50 s
The speed of sound in air is [tex]v_s = 343 \ m/s[/tex]
The speed of sound in the concrete is [tex]v_c = 3000 \ m/s[/tex]
Generally the distance where the collision occurred is mathematically represented as
[tex]d = v * t[/tex]
Now from the question we see that d is the same for both sound waves
So
[tex]v_c t = v_s * t_1[/tex]
Now
So [tex]t_1 = k + t[/tex]
[tex]v_c t = v_s * (t+ k)[/tex]
=> [tex]3000 t = 343* (t+ 0.50)[/tex]
=> [tex]3000 t = 343* (t+ 0.50)[/tex]
=> [tex]t = 0.0645 \ s[/tex]
So
[tex]d = 3000 * 0.0645[/tex]
[tex]d = 193.6 \ m[/tex]
On a separate sheet of paper, tell why scientists in different countries can easily compare the amount of matter in similar objects in their countries
Answer: no u
Explanation: no u
A car is travelling west at 22.2 m/s when it accelerated for 0.80 s to the west at 2.68 m/s2. Calculate the car's final velocity. Show all your work.
Answer:
24.34 m/s
Explanation:
recall that one of the equations of motions takes the form:
v = u + at
where,
v = final velocity
u = initial velocity (given as 22.2 m/s)
a = acceleration (given as 2.68m/s²)
t = time elapsed during acceleration (given as 0.80s)
since we are told that the the acceleration is in the direction of the intial velocity, we can simply substitute the known values into the equation above:
v = u + at
v = 22.2 + (2.68) (0.8)
v = 24.34 m/s
Rank the ultraviolet, infrared, and visible regions of theelectromagnetic spectrum in terms of lowest to highest energy,frequency, and wavelength.
Energy: < <
Frequency: < <
Wavelength: <
Answer:
1. Energy: ultraviolet>> visible> infrared
2. Frequency: ultraviolet>> visible > infrared
3. Wavelength: infrared >> visible > ultraviolet
Explanation:
Electromagnetic waves are a class of waves that do not require material medium for their propagation, and travel at the same speed. They are arranged with respect to either their decreasing wavelength or increasing frequency to form a spectrum called an electromagnetic spectrum.
Comparing the energy, frequency and wavelength of ultraviolet, infrared and visible regions, it can be deduced that:
1. Energy: ultraviolet has the highest energy, then followed by visible, and infrared has the lowest energy.
i.e energy: ultraviolet>> visible> infrared
2. Frequency: ultraviolet radiation has the highest frequency, visible region has a greater frequency than that of infrared.
i.e frequency: ultraviolet>> visible > infrared
3. Wavelength: infrared radiation has the highest wavelength, followed by visible region, and ultraviolet radiation has the lowest wavelength.
i.e wavelength: infrared >> visible > ultraviolet
In terms of lowest to the highest energy,frequency, and wavelength is;
Energy: infrared > visible light > ultraviolet
Frequency: infrared > visible light > ultraviolet
Wavelength: ultraviolet > visible light > infrared
The electromagnetic spectrum:
The electromagnetic spectrum is made up of all the electromagnetic waves (ultraviolet, infrared, and visible) arranged according to their energy,frequency, and wavelength.
The ultraviolet: This wave is seen in the sunlight and is made up of wavelength of 10nm to 400nm. A frequency of [tex]10^{16}[/tex](Hz).Infrared wave: They are invisisble to the human eye but can be felt as heat. It has frequency of [tex]10^{12}[/tex]Hz and a wavelength of 780nm to 1mm.Visible light: This is part of the electromagnetic wave that the eye can view. It has frequency of [tex]10^{15}[/tex]Hz and a wavelength of 380 to 700nm.Learn more about electromagnetic spectrum here:
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y=k/x, x is halved.
what happens to the value of y
Answer:
y is doubled
Explanation:
If x is halved, that means the value is doubled. Here is an exmaple:
y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.
At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.10 ✕ 10−5 T
Answer:
The speed of the proton is 4059.39 m/s
Explanation:
The centripetal force on the particle is given by;
[tex]F = \frac{mv^2}{r}[/tex]
The magnetic force on the particle is given by;
[tex]F = qvB[/tex]
The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.
[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]
where;
r is the radius of the circular path moved by both electron and proton;
⇒For electron;
[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]
⇒For proton
The speed of the proton is given by;
[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]
Therefore, the speed of the proton is 4059.39 m/s
A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire
Answer:
The direction of the force is towards the East.
Explanation:
Using the right hand rule, the force on the current carrying conductor is east.
In the right hand rule, if the hand is held with the fingers pointed parallel to the palm representing the magnetic field, and the thumb held at right angle to the rest of the fingers representing the direction of the current, then the palm will push in the direction of the force.
In this case, the thumb is pointing downwards, with the fingers pointing north away from the body in the direction of the earth's magnetic field, the palm will push east.
Two long straight wires carry currents perpendicular to the xy plane. One carries a current of 50 A and passes through the point x = 5.0 cm on the x axis. The second wire has a current of 80 A and passes through the point y = 4.0 cm on the y axis. What is the magnitude of the resulting magnetic field at the origin?
Answer:
450 x10^-6 T
Explanation:
We know that the magnetic of each wire is derived from
ByB= uoi/2pir
Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)
= 0.2 x 10^ -3T
B2=
4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)
= 0.4 x 10^ -3T
So
(Bnet)² = (Bx)² + ( By)²
= (0.2² + 0.4²)mT
= 450 x10^-6T
The magnitude of magnetic field at the origin is required.
The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]
x = Location at x axis = 5 cm
y = Location at y axis = 4 cm
[tex]I_x[/tex] = Current at the x axis point = 50 A
[tex]I_y[/tex] = Current at the y axis point = 80 A
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]
Magnitude of the magnetic field is given by
[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]
Finding the net magnetic field using the Pythagoras theorem
[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]
The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]
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A particle undergoes damped harmonic motion. The spring constant is 100 N/m, the damping constant is 8.0 x 10-3 kg.m/s, and the mass is 0.050 kg. If the particle starts at its maximum displacement, x = 1.5 m, at time t = 0. What is the amplitude of the motion at t = 5.0 s?
Answer:
The amplitude [tex]A(5) = 1 \ m[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 100 \ N/m[/tex]
The damping constant is [tex]b = 8.0 *10^{-3} \ kg \cdot m/s[/tex]
The mass is [tex]m = 0.050 \ kg[/tex]
The maximum displacement is [tex]A_o = 1.5 \ m \ at t = 0[/tex]
The time considered is t = 5.0 s
Generally the displacement(Amplitude) of damped harmonic motion is mathematically represented as
[tex]A(t) = A_o * e ^{ - \frac{b * t}{2 * m} }[/tex]
substituting values
[tex]A(5) = 1.5 * e ^{ - \frac{ 8.0 *10^{-3} * 5}{2 * 0.050} }[/tex]
[tex]A(5) = 1 \ m[/tex]
1.) When the acceleration is zero, what can you say about the velocity of an object?
Answer:
it is either constant or zero
Explanation:
a car moves for 10 minutes and travels 5,280 meters .What is the average speed of the car?
Answer:use the formular distance over time i.e distance/time. Make sure to convert the distance from metres to kilometers and time from minutes to hours .
Explanation:
The average speed of the car is 31,680 meters per hour.
To calculate the average speed of the car, you need to divide the total distance traveled by the time it took to travel that distance.
Given:
Time taken (t) = 10 minutes = 10 minutes × (1 hour / 60 minutes) = 10/60 hours = 1/6 hours
Distance traveled (d) = 5,280 meters
Average Speed (v) = Distance (d) / Time (t)
Average Speed (v) = 5280 meters / (1/6) hours
To simplify, when you divide by a fraction, it's equivalent to multiplying by its reciprocal:
Average Speed (v) = 5280 meters × (6/1) hours
Average Speed (v) = 31,680 meters per hour
Hence, the average speed of the car is 31,680 meters per hour.
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The Moon orbits Earth in a nearly circular orbit (mean distance is 378,000 km ). The moon Charon orbits Pluto in a nearly circular orbit as well (mean distance is 19,600 km ).
Earth Moon Pluto Charon
Mass (kg) 5.97 x 10^24 0.07342 x 10^24 0.0146 x 10^24 0.00162 x 10^24
Equatorial radius (km) 6378.1 1738.1 1185 604
Which object exhibits the longest orbital period? Hint: perform order of magnitude analysis.
a. Moon around Earth
b. Charon around Pluto
c. About the same for both
Answer:
a. Moon around Earth.
Explanation:
Charon orbit takes around 6.4 earth days to complete its orbit. Charon does not rises or sets, it hovers over same spot around the Pluto. The same side of Charon faces the Pluto, this is called Tidal Locking.
The moon orbit takes around 27 days to complete its orbit. The moon has different sides that are faced with sun which creates light or dark face of moon on the earth. Moon has 384,400 km distance from the earth.
The object that should exhibit the longest orbital period is option a. Moon around Earth.
What is Charon's orbit?Charon's orbit takes around 6.4 earth days to finish its orbit. Charon does not rise or sets, it hovers over similar spot around Pluto. The same side of Charon faces Pluto, this we called Tidal Locking. Here the moon orbit should take approx 27 days to finish its orbit. The moon has various sides that are faced with the sun which developed the light or dark face of the moon on the earth. Also, Moon has 384,400 km distance from the earth.
learn more about orbit here: https://brainly.com/question/25404554
A piano string having a mass per unit length equal to 4.80 ✕ 10−3 kg/m is under a tension of 1,300 N. Find the speed with which a wave travels on this string.
Answer:
Velocity of wave (V) = 5.2 × 10² m/s
Explanation:
Given:
Per unit length mass (U) = 4.80 × 10⁻³ kg/m
Tension (T)= 1,300 N
Find:
Velocity of wave (V)
Computation:
Velocity of wave (V) = √T / U
Velocity of wave (V) = √1300 / 4.80 × 10⁻³
Velocity of wave (V) = √ 270.84 × 10³
Velocity of wave (V) = 5.2 × 10² m/s
Approximating the eye as a single thin lens 2.70 cmcm from the retina, find the focal length of the eye when it is focused on an object at a distance of 265 cmcm
Answer:
0.37 cm
Explanation:
The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.
The object is at a distant of 265 cm to the lens of the eye.
From lens formula,
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
where: f is the focal length, u is the object distance and v is the image distance.
Thus, u = 265.00 cm and v = 2.70 cm.
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{265}[/tex] + [tex]\frac{27}{10}[/tex]
= [tex]\frac{10+7155}{2650}[/tex]
[tex]\frac{1}{f}[/tex] = [tex]\frac{7165}{2650}[/tex]
⇒ f = [tex]\frac{2650}{7165}[/tex]
= 0.37
The focal length of the eye is 0.37 cm.
