Answer:
The induced emf is [tex]\epsilon = 0.0280 \ V[/tex]
Explanation:
From the question we are told
The diameter of the loop is [tex]d = 6.7 cm = 0.067 \ m[/tex]
The magnetic field is [tex]B = 1.27 \ T[/tex]
The time taken is [tex]dt = 0.16 \ s[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{\Delta \phi}{dt}[/tex]
Where N = 1 given that it is only a circular loop
[tex]\Delta \phi = \Delta B * A[/tex]
Where [tex]\Delta B = B_f - B_i[/tex]
where [tex]B_i[/tex] is 1.27 T given that the loop of wire was initially in the magnetic field
and [tex]B_f[/tex] is 0 T given that the loop was removed from the magnetic field
Now the area of the of the loop is evaluated as
[tex]A = \pi r^2[/tex]
Where r is the radius which is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.067}{2}[/tex]
[tex]r = 0.0335 \ m[/tex]
So
[tex]A = 3.142 * (0.0335)^2[/tex]
[tex]A = 0.00353 \ m^2[/tex]
So
[tex]\Delta \phi = (0- 127)* (0.00353)[/tex]
[tex]\Delta \phi = -0.00448 Weber[/tex]
=> [tex]\epsilon = - 1 * \frac{-0.00448}{0.16}[/tex]
=> [tex]\epsilon = 0.0280 \ V[/tex]
A father and his young son get on a teeter-totter. The son sits 2 m fromthe center, but the father has to sit closer to balance. Where does the father have to sit to balance the teeter-totter if he weighs 4 times as much as his son?
Answer:
The distance of the father from the center is [tex]d_f = \frac{1}{2} \ m[/tex]
Explanation:
From the question we are told that
The distance of the son from the center is [tex]d_s = 2 \ m[/tex]
Let the mass of the son be [tex]m_s[/tex]
then the mass of the father is [tex]m_f = 4m_s[/tex]
Now for the teeter-totter to be balanced the torque due to the weight of the father must be equal to the torque due to the weight the son, this is mathematically represented as
[tex]\tau_s = \tau_f[/tex]
Where [tex]\tau_s[/tex] is the torque of the son which is mathematically represented as
[tex]\tau_ s = m_s * d_s * g[/tex]
while [tex]\tau_f[/tex] is the torque of the father which is mathematically represented as
[tex]\tau_f = m_f * d_f * g[/tex]
=> [tex]\tau_f = 4 m_s * d_f * g[/tex]
So
[tex]4 m_s * d_f * g = m_s * d_s * g[/tex]
substituting values
[tex]4 * d_f * = 2[/tex]
=> [tex]d_f = \frac{1}{2} \ m[/tex]
Two spheres A and B of negligible dimensions and masses 1 kg and √3 kg respectively, are supported on the smooth circular surface, fixed to the ground with a centre O and radius of 0.1m. The spheres are joined by the cord shown in length π/20 m; determine the angles α and β corresponding to the position of equilibrium of the spheres with respect to the vertical passing through O.
Answer:
α = π/3
β = π/6
Explanation:
Use arc length equation to find the sum of the angles.
s = rθ
π/20 m = (0.1 m) (α + β)
π/2 = α + β
Draw a free body diagram for each sphere. Both spheres have three forces acting on them:
Weight force mg pulling down,
Normal force N pushing perpendicular to the surface,
and tension force T pulling tangential to the surface.
Sum of forces on A in the tangential direction:
∑F = ma
T − m₁g sin α = 0
T = m₁g sin α
Sum of forces on B in the tangential direction:
∑F = ma
T − m₂g sin β = 0
T = m₂g sin β
Substituting:
m₁g sin α = m₂g sin β
m₁ sin α = m₂ sin β
(1 kg) sin α = (√3 kg) sin (π/2 − α)
1 sin α = √3 cos α
tan α = √3
α = π/3
β = π/6
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the right when the driver turns on the motor. The boat speeds up for 6.0\,\text s6.0s6, point, 0, start text, s, end text with an acceleration of 4.0\,\dfrac{\text m}{\text s^2}4.0 s 2 m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction leftward.
The question is incomplete. Here is the entire question.
A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?
Answer: Δx = - 42m
Explanation: The jetboat is moving with an acceleration during the time interval, so it is a linear motion with constant acceleration.
For this "type" of motion, displacement (Δx) can be determined by:
[tex]\Delta x = v_{i}.t + \frac{a}{2}.t^{2}[/tex]
[tex]v_{i}[/tex] is the initial velocity
a is acceleration and can be positive or negative, according to the referential.
For Referential, let's assume rightward is positive.
Calculating displacement:
[tex]\Delta x = 5(6) - \frac{4}{2}.6^{2}[/tex]
[tex]\Delta x = 30 - 2.36[/tex]
[tex]\Delta x[/tex] = - 42
Displacement of the boat for t=6.0s interval is [tex]\Delta x[/tex] = - 42m, i.e., 42 m to the left.
Types of friction in physics
-- static friction
-- kinetic friction
-- fluid friction
-- sliding friction
-- air resistance
-- drag
-- professional debate
The unstretched rope is 20 meters. After getting dunked a few times the 80 kg jumper comes to rest above the water with the rope now stretched to 30 meters. What is the maximum length of the rope in meters when the jumper is being dunked?
Answer:
Therefore maximum stretch is y2 = 32.36 m
Explanation:
In this problem let's use the initial data to find the string constant, let's apply Newton's second law when in equilibrium
[tex]F_{e}[/tex] - W = 0
k Δx = mg
k = mg / Δx
k = 80 9.8 / (30-20)
k = 78.4 N / m
now let's use conservation of energy to find the velocity of the body just as the string starts to stretch y = 20 m
starting point. When will you jump
Em₀ = U = mg y
final point. Just when the rope starts to stretch
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
mg y = ½ m v²
v = √ 2g y
v = √ (2 9.8 20)
v = 19.8 m / s
now all kinetic energy is transformed into elastic energy
starting point
Em₀ = K = ½ m v²
final point
Em_{f} = [tex]K_{e}[/tex] + U = ½ k y² + m g y
Emo = Em_{f}
½ m v² = ½ k y² + mgy
k y² + 2 m g y - m v² = 0
we substitute the values and solve the quadratic equation
78.4 y² + 2 80 9.8 y - 80 19.8² = 0
78.4 y² + 1568 y - 31363.2 = 0
y² + 20 y - 400 = 0
y = [- 20 ±√ (20 2 +4 400)] / 2
y = [-20 ± 44.72] / 2
the solutions are
y₁ = 12.36 m
y₂ = 32.36 m
These solutions correspond to the maximum stretch and its rebound.
Therefore maximum stretch is y2 = 32.36 m
Why do we use semiconductor instead of metal in thermopile?
Answer:
Metal or conductors , what they do is that they allow full flow of current that is conduction is due to free electrons only and there is literally no gap between valence and conduction band,so free electrons can easily jump into conduction band from valence band.
And now lets talk about insulators,what they do is that they don’t allow any current to flow i.e they act as strong dielectric,and gap between valence band and conduction band is so big that free electron can never come into conduction band from valence band ever if they try is for eternity…
But semiconductors can act both as an insulator as well as a conductor based on the voltage input. Hence, there is a possibility to control the current flow in semiconductors , so they don’t just relax and let the current pass by, they can control it, and that is why you can design logic circuits with it.
But as the temperature increases free electrons from valence band of insulators can jump to conduction band and can cause a little conductivity, and then the insulator will act as a semi-conductor.
Water flows at speed v in a pipe of radius R. At what speed does the water flow through a constriction in which the radius of the pipe is R/3
Answer:
v₂ = 9 v
Explanation:
For this exercise in fluid mechanics, let's use the continuity equation
v₁ A₁ = v₂ A₂
where v is the velocity of the fluid, A the area of the pipe and the subscripts correspond to two places of interest.
The area of a circle is
A = π R²
let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint
In this case v₁ = v and the area is
A₁ = π R²
in the second point
A₂= π (R / 3)²
we substitute in the continuity equation
v π R² = v₂ π R² / 9
v = v₂ / 9
v₂ = 9 v
If a vacuum pump reduces the pressure of a gas to 1.0 x 10-6 atm, what is the pressure expressed in millimeters of mercury
Answer:
[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.
Explanation:
According to current SI unit conversions, 1 atmosphere is equal to 760 millimeters of mercury. The current pressure is determined by simple rule of three:
[tex]p = \frac{760\,mm\,Hg}{1\,atm} \times (1\times 10^{-6}\,atm)[/tex]
[tex]p = 7.6\times 10^{-4}\,mm\,Hg[/tex]
[tex]1.0\times 10^{-6}[/tex] atmospheres are equivalent to [tex]7.6\times 10^{-4}[/tex] millimeters of mercury.
Which of
of
these
following material is
used as fuse material?
carbon,
silver
Copper
Aluminium
The provided question is not correct as, there is more than one options are correct, however the explaining every correct option -
Answer:
The correct answer are - silver, copper and aluminium all three used as fuse material.
Explanation:
A safety device in any electric circuit of that prevents the electric system in case of short circuit by breaking the connection of electric system or circuit termed as the Fuse or fuse element. Normally the fuse are made up of wire or element of material that are low in melting point and high in resistance.
Zinc, lead, tin, silver, copper, aluminium, and alloy of tin and alloy are used as fuse element or material for their low melting point and high resistance these are easily breaks the electric path in case of short circuit.
A magnetic field near the floor points down and is increasing. Looking down at the floor, does the non-Coulomb electric field curl clockwise or counter-clockwise?
a. clockwiseb. counter-clockwise c. no curly E
Answer:
when a magnetic field near the floors points down and is increasing then the electric field curl (a) clockwise.
Explanation:
The magnetic field this is the area that is around a magnet which there is presence of magnetic force. The Moving electric charges can create magnetic fields. we say In physics, that the magnetic field is a field that passes through space and which makes a magnetic force move electric charges.
The Non-coulomb electric field curls ; ( B ) counterclockwise
Non-coulomb electric field also known as induced EMF is the Negative time rate of change of a magnetic flux in a closed loop through the loop. Non-coulomb electric field is expressed as ; Fnc = qEnc
Given that the magnetic field points downwards and the value of the electric field ( ε ) is increasing ( i.e. ε > 0 ) The direction of the non-coulomb electric field will curl in a counter-clockwise direction.
Hence we can conclude that The Non-coulomb electric field curls in a counterclockwise direction.
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The source of sound moves away from the listener.The listener has the impression that the source is lower in pitch. Why?
When the source is moving away from the observer the velocity of the source is added to the speed of light. This increases the value of the denominator, decreasing the value of the observed frequency. Frequency corresponds to pitch or tone; a lower observed frequency will result in a lower observed pitch.
Ratio of the speed of light in a vacuum to the speed of light in a medium Rule for how light is refracted at the boundary between two materials Process that occurs when the angle of incidence is greater than the critical angle
Answer:
TOTAL INTERNAL REFLECTION
Explanation:
Retraction is defined as the change in the direction of light rays as it moves from less dense medium to a denser medium.
For us to have a critical angle, the ray must be passing from the denser medium to the less dense medium. As the angle of refraction in the less dense medium is increasing, the angle of incidence in the less dense medium also increases. A point will reach when the refracted ray will be parallel to the interface i.e angle of refraction is 90°, the angle of incidence at this point is known as the critical angle. If the angle of refraction keeps increasing further, it will get to a point when the refracted ray becomes reflected into the denser medium. At this stage we say that the ray is internally reflected and this is the point when the angle of incidence is greater than the critical angle.
Hence it can be concluded that the process that occurs when the angle of incidence is greater than the critical angle is called TOTAL INTERNAL REFLECTION
Infrared and ultraviolet waves have different frequencies.
Both types of wave can have harmful effects on human beings.
Describe the harmful effects of infrared and ultraviolet waves, relating them to the frequencies of the waves.
Answer:
For infrared and ultraviolet waves have different frequencies. Both types of wave can have harmful effects on human beings. Describe the harmful effects of infrared and ultraviolet waves, relating them to the frequencies of the waves. Medical studies indicate that prolonged IR exposure can lead to lens, cornea and retina damage, including cataracts, corneal ulcers and retinal burns, respectively. To help protect against long-term IR exposure, workers can wear products with IR filters or reflective coatings.When you look at the EM spectrum, UV waves are quite a bit smaller in wavelength than infrared, and x-rays/gamma rays are even smaller. Therefore, UV waves are probably causing more harm than infrared waves, and x-rays/gamma rays are probably doing even more damage.
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Infrared and ultraviolet waves have different frequencies. Infrared waves have lower frequencies and longer wavelengths, while ultraviolet waves have higher frequencies and shorter wavelengths.
Harmful effects of Infrared waves:
Infrared waves have lower frequencies and are often associated with heat radiation. Prolonged exposure to intense infrared radiation can lead to thermal burns and damage to the skin and eyes. Infrared radiation can also cause dehydration and overheating of the body, especially in hot environments. While infrared radiation is not as harmful as ultraviolet radiation, excessive exposure can still lead to health issues.
Harmful effects of Ultraviolet waves:
Ultraviolet waves have higher frequencies and shorter wavelengths, making them more energetic than infrared waves. UV radiation from the sun is a well-known harmful agent. Short-term exposure to intense UV radiation can cause sunburn, skin redness, and eye irritation. Long-term exposure to UV radiation can lead to more serious health problems such as skin aging, cataracts, and an increased risk of skin cancer. UV radiation can also damage DNA in skin cells, leading to mutations and potential carcinogenesis.
It is essential to protect ourselves from both infrared and ultraviolet waves to prevent harmful effects. Using sunscreen and wearing protective clothing can help shield the skin from UV radiation. Limiting exposure to intense sources of infrared radiation, such as hot objects or infrared heaters, can help reduce the risk of thermal burns and overheating. Understanding the differences in the frequencies of these waves allows us to implement appropriate safety measures and protect our health.
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Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long. How long is the other pipe?
Answer:
The length of the longer pipe is L = 2.30 m
Explanation:
Given that:
Two open organ pipes, sounding together, produce a beat frequency of 8.0 Hz . The shorter one is 2.08 m long.
How long is the other pipe?
From above;
The formula for the frequency of open ended pipes can be expressed as:
[tex]f = \dfrac{nv}{2L}[/tex]
where n = 1 ( since half wavelength exist between those two pipes)
v = 343 m/s and L = 2.08 m
Thus, the shorter pipe produces a frequency of :
[tex]f = \dfrac{1*343}{2*2.08}[/tex]
[tex]f = \dfrac{343}{4.16}[/tex]
[tex]f =82.45 \ Hz[/tex]
Also; we know that the beat frequency was given as 8.0 Hz
Then,
The lower frequency of the longer pipe = ( 82.45 - 8.0 )Hz
The lower frequency of the longer pipe = 74.45 Hz
Finally;
From the above equation; make Length L the subject of the formula. Then,
The length of the longer pipe is L = [tex]\dfrac{nv}{2f}[/tex]
The length of the longer pipe is L = [tex]\dfrac{1*343}{2*74.45}[/tex]
The length of the longer pipe is L = [tex]\dfrac{343}{148.9}[/tex]
The length of the longer pipe is L = 2.30 m
A 6.0 cm high object is placed 30.0 cm in front of a convex mirror. The radius of curvature of the mirror is 20.0 cm. How high is the image? Is it upright or inverted?
Photons — particles of light — stream toward the smooth pane of glass and bounce off it. The image of everything in front of the mirror is reflected backward, retracing the path it traveled to get there. Nothing is switching left to right or up-down. Instead, it's being inverted front to back.
When an object is placed in front of the convex mirror, the image is formed behind the mirror. The height of the image is 1.5 cm. The image is upright.
What is Convex mirror?Convex mirror is the spherical type of mirror. It has positive focal length. It forms the images by the virtual meeting of the rays coming from the object.
By the lens maker formula,
1/f = 1/u +1/v
where image distance is v, the object distance is 30 cm and the focal length of mirror, f = Radius of curvature /2
f = 20/2 = 10 cm
Substitute the values, we get the image distance as
1/10 = 1/-30 +1/v
v = 7.5 cm (on right of lens)
From the definition,
m = -image distance /object distance = image height / object height
-(7.5) /30 = hi / 6
hi = 1.5 cm
Thus, the height of the image is 1.5 cm. The image is upright.
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A quantity of an ideal gas is compressed to half its initial volume. The process may be adiabatic, isothermal or occurring at constant pressure. Rank those three processes in order of the work required of an external agent, least to greatest. A : adiabatic, isothermal, constant pressure B : constant pressure, isothermal, adiabatic C : adiabatic, constant pressure, isothermal D : isothermal, adiabatic, constant pressure E : constant pressure, adiabatic, isothermal
Answer:
B. constant pressure, isothermal, adiabatic
Explanation:
A quantity of an ideal gas is compressed to half its initial volume.
The process may be adiabatic, isothermal or occurring at constant pressure.
Adiabatic-constant heat
Constant pressure or isobaric
Isothermal or constant temperature
An external agent is a system that does work on a system or a machine.
This external agent applies force , or changes the state of the body it is acting on.
In order of the work required of an external agent, least to greatest
The following processes will be arranged.
constant pressure, isothermal, adiabatic
an ideal gas is confined to a container with adjustable volume. the number of moles, n, and temperature, t, are constant. by what factor will the volume change if pressure increase by a factor of 5.1
Answer:
The volume will decrease by a factor of 10/51.
Explanation:
Hello,
In this case, since both moles and temperature remain constant, we can use the Boyle's law that relates the volume and pressure as an inversely proportional relationship:
[tex]P_1V_1=P_2V_2[/tex]
Thus, since the pressure increases by a factor of 5.1 (statement), we have:
[tex]P_2=5.1P_1[/tex]
Thus, the final volume is:
[tex]V_2=\frac{P_1V_1}{P_2} =\frac{P_1V_1}{5.1P_1}\\\\V_2=\frac{10}{51}V_1[/tex]
It means that the volume will decrease by a factor of 10/51.
Regards.
The intensity of the waves from a point source at a distance d from the source is I. What is the intensity at a distance 2d from the source?
Answer:
The intensity at distance 2d from source is [tex]I_1 = \frac{1}{4} * I[/tex]
Explanation:
From the question we are told that
The distance of the wave from point source is d
The intensity is [tex]I[/tex]
The distance we are considering is 2d
Generally the intensity of a wave is mathematically represented as
[tex]I = \frac{ P }{\pi d^2 }[/tex]
Here P is power of point source
Now when d = 2d
[tex]I_1 = \frac{ P }{\pi (2d)^2 }[/tex]
[tex]I_1 = \frac{ 1 }{4 } * \frac{ P }{\pi d^2 }[/tex]
=> [tex]I_1 = \frac{1}{4} * I[/tex]
The intensity at a distance 2d from the source is equal to [tex]I'=\frac{I}{4}[/tex]
Given the following data:
Distance = dIntensity = ITo determine the intensity at a distance 2d from the source:
Mathematically, the intensity of a wave is given by the formula:
[tex]I=\frac{P}{\pi d^2}[/tex]
Where:
I is the intensity of a wave.P is the power.d is the distance.Since the distance is doubled (2d), we have:
Let the new intensity be [tex]I'[/tex][tex]I'=\frac{P}{\pi (2d)^2}\\\\I'=\frac{P}{4\pi (d)^2}\\\\I'=\frac{1}{4} \times \frac{P}{\pi (d)^2}\\\\I'=\frac{1}{4} \times I\\\\I'=\frac{I}{4}[/tex]
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Question 5 of 10
Heat is being transferred through currents within a liquid. When will this heat
transfer mostly end?
O A. When the substance changes state and becomes a gas
O B. When the entire liquid is a single temperature
O C. When the substance is very hot on top and cold beneath
O D. When the particles stop bumping into each other
SUBMIT
Answer:
When the entire liquid is a single temperature
Explanation:
When a liquid is heated, a convection current is set up. Convection is the movement of
fluid particles in response to a temperature gradient.
When you start heating a liquid, the particles near the base of the heating vessel increase in temperature, become less dense and rise upwards while the denser particles move downwards. This convection current will continue until an equilibrium temperature is obtained throughout the liquid.
Polarized light passes through a polarizer. If the electric vector of the polarized light is horizontal what, in terms of the initial intensity I0, is the intensity of the light that passes through a polarizer if the polarizer is tilted 22.5° from the horizontal?
Answer: I0*0.853
Explanation:
Ok, the Malus's law says that:
If you have light polarized along a given line with an intensity I0, and it passes through a polaroid which axis of polarization forms an angle θ with respect to the polarization of the light, then the intensity of the resulting beam is:
I(θ) = I0*cos^2(θ)
For example, if the axis of the polaroid is exactly the same as the axis of polarization of the light beam that will impact it, then we have θ = 0°, and the equation above says that the intensity of the beam will not change.
In this particular case, we have that the intensity of the light is I0, and the angle is θ = 22.5°
Then:
I(22.5°) = I0*cos^2(22.5°) = I0*0.853
4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest point? (b) How high does the ball rise? (c) How long after projection will the ball have a velocity of 10m/s upward? A velocity of 10m/s downward? (d) When is the displacement of the ball zero? (e) When is the magnitude of the ball’s velocity equal to half its velocity of projection? (f) When is the ball‘s displacement equal to half the maximum height to which it rises? (g) What is the magnitude and direction of the acceleration while the ball is moving upward? While moving downward? While at the highest point?
All the answers are:
a) The time that will it take to rise to the highest point is 3.06 seconds.
b) The ball will rise to a height of 45.87 meters.
c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.
The time when the ball has 10 m/s downward is 1.02 seconds.
d) The displacement of the ball will be zero at 6.12 seconds.
e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.
f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.
g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.
Let's calculate the values for each case.
a) At the highest point, the final velocity is 0, so we can use the following equation.
[tex]v_{f}=v_{i}-gt[/tex] (1)
Where:
v(i) is the initial velocityv(f) is the final velocityg is the acceleration due to gravity (9.81 m/s²)We know that v(i) = 30 m/s.
[tex]0=30-9.81t[/tex]
Solve it for t:
[tex]t=3.06\: s[/tex]
Hence, the time is 3.06 s.
b) At the highest point, the final velocity is 0, so we can use the following equation.
[tex]v_{f}^{2}=v_{i}^{2}-2gh[/tex] (2)
[tex]0=v_{i}^{2}-2gh[/tex]
We know that the initial velocity is 30 m/s.
[tex]0=30^{2}-2gh[/tex]
Solving it for h we have:
[tex]h=\frac{30^{2}}{2*9.81}[/tex]
[tex]h=45.87 \: m[/tex]
Then, the height is 45.87 m.
c) Using equation (1) we can find the time (t).
[tex]10=30-(9.81t)[/tex]
So, the time elapsed to get 10 m/s is:
[tex]t_{upward}=2.04\: s[/tex]
We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.
[tex]t_{upward}=2.04+t[/tex]
[tex]t=1.02\: s[/tex]
This is the time when the ball has 10 m/s downward.
Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.
d) It will be when the ball returns to the ground.
[tex]t=2t_{upward}[/tex]
[tex]t=2*3.06[/tex]
[tex]t=6.12\: s[/tex]
The displacement will be zero after 6.12 s.
e) Here we need to find the time when v(f) is 15 m/s
[tex]15=30-gt[/tex]
[tex]t=\frac{15}{9.81}[/tex]
[tex]t=1.53\: s[/tex]
The time when the v(f) is 15 m/s is 1.53 s.
f) Here, we need to find t when h = 45.87/2 m = 22.94 m
We can use the next equation:
[tex]h=v_{i}t-0.5gt^{2}/tex]
[tex]22.94=30t-0.5*9.81*t^{2}/tex]
Solving this quadratic equation, t will be:
[tex]t=0.90\: s/tex]
Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.
g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.
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Describe and name the different types of collision. In which are the linear momentum and kinetic energy conserved
Answer:
1. Elastic collision
2. Inelastic collision
Explanation:
Elastic collision: collision is said to be elastic if total kinetic energy is not conserved and if there is a rebound after collision
the collision is described by the equation bellow
[tex]m1U1+ m2U2= m1V1+m2V2[/tex]
Inelastic collision: this type of collision occurs when the total kinetic energy of a body is conserved or when the bodies sticks together and move with a common velocity
the collision is described by the equation bellow
[tex]m1U1+ m2U2= V(m1+m2)[/tex]
A force cannot exist without an agent and a system.
True
False
Answer:
true
Explanation:
forces require an agent
you should always be able to identify what (the agent) is producing the force
a model car moves round a circular path of radius 0.3m at 2 revolutions per secs what is its angular speed, the period of the car and the speed of the car
Answer:
a) T = 0.5 s
b) v = 1.2π m/s ≈ 3.77 m/s
Explanation:
It makes two revolutions in one second so makes one revolution in ½ second
circumference of the circle is
C = 2πr = 0.6π m
which it traverses in one time period
0.6π m / 0.5 s = 1.2π m/s
To solve this, we must be knowing each and every concept related to speed and its calculations. Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.
What is speed?Speed may be defined as the distance traveled by an item in the amount of time it requires to travel that distance. In other words, it measures how rapidly an item travels but does not provide direction.
Speed may be calculated in Science. The speed equation is a scientific formula that is used to calculate various types of speed.
Mathematically, the formula for speed can be given as
speed= distance/time
Values that are given
Time period= 0.5 s
Circumference = 2πr = 0.6π m
substituting all the given values in the above equation, we get
speed =0.6π m / 0.5 s
On calculations, we get
= 1.2π m/s
=3.77 m/s
Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.
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In the 25 ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity 2500 W/m2 at the floor of the facility.
A) Find the average radiation pressure (in pascals) on a totally absorbing section of the floor.B) Find the average radiation pressure (in atmospheres) on a totally absorbing section of the floor.C) Find the average radiation pressure (in pascals) on a totally reflecting section of the floor.D) Find the average radiation pressure (in atmospheres) on a totally reflecting section of the floor.
Answer:
a) 8.33 x 10^-6 Pa
b) 8.23 x 10^-11 atm
c) 1.67 x 10^-5 Pa
d) 1.65 x 10^-10 atm
Explanation:
Intensity of the light [tex]I[/tex] = 2500 W/m^2
speed of light [tex]c[/tex] = 3 x 10^8 m/s
a) we know that the pressure for for a totally absorbing surface is given as
[tex]P_{abs}[/tex] = [tex]I/c[/tex] = 2500/(3 x 10^8) = 8.33 x 10^-6 Pa
b) 1 atm = 101325 Pa
[tex]P_{abs}[/tex] = (8.33 x 10^-6)/101325 = 8.23 x 10^-11 atm
c) for a totally reflecting surface
[tex]P_{ref}[/tex] = [tex]2I/c[/tex] = twice the value for totally absorbing
[tex]P_{ref}[/tex] = 2 x 8.33 x 10^-6 = 1.67 x 10^-5 Pa
d) 1 atm = 101325 Pa
[tex]P_{ref}[/tex] = 2 x 8.23 x 10^-11 = 1.65 x 10^-10 atm
Nine tree lights are connected inparallel across 120-V potential difference. The cord to the wall socket carries a current of 0.43 A. Required:a. Detrmine the resistance of one of the bulbs.b. What would the current be if the bulbs were connected in series?
Answer:
a)3000ohm
b)4.44mA
Explanation:
a) we were given a Nine tree lights connected inparallel across 120-V potential difference, since the resistor are in parallel we use the expresion below
1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉
But according to ohm'law which can be expressed below
V=IR
R=V/I
R(total)= 120/0.36
= 333.33ohm
1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉
R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉
1/R(total)=9/R
1/333.33= 9/R
R= 3000ohm
Therefore, the resistance is 3000ohm
b)the bulbs were connected in series here, then for series connection we use below expression
R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉
R(total)=9R
= 9*3000
=27000ohm
I=VR
I=V/R
I= 120/27000
= 4.44*10⁻³A
4.44mA
Therefore, the current is 4.4mA
The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal contact and the combination is thermally insulated. The final temperature of both objects is
Answer:
The final temperature of both objects is 400 K
Explanation:
The quantity of heat transferred per unit mass is given by;
Q = cΔT
where;
c is the specific heat capacity
ΔT is the change in temperature
The heat transferred by the object A per unit mass is given by;
Q(A) = caΔT
where;
ca is the specific heat capacity of object A
The heat transferred by the object B per unit mass is given by;
Q(B) = cbΔT
where;
cb is the specific heat capacity of object B
The heat lost by object B is equal to heat gained by object A
Q(A) = -Q(B)
But heat capacity of object B is twice that of object A
The final temperature of the two objects is given by
[tex]T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}[/tex]
But heat capacity of object B is twice that of object A
[tex]T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K[/tex]
Therefore, the final temperature of both objects is 400 K.
Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. What is the effective "spring constant" of this simple harmonic motion?
Express your answer to three significant digits and include the appropriate units.
We have that the spring constant is mathematically given as
[tex]k=2.37*10^{11}N/m[/tex]
Generally, the equation for angular velocity is mathematically given by
[tex]\omega=\sqrt{k}{m}[/tex]
Where
k=spring constant
And
[tex]\omega =\frac{2\pi}{T}[/tex]
Therefore
[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]
Hence giving spring constant k
[tex]k=m((\frac{2 \pi}{T})^2[/tex]
Generally
Mass of earth [tex]m=5.97*10^{24}[/tex]
Period for on complete resolution of Earth around the Sun
[tex]T=365 days[/tex]
[tex]T=365*24*3600[/tex]
Therefore
[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]
[tex]k=2.37*10^{11}N/m[/tex]
In conclusion
The effective spring constant of this simple harmonic motion is
[tex]k=2.37*10^{11}N/m[/tex]
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Krishna and Seldon now try a homework problem. A policeman sitting in his unmarked police car sees an approaching motorcyclist go through a red light two blocks away. He turns on his siren at a frequency of 1000 Hz as the motorcyclist heads directly toward him at 61 mph (27.27 m/s). What frequency does the motorcyclist hear? (Enter your answer to at least the nearest integer. Assume the speed of sound in air is 331 m/s.) Hz What frequency does the motorcyclist hear when stopped with the police car approaching at 61 mph (27.27 m/s)? (Enter your answer to at least the nearest integer. Assume the speed of sound in air is 331 m/s.) Hz
Answer:
Explanation:
We shall apply formula of Doppler's effect
Here source is fixed and observer is approaching the source
f = f₀ x [(V + v ) / V ]
f₀ is original and f is apparent frequency , V is velocity of sound and v is velocity of motorcyclist .
f = 1000 x [(331 + 27.27 ) / 331 ]
= 1082 .4 Hz
This is the frequency heard by motorcyclist .
When police car is approaching him when he is stopped
f = f₀ x [V /(V - v ) ]
v is velocity of police car .
= 1000 x 331 / (331 - 27.27)
= 1090 Hz
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