Answer:
The average emf induce is [tex]V = 2.625 * 10^{-5} \ V[/tex]
Explanation:
From the question we are told that
The radius of the coil is [tex]r = 0.30 \ m[/tex]
The number of turns is [tex]N = 420 \ turns[/tex]
The frequency of the transition radio wave is [tex]f = 1.3\ MHz = 1.3 *10^{6} Hz[/tex]
The magnetic field is [tex]B_,{max} = 1.7 * 10^{-13} \ T[/tex]
The time taken for the magnetic field to go from zero to maximum is [tex]\Delta T = \frac{T}{4}[/tex]
The period of the transmitted radio wave is [tex]T = \frac{1}{f}[/tex]
So
[tex]\Delta T = \frac{T}{4} = \frac{1}{4 f}[/tex]
The potential difference can be mathematically represented as
[tex]V = NA (\frac{\Delta B}{\Delta T} )[/tex]
[tex]V = NA ([B_{max} - B_{min} ] * 4f)[/tex]
Where [tex]B_{min} = 0T[/tex]
substituting values
[tex]V = 420 * (\pi *(0.30)^2) * (1.7 *10^{-13} * 4 * 1.3 *10^{6})[/tex]
[tex]V = 2.625 * 10^{-5} \ V[/tex]
At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to help people avoid collisions. The magnitude of the mirror's radius of curvature is 0.560 m.
A) Locate the image of a patient10.6m from the mirror. B) Indicate whether the image is upright or inverted.C) Determine the magnification of the image.
Answer:
Explanation:
For a convex mirror, the value of its image distance and its focal length are negative.
using the mirror formula 1/f = 1/u+1/v
f is the focal length = Radius of curvature/2 = 0.560/2
f= 0.28m
u is the object distance = 10.6m
v is the position of the image = ?
On substitution;
1/0.28 = 1/10.6 + 1/-v
3.57 = 0.094 - 1/v
3.57 - 0.094 = -1/v
3.476 = -1/v
v = -1/3.476
v = -0.2877m
B) Since the image distance is negative, this means that the image is an upright and a virtual image. All Upright images has their image distance to be negative.
C) Magnification = Image distance/object distance
Magnification = 0.2877/10.6
Magnification = 0.0271
Which statement BEST explains the relationship between voltage, current, and power?
A. If voltage increases and everything else remains constant, then power will increase.
B. If voltage increases and everything else remains constant, then power will decrease.
C. If current decreases and everything else remains constant, then power will increase.
D. Voltage and power are inversely related.
A) In the figure below, a cylinder is compressed by means of a wedge against an elastic constant spring = 12 /. If = 500 , determine what the minimum compression in the spring will be so that the pad does not move. Disregard the weight of the blocks and . The coefficient of friction between and the pad and between the floor and the pad is s = 0.4. Consider that the friction between the cylinder and the vertical walls is negligible
Answer: 4.08 cm.
B) Determine the lowest force required to lift the weight of 750 . The static coefficient of friction between and and between and is s= 0.25, and between and is 's = 0.5. Disregard the weight of the shims and .
Answer : 1095.4 N.
Explanation:
A) Draw free body diagrams of both blocks.
Force P is pushing right on block A, which will cause it to move right along the incline. Therefore, friction forces will oppose the motion and point to the left.
There are 5 forces acting on block A:
Applied force P pushing to the right,
Normal force N pushing up and left 10° from the vertical,
Friction force Nμ pushing down and left 10° from the horizontal,
Reaction force Fab pushing down,
and friction force Fab μ pushing left.
There are 2 forces acting on block B:
Reaction force Fab pushing up,
And elastic force kx pushing down.
(There are also horizontal forces on B, but I am ignoring them.)
Sum of forces on A in the x direction:
∑F = ma
P − N sin 10° − Nμ cos 10° − Fab μ = 0
Solve for N:
P − Fab μ = N sin 10° + Nμ cos 10°
P − Fab μ = N (sin 10° + μ cos 10°)
N = (P − Fab μ) / (sin 10° + μ cos 10°)
Sum of forces on A in the y direction:
N cos 10° − Nμ sin 10° − Fab = 0
Solve for N:
N cos 10° − Nμ sin 10° = Fab
N (cos 10° − μ sin 10°) = Fab
N = Fab / (cos 10° − μ sin 10°)
Set the expressions equal:
(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)
Cross multiply:
(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)
Distribute and solve for Fab:
P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)
P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)
Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)
Sum of forces on B in the y direction:
∑F = ma
Fab − kx = 0
kx = Fab
x = Fab / k
x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))
Plug in values and solve.
x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))
x = 0.0408 m
x = 4.08 cm
B) Draw free body diagrams of both blocks.
Force P is pushing block A to the right relative to the ground C, so friction force points to the left.
Block A moves right relative to block B, so friction force on A will point left. Block B moves left relative to block A, so friction force on B will point right (opposite and equal).
Block B moves up relative to the wall D, so friction force on B will point down.
There are 5 forces acting on block A:
Applied force P pushing to the right,
Normal force Fc pushing up,
Friction force Fc μ₁ pushing left,
Reaction force Fab pushing down and left 15° from the vertical,
and friction force Fab μ₂ pushing up and left 15° from the horizontal.
There are 5 forces acting on block B:
Weight force 750 n pushing down,
Normal force Fd pushing left,
Friction force Fd μ₁ pushing down,
Reaction force Fab pushing up and right 15° from the vertical,
and friction force Fab μ₂ pushing down and right 15° from the horizontal.
Sum of forces on B in the x direction:
∑F = ma
Fab μ₂ cos 15° + Fab sin 10° − Fd = 0
Fd = Fab μ₂ cos 15° + Fab sin 15°
Sum of forces on B in the y direction:
∑F = ma
-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0
Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750
Substitute:
(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750
Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750
Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750
Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)
Sum of forces on A in the y direction:
∑F = ma
Fc + Fab μ₂ sin 15° − Fab cos 15° = 0
Fc = Fab cos 15° − Fab μ₂ sin 15°
Sum of forces on A in the x direction:
∑F = ma
P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0
P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁
Substitute:
P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁
P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°
P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)
First, find Fab using the given values.
Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)
Fab = 1151.9 N
Now, find P.
P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)
P = 1095.4 N
A swimmer heading directly through a 200m wide river reaches the opposite shore in 6 min 40s. She is washed downstream 480 m. How fast can you swim in calm water?
Answer :v=480m400s=1.2ms
2002+4802=H2
The hypotenuse H=520m
A quicker way to get the length of the hypotenuse is to recognize that this is a simple 5–12–13 triangle where the sides are multiples of 5, 12, and 13:
5(40) = 200m, 12(40)= 480m, 13(40)= 520m
We know that the swimmer travelled 520 m in 400 seconds, so her average speed was:
VR=520m400sec= 1.3ms
hope i got it right!! xx
Explanation:
Which of the followings is true about EMF?
a. an induced emf is caused by a changing magnetic flux.
b. an emf can only be induced in a conducting loop by moving the loop through an area that has a constant magnetic field.
c. an induced emf can be observed by measuring the current that is created.
d. an induced emf and conventional induced current are in opposite directions.
Answer:
a. TRUTH
b. FALSE
c. TRUTH
d. FALSE
Explanation:
The emf (electromagnetic force) is generated in a loop or solenoid by the change in the magnetic flux in a closed conductor path (for example, a wire).
This can be noted in the following formula, which is known as the Lenz's law:
[tex]emf=-N\frac{d\Phi_B}{dt}=-N\frac{d(AB)}{dt}[/tex] (1)
Then, the change, in time, of the area of the conductor, or the change in the magnitude of the magnetic field, the induced emf acquires different values. Furthermore, the loops have a resistance, then, a current can be measured when an emf is induced.
Based on this information you have:
a. an induced emf is caused by a changing magnetic flux. TRUTH
b. an emf can only be induced in a conducting loop by moving the loop through an area that has a constant magnetic field. FALSE
c. an induced emf can be observed by measuring the current that is created. TRUTH
d. an induced emf and conventional induced current are in opposite directions. TRUTH (the minus sing in the equation (1) )
A large box containing your new computer sits on the bed of your pickup truck. You are stopped at a red light. When the light turns green, you stomp on the gas and the truck accelerates. To your horror, the box starts to slide toward the back of the truck. Draw clearly labeled free-body diagrams for the truck and for the box. Indicate pairs of forces, if any, that are third-law action–reaction pairs. (The horizontal truck bed is not frictionless.)
Answer:
The description of that same situation has been listed throughout the explanation segment below.
Explanation:
When another huge box or container containing your new machine or device sits on someone's pick-up truck's bed, the third low portion of the operation response force. This same friction force of the box mostly on the truck bed as well as the friction force including its truck bed on either the box from either the immune response pair.So that the above seems to be the right answer.
Scenario 2: Use the following information to answer questions 3 and 4:
Your client, Jim, is interested in weight control. He weighs 75kg.
3. If Jim walks 3.3 mph (0% grade), how long must he walk to expend 300 kcal total?
A. 52 min
B. 42 min
C. 65 min
D. 99 min
4. If Jim exercises at an intensity of 6 kcal/min, what is the leg ergometer work rate?
A. 47 watts
B. 90 watts
C. 61 watts
D. 71 watts
Answer:
A. 52 min
.A. 47 watts
Explanation:
Given that;
jim weighs 75 kg
and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.
Using the following relation to determine the amount of calories burned per minute while walking; we have:
[tex]\dfrac{MET*weight (kg)*3.5}{200}[/tex]
here;
MET = energy cost of a physical activity for a period of time
Obtaining the data for walking with a speed of 3.3 mph From the standard chart for MET, At 3.3 mph; we have our desired value to be 4.3
However;
the calories burned in a minute = [tex]\dfrac{4.3*75 (kg)*3.5}{200}[/tex]
= 5.644
Therefore, for walking for 52 mins; Jim burns approximately 293.475 kcal which is nearest to 300 kcal.
4.
Given that:
mass m = 75 kg
intensity = 6 kcal/min
The eg ergometer work rate = ??
Applying the formula:
[tex]V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7[/tex]
where ;
[tex]V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}[/tex]
[tex]V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}[/tex]
[tex]V_O_2 ( intensity ) = 0.0012[/tex]
∴[tex]0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min[/tex]
Converting to watts;
Since; 6.118kg-m/min is = 1 watt
Then 291.66 kgm /min will be equal to 47.67 watts
≅ 47 watts
Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section), and the power fit of your Trendline equation,calculate the drag coefficient. Solve for it first (see video) and then plug in the values.
Answer:
The drag coefficient is [tex]D_z = 1.30512[/tex]
Explanation:
From the question we are told that
The density of air is [tex]\rho_a = 1.21 \ kg/m^3[/tex]
The diameter of bottom part is [tex]d = 0.15 \ m[/tex]
The power trend-line equation is mathematically represented as
[tex]F_{\alpha } = 0.9226 * v^{0.5737}[/tex]
let assume that the velocity is 20 m/s
Then
[tex]F_{\alpha } = 0.9226 * 20^{0.5737}[/tex]
[tex]F_{\alpha } = 5.1453 \ N[/tex]
The drag coefficient is mathematically represented as
[tex]D_z = \frac{2 F_{\alpha } }{A \rho v^2 }[/tex]
Where
[tex]F_{\alpha }[/tex] is the drag force
[tex]\rho[/tex] is the density of the fluid
[tex]v[/tex] is the flow velocity
A is the area which mathematically evaluated as
[tex]A = \pi r^2 = \pi \frac{d^2}{4}[/tex]
substituting values
[tex]A = 3.142 * \frac{(0.15)^2}{4}[/tex]
[tex]A = 0.0176 \ m^2[/tex]
Then
[tex]D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }[/tex]
[tex]D_z = 1.30512[/tex]
A tank circuit consists of an inductor and a capacitor. Give a simple explanation for why the magnetic field in the induc- tor is strongest at the moment that the separated charge in the capacitor reaches zero.
Answer:
If you pull a permanent magnet rapidly away from a tank circuit, what is likely to happen in that circuit?
Charge will oscillate in the tank's capacitor and inductor.
Explanation:
02
Blue light has a frequency of about 7.5 x 1014 Hz. Calculate the energy, in Joules, of a single photon associated with this frequency
Answer:
49.725× 10^-24J
Explanation:
The Energy associated with a Photon us defined as;
E = hf
Where h is Planck's constant = 6.63× 10^-34m2kg/s
f is the frequency= 7.5 x 10^14 Hz
Hence
E = 6.63× 10^-34 × 7.5 x 10^14 =49.725× 10^-24J
A mass m at the end of a spring vibrates with a frequency of 0.72 Hz . When an additional 700 g mass is added to m, the frequency is 0.64 Hz . Part A What is the value of m? Express your answer using two significant figures.
Answer:
The value of m is 2635.294 grams.
Explanation:
Let suppose that mass-spring system has a simple harmonic motion, to this respect the formula for frequency is:
[tex]f = \frac{\omega}{2\pi}[/tex]
Where [tex]\omega[/tex] is the angular frequency, measured in radians per second.
For a mass-spring system under simple harmonic motion, the angular frequency is:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
Where:
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m[/tex] - Mass, measured in kilograms.
The following equation is obtained after replacing angular frequency in frequency formula:
[tex]f = \frac{1}{2\pi}\cdot \sqrt{\frac{k}{m} }[/tex]
As this shows, frequency is inversely proportional to the square root of mass. Hence, the following relationship is deducted:
[tex]f_{1}\cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{2}}[/tex]
If [tex]m_{2} = m_{1} + 700\,g[/tex], [tex]f_{1} = 0.72\,hz[/tex] and [tex]f_{2} = 0.64\,hz[/tex], the resulting expression is simplified and then initial mass is found after clearing it:
[tex]f_{1} \cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{1}+700\,g}[/tex]
[tex]f_{1}^{2} \cdot m_{1} = f_{2}^{2}\cdot (m_{1} + 700\,g)[/tex]
[tex]\left(\frac{f_{1}}{f_{2}} \right)^{2}\cdot m_{1} = m_{1} + 700\,g[/tex]
[tex]\left[\left(\frac{f_{1}}{f_{2}}\right)^{2} - 1\right]\cdot m_{1} = 700\,g[/tex]
[tex]m_{1} = \frac{700\,g}{\left(\frac{f_{1}}{f_{2}} \right)^{2}-1}[/tex]
[tex]m_{1} = \frac{700\,g}{\left(\frac{0.72\,hz}{0.64\,hz} \right)^{2}-1}[/tex]
[tex]m_{1} = 2635.294\,g[/tex]
The value of m is 2635.294 grams.
World religions: Shinto
Most Shinto rituals are tied to
A) worshiping the kami.
B) the life-cycle of humans and the seasonal cycles of nature.
C) forgiveness of sins.
D) preparing for the afterlife.
A 148 g ball is dropped from a tree 11.0 m above the ground. With what speed would it hit the ground
Answer:
14.68m/s
Explanation:
As per the question, the data provided is as follows
Mass = M = 0.148 kg
Height = h = 11 m
Initial velocity = U = 0 m/s
Final velocity = V
Gravitational force = F
Mass = M
Based on the above information, the speed that hit to the ground is
As we know that
Work to be done = Change in kinetic energy
[tex]F ( S) = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]
[tex]M g h = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]
[tex]g h = (\frac{1}{2} ) ( V^2 - U^2 )[/tex]
[tex]V^2 - U^2 = 2gh[/tex]
[tex]V^2 - 0 = 2gh[/tex]
[tex]V = \sqrt{2 g h}[/tex]
[tex]= \sqrt{2\times9.8\times11}[/tex]
= 14.68m/s
What type of device forms images by changing the speed at which light travels?
Answer:
A lens
Explanation:
A lens forms images when light passes Through it bending the rays of in the process.A phenomena called refraction and the speed of light changes in the process because it enters a medium since it's wavelength is reduced.
The type of device that forms images by changing the speed at which light travels is the lens.
What is refraction through the lens?
A lens bends a light beam at an aimed perspective and converges or diffuses bundles of rays by taking benefit of refraction taking vicinity while the mild travels from air into glass or plastic. For that purpose, the aspect geometry of a lens has a spherical parent, which can be kind of divided into sorts.
A lens bends a mild beam at an aimed perspective and converges or diffuses bundles of rays through taking gain of refraction taking area whilst the mild travels from air into glass or plastic. For that motive, the facet geometry of a lens has a round parent, which may be kind of divided into sorts.
Learn more about the speed of light here:-https://brainly.com/question/104425
#SPJ2
Michelson and Morley's experiment is widely considered to have been:______
a. a success because it detected a shift in the interference pattern.
b. a failure because it detected a shift in the interference pattern.
c. a success because it did not detect a shift in the interference pattern.
d. a failure because it did not detect a shift in the interference pattern.
e. lacking the necessary precision to determine a shift in the interference pattern.
Answer:
The correct answer is option (c) a success because it did not detect a shift in the interference pattern.
Explanation:
In Michelson and Morley experiment it was considered to be successful.
They both found out that the experiment that was carried out was not a failure since it did not detect any shift in the interference pattern.
With this findings it was widely regarded as correct and precise.
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 165 cmcm , but its circumference is decreasing at a constant rate of 14.0 cm/scm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.800 TT , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
(a) Find the emf induced in the loop at the instant when 9.0 s have passed.
(b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Answer:
(a) emf = 1.18 mV
(b) counter-clockwise sense
Explanation:
(a) The induced emf is given by the following formula:
[tex]emf=-\frac{d\Phi_B}{dt}[/tex] (1)
where:
ФB: magnetic flux = AB = (area of the loop)*(magnitude of the magnetic field)
A = πr^2
B = 0.800 T
You replace the expression for the magnetic flux in the equation (1):
[tex]emf=-B\frac{\Delta A}{\Delta t}=-B\frac{A_2-A_1}{t_2-t_1}[/tex]
A1: initial area
A2: final area
t2-t1: time interval = 9.0s
Then you have to calculate the change in the area of the loop, by using the information about the circumference of the loop. First you calculate the radius of the loop for a circumference of 165 cm = 1.65m
[tex]s=1.65m=2\pi r\\\\r=\frac{1.65m}{2\pi}=0.262m[/tex]
You calculate the initial area A1:
[tex]A_1=\pi (0.262m)^2=0.215m^2[/tex]
After 9.0 second the circumference will be:
[tex]s'=1.65m-0.14\frac{m}{s}(9.0s)=0.39m[/tex]
the new radius and the final area is:
[tex]r=\frac{0.39m}{2\pi}=0.062m[/tex]
[tex]A_2=\pi(0.062m)^2=0.012m^2[/tex]
Finally, you replace in the equation (1):
[tex]emf=-(0.800T)\frac{0.012m^2-0.215m^2}{9.0s}=1.8*10^{-3}V=1.8mV[/tex]
The induced emf in the circular loop is 1.18mV
(b) The induced emf generates an electric current, which produces a magnetic field that is opposite to the direction of the constant magnetic field of 0.800T. Due to this magnetic field point into the loop. The current has to have a direction in a counter-clockwise sense.
You rub a balloon on your head and it becomes negatively charged. The balloon will be most attracted to what?
Answer:
To things that are positive charged
An airplane takes off a runway at a constant speed of 49m/s at constant angle 30 to the horizontal
Complete Question
An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters ) is the airplane above the ground 13 seconds after takeoff?
Answer:
The height is [tex]H = 318.5 \ m[/tex]
Explanation:
From the question we are told that
The speed at which the plane takes off is [tex]u = 49 \ m/s[/tex]
The angle at which it takes off is [tex]\theta = 30 ^o[/tex]
The time taken is [tex]t = 13 s[/tex]
The vertical distance traveled is mathematically represented as
[tex]H = u sin \theta t[/tex]
Substituting values
[tex]H = (49) * sin (30) *13[/tex]
[tex]H = 318.5 \ m[/tex]
Someone please helppppppp!!!!!
Space-faring astronauts cannot use standard weight scales (since they are constantly in free fall) so instead they determine their mass by measuring the period of oscillation when sitting in a chair connected to a spring. Suppose a chair is connected to a spring with a spring constant of 600 N/m. If the empty chair oscillates with a period of 0.9s, what is the mass of an astronaut who oscillates with a period of 2.0 s while sitting in the chair
Answer:
ma = 48.48kg
Explanation:
To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:
[tex]T=2\pi\sqrt{\frac{m_c}{k}}[/tex] (1)
mc: mass of the chair
k: spring constant = 600N/m
T: period of oscillation of the chair = 0.9s
You solve the equation (1) for mc, and then you replace the values of the other parameters:
[tex]m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg[/tex] (2)
Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:
T': period of chair when the astronaut is sitting = 2.0s
M: mass of the astronaut plus mass of the chair = ?
[tex]T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg[/tex] (3)
Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :
[tex]m_a=M-m_c=60.79kg-12.31kg=48.48kg[/tex]
The mass of the astronaut is 48.48 kg
wha is amplitde in sound
Answer:
The number of molecules displaced in a vibration makes the amplitude of a sound.
g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber, and from the bottom hangs a 10 gram lead weight. The density of lead is 11.3 g/cm3. What fraction of the bobber's volume is submerged, as a percent of the total volume
Answer:
Explanation:
total weight acting downwards
= 3g + 10g
13 g
volume of lead = 10 / 11.3 = .885 cm³
Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g
Buoyant force on lead = .885 x 1 x g
total buoyant force = vg + .885 g
For floating
vg + .885 g = 13 g
v = 12.115 cm³
total volume of bobber
= 4/3 x 3.14 x 2³
= 33.5 cm³
fraction of volume submerged
= 12.115 / 33.5
= .36
= 36 %
The fraction of the bobber's volume submerged as a percent of the total volume is 36.2 %.
The given parameters;
diameter of the bobber, d = 4 cmmass of the bobber, m = 3 gmass of the lead, m = 10 gdensity of the lead, ρ = 11.3 g/cm³The volume of the bobber is calculated as follows;
[tex]V = \frac{4}{3} \pi \times r^3\\\\V = \frac{4}{3} \pi \times (2)^3\\\\V = 33.52 \ cm^3[/tex]
The buoyant force experienced by the bobber due to the volume submerged is calculated as follows;
[tex]F _b= \rho Vg\\\\F_b = 1 \times V \times g\\\\F_b = Vg[/tex]
The volume of the lead is calculated as follows;
[tex]V = \frac{mass}{density} \\\\V = \frac{10}{11.3} \\\\V = 0.885 \ cm^3[/tex]
The buoyant force experienced by the lead due to the volume submerged is calculated as follows
[tex]F_b = \rho Vg\\\\F_b = 0.885 g[/tex]
The total buoyant force is calculated as;
[tex]Vg + 0.885g = (3+ 10)g\\\\g(V + 0.885) = 13g\\\\V+ 0.885 = 13\\\\V = 13 -0.885\\\\V = 12.12 \ cm^3[/tex]
The fraction of the bobber's volume submerged as a percent of the total volume is calculated as follows;
[tex]= \frac{12.12}{33.52} \times 100\%\\\\= 36.2 \ \%[/tex]
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Tech A says that as engines gain miles, the spark plug gap increases, which raises the ignition system’s available voltage. Tech B says that misfire occurs when required voltage is higher than available voltage. Who is correct? Group of answer choices
Answer: Tech A is correct
Explanation:
Every vehicle has ignition system and without this system,it will not work. The battery of everything vehicle contain energy that start the vehicle and ignore it to start working. Electrical current move from the vehicle's battery and get to the induction coil, the induction coil increases the voltage in it so that the plug will be ignited. The spark plugs produce fire. The spark plug is connected to the ignition system. Once voltage is produced from the induction coil, electrical impulses move from induction coil to insulated plug wires. The spark plug need a very high voltage from the small voltage battery. Once the high voltage exceed the dielectric strength of the gases, spark jump the gap between the plug's fire end.
The Nardo ring is a circular test track for cars. It has a circumference of 12.5km. Cars travel around the track at a constant speed of 100km/h. A car starts at the easternmost point of the ring and drives for 15 minutes at this speed.
1. What distance, in km, does the car travel?
2. What is the magnitude of the car's displacement, in km, from its initial position?
3. What is the speed of the car in m/s?
Answer:
1. 25 Km
2. zero
3. 27.7 m/s
Explanation:
Data provided in the question:
Circumference of the track = 12.5 km
Speed of the car = 100 Km/h
Time for which car travels = 15 minutes = [tex]\frac {15}{60}[/tex] hr
Now,
1. Distance traveled = Speed × Time
= 100 × [tex]\frac{15}{60}[/tex]
= 25 Km
2. The distance traveled is 2 times the circumference of the track (i.e 2 × 12.5 = 25 Km)
Which means that the car is again at the initial position
Therefore, The displacement is zero.
3. Speed of car in Km/hr = 100 Km/h
now,
1 Km = 1000 m
1 hr = 3600 seconds
therefore,
100 Km/h = [tex]100\times\frac{1000}{3600}[/tex] m/s
= 27.7 m/s
Hence, the speed of car in m/s = 27.7
As the temperature of a medium increases, the speed of the sound wave ....
Answer:
Increases
Explanation:
Due to an increase in temperature, molecules within the medium will vibrate more vigorously, meaning that the rate of chemical reactions generally increases with temperature due to an increase in kinetic energy. Because sound is a form of kinetic energy, it is safe to assume that the speed of sound waves increases with temperature.
Answer:
A- increases because The particles bump into each other more often.
Explanation:
Just took the test
A constant force applied to object A causes it to accelerate at 5 m/s2. The same force applied to object B causes an acceleration of 3 m/s2. Applied to object C, it causes an acceleration of 7 m/s2.
A. Which object has the largest mass?B. Which object has the smallest mass?C. What is the ratio of mass A to mass B?
Answer:
(A) object B has the largest mass because it has the least acceleration
(B) object C has the smallest mass because it has the largest acceleration
(C) mass A : mass B = 3 : 5
Explanation:
Given;
acceleration of object A = 5 m/s²
acceleration of object B = 3 m/s²
acceleration of object C = 7 m/s²
A constant force, F
According to Newton's second law of motion;
F = ma
m = F / a
Mass of object A:
m = F / 5
Mass of object B:
m = F / 3
Mass of object C:
m = F / 7
(A). Which object has the largest mass:
object B has the largest mass because it has the least acceleration
(B). Which object has the smallest mass:
object C has the smallest mass because it has the largest acceleration
(C). What is the ratio of mass A to mass B;
mass A = F / 5
mass B = F / 3
[tex]mass \ A : \ mass \ B = \frac{F}{5} : \frac{F}{3} \\\\\frac{mass \ A}{mass \ B} = \frac{F}{5} * \frac{3}{F}= \frac{3}{5} \\\\mass \ A : \ mass \ B = 3: 5[/tex]
A. The Object B has largest mass.
B. The Object A has smallest mass.
C. The ratio of mass A to mass B is, [tex]\frac{3}{5}[/tex]
Newton second law of motion:The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
[tex]F=ma\\\\m=\frac{F}{a}[/tex]
For constant force, mass is inversely proportional to acceleration of object.Given that, acceleration of object A is [tex]5m/s^{2}[/tex] and object B is [tex]3m/s^{2}[/tex]Thus, Object B has largest mass.Object A has smallest mass.the ratio of mass A to mass B is,[tex]\frac{m_{A}}{m_{B}} =\frac{a_{B}}{a_{A}} =\frac{3}{5}[/tex]
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Carbon is added to iron to make steel. Steel is stronger than either carbon or iron by itself.
What does this example show?
Answer:
This example shows that alloys are stronger than either of it's parent materials by themselves.
Explanation:
Since carbon is added to iron to make steel, it means steel is an alloy of iron and carbon.
This is because an alloy is a mixture of two or more elements, where at least one element is a metal.
Now, steel is stronger than either carbon or or iron by itself because Steel contains atoms of other elements including carbon and iron. These atoms have different sizes to iron carbon atoms, so they distort the layers of atoms in the pure iron and carbon. This means that a greater force is required for the layers to slide over each other in steel, so steel is harder than pure iron.
how does the statement " silence is golden " relate to ethics in communicating at the workplace.?
Answer:
Being silent most of the time is a good virtue under certain circumstances and environment. It is always advisable to remain quite silent and not be too quick to respond to situations or issues so as to avoid making and saying wrong words.
The ethics in a workplace involves communicating with others with less amount of talking as possible and more of body languages and signs. This is because the workplace is meant to be a serene place.
A uniformly charged ring of radius 10.0 cm has a total charge of 71.0 μC. Find the electric field on the axis of the ring at the following distances from the center of the ring. (Choose the x-axis to point along the axis of the ring.)
(a) 1.00 cm
What is the general expression for the electric field along the axis of a uniformly charged ring? i MN/C
(b) 5.00 cm
i MN/C
(c) 30.0 cm
i MN/C
(d) 100 cm
i MN/C
Answer:
General Expression: E = kql/(l² + r²)^(3/2)
(a) 6.3 MN/C
(b) 22.8 MN/C
(c) 6.1 MN/C
(d) 0.63 MN/C
Explanation:
The general expression for electric field along axis of a uniformly charged ring is:
E = kqL/(L² + r²)^(3/2)
where,
E = Electric Field Strength = ?
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q = Total Charge = 71 μC = 71 x 10⁻⁶ C
L = Distance from center on axis
r = radius of ring = 10 cm = 0.1 m
(a)
L = 1 cm = 0.01 m
Therefore,
E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.01 m)/[(0.01 m)² + (0.1 m)²]^(3/2)
E = (6390 N.m³/C)/(0.00101 m³)
E = 6.3 x 10⁶ N/C = 6.3 MN/C
(b)
L = 5 cm = 0.05 m
Therefore,
E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.05 m)/[(0.05 m)² + (0.1 m)²]^(3/2)
E = (31950 N.m³/C)/(0.00139 m³)
E = 22.8 x 10⁶ N/C = 27.4 MN/C
(c)
L = 30 cm = 0.3 m
Therefore,
E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.3 m)/[(0.3 m)² + (0.1 m)²]^(3/2)
E = (191700 N.m³/C)/(0.03162 m³)
E = 6.1 x 10⁶ N/C = 6.1 MN/C
(d)
L = 100 cm = 1 m
Therefore,
E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(1 m)/[(1 m)² + (0.1 m)²]^(3/2)
E = (639000 N.m³/C)/(1.015 m³)
E = 0.63 x 10⁶ N/C = 0.63 MN/C
A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?
Answer:
(a) q = 2.357 x 10⁻⁵ C
(b) Φ = 2.66 x 10⁶ N.m²/C
Explanation:
Given;
diameter of the sphere, d = 1.1 m
radius of the sphere, r = 1.1 / 2 = 0.55 m
surface charge density, σ = 6.2 µC/m²
(a) Net charge on the sphere
q = 4πr²σ
where;
4πr² is surface area of the sphere
q is the net charge on the sphere
σ is the surface charge density
q = 4π(0.55)²(6.2 x 10⁻⁶)
q = 2.357 x 10⁻⁵ C
(b) the total electric flux leaving the surface of the sphere
Φ = q / ε
where;
Φ is the total electric flux leaving the surface of the sphere
ε is the permittivity of free space
Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)
Φ = 2.66 x 10⁶ N.m²/C
