Answer:I know the answer for B cus I’m doing the same problem. For B, you would only take the coefficient of friction given and then multiply it by the Normal Force, which in this case is the same as the Gravitational Force.
Explanation:
A diesel engine lifts the 225 kg hammer of a pile driver 20 m in 5 seconds. How much work is done on
the hammer? What is the power?
Answer:
a. Workdone = 44100 Joules
b. Power = 8820 Watts.
Explanation:
Given the following data:
Mass = 225kg
Distance = 20m
Time = 5 seconds
To find the workdone;
Workdone = force * distance
But force = mg
We know that acceleration due to gravity is equal to 9.8m/s²
Force = 225*9.8 = 2205N
Substituting the values into the equation, we have;
Workdone = 2205 * 20
Workdone = 44100 Joules
b. To find the power;
Power = workdone/time
Power = 44100/5
Power = 8820 Watts.
The tray dispenser in your cafeteria has broken and is not repairable. The custodian knows that you are good at design-ing things and asks you to help him build a new dispenser out of spare parts he has on his workbench. The tray dispenser supports a stack of trays on a shelf that is supported by four springs, one at each corner of the shelf. Each tray is rectangu-lar, with dimensions 45.3 cm by 35.6 cm. Each tray is 0.450 cm thick and has a mass of 580 g. The custodian asks you to design a new four-spring dispenser such that when a tray is removed, the dispenser pushes up the remaining stack so that the top tray is at the same position as the just-removed tray was. He has a wide variety of springs that he can use to build the dispenser. Which springs should he use
Answer:
you have to find 4 spring with this elastic constant k = 316 N / m
Explanation:
In this case for the design of the dispenser the four springs are placed in the four corner at the bottom, therefore we can use the translational equilibrium relationship
4 F_e -W = 0
where the elastic force is
F_e = k x
we substitute
4 kx = mg
k = [tex]\frac{mg}{4x}[/tex]
Each tray has a thickness of x = 0.450 cm = 0.450 10⁻² m, this should be the elongation of the spring so that when the tray is in position it will remain fixed.
let's calculate
k = [tex]\frac{0.580 \ 9.8}{4 \ 0.450 \ 10^{-2} }[/tex]
k = 3.1578 10² N / m
k = 316 N / m
therefore you have to find 4 spring with this elastic constant
what are the types of energy sources based on
time of replacement ? write down their names
Answer:
solar energy
wind power
geothermal energy
hydraulic power
biomass energy
energy storage
(That's all I know).
Imagin you have mixed together some sand and salt Based on the venn diagram this mixture would be placed where
Answer:
a
Explanation:
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help please i will mark brainlist!!!
Answer:
.50 M
Explanation:
5*.50=2.5 + 2*.25=.5 = 3n
6*.50= 3N
Final answer is .50M
A spring with a constant of 76 N/m is extended by 0.9 m. How much energy is stored in the extended spring?
Answer:
[tex]E=30.78\ J[/tex]
Explanation:
The force constant of the spring, k = 76 N/m
The extension in the spring, x = 0.9 m
We need to find the energy is stored in the extended spring. The energy stored in the spring is given by :
[tex]E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 76\times (0.9)^2\\\\E=30.78\ J[/tex]
So, 30.78 J of energy is stored in the spring.
What is a black hole's escape velocity?
The simplest definition of a black hole is an object that is so dense that not even light can escape its surface. If we squished the Earth's mass into a sphere with a radius of 9 mm, the escape velocity would be the speed of light. Just a wee-bit smaller, and the escape velocity is greater than the speed of light.
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A 50kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3m/s. What was the force acting on the mass?
Answer:
75N
Explanation:
a = v/t = 3/2
F = ma = 50(3/2) = 75
Which phrase describes velocity?
u
A. A quantity with direction only
B. A quantity with magnitude only
C. A quantity with no units
D. A quantity with magnitude and direction
SUBMI
What formula could be used to find distance if you know the speed an the time
Answer: d = st
Explanation:
We know that the distance is equal to the rate (speed) times the time
d = st
1. If a wave has a wavelength of 5.5m and a frequency of 45hz, what is its speed?
Answer:
By using the most simple velocity equation, velocity = distance / time, meaning the speed would be 247.5 meters per second.
List down the types of centripetal force?
Answer: Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force.
Answer:
roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge
Explanation:
Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1, the ball bounces off a cement floor, and in Case 2, the ball bounces off a piece of stretchy rubber. Two cases of a ball dropping and bouncing off of a surface. In case one, the ball approaches a cement floor, comes in contact, momentarily comes to rest, and then bounces off with a velocity smaller than the approach velocity. In case 2, the ball approaches a piece of stretchy rubber. When the ball comes in contact with the rubber, it deforms the rubber and comes momentarily to rest. The ball bounces off again with a velocity smaller than the approach velocity, and the rubber regains its original shape. In both cases, the balls are dropped from the same initial height and reach the same final height. In which case is the magnitude of the ball's change in momentum the greatest
Answer:
the impulse must be the same in these two cases F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
Explanation:
For this exercise we use the relationship between momentum and momentum
I = Δp
F t = m v_f - m v₀
To know the speed we use the conservation of energy
starting point. Highest point
Em₀ = U = m g h
fincla point. Just before the crash
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
we substitute in the impulse relation
F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases
A simple pendulum is used to measure gravity using the following theoretical equation,TT=2ππ�LL/gg ,where L is the length of the pendulum, g is gravity, andT is the period of pendulum.Twenty measurements of T give a mean of 1.823 seconds and a standard deviation of 0.0671 s. The device used to measure time has a resolution of 0.02 s. The pendulum length is measured once to be 0.823 m (with a scale having a resolution of 0.001 m). Determine the value of g and its uncertainty (assume 90% confidence where necessary). You may use any method of uncertainty propagation that we covered in class.
Answer:
g ±Δg = (9.8 ± 0.2) m / s²
Explanation:
For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use
T = [tex]2\pi \sqrt{ \frac{L}{g} }[/tex]
T² = [tex]4\pi ^2 \frac{L}{g}[/tex]4pi2 L / g
g = [tex]4\pi ^2 \frac{L}{T^2}[/tex]
They indicate the average time of 20 measurements 1,823 s, each with an oscillation
let's calculate the magnitude
g = [tex]4\pi ^2 \frac{0.823}{1.823^2}[/tex]4 pi2 0.823 / 1.823 2
g = 9.7766 m / s²
now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation
for the period
T = t / n
ΔT = [tex]\frac{dT}{dt}[/tex] Δt + [tex]\frac{dT}{dn}[/tex] ΔDn
In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently
ΔT = Δt / n
ΔT = Δt
now let's look for the uncertainty of g
Δg = [tex]\frac{dg}{dL}[/tex] ΔL + [tex]\frac{dg}{dT}[/tex] ΔT
Δg = [tex]4\pi ^2 \frac{1}{T2}[/tex] ΔL + 4π²L (-2 T⁻³) ΔT
a more manageable way is with the relative error
[tex]\frac{\Delta g}{g} = \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}[/tex]
we substitute
Δg = g ( \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}DL / L + ½ Dt / T)
the error in time give us the stanndard deviation
let's calculate
Δg = 9.7766 ([tex]\frac{0.001}{0.823} + \frac{1}{2} \ \frac{0.671}{1.823}[/tex])
Δg = 9.7766 (0.001215 + 0.0184)
Δg = 0.19 m / s²
the absolute uncertainty must be true to a significant figure
Δg = 0.2 m / s2
therefore the correct result is
g ±Δg = (9.8 ± 0.2) m / s²
A spring is stretched 5 mm by a force of 125 N. How much will the spring stretch
when 250 N force is applied?
Answer:
10 mm
Explanation:
We'll begin by calculating the spring constant of the spring. This can be obtained as follow:
Extention (e) = 5 mm
Force (F) = 125 N
Spring constant (K) =?
F = Ke
125 = K × 5
Divide both side by 5
K = 125 / 5
K = 25 N/mm
Finally, we shall determine how much the spring will stretch when a 250 N force is applied. This can be obtained as follow:
Force (F) = 250 N
Spring constant (K) = 25 N/mm
Extention (e) =?
F = Ke
250 = 25 × e
Divide both side by 25
e = 250 / 25
e = 10 mm
Thus, the spring will stretch 10 mm when a 250 N force is applied.
A painter sits on a scaffold that is connected to a rope passing over a pulley. The other end of the rope rests in the hands of the painter who wants to lift the scaffold. She plans to pull downward on the loose end of the rope, thinking that the scaffold will then rise vertically with her along for the ride. The scaffold has a mass of 52 kg, and her mass is 63 kg. The painter pulls downward on the rope with a force of 600.0 N, while she and the scaffold are hanging from the other end above the ground.
Required:
a. What is the net acceleration on the system consisting of the painter and the scaffold?
b. What is the magnitude of the normal force exerted on the painter by the scaffold?
Solution :
a). From Newtons second law,
F = ma
The total tension force is 2T.
∴ 2T - (m + M)g = (m+ M)a
Then
[tex]$a=\frac{2T-(m+M)g}{m+M}$[/tex]
[tex]$a=\frac{2\times 600-(52+63)9.8}{52+63}$[/tex]
[tex]$=0.63 \ m/s^2$[/tex]
b). From the person,
F = ma
T - Mg + N = Ma
or N = Ma + Mg - T
= (63 x 9.8) + (52 x 9.8) - 600
= 617.4 + 509.6 - 600
= 527 N
if a body of mass m is placed on earth ,what is the amount of potential energy possessed by it (g:-9.8m/s
Answer:
mgh
Explanation:
Assume the height of the body is 1.8m.
The gravity?of the body is G=mg
the height of the gravity center is about 0.9m
E=mgh
=m*9.8m/s*0.9m
= 8.82mJ
A particle move in the xy plane so that its position vector r=bcosQi +bsinQj+ ctk, where b, Q and c are constants. show that the partial move with constant speed.
Answer:
The speed of this particle is constantly [tex]c[/tex].
Explanation:
Position vector of this particle at time [tex]t[/tex]:
[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].
Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:
[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].
Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].
Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:
[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].
The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :
[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].
Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)
What is the maximum centripetal acceleration experienced by a person standing still on the surface of the Earth? Where must they be located?
Answer:
The person must be located in the Equator Line. The maximum centripetal acceleration experienced by a person is 0.0337 meters per square second.
Explanation:
Physically speaking, the centripetal acceleration ([tex]a_{r}[/tex]), measured in meters per square second, experienced by a person is defined by the following expression:
[tex]a_{r} = \omega^{2}\cdot r[/tex] (1)
Where:
[tex]\omega[/tex] - Angular speed of the Earth, measured in radians per second.
[tex]r[/tex] - Distance perpendicular to the rotation axis, measured in meters.
Since rotation axis passes through poles and distance described above is directly proportional to centripetal acceleration. The person must be located in the Equator Line, which is equivalent to the radius of the planet.
In addition, the angular speed of the Earth can be calculated in terms of its period ([tex]T[/tex]), measured in seconds:
[tex]\omega = \frac{2\pi}{T}[/tex] (2)
If we know that [tex]r = 6.371\times 10^{6}\,m[/tex] and [tex]T = 86400\,s[/tex], then the maximum centripetal acceleration experienced by a person is:
[tex]a_{r} = \left(\frac{2\pi}{86400\,s} \right)^{2}\cdot (6.371\times 10^{6}\,m)[/tex]
[tex]a_{r} = 0.0337\,\frac{m}{s^{2}}[/tex]
The maximum centripetal acceleration experienced by a person is 0.0337 meters per square second.
The person standing still on the surface of the earth must be located in the equator line
Recall: the the centripetal acceleration at the Equator is about 0.03 m/s2.
This then means that the maximum centripetal acceleration of a person standing in the equator line is 0.03 m/s2
What is meant by maximum centripetal acceleration?The maximum centripetal acceleration as the name implies is the maximum speed of a body or object in a circular path
Learn more about centripetal acceleration:
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A liquid fueled rocket is red on a test stand. The rocket nozzle has an exit diameter of 30 cm and the combustion gases leave the nozzle at a velocity of 3800 m/s and a pressure of 100 kPa, which is the same as the ambient pressure. The temperature of the gases in the combustion area is 2400 C. Find (a) the temperature of the gases at the nozzle exit plane, (b) the pressure in the combustion area, and (c) the thrust developed. Assume that the gases have a speci c heat ratio of 1.3, and a molar mass of 9. Assume that the ow in the nozzle is isentropic.
Answer:
1. Temperature= 869.35 K
2. Pressure of combustion = 12994.043 kpa
3. Thrust = 127x10⁶N
Explanation:
this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.
1.
The temperature = (273+2400k) - (3800)²/2(4003)
= 2673 - 14440000/8006
= 2673 - 1803.65
= 869.35 K
Approximately 869.4K
2. We first get mach number
= 3800/√1.3(923.8)(869.35)
= 3800/1021.78
= 3.719
Pressure = 100kpa[1+2.07464415]^1.3/0.3
= 12995.043kpa
C. Thrust
Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)
= 12678.621
= 126.781 kN
Thrust is approximately 127kN = 127x10⁶N
If the diameter of a moose eye is 40 mm, what is the total refractive power of the anterior portion of the eye?
Answer:
-the ratio of the speed of light
in air to the speed of light in the substance.
-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.
-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5
Explanation:
1. A block with mass 20 kg is
sliding up a plane (Ukinetic=0.3,
inclined at 10°) at a speed of
2 m/s to the right (positive
X-direction). How far does it
go up along the plane before
it comes to rest momentarily?
Answer: 0.435 m
Explanation:
Given
mass m=20 kg
initial speed u=2 m/s
coefficient of kinetic friction [tex]\mu_k=0.3[/tex]
deceleration which opposes the motion is given by
[tex]\Rightarrow a=g\sin \theta+\mu_kg\cos \theta\\\Rightarrow a=g(\sin \theta +\mu_k\cos \theta)[/tex]
[tex]\Rightarrow a=9.8(\sin 10^{\circ}+0.3\times \cos 10^{\circ})\\\Rightarrow a=4.59\ m/s^2[/tex]
using [tex]v^2-u^2=2as[/tex]
[tex]\Rightarrow s=\dfrac{2^2}{2\times 4.59}=0.435\ m[/tex]
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25 cm and the point-like object with charge q2 = −2.14 µC is located at x2 = −1.80 cm.
A) Determine the total electric potential (in V) at the origin.
B) Determine the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
Answer:
a) the total electric potential is 2282000 V
b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
Explanation:
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
so
Electric potential at p in the diagram 1 below is;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we know that; Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)
the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
r1² = 0.015² + 0.0125²
r1 = √[ 0.015² + 0.0125² ]
r1 = √0.00038125
r1 = 0.0195
Also
r2² = 0.015² + 0.018²
r2 = √[ 0.015² + 0.018² ]
r2 = √0.000549
r2 = 0.0234
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
a) The total electric potential is 2282000 V
b) The total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
What is electric potential?The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to a specific point in an electric field.
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
Electric potential at p in diagram 1 below is;
[tex]V_P=V_1+V_2[/tex]
[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]
we know that; the Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)The total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
[tex]r_1^2=0.015^2+0.0125^2[/tex]
[tex]r_1 = \sqrt{[ 0.015^2 + 0.0125^2 ][/tex]
[tex]r_1 = \sqrt{0.00038125}[/tex]
[tex]r_1 = 0.0195[/tex]
Also
[tex]r_2^2 = 0.015^2 + 0.018^2[/tex]
[tex]r_2 = \sqrt{0.015^2 + 0.018^2}[/tex]
[tex]r_2 = \sqrt{0.000549[/tex]
[tex]r_2 = 0.0234[/tex]
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
To know more about electric potential follow
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what is the effect of divorce on females?
Answer:
Numerous studies have shown that the economic costs of divorce fall more heavily on women. After separation, women experience a sharper decline in household income and a greater poverty risk (Smock 1994; Smock and Manning
Answer:
sadness and stress...................
You are standing on the bottom of a lake with your torso above water. Which statement is correct?
a. You feel a buoyant force only when you momentarily jump up from the bottom of the lake.
b. There is a buoyant force that is proportional to the weight of your body below the water level.
c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.
d. There is no buoyant force on you since you are supported by the lake bottom.
Answer:
c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.
Explanation:
Buoyancy can be defined as a force which is created by the water displaced by an object.
Simply stated, buoyancy is directly proportional to the amount of water that is being displaced by an object.
Hence, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up.
The buoyancy of an object is given by the formula;
[tex] Fb = pgV [/tex]
[tex] But, \; V = Ah [/tex]
[tex] Hence, \; Fb = pgAh [/tex]
Where;
Fb = buoyant force of a liquid acting on an object.
g = acceleration due to gravity.
p = density of the liquid.
v = volume of the liquid displaced.
h = height of liquid (water) displaced by an object.
A = surface area of the floating object.
The unit of measurement for buoyancy is Newton (N).
In this scenario, you are standing on the bottom of a lake with your torso above water. Thus, there is a buoyant force that is proportional to the volume of your body that is below the level of the water.
We assume the foam plate has a positive charge when rubbed with paper towels.
Lift the pan away from the charged plate using the styrofoam cup. Briefly touch the rim of the pan to neutralize it. Place the neutralized pan on the plate and observe the tape rise. When the pan is on the plate, the rim of the plate has a _____________. This means that the pan base is ________________ charged because the net charge on the pan is __________. You know that this must be the case because as you lift the pan with the cup away from the plate, the tape on the rim goes down.
Answer:
POSITIVE CHARGE, NEGATIVE CHARGE, ZERO
Explanation:
To solve this completion exercise, we must remember that charges of the same sign repel each other and in a metallic object (frying pan) the charge is mobile.
Let's analyze the situation when we touch the pan, the charges are neutralized, therefore when we bring the pan to the plate that has a positive charge, it attracts the mobile negative charges in the pan, until it is neutralized, therefore on the opposite side of the pan. pan (edge with a glued tape) is left with a positive charge; therefore the edge and the tape, which is very light, have positive charges and repel each other.
We must assume that the frying pan is insulated so that the net charge is zero, since the induction process.
Consequently the words to complete the sentence are
When the pan is on the plate, the edge of the plate has a _POSITIVE CHARGE_____.
This means that the base of the container is loaded NEGATIVE CHARGE_____ because the net charge of the container is ___ZERO_
Consider a uniformly charged sphere of total charge Q and radius R centered at the origin. We want to find the electric field inside the sphere (r
Answer:
Hello your question is incomplete attached below is the complete question
answer :
Total charge enclosed within the sphere : [tex]\frac{q_{r1} }{4\pi e_{0}R^3 } . r[/tex]
Total charge enclosed outside the sphere : [tex]\frac{q}{4\pi e_{0}r^2 } .r[/tex]
Explanation:
Given data:
Total charge of a uniformly charged sphere = Q
radius = R
first step : find the electric field inside and outside the uniformly charged sphere
2nd step : determine the total charge enclosed within and outside the sphere
make a sketch of the uniformly charged sphere
Attached below is a detailed solution
A 680 nm laser illuminates a double slit apparatus with a slit separation distance of 7.83 μm. On the viewing screen, you measure the distance from the central bright fringe to the 2nd bright fringe to be 88.2 cm. How far away (in meters) is the viewing screen from the double slits?
Answer:
L = 5.08 10⁻¹ m = 50.8 cm
Explanation:
In a double slit experiment for constructive interference is given by
d sin θ = m λ
let's use trigonometry to find a relationship with the distance
tan θ = y / L
these experiments are very small angles
tan θ = [tex]\frac{sin \ \theta}{cos \ \theta}[/tex] = sin θ
when substituting
sin θ = y / L
substituting in the first equation
d y / L = m λ
L = [tex]\frac{d \ y}{m \ \lambda}[/tex]
let's calculate
L = [tex]\frac{ 0.783 \ 10^{-6} \ 88.2 \ 10^{-2} }{2 \ 680 \ 10^{-9}}[/tex]
L = 5.08 10⁻¹ m
The distance of viewing screen from the double slits will be L = 5.08 *10⁻¹ m = 50.8 cm
What is constructive interference?
This position, where the resulting wave is larger than either of the two original, is called constructive interference.
In a double slit experiment for constructive interference is given by
d sin θ = m λ
let's use trigonometry to find a relationship with the distance
[tex]tan\theta=\dfrac {y}{L}[/tex]
these experiments are very small angles
tan θ = = sin θ
when substituting
[tex]sin\theta = \dfrac{y}{l}[/tex]
substituting in the first equation
[tex]\dfrac{dy}{L}=m\lambda[/tex]
[tex]L=\dfrac{dy}{L\lambda}[/tex]
let's calculate
[tex]L=\frac{0.783\times 10^{-6}\times 88.2\times 10^{-2}}{2.680\times 10^{-9}}[/tex]
L = 5.08 10⁻¹ m
Hence the distance of viewing screen from the double slits will be L = 5.08 10⁻¹ m = 50.8 cm
To know more about constructive interference, follow
https://brainly.com/question/1346741
what is the relation of pressure of a liquid with its depth and density?
Answer:
★ Pressure and depth have a directly proportional relationship. This is due to the greater column of water that pushes down on an object submerged. Conversely, as objects are lifted, and the depth decreases, the pressure is reduced.
Explanation:
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A freight train has a mass of [02] kg. The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 km/h?
Answer:
t = 300.3 seconds
Explanation:
Given that,
The mass of a freight train, [tex]m=1.01\times 10^7\ kg[/tex]
Force applied on the tracks, [tex]F=7.5\times 10^5\ N[/tex]
Initial speed, u = 0
Final speed, v = 80 km/h = 22.3 m/s
We need to find the time taken by it to increase the speed of the train from rest.
The force acting on it is given by :
F = ma
or
[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.01\times 10^7\times (22.3-0)}{7.5\times 10^5}\\\\t=300.3\ s[/tex]
So, the required time is 300.3 seconds.