Answer:
Before providing an answer to the question, the values for acceleration given in questions A and B were written twice. So correction would go like this: For (a) when the truck accelerates at 2.20m/s2 northward, and for (b) when it accelerates at 3.40m/s2 southward.
The answer:
(a) 88N, northward.
(b) 78.4, southward.
Explanation:
(a) Maximum frictional force acting on the packing case= (coefficient of static friction) X (Normal force)
Normal force = mass X acceleration due to gravity
Maximum static frictional force acting on the packing case = (coefficient of static friction) X (mass of packing case X acceleration due to gravity)
Maximum static frictional force = (0.30) X (40.0-kg) X (9.8m/s 2) = 117.6N
While Reaction force acting on the packing case = (mass of packing case) x (acceleration generated by the pickup truck)
Reaction force acting on the case = (40.0-kg) X (2.20m/s2) = 88N
With these values, one can conclude that the packing case is at rest since the reaction force of the case acting in the opposite direction is lesser than the frictional force. Making the magnitude and direction of the friction force acting on the case still move northward, and the static frictional force acting on equals the reaction force.
The answer is 88N, northward.
(b) Here too, we need to still compare the reaction force with the value of the already determined Maximum static frictional force (117.6N) above. This is necessary to know the frictional force between the pickup truck"s floor and the packing case.
Reaction force acting on the case when acceleration is 3.40m/s2 = (40.0-kg) X (3.40m/s2) = 136 N
We can conclude that the reaction force (136 N) is greater than the maximum static frictional force (117.6N), suggesting that the packing case is in motion and the frictional force is no longer static.
This means a kinetic force is now acting on the pickup truck"s floor causing the packing case to also move. This kinetic force can be calculated as:
kinetic force = (coefficient of kinetic friction) X (mass of packing case X acceleration due to gravity)
= (0,20) X (40.0-kg) X (9.8m/s 2) = 78.4N
A rocket, with a mass of 5100 kg, has an engine that provides a net upward force of 8.0 x 10^5 N. It starts from rest and reaches a maximum speed of 900 m/s. How long does it take to reach that maximum velocity?
Answer:
5.7375 seconds
Explanation:
The computation of the time required to reach that maximum velocity is shown below:
Given that
Mass = m = 5100 kg
Net upward force F = 8 × times 10^5 N
Initial speed = V_i = 0
Maximum speed = V = 900 m.s
Based on the above information
Impluse J = m(V - V_i)
= 5100 (900 - 0)
= 459 × 10^4 kg m.s
As we know that
J = FT
So
T = J ÷ F
= (459 × 10^4) ÷ (8 × 10^5)
= 5.7375 seconds
If you have a 0.125 kg lead piece at
20.0°C, how much heat must you
add to melt it? (Remember, you
must warm it to its melting point
first.)
Material
Lead
Melt Pt (°C)
327
L (1/kg)
2.32.104
Boil Pt (°C) Lv (1/kg)
1750 8.59.105
c (1/(kg*c)
128
(Unit = J)
Answer:
7,812 J
Explanation:
Using the relation:
Q = mcΔθ
Q = quantity of heat
C = specific heat capacity of lead
Δθ = temperature change (T2 - T1)
M = mass of substance
Q = mass * specific heat * Δθ
Q = 0.125kg * 128 * (327 – 20)
Q = 0.125 * 128 * 307
Q = 4912 J
For melting:
Q = mass * Hf
0.125 * (2.32 * 10^4)
= 2,900 J
Total = 4,912 J + 2,900 J = 7,812 J
39. What is the change in momentum for a 5,000 kg ship in
outer space that experiences no net force over a 1 hr
period?
Answer:
Change in momentum is zero.
Explanation:
The following data were obtained from the question:
Mass (m) = 5000 kg
Time (t) = 1 h
Net force (F) = 0
Change in momentum =?
Force = Rate of change of momentum
0 = change in momentum
Change in momentum = 0
We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.
The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is [01] m. (a) What is the force (N) that the track must exert on the car? (positive is up) (b) What must be the force (N) that the car exerts on a 61 kg passenger?
This question is incomplete, the complete question is;
The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is 15.8 m.
(a) What is the force (N) that the track must exert on the car? (positive is up)
(b) What must be the force (N) that the car exerts on a 61 kg passenger?
Answer:
a) the force (N) that the track must exert on the car is -6139.14 N
b) the force (N) that the car exerts on a 61 kg passenger is -1170.27 N
Explanation:
Given the data in the question;
Let N represent the force that the track must exerted on the car
Net force on the car Fnet = Mg + N
so
M × a = Mg + N
N = Ma - Mg
N = Ma - M(v²/R)
we substitute
N = (320kg × 9.8m/s²) - ( 320 × ((21.4m/s)² / 15.8 m) )
N = 3136 - ( 320 × 28.9848 )
N = 3136 - 9275.136
N = -6139.14 N
Therefore, the force (N) that the track must exert on the car is -6139.14 N
b) What must be the force (N) that the car exerts on a 61 kg passenger?
Let N represent the force that the car exerts on 61kg passengers
so
Net force of passengers Fnet = mg + N
Ma = Mg + N
N = Ma - Mg
N = Ma - M(v²/R)
N = (61kg × 9.8m/s²) - ( 61 × ((21.4m/s)² / 15.8 m) )
N = 597.8 - ( 61 × 28.9848)
N = 597.8 - 1768.0728
N = -1170.27 N
Therefore, the force (N) that the car exerts on a 61 kg passenger is -1170.27 N
The centripetal force of the track on the car moving in the circular path is [tex]1.465 \times 10^6 \ N[/tex].
The force (N) that the car exerts on a 61 kg passenger is 597.8 N.
Centripetal force of the track
The centripetal force of the track on the car moving in the circular path is calculated as follows;
[tex]F_c = \frac{mv^2}{r}\\\\ F_c = \frac{320 \times 21.4^2}{0.1} \\\\F_c = 1.465 \times 10^6 \ N[/tex]
Normal force of the passengerThe force (N) that the car exerts on a 61 kg passenger is equal to the force the passenger exerts on the car based on Newton's third law of motion.
F = mg
F = 61 x 9.8
F = 597.8 N
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Cameron is standing on the edge of a 60 m high cliff. He horizontally throws a football
with an initial velocity of 10 m/s. How long does it take for the football to hit the
ground?
Answer:34.6 m/s
Explanation: It is asking how long meaning the answer is in time
Albert Bandura emphasized the idea of __________, which is the belief one has in one’s own ability to succeed. A. operant conditioning B. determinism C. self-efficacy D. self-worth
Answer:
C
Explanation:
Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.
What is Self efficacy?
A person's self-efficacy relates to their confidence in their ability to carry out the behaviors required to achieve particular performance goals (Bandura, 1977, 1986, 1997).
The belief in one's capacity to exercise control over one's own motivation, behavior, and social environment is known as self-efficacy. The goals for which people strive, the amount of effort put out to obtain goals, and the possibility of achieving particular levels of behavioral performance are all influenced by these cognitive self-evaluations.
Self-efficacy beliefs, unlike conventional psychological notions, are anticipated to change according to the operating domain and the environment in which an action occurs.
Therefore, Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.
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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
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.
because there the sperm and eggs are combining together to produce
so thats why they look alike
Answer:
yes. that is how a baby is conceived.
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]
if a current of 5A flows for 2minutes, find the quantity of electricity transfered
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.
A soccer ball was kicked over the edge of a wall and traveled 35 m horizontally at a speed of 5.6m/s. Calculate the vertical height of the wall.
Answer:
Are you sure it was soccer ball? Or meine hearts
Explanation:
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
your "A" never changes, while your "Y" changes depending on strength of gravity. is it mass or weight?
Answer:
A - mass. B - Weight
Explanation:
This is because weight varies with the strength of gravity. Mass is just the amount of matter in an object
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
The sides of a right triangle that has any given vector for the hypotenuse are called _____
A. Scalar
B. Component
C. Resultant
D. Vector
Answer:
They are the resultant vector.
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.
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:
A 450.0 kg roller coaster is traveling in a circle with radius 15.0m. Its speed at point A is 28.0m/s and its speed at point B is 14.0 m/s. At point A the cart is already moving with circular motion. a) Draw free bodydiagramsfor the cartatpointsAand B(two separate free body diagrams). b) Calculate the acceleration of the cartat pointsAandB(magnitude and direction). c) Calculate the magnitude of the normal force exerted by the trackson the cartat point A. d) Calculate the magnitude of the normal force exerted by the tracks on the cart at point B.
Answer:
b) a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N
Explanation:
a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle
b) Let's start at point A
Let's use that the acceleration is centripetal
a = v² / r
let's calculate
a = 28² / 15.0
a = 52.26 m / s²
as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards
Point B
a ’= 142/15
a ’= 13.06 m / s²
in this case the acceleration is vertical downwards
c) The values of the normal force
point A
let's use Newton's second law
∑ F = m a
N- W = m a
N = mg + ma
N = m (g + a)
N = 450.0 (9.8 + 52.25)
N = 2.79 10⁴ N
d) Point B
-N -W = m (-a)
N = ma -m g
N = m (a-g)
N = 450.0 (14.0 - 9.8)
N = 1.89 10³ N
.You have always been impressed by the speed of the elevators in your apartment building. You wonder about the maximum acceleration for these elevators during normal operation, so you decide to measure it by using your bathroom scale. While the elevator is at rest on the ground floor,you get in, put down your scale, and stand on it. The scale reads 50 kg. You continue standing on the scale when the elevator goes up, carefully watching the reading. During the trip to the 10th floor, the greatest scale reading was
Answer:
5.51 m/s^2
Explanation:
Initial scale reading = 50 kg
assume the greatest scale reading = 78.09 kg
Determine the maximum acceleration for these elevators
At rest the weight is = 50 kg
Weight ( F ) = mg = 50 * 9.81 = 490.5 N
At the 10th floor weight = 78.09 kg
Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N
F = change in weight
Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)
50 * a = 275.61
Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2
Please help 25 points!
Three waves with frequencies of 1 Hertz (Hz), 3 Hz, and 9Hz travel at the same speed. Which of the following statements is correct?
A. The 1 Hz wave contains the most energy.
B. The crests of all three waves are of equal height.
C. The wavelength of the 9Hz wave is three times that of the 3 Hz wave.
D. The 1 Hz wave has the longest wavelength.
Answer:
B
Explanation:
The crest of all three waves are of equal height
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...................
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.)
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
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A fan has four identical, symmetrically placed blades. The blades are rotating clockwise at twenty revolutions per second.
A) What is the smallest time interval between stroboscope flashes that will make the fan blades appear motionless?
B) What is the highest frequency (in flashes per second) at which a stroboscope will make the
fan blades appear to stand still? Show your calculation.
C) The same questions as (a) and (b), but someone has put a yellow dot on one blade, and now you want the yellow dot to appear to be standing still. Explain, and show your calculation.
D) Now the stroboscope is set for nineteen flashes per second, and the yellow dot appears to be slowly rotating. Which direction does it appear to rotate, clockwise or counterclockwise? Explain, and show your calculation.
E) The same as (d), but the stroboscope is set for twenty-one flashes per second. Explain, and show your calculation.
Answer:
A) t = 1.249 10⁻² s, B) f = 80 Hz, C) f = 20 Hz,
D) slowly advancing an angle of approximately Δθ = 0.05 rad each flash
E) In each flash it seems to go backward an angle of Δθ = -0.05 rad
Explanation:
A) To make it appear that the blades are immobile, it implies that every time the light turns on, a blade should be in the same position, therefore, as we have 4 blades, they must rotate an angle of 2π/4,
θ = π / 2
θ = 1.57 rad
taking the angle let's use the endowment kinematics relations
θ = w₀ t + ½ α t²
in general the fans rotate at constant speed α= 0
θ = w₀ t
t = θ / w₀
let's reduce the magnitudes to the SI system
w₀ = 20 rev / s (2π rad / 1rev) = 125.66 rad / s
let's calculate
t = 1.57 / 125.66
t = 1.249 10⁻² s
B) the fastest speed for the blades to rotate is when one blade of a complete turn , we use the relationship between the fecuance and the period
f = 1 / T
f = 1 / 1.25 10⁻²
f = 80 Hz
C) we have two possibilities:
* a yellow dot is placed on each sheet
In this case the angular velocity of the blade is the same at all points, therefore the results obtained should not change
* a yellow dot is placed on a single sheet.
Here for the point to remain fixed the angle of rotation must be
θ= 2π rad
the time is
t = 2π / 125.66
t = 5 10⁻² s
the maximum frequency is
f = 1/5 10⁻²
f = 20 Hz
D) The copy strobe rotates at f = 19 Hz, the time between each flash is
t = 1/19
t = 5.26 10⁻² s
this time is higher, so the angle turned is large
θ = w t
θ = 125.66 5.26 10⁻²
θ = 6.61 rad
the relationship between this angle and the angle of a circle is
θ = 1,052
We can see that it is this time the blade rotates 1 complete turns, for this the position of the blade changes us, for the other 0.052 rad the blade rotates a little more than the circumference therefore it seems that it is slowly advancing an angle of approximately
Δθ = 0.05 rad each flash
E) in this case changes the flash speed
t = 1/21
t = 4.76 10⁻² s
the angle rotated is
θ = 125.66 4.76 10⁻²
θ = 5.984 rad
θ / 2π = 0.95
in that case, the blade did not complete the turn, therefore in each flash it seems to go backward an angle
Δθ = -0.05 rad
Which statement best compares potential and kinetic energy?
O Objects always have more potentiał energy than kinetic energy.
O Kinetic energy increases and potential energy decreases when the velocity of an object increases
O Only potential energy decreases when an object's height increases.
O Objects always have more kinetic energy than potential energy.
Answer:
Kinetic energy increases and potential energy decrease when velocity of an object increase.
QUCIK!! SOMEONE PLEASE HELP! I’LL MARK BRAINLIEST!!
Answer:
A. v = √2gh
B. No! The final velocity does not depend on the mass of the car.
C. Yes! the final velocity depends on the steepness of the hill
D. 3.28 m/s
Explanation:
A. Determination of the final velocity.
½mv² = mgh
Cancel out m
½v² = gh
Cross multiply
v² = 2gh
Take the square root of both side
v = √2gh
B. Considering the formula obtained for the final velocity i.e
v = √2gh
We can see that there is no mass (m) in the formula.
Thus, the final velocity does not depend on the mass of the car.
C. Considering the formula obtained for the final velocity i.e
v = √2gh
We can see that there is height (h) in the formula.
Thus, the final velocity depends on the steepness of the hill
D. Determination of the final velocity.
Height (h) = 0.55 m
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) =?
v = √2gh
v = √(2 × 9.8 × 0.55)
v = √10.78
v = 3.28 m/s
Question 3 (5 points)
Yissel was going to be late to Mr. Scharff's science class. Just as the bell was about to ring. Vissel ran the last little bit of the hallway at 2.47
meters/second for 8 seconds to beat the bell. How far away was Yissel from Mr. Scharff's classroom when she started to run?
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: