Equipotential lines are usually shown in a manner similar to topographical contour lines, in which the difference in the value of consecutive lines is constant. Clear the equipotential lines using the Erase button on the voltage tool. Place the first equipotential line 1 m away from the charge. It should have a value of roughly 9 V. Now, produce several additional equipotential lines, increasing and decreasing by an interval of 3 V (e.g., one with 12 V, one with 15 V, and one with 6 V). Don't worry about getting these exact values. You can be off by a few tenths of a volt. Which statement best describes the distribution of the equipotential lines?
1. The equipotential lines are closer together in regions where the electric field is weaker.
2. The equipotential lines are closer together in regions where the electric field is stronger.
3. The equipotential lines are equally spaced. The distance between each line is the same for all adjacent lines.
Answer:
B or 2
Explanation:
A hollow sphere is attached to the end of a uniform rod. The sphere has a radius of 0.64 m and a mass of 0.48 kg. The rod has a length of 1.78 m and a mass of 0.50 kg. The rod is placed on a fulcrum (pivot) at X = 0.34 m from the left end of the rod.
(a) Calculate the moment of inertia (click for graphical table) of the contraption around the fulcrum. kg m2
(b) Calculate the torque about the fulcrum, using CCW as positive. N.m
(c) Calculate the angular acceleration of the contraption, using CCW as positive. rad/s2
(d) Calculate the linear acceleration of the right end of the rod, using up as positive. m/s2
The image of this hollow sphere and uniform rod is missing, so i have attached it.
Answer:
A) J = 0.7443 kg•m²
B) T = 1.9169 N•m CCW
C) α = 2.5754 rad/s²
D) a = 3.966 m/s²
Explanation:
A) The moment of inertia J of the contraption around the fulcrum is given by the formula;
J = Jℓ + Jr
Let's calculate Jℓ
Jℓ = [((0.34²/3) × 0.50 × 0.34)/1.78] + (0.48 × (0.34 + 0.64)²)
Jℓ = 0.4647 kg•m²
Now, let's Calculate Jr
Jr = ((1.78 - 0.34)²/3) × ((1.78 - 0.34)/1.78) × 0.50
Jr = 0.2796 kg•m²
Thus;
J = 0.4647 + 0.2796
J = 0.7443 kg•m²
(b) Using CCW as positive, Torque in Nm is calculated as;
T = Tℓ - Tr
Let's calculate Tℓ
Tℓ = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81
Tℓ = 4.7739 N•m CCW
Now, let's Calculate Tr;
Tr = [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81
Tr = 2.857 N•m CW
Thus;
T = 4.7739 - 2.857
T = 1.9169 N•m CCW
(c) The angular acceleration α of the contraption, using CCW is gotten from the formula;
α = T/J
α = 1.9169/0.7443
α = 2.5754 rad/s²
(d) The linear acceleration a of the right end of the rod, using up as positive is given by;
a = α*(1.78 - 0.34)
a = 2.5754 × 1.54
a = 3.966 m/s²
A) the moment of inertia of the contraption is 0.7443 kgm²
B) The torque about the fulcrum is 1.9169 Nm
C) Angular acceleration of the contraption is 2.5754 rad/s²
D) The linear acceleration of the contraption is 3.966 m/s²
Moment of inertia:(A) The moment of inertia I of the contraption around the fulcrum is given by :
[tex]I = [(0.34^2/3) \times 0.50 \times 0.34)/1.78 + (0.48 \times (0.34 + 0.64)^2)] + [(1.78 - 0.34)^2/3) \times (1.78 - 0.34)/1.78) \times 0.50][/tex]
I = 0.4647 + 0.2796
I = 0.7443 kgm²
(B) Using CCW as positive, Torque in Nm is given by;
T = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81 - [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81
T = 4.7739 - 2.857
T = 1.9169 Nm
(C) The angular acceleration (α) of the contraption is given by:
α = T/I
since, torque is defined as T = Iα
α = 1.9169/0.7443
α = 2.5754 rad/s²
(D) The linear acceleration (a) of the right end of the rod
a = αr
where r is the distance from the pivot
a = α × (1.78 - 0.34)
a = 2.5754 × 1.54
a = 3.966 m/s²
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2. A test reveals that 150 J of work is required to lift an object 3 m at a
constant speed. What is the weight of the object?
0 50 N
25N
55N
O 75N
Answer:
50N
Explanation:
W=Fd
150=F(3)
50N=F
Name one similarity and one difference between a set and a bump in volleyball??
One similarity is the use of physical body whereas one difference is that one is exercise and the other is a sport.
One similarity and one differenceOne similarity between a set and a bump in volleyball is the movement and use of legs and hands.
Whereas one difference between a set and a bump in volleyball is that completing several reps of a particular exercise in a row is called a set while on the other hand, the basic pass in volleyball is known as bump.
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A runner is moving at a speed of 20 m/s. How much distance would they cover in 10 seconds?
Answer:
200 meters
Explanation:
20 x 10 = 200
HELP!!! how does gravity affect how objects fall to the ground
Answer:
c
Explanation:
Answer:
When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion. Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls.
Explanation:
In one type of mass spectrometer, ions having the same velocity move through a uniform magnetic field. The spectrometer is being used to distinguish 12C and 14C ions. The 12C ions move in a circle of diameter 45.4 cm. Use these atomic mass values: 12C, 12.0 u; 14C, 14.0 u.
Answer:
r = 0.5297 m
Explanation:
In this exercise we use Newton's second law where the force is magnetic
F = ma
centripetal acceleration
a = v² / r
F = q v x B = q v B sin θ
where the angle between the velocity and the magnetic field is 90º, therefore the sin 90 = 1
we substitute
q v B= m v² / r
r = [tex]\frac{m v^2 }{qv B}[/tex]
the mass of each isotope is
12C
m12 = 6 m_proton + 6 m_neutrons
m12 = (6 1,673 +6 1,675) 10⁻²⁷
m12 = 20.088 10-27 kg
14C
m14 = 6 m_proton + 8 m_neutron
m14 = (6 1,673 + 8 1,675) 10-27
m14 = 23,438 10⁻²⁷ kg
in the exercise they indicate that the velocity of the two particles is the same, therefore with the initial data we can calculate the parameters that do not change in the experiment.
[tex]\frac{v}{qB} = \frac{r}{m_{12}}[/tex]
v / qB = 0.454 / 20.088 10⁻²⁷
v / qb = 2.26 10²⁵
this quantity remains constant, let's use the other data to calculate the radius
r = 23.438 10⁻²⁷ 2.26 10²⁵
r = 5.297 10⁻¹ m
r = 0.5297 m
A large box slides across a frictionless surface with a velocity of 12 m/s and a mass of 4
kg, collides with a smaller box with a mass of 2 kg that is stationary. The boxes stick
together. What is the velocity of the two combined masses after collision?
8 m/s
O m/s
12 m/s
4 m/s
us 12:18
What is used to measure the radiation?
Answer:
geiger counter with geiger mueller
uniform solid sphere has a mass of 1.765 kg and a radius of 0.854 m.a. Find the torque required to bring the sphere from rest to an angular velocity of 317 rad/s, clockwise, in 15.5 s.b. What magnitude force applied tangentially at the equator would provide the needed torque
Answer:
a) the torque required is 10.53 N-m
b) The magnitude force applied tangentially is 12.33 N
Explanation:
Given the data in the question;
mass m = 1.765 kg
radius r = 0.854 m
first we calculate the moment of inertia;
[tex]I[/tex] = [tex]\frac{2}{5}[/tex]mr²
we substitute
[tex]I[/tex] = [tex]\frac{2}{5}[/tex] × 1.765 × (0.854)²
[tex]I[/tex] = 0.514897 kg.m²
a)
Find the torque required to bring the sphere from rest to an angular velocity of 317 rad/s, clockwise, in 15.5 s
ω[tex]_{initial[/tex] = 0
ω[tex]_{final[/tex] = 317 rad/s
t = 15.5 s
we know that; ω[tex]_{final[/tex] = ω[tex]_{initial[/tex] + ∝t
so we substitute
317 = 0 + ∝(15.5)
∝ = 317 / 15.5
∝ = 20.4514 rad/s²
so
ζ = [tex]I[/tex] × ∝
we substitute
ζ = 0.514897 × 20.4514
ζ = 10.53 N-m
Therefore, the torque required is 10.53 N-m
b)
What magnitude force applied tangentially at the equator would provide the needed torque.
ζ = F × r
we substitute
10.53 = F × 0.854
F = 10.53 / 0.854
F = 12.33 N
Therefore, magnitude force applied tangentially is 12.33 N
1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into the water and the level rises to 12 cm.
(a) What is the volume of water displaced by the rock?
(b) What is the volume of the rock?
(c) Calculate the density of the rock
Answer:
(a) The volume of water is 100 cm³
(b) The volume of the rock is 20 cm³
(c) The density of the rock is 30 g/cm³
Explanation:
The given parameters of the perspex box are;
The area of the base of the box, A = 10 cm²
The initial level of water in the box, h₁ = 10 cm
The mass of the rock placed in the box, m = 600 g
The final level of water in the box, h₂ = 12 cm
(a) The volume of water in the box, 'V', is given as follows;
V = A × h₁
∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³
The volume of water in the box, V = 100 cm³
(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the water, V, is given as follows;
[tex]V_T[/tex] = [tex]V_r[/tex] + V
[tex]V_T[/tex] = A × h₂
∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³
The total volume, [tex]V_T[/tex] = 120 cm³
The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V
∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³
The volume of the rock, [tex]V_r[/tex] = 20 cm³
(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)
∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³
A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 6.9 s later. How high is the cliff?
What is the acceleration of a 4,000 kg car pushed with a
force of 12,000 N?
Answer:
3 m/s
Explanation:
A= F/m
12,000/ 4000 = 3
Answer:
3 m/s^2
Explanation:
The equation you have to use is F=ma because the problem is a Newton's 2nd law problem.
Our known values are:
F ( Force ) = 12,000 N
m ( mass ) = 4,000 kg
a ( acceleration ) = ?
Now we plug in the known values into the equation and solve
F=ma
12,000=4,000a
We have to divide 4,000 by both sides to isolate the a value
12,000/4,000=4,000/4,000a
The 4,000s on the right of the equation cancel.
And 12,000 divided by 4,000 equals 3
The acceleration (a) is 3 meters per second squared (m/s^2)
Next, check to make sure 3 does work by plugging it back into the equation.
12,000=4,000*3
12,000=12,000 ✔
As you can see, the acceleration will be 3 m/s^2
A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is
the cat's displacement during this motion?
A. 6m
B. 2.5 m
C. 8.5 m
D. 4.5 m
20
А
B
D
Answer:
see that the correct one is B
Explanation:
To solve this exercise let us use the kinematic relations
v² = v₀² - 2 a x
as they indicate that the car stops, therefore the final speed is yield v = 0
x = v₀² / 2a
let's calculate
x = 2²/(2 0.8)
x = 2.5 m / s²
When reviewing the answers we see that the correct one is B
A window air conditioner that consumes 2 kW of electricity when running and has a coefficient of performance of 3 is placed in the middle of a room, and is plugged in. The rate of cooling or heating this air conditioner will provide to the air in the room when running is:
Answer: Q= QH - QL
= W + QL - QL
= W
= 1 kj/8
Explanation:
Since the net rate is positive the room will be heated.
TRUE OR FALSE
2 QUESTIONS
please HELP ASAP
1. false
2. true
I hope this helps ^-^
The mass of the Moon is 7.3x1022 kg and its radius is 1738 km. What is the strength of the gravitational field on the
surface of the Moon? (Do all required steps)
Answer:
1.61 N/kg
Explanation:
Take the universal gravitational constant G as 6.67 × 10^(-11) Nm²/kg²
The required gravitational field strength
= 6.67 × 10^(-11) × 7.3 × 10^(22) / (1738000)²
= 1.61194103 N/kg
= 1.61 N/kg (corr. to 3 sig. fig.)
A block and a ball have the same mass and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping. 1)Which one makes it furthest up the ramp
Answer:
Both.
Explanation:
Given that a block and a ball have the same mass and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping. 1)Which one makes it furthest up the ramp ?
Since both of them have the same mass and the same initial velocity, then, they will both have the same kinetic energy.
That is,
K.E = 1/2mv^2
Friction is a force that opposes motion. And since the frictional force is zero,
Both of them will accelerate from Newton's law.
F = ma
We can therefore conclude that both of them will make it further up the ramp.
which one is odd copper,plastic,rubber
Answer:
It's plastic.
trust me it's plastic, i've rad it somewhere.
All of them have something that's not like the others.
-- Rubber is the only one on the list that has two repeated letters.
-- Plastic is the only one on the list thagt has no repeated letters.
-- Plastic is the only one on the list that has no 'r' in its name.
-- Copper is the only one on the list that is an element, not a compound.
-- Copper is the only good electrical conductor on the list.
-- Plastic is the only one on the list with more than six letters in its name.
-- Rubber is the only one on the list with no 'p' in its name.
-- Plastic is the only one on the list that doesn't end in "-er".
someone please help I can mark brainless
what determines the magnification of an imagev
measure:what the current values of
Answer:
The magnification of an image is equal to the ratio of the image height to the object height.
We can not hear Infrasound. Why is that?
Answer :
Last choice
Answer:
its frequency is too low
PLEASE CLICK ON THIS IMAGE I NEED HELP
Answer:
Explanation:
you can say the law of superpoition can tell us that each rock layer is older than the one above it. So, the relative age of the rock or fossil in the rock or fossil in the rock is older if it is farther down in the rock layers. hope it helps
Scientific work is currently under way to determine whether weak oscillating magnetic fields can affect human health. For example, one study found that driv- ers of trains had a higher incidence of blood cancer than other railway workers, possibly due to long expo- sure to mechanical devices in the train engine cab. Consider a magnetic field of magnitude 1.00 3 1023 T, oscillating sinusoidally at 60.0 Hz. If the diameter of a red blood cell is 8.00 mm, determine the maximum emf that can be generated around the perimeter of a cell in this field.
Answer:
fem = 7.58 10⁻⁵ V
Explanation:
For this exercise we use Faraday's law
fem = [tex]- \frac{d \Phi _B}{dt}[/tex]
the magnetic flux is
Ф_B = B. A = B A cos θ
Tje bold are vectros. Suppose the case where the normal to the surface of the red blood cell is parallel to the field, therefore the angle is zero and the cos 0 = 1
The red blood cell area is
A =π r²
indicate that the diameter is r = 8.00 mm = 8.00 10⁻³ m
the magnetic field has a frequency of f=60 Hz, and B₀ = 1.00 10⁻³T, therefore we can write it
B = B₀ sin (wt) = B₀ sin( 2π f t)
we substitute
fem = - A dB / dt
fem = - A B₀ [tex]\frac{ d (sin ( 2\pi f t)}{dt}[/tex]
fem = - π r² Bo (2πf cos 2πft)
the maximum electromotive force occurs when the function is ±1
fem = 2 π² r² B₀ f
let's calculate
fem = 2π² (8.00 10⁻³)² 1.00 10⁻³ 60
fem = 7.58 10⁻⁵ V
The lens inside the human eye is convex.
True
False
Answer: True
Explanation: "The lens is a transparent biconvex structure in the eye that, along with the cornea, helps to refract light to be focused on the retina."
How long would it take a machine to do 5.000
joules of work if the power rating of the machine
is 100 watts?
A. 5,000 sec
B.
50 sec
C.
10 sec
D. 0.2 sec
Answer : B
Answer:
a
Explanation:
How long would it take a machine to do 5.000
joules of work if the power rating of the machine
is 100 watts?
The given machine will take 50 s to complete the work. the power is the rate of performing work.
What is power?It can be defined as the rate of performing work. It can also be written as the amount of work divided by the time it takes to complete the work.
[tex]p = \dfrac wt[/tex]
So
[tex]t = \dfrac w p[/tex]
Where,
[tex]p[/tex] - power = 100 watt = 100 J/s
[tex]t[/tex] - time = ?
[tex]w[/tex]- work = 5000 J
Put the values in the formula,
[tex]t = \dfrac{ 5000 \rm \ J} {100 \rm \ J/s}\\\\t = 50 \rm \ s[/tex]
Therefore, the given machine will take 50 s to complete the work.
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A train travels at a speed of 70km/h and travels a distance of 630 km. How long did it take the train to complete its journey?
Answer:
9 hr
Explanation:
Speed = distance/time
Let the time taken by the train be t.
=> 70 = 630/t
=> t = 630/70
=> t = 9
it take 9 hr to complete its journey
Answer:
9 hours
Explanation:
Using the formula to solve motion problems: d=r*t (Distance = Rate * Time)
Given:
Distance = 630 km
Rate (Speed) = 70 km/h
Time = Distance / Time
Time = 630/70
= 9
Section of hollow pipe and a solid cylinder have the same radius, mass, and length. They
both rotate about their long central axes with the same angular speed. Which object has the
higher moment of inertia? (a) the hollow pipe (b) the solid cylinder (c) they have the same
rotational kinetic energy (d) impossible to determine
Answer:
Option B (The hollow pipe has the higher angular momentum)Explanation:
Angular momentum = [tex]I*w[/tex]
I = Moment of inertia
w = Angular velocity
Hollow pipe can be considered as hollow cylinder
Moment of Inertia of Hollow cylinder = [tex]Mr^2[/tex]
Moment of Inertia of solid cylinder= [tex]\frac{Mr^2}{2}[/tex]
Clearly Moment of inertia of hollow pipe is greater which states that Angular momentum of it will be greater.
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Use the worked example above to help you solve this problem. An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force on the sled by pulling on the rope. Suppose the coefficient of kinetic friction between the loaded sled and snow is 0.200.
(a) The Eskimo pulls the sled 5.90 m, exerting a force of 1.10 102 N at an angle of θ = 0°. Find the work done on the sled by friction, and the net work. Wfric = Correct: Your answer is correct. . J Wnet = Correct: Your answer is correct. . J
(b) Repeat the calculation if the applied force is exerted at an angle of θ = 30.0° with the horizontal. Wfric = J Wnet = J
Answer:
(a)
W_friction = 98.1 J
W_net = 550.9 J
(b)
W_friction = 98.1 J
W_net = 463.95 J
Explanation:
(a)
First, we will calculate the work done by friction:
[tex]W_{friction} = \mu R = \mu W = \mu mg\\W_{friction} = (0.2)(50\ kg)(9.81\ m/s^2)\\[/tex]
W_friction = 98.1 J
Now, the work done by Eskimo will be:
[tex]W_{Eskimo} = FdCos\theta\\W_{Eskimo} = (110\ N)(5.9\ m)Cos\ 0^o\\[/tex]
W_Eskimo = 649 J
So, the net work will be:
W_net = W_{Eskimo} - W_{friction}
W_net = 649 J - 98.1 J
W_net = 550.9 J
(b)
First, we will calculate the work done by friction:
[tex]W_{friction} = \mu R = \mu W = \mu mg\\W_{friction} = (0.2)(50\ kg)(9.81\ m/s^2)\\[/tex]
W_friction = 98.1 J
Now, the work done by Eskimo will be:
[tex]W_{Eskimo} = FdCos\theta\\W_{Eskimo} = (110\ N)(5.9\ m)Cos\ 30^o\\[/tex]
W_Eskimo = 562.05 J
So, the net work will be:
W_net = W_{Eskimo} - W_{friction}
W_net = 562.05 J - 98.1 J
W_net = 463.95 J
A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 38.0 ∘ above the horizontal. The glider has mass 9.00×10−2 kg. The spring has 590 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.70 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring.
Required:
a. What distance was the spring originally compressed?
b. When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?
Answer:
x = 0.056 m
ΔKE = 0.489 J
Explanation:
Given that
Angle, θ = 38°
Length, L = 1.7 m
Mass, m = 0.09 kg
Spring constant, K = 590 N/m
If we use the Work-Energy theorem, then we know that Potential Energy, PE = Kinetic Energy, KE
This is mathematically written as
1/2kx² = mgH
The height, H we can get by using the relation
H = L.Sinθ
H = 1.7 * Sin 38
H = 1.7 * 0.6157
H = 1.047 m
Next, we use the Work-Energy theorem
1/2kx² = mgH
1/2 * 590 * x² = 0.09 * 9.8 * 1.047
295 * x² = 0.9234
x² = 0.9235 / 295
x² = 0.00313
x = √0.00313
x = 0.056 m
If the spring loses contact at x = 0.056, definitely, it will also lose contact at x = 0.8
Then we use the formula
ΔKE = mg(H - H1)
ΔKE = mg(xsinθ - x2.sinθ)
Where, x = 1.7 , x2 = 0.8
ΔKE = 0.09 * 9.8 (1.7 * sin 38 - 0.8 * sin 38)
ΔKE = 0.882(1.047 - 0.493)
ΔKE = 0.882 * 0.554
ΔKE = 0.489 J