Answer:
The magnetic field is 1.16 T.
Explanation:
speed, v = 10% of speed of light = 3 x 10^7 m/s
diameter, d = 54 cm
radius, r = 0.27 m
charge, q = 1.6 x 10^-19 C
mass, m = 1.67 x 10^-27 kg
Let the magnetic field is B.
The centripetal force is balanced by the magnetic force.
[tex]qvB=\frac{mv^2}{r}\\\\B =\frac{mv}{qr}\\\\B =\frac{1.67\times 10^{-27}\times 3\times 10^{7}}{1.6\times 10^{-19}\times 0.27}\\\\B =1.16 T[/tex]
A 15 000-N car on a hydraulic lift rests on a cylinder with a piston of radius 0.20 m. If a connecting cylinder with a piston of 0.040-m radius is driven by compressed air, what force must be applied to this smaller piston in order to lift the car
Answer:
the force applied to the smaller piston is 600 N
Explanation:
Given;
weight of the car, F = 15,000 N
radius of the lager piston, R = 0.2 m
radius of the smaller piston, r = 0.04 m
let the force applied to the smaller piston = f
The pressure applied on both piston is constant;
[tex]P = \frac{F}{A} = \frac{f}{a} \\\\\frac{F}{R^2} = \frac{f}{r^2} \\\\f = \frac{F\times r^2}{R^2} = \frac{15,000 \times (0.04)^2}{(0.2)^2} = 600 \ N[/tex]
Therefore, the force applied to the smaller piston is 600 N
A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 2.12 m/s and observes that it takes 1.5 s to reach the water.
(a) List the knowns in this problem.
(b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.
Each year 500 runners run up the stairs to the 86th floor of the Empire State Building in New York City. There are 1576 steps and each step is 0.241 m high. In 2003, Australian Paul Crake (20-29 age group) set the overall record by reaching the 86th floor in 9:33. His mass was 70.0 kg. Question 2 HomeworkUnanswered What was Paul Crake's power output during this climb
Answer:
The power is 465.44 W.
Explanation:
mass, m = 70 kg
number of steps, n = 1576
height of each step, h = 0.241 m
time taken, t = 9.33 min= 9.33 x 60 s
The power is given by the rate of doing work.
W = n m g h
W = 1576 x 70 x 9.8 x 0.241
W = 260553.776 J
The power is given by
[tex]P = \frac{W}{t}\\\\P = \frac{260553.776}{9.33\times 60}\\\\P = 465.44 W[/tex]
Lauren pushed a crate of mass 28.9 kg a distance of 2.7 meters along a horizontal surface. On that part of the surface, the crate could slide with negligible friction. Lauren exerted a constant force of 121 newtons for 2.7 meters. The crate then slid down an inclined plane of height 1.8 meters, also with negligible friction. As the crate slid down the plane, the only significant forces on it were the normal force from the plane and gravity (with g = 9.81 meters per second squared). At the bottom of the incline, the crate began sliding along a horizontal surface with ordinary kinetic friction. The coefficient of friction between this surface and the crate was 0.41. This surface ended at a vertical wall after a distance of d2 = 5.2 meters but the crate did not slide that far. Attached to the vertical wall was a long ideal spring with length dy = 3.4 meters and a spring constant of 154 newtons per meter.
Required:
Calculate the distance that the crate compressed the spring before coming to rest.
Answer:
3.034 m
Explanation:
From the law of conservation of energy, the energy at the top of the incline equals the energy at the bottom of the incline since at the top of the incline, the horizontal surface is frictionless and along the incline there is no friction.
The work done in moving the crate a distance, d = 2.7 m with a force of F = 121 N to the top of the incline is W = Fd = 121 N × 2.7 m = 326.7 J.
From work-kinetic energy principles, this work W = kinetic energy of the crate at the top of the incline, K₁.
Now, the total mechanical energy at the top of the incline, E equals the total mechanical energy at the bottom of the incline E' since there is no friction along the incline.
So, E = E'
U₁ + K₁ = U₂ + K₂ where U₁ = potential energy of crate at top of incline = mgh where m = mass of crate = 28.9 kg, g = acceleration due to gravity = 9.8 m/s², h = height of incline = 1.8 m, K₁ = kinetic energy of crate at top of incline = 326.7 J, U₂ = potential energy of crate at bottom of incline = 0 J(since it is at an elevation h = 0) and K₂ = kinetic energy of crate at bottom of incline
So, substituting the values of the variables into the equation, we have
U₁ + K₁ = U₂ + K₂
mgh + K₁ = U₂ + K₂
28.9 kg × 9.8 m/s² × 1.8 m + 326.7 J = 0 J + K₂
509.796 J + 326.7 J = K₂
K₂ = 836.496 J
K₂ ≅ 836.5 J
Now since the vertical wall is a distance d2 away and the long ideal spring has a length dy = 3.4 m, let x be the compression of the spring. So, the distance moved by the crate is thus D = d2 - dy - x.
Now, the change in kinetic energy of the crate ΔK equals the work done by friction and that done by the spring W.
So ΔK = -W (from work-kinetic energy principles)
Let W' = work done by friction = μmgD where μ = coefficient of kinetic friction between surface and crate = 0.41, m = mass of crate = 28.9 kg, g = acceleration due to gravity = 9.8 m/s² and D = distance moved by crate = D = d2 - dy - x = 5.2 m - 3.4 m - x = 1.8 - x
So, W' = μmgD
W' = 0.41 × 28.9 kg × 9.8 m/s² (1.8 - x)
W' = 116.12(1.8 - x)
W' = 2090.16 - 116.12x
The work done by the spring W" = 1/2k(x₀² - x²) where k = spring constant = 154 N/m, x₀ = initial spring length = dy = 3.4 m and x = final spring compression.
So, W" = 1/2k(x₀² - x²)
W" = 1/2 × 154 N/m[(3.4 m)² - x²]
W" = 77 N/m[11.56 m² - x²]
W" = 890.12 - 77x²
So, W = W' + W"
W = 2090.16 - 116.12x + 890.12 - 77x²
W = 2980.28 - 116.12x - 77x²
Since the crate stops, final kinetic energy K₃ = 0. So, ΔK = K₃ - K₂ = 0 - 836.5 J = -836.5 J
Also, ΔK = -W
-836.5 = -(2980.28 - 116.12x - 77x²)
836.5 = 2980.28 - 116.12x - 77x²
77x² + 116.12 -2980.28 + 836.5 = 0
77x² + 116.12x -2143.78 = 0
dividing through by 77, we have
x² + 1.508x -27.841 = 0
Using the quadratic formula to find x, we have
[tex]x = \frac{-1.508 +/-\sqrt{1.508^{2} - 4 X 1 X (-27.841)} }{2 X 1.508} \\x = \frac{-1.508 +/-\sqrt{2.274064 + 111.364} }{3.016} \\x = \frac{-1.508 +/-\sqrt{113.638064} }{3.016} \\x = \frac{-1.508 +/- 10.66}{3.016} \\x = \frac{-1.508 - 10.66}{3.016} or x = \frac{-1.508 + 10.66}{3.016} \\x = \frac{-12.168}{3.016} or x = \frac{9.152}{3.016} \\x = -4.03 or 3.034[/tex]
x = -4.03 or 3.034
Since the compression of the spring is positive, we choose x = 3.034
So, the crate compresses the spring 3.034 m
A jogger travels a route that has two parts. The first is a displacement ->A of 2.20 km due south, and the second involves a displacement ->B that points due east.
(a) The resultant displacement ->A + ->B has a magnitude of 3.81 km.
What is the magnitude of B?
______ km
What is the direction of A + B relative to due south?
_____° west of south or east of south?
(b) Suppose that A - B had a magnitude of 3.81 km. What then would be the magnitude of B and what is the direction of A - B relative to due south?
Answer:
a) B = 3.11 km. θ= 54.7º E of S
b) B = 3.11 km θ= 54.7º W of S
Explanation:
a)
Since we know the value of the total displacement, and the value of the displacement A and its direction, we can find the magnitude of B just applying the Pythagorean Theorem, as follows:[tex]C=\sqrt{(2.2km)^{2} + B^{2} } = 3.81 Km (1)[/tex]
Solving for B, the only unknown, we get:[tex]B=\sqrt{(3.81km)^{2} -(2.2km)^{2} } = 3.11 Km (2)[/tex]
Now, applying some simple trig, we can find the angle that (A+B) makes with the S axis, as follows:[tex]\theta = arc tg (\frac{B}{A} )= arc tg ( \frac{3.11}{2.2} )= arctg (1.414) = 54.7 deg (3)[/tex]
Since it's a positive number, applying the convention that the positive angles are measured counterclockwise, this means that this angle is measured East of South.b)
If the magnitude of A-B is the same than the one for A+B, this means that the magnitude of B remains the same, i.e. 3.11 Km.But if we do graphically A-B, as it is the same as adding A + (-B), we find that the angle of A-B is different to the one in A+B, even the magnitudes of both displacements are the same.In this case, B is a negative number, because it's a displacement due west.So, applying the same trig that for a) we can find the angle that (A-B) makes with the S Axis, as follows:[tex]\theta = arc tg (\frac{-B}{A} )= arc tg ( \frac{-3.11}{2.2} )= arctg (-1.414) = -54.7 deg (4)[/tex]
So, since it is negative, it's measured clockwise from the S axis, so it's 54.7º W of S.Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if the system it is rotated about (a) one end of the rod and (b) the center of the rod?
Explanation:
a) [tex]I=\displaystyle \sum_{i}m_ir_i^2[/tex]
where [tex]r_i[/tex] is the distance of the mass [tex]m_i[/tex] from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,
[tex]I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2[/tex]
b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,
[tex]\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2[/tex]
Help please help I don’t want to fail please
Answer:
B
Explanation:
. Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 15 m/s, and the mass of rain per second striking the roof is 0.071 kg/s. (a) Assuming the drops come to rest after striking the roof, find the average force exerted by the rain on the roof.
Answer:
the average force exerted by the rain on the roof is 1.065 N
Explanation:
Given;
speed of the rainfall, v = 15 m/s
the mass of the rate of the rainfall, m' (m/t) = 0.071 kg
Let the average force exerted by the rain on the roof = F
The average force exerted by the rain on the roof is calculated by Newton's second law of motion;
[tex]F = ma = m \frac{v}{t} = \frac{m}{t} \times v \\\\Recall, \ m' = \frac{m}{t} \\\\F = m'\times v\\\\F = 0.071 \times 15\\\\F = 1.065 \ N[/tex]
Therefore, the average force exerted by the rain on the roof is 1.065 N
What does it mean when work is positive?
O Velocity is greater than kinetic energy.
O Kinetic energy is greater than velocity.
O The environment did work on an object.
O An object did work on the environment.
Answer:
O The environment did work on an object
Explanation:
Used car lots and metal scrapyards use electromagnets to move pieces of metal from one place to another. If the electromagnet/crane operator needed to drop a car to a new space, what would he need to do for this to happen? A) Lift the electromagnet farther away from the pile of cars. B) Reverse the magnetic field by switching the poles of the core. C) Turn off the magnetic field by switching off the electric current. The magnetic field can be decreased by placing it close to another electromagnet
Answer:
C) Turn off the magnetic field by switching off the electric current.
PLEASE MARK THIS BRAINLIEST ❤️
three people are standing by maintaining social distancing i.e 6 feet from each other. A doctor checks one person first then go to the other standing towards the east and then third person towards north. what will be the resultant displacement covered the doctor?
Answer:
8.5 feet.
Explanation:
A sketch of the position of the three people gives a right angled triangle. The hypotenuse of the triangle gives the resultant displacement, while the two other sides are 6 feet each.
The resultant displacement, R, is the overall displacement covered by the doctor. This can be determined by;
R = [tex]\sqrt{d_{1} ^{2} + d_{2} ^{2} }[/tex]
where: [tex]d_{1}[/tex] = 6 feet and [tex]d_{2}[/tex] = 6 feet.
Thus,
R = [tex]\sqrt{6^{2} + 6^{2} }[/tex]
= [tex]\sqrt{72}[/tex]
R = 8.49
The resultant displacement covered by the doctor is 8.5 feet.
16 km
14 km
An ambulance traveled on roads from the hospital
14 kilometers east, then 16 kilometers north to
reach an accident. If the ambulance traveled for
0.40 hours, what was its average speed?
12 kilometers/hour
O 30 kilometers/hour
O 75 kilometers/hour
10 kilometers/hour
Answer:
O 75 kilometers/hour
Explanation:
v = distance/time
v = 30km/0.4h
v = 75km/h
WOODHAVEN MI AND SAGINAW MI
In a separate location, take notes from the sources you have identified. This will take place over two or more days. While taking notes, consider using these reading strategies. From your notes, select one piece of evidence describing a climate similarity or difference that you discovered between these two cities. Write it in the space provided.
Answer:
LOL PUEDO HABLAR CON
Explanation:
Answer:
Whats the anwser/????
Explanation:
Ahmed is pushing a 4 Kg box to the right and Rashid is Pushing it to the right as well with a force of 12 N , the box accelerates by 5 m/s^2. What is the Force that is applied by Ahmed
Answer:
8 N
Explanation:
Applying,
(F'+F) = ma............... Equation 1
Where F' = Amhed's force, F = Rashid's force, m = mass of the box, a = acceleration of the box.
From the question,
Given: F = 12 N, m = 4 kg, a = 5 m/s²
Substitute these values into equation 1
(F'+12) = 4×5
(F'+12) = 20
F' = 20-12
F' = 8 N.
Hence Ahmed's force is 8 N
The total output power of an AM transmitter that is being operated at 50% modulation is measured to be 1800 watts. What is the carrier power?
a. 1440 watts
b. 1600 watts
c. 900 watts
d. 2025 watts
Answer:
b. 1600 watts
Explanation:
Modulation is the process of impressing a low frequency signal unto a high frequency signal so a to transmit over long distance.
If the modulation index (m) for an AM wave is less than 1, modulation would occur with no distortion, if m = 1, modulation occurs but if m > 1, it leads to over modulation and loss in AM envelope.
The total output power of an AM is the sum of the carrier power, upper side band power and lower side band power.
Total output power of an AM [tex](P_{AM})=(\frac{2+m^2}{2} )P_c\\[/tex]
Pc is the carrier power. Given that modulation (m) = 50% = 0.5:
[tex]P_{AM}=\frac{2+m^2}{2} P_c\\\\1800=\frac{2+0.5^2}{2} P_c\\\\1800=1.125P_c\\\\P_c=1600\ watts[/tex]
5. How much heat is generated when an
electrical coil rated 200W is used to heat a
given mass of water for 1hour
Answer:
I HOPE THIS IS CORRECT
Explanation:
Power of water =2 kw=2000w
Mass of water =200kg
difference in temperature ΔT=70−10=60oC
Concept
energy required to heat the water = energy given by water in time t=pt
energy required to increase tempeature of water by 60oC,Q=msΔT
S= specific heat =4200J/kgoC
pt=msΔT
2000×t=200×4200×60
t=25200
or t=25.2×103sec.
The quantity of heat generated is 720,000 Joules or 720 kiloJoules.
Given the following data:
Time = 1 hourPower = 200 WattsTo find the quantity of heat generated when the electrical coil is used:
Note: the quantity of heat generated is energy.
First of all, we would convert the time in hour to seconds.
Conversion:
1 hour = 3600 seconds
Mathematically, the energy of a device is given by the formula;
[tex]Energy = power[/tex] × [tex]time[/tex]
Substituting the values into the formula, we have;
[tex]Energy = 200[/tex] × [tex]3600[/tex]
Energy = 720,000 Joules or 720 kiloJoules.
Therefore, the quantity of heat generated is 720,000 Joules or 720 kiloJoules.
Read more: https://brainly.com/question/23153766
Two long straight wires lie parallel to each other 2.00 cm apart, and carry equal and opposite currents of 145 A. What is the strength of the magnetic field they produce, halfway between them
Answer:
B = 2.9 mT
Explanation:
Given that,
The distance between the wires, r = 2 cm
Distance halfway between them d = 1 cm = 0.01 m
The current in the wires, I = 145 A
We need to find the strength of the magnetic field they produce, halfway between them. The formula for the magnetic field in the wire is given by :
[tex]B=\dfrac{\mu_o I}{2\pi d}\\\\B=\dfrac{4\pi \times 10^{-7}\times 145}{2\pi \times 0.01}\\\\B=2.9\ mT[/tex]
So, the required magnetic field is 2.9 mT.
Can someone help ASAP pleasee
Answer:
C.
Explanation:
The x axis is the origin and the degrees move counter clockwise from there so a 90° flip would be answer C.
During which radioactive decay process does a neutron change into a proton?
A. Alpha decay
B. Gamma decay
C. Beta decay (positron)
D. Beta decay (electron)
A neutron changes into a proton during a radioactive decay process called beta decay (positron), which is option D.
What is beta decay?A beta decay in physics is a nuclear reaction in which a beta particle (electron or positron) is emitted.
A positron is an electron with a positive charge.
During a beta decay, a neutron in the nucleus of the radioactive material suddenly changes into a proton, causing an increase in the atomic number of an element.
Learn more about beta decay at: https://brainly.com/question/25455333
#SPJ1
What speed will the body have 3 seconds after the beginning of free fall?
a.
20 m/s
b.
10 m/s
c.
30 m/s
d.
1 m/s
Answer:
option C
30m/s
Explanation:
preciesly 29.4
Answer:
The answer would be B. 30 m/s
Explanation:
Acceleration due to gravity is 9.8m/s per second.
So 9.8 × 3 = 29.4
closest to 29.4 is 30
A 5.0 Ω resistor is hooked up in series with a 10.0 Ω resistor followed by a 20.0 Ω resistor. The circuit is powered by a 9.0 V battery. Draw a labeled circuit diagram for the circuit described using correct symbols. Calculate the equivalent resistance. Calculate the voltage drop across each resistor in the circuit.
(a) Attached to the response as Figure 1.
(b) 35.0Ω
(c) Across 5.0Ω = 1.3V
Across 10.0Ω = 2.6Ω
Across 20.0Ω = 5.2Ω
Explanation:(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.
(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e
R = R₁ + R₂ + R₃ -------------------(i)
Where;
R₁ = 5.0 Ω
R₂ = 10.0 Ω
R₃ = 20.0 Ω
Substitute these values into equation (i) as follows;
∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω
∴ R = 35.0 Ω
Therefore, the equivalent resistance is ∴ R = 35.0Ω
(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;
i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.
ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;
V = I R ------------------(ii)
Where;
V = Voltage supplied to the circuit = 9.0V
I = Current through the circuit
R = Resistance of the equivalent resistor = 35.0Ω
Substitute these values into equation (ii)
9.0 = I x 35.0
I = [tex]\frac{9.0}{35.0}[/tex]
I = 0.26A
This is also the current flowing through each of the resistors separately.
iii. Calculate the voltage drop across
1. 5.0 Ω resistor
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 5.0Ω resistor
I = current through the 5.0Ω resistor = 0.26A
R = resistance of the 5.0Ω resistor = 5.0Ω
=> V = 0.26 x 5.0
=> V = 1.3V
2. 10.0 Ω resistor
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 10.0Ω resistor
I = current through the 10.0Ω resistor = 0.26A
R = resistance of the 10.0Ω resistor = 10.0Ω
=> V = 0.26 x 10.0
=> V = 2.6V
3. 20.0 Ω resistor
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 20.0Ω resistor
I = current through the 20.0Ω resistor = 0.26A
R = resistance of the 20.0Ω resistor = 10.0Ω
=> V = 0.26 x 20.0
=> V = 5.2V
17
Một hộ gia đình sử dụng điện với dòng điện trung bình là 6 A. Tính điện năng tiêu thụ và tiền điện mà hộ gia đình này phải trả hàng tháng (thuế VAT là
10%). Biết trung bình mỗi ngày sử dụng điện 12 giờ; giá tiền điện cho trong bảng sau (chưa có VAT):
Answer:
English please!!!!!!!!!!!!!!!!!!!!!!!
what is entrapersonal environment
Answer:
The interpersonal environment is considered to be a subset of the organizational environment – defined as the employee’s perception of the practices, policies, and processes of an organization
Explanation:
who is the biggest man in the world
Answer:
Sultan Kösen
here is a pic
examples of frictional forces grade 8
Answer:
Dragging a Chair, Climbing, Rubbing Hands
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. We assume the upward direction to be positive, and the downward direction to be negative.
(a) How long are her feet in the air?(b) What is her highest point above the board?(c) What is her velocity when her feet hit the water?
Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
[tex]2gh = v_f^2 - v_i^2[/tex]
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
[tex](2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\[/tex]
h = 0.82 m
Now, for the time in air during upward motion we use first equation of motion:
[tex]v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s[/tex]
(c)
Now we will consider the downward motion and use the third equation of motion:
[tex]2gh = v_f^2-v_i^2[/tex]
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
[tex]2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\[/tex]
vf = 7.17 m/s
Now, for the time in air during downward motion we use the first equation of motion:
[tex]v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s[/tex]
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
t = 1.14 s
what is the friction
Explanation:
Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are several types of friction: Dry friction is a force that opposes the relative lateral motion of two solid surfaces in
Assume you are going to race the three objects (hollow sphere, disk and ring used in Experiment 8.2) by releasing them from rest at the top of an inclined plane.Which object do you expect to reach the bottom of the inclined plan first? Why?
Answer:
a. The disk
b. Because it has the smallest rotational inertia
Explanation:
a. Which object do you expect to reach the bottom of the inclined plan first?
I would expect the disk to reach the bottom first.
b. Why?
This is because the disk has the smallest rotational inertia.
The rotational inertial of the hollow sphere, disk and ring are 2/3MR², 1/2MR² and MR² respectively.
Since the three objects are rolling from the same height, they have the same mechanical energy.
But, since the disk has the smallest rotational inertia, it would have the smallest rotational kinetic energy and largest translational kinetic energy. The disk's smaller rotational kinetic energy will cause to rotate less but translate more than the other objects and thus reach the bottom first.
The object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.
What is a moment of inertia?Moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.
There are three objects, hollow sphere, disk and ring.
The moment of inertia of the hollow sphere object is given as,[tex]I=\dfrac{2}{3}mr^2[/tex]
The moment of inertia of the ring is,[tex]I=mr^2[/tex]
The moment of inertia of the disk is,[tex]I=\dfrac{1}{2}mr^2[/tex]Here, (m) is the mass and (r) is the radius of the object.
These three objects are going to race by releasing from rest at the top of an inclined plane to the bottom of the plane.
As moment of inertia is the force which acts in the opposite direction of the force of angular acceleration acting on the body.
Thus the less the value of inertia will result in less the time required to reach at the bottom of the inclined plane.
Hence, the object which is expected to reach the bottom of the inclined plan first is the disk, as it has the lowest rotational inertia.
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A set of data was collected measuring the length of a sheet of steel using three different instruments. The actual length of the sheet is 2.00 m. Which instrument is the most accurate in its measurements
Accuracy of measurement is the degree of how close the measurement is, to the actual measurement.
For a measurement of 2.00 m, 1.99 m is more accurate because it is closer to 2.00 m than 1.95 m.
The set of data is not given. So, I will give a general explanation.
For a measurement to be the accurate, the measurement must be close to the actual measurement, and in some cases, the measurement should be approximated to the actual measurement.
Given that:
[tex]Length =2.00m[/tex]
An example of an accurate measurement is:
[tex]Length =1.95m[/tex]
Why? The reasons are
1.95 m is close to 2.00 m1.95 m is approximately 2.00 mAnother example is:
[tex]Length = 1.99m[/tex]
1.99 m is also an accurate measurement because of the reasons above.
However, 1.99 m is more accurate because it is closer to 2.00 m than 1.95 m
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Find the tension in cord A in the figure if the weight of the suspended object is 250 N. the cord is at a 30 degree angle
366 N
183 N
276 N
224 N
Answer:
276 your welcome give me brainliest
