Answer:
The correct answer is - it is an error in the observation.
Explanation:
Backlash error is the error in the motion that takes place during shifting the direction of gears. A backlash error is an error in the observation which occurs due to the wear and tear of threads of the screw observed that takes place at the time of reversing the direction of rotation of the thimble where the tip of the screw does not move in the opposite direction but remains stationary for a part of the rotation.
Avoid backlash error:
While taking measurements, the screw should be rotated in one direction only.
(A)
(B)
(C)
(D)
Which graph is a quadratic graph?
OA graph A
OB. graph B
OC graph
OD. graph D
A copper (Cu) atom has 29 protons, 34 neutrons, and 29 electrons. What is
copper's atomic number?
O
A. 63
B. 29
C. 58
W
.
D. 34
A superconductor performs best at ______.
Extremely high temperatures
Room temperature
Very cold temperatures
Any temperature
Answer:
c
Explanation:
Superconductors perform best at very cold temperatures because resistivity of this kind of materials decays drastically with temperature. Chromium is most likely to be the best conductor of electricity because it belongs to the Transition Metal group of the periodic table
Long Shore Drift is the ...
A. landforms along the shoreline of an ocean or sea
B. submerged or partly exposed ridge of sand along an ocean front
C. accumulation of sand grains along a beach front
D. movement of material along a coast by waves that approach at an angle to the shore.
Answer: D. movement of material along a coast by waves that approach at an angle to the shore.
Explanation:
Longshore drift is also referred to as the littoral druft and it means the sediment that is moved by the longshore current.
Longshore drift is the movement of material along a coast by waves which approach at an angle to the shore but then recede down the beach.
Therefore, the correct option is D.
A person pushes a 15.7-kg shopping cart at a constant velocity for a distance of 25.9 m on a flat horizontal surface. She pushes in a direction 23.7 ° below the horizontal. A 32.7-N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.
Answer:
a) F = 35.7 N, b) W = 846.7 J, c) W = - 846.9 J, d) W=0
Explanation:
a) For this exercise let's use Newton's second law, let's set a reference frame with the x-axis horizontally
let's break down the pushing force.
cos (-23,7) = Fₓ / F
sin (-237) = F_y / F
Fₓ = F cos 23.7 = F 0.916
F_y = F sin (-23.7) = - F 0.402
Y axis
N- W - F_y = 0
N = W + F 0.402
X axis
Fₓ - fr = 0
F 0.916 = fr
F = fr / 0.916
F = 32.7 / 0.916
F = 35.7 N
It is asked to calculate several jobs
b) the work of the pushing force
W = fx x
W = 35.7 cos 23.7 25.9
W = 846.7 J
c) friction force work
W = F x cos tea
friction force opposes movement
W = - fr x
W = - 32.7 25.9
W = - 846.9 J
d) The work of the force would gravitate, as the displacement and the force of gravity are at 90º, the work is zero
W = 0
a 2kg block of wood starts at rest and slides down a ramp. Its initail height is 12m. if the final velocity of the block is 13m/s, determine the energy of this system that has been turned into heat
Answer:
E = 66.44 J
Explanation:
From the law of conservation of energy:
Total Mechanical Energy at Start = Total Mechanical Energy at the End
Potential Energy at Start = Kinetic Energy at End + Energy Lost
[tex]mgh = \frac{1}{2} mv^2 + E\\\\E = mgh - \frac{1}{2} mv^2\\\\[/tex]
where,
E = Energy turned into heat = ?
m = mass of block = 2 kg
g = acceleration due to gravity = 9.81 m/s²
h = height = 12 m
v = final speed = 13 m/s
Therefore,
[tex]E = (2\ kg)(9.81\ m/s^2)(12\ m)-\frac{1}{2} (2\ kg)(13\ m/s)^2\\\\E = 235.44\ J - 169\ J\\[/tex]
E = 66.44 J
As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force
acting on it is equal to (41 - 9j) N. If the speed of the object at the initial position is 4.0 m/s,
what is its kinetic energy at its final position?
Answer:
Answer:
v_f = 10.38 m / s
Explanation:
For this exercise we can use the relationship between work and kinetic energy
W = ΔK
note that the two quantities are scalars
Work is defined by the relation
W = F. Δx
the bold are vectors. The displacement is
Δx = r_f -r₀
Δx = (11.6 i - 2j) - (4.4 i + 5j)
Δx = (7.2 i - 7 j) m
W = (4 i - 9j). (7.2 i - 7 j)
remember that the dot product
i.i = j.j = 1
i.j = 0
W = 4 7.2 + 9 7
W = 91.8 J
the initial kinetic energy is
Ko = ½ m vo²
Ko = ½ 2.0 4.0²
Ko = 16 J
we substitute in the initial equation
W = K_f - K₀
K_f = W + K₀
½ m v_f² = W + K₀
v_f² = 2 / m (W + K₀)
v_f² = 2/2 (91.8 + 16)
v_f = √107.8
v_f = 10.38 m / s
in a Mercury thermometer the level of Mercury Rises when its bulb comes in contact with a hot object what is the reason for this rise in the level of Mercury
Answer:
because thermometric liquid readily expands on heating or contracts on cooling even for a small difference in the temperature of the body.
1. What did you observe about the magnitudes of the forces on the two charges? Were they the same or different? Does your answer depend on whether the charges were of the same magnitude or different? How does this relate to Newton’s 3rd law?
Answer:
Following are the solution to the given question:
Explanation:
Its strength from both charges is equivalent or identical. The power is equal. And it is passed down
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Therefore, the extent doesn't rely on the fact that charges are the same or different. Newton's third law complies with Electrostatic Charges due to a couple of charges. They are similar in magnitude, and they're in the other way.
[tex]|F_{12}| = |F_{21}|[/tex]
A 0.20-kg block and a 0.25-kg block are connected to each other by a string draped over a pulley that is a solid disk of inertia 0.50 kg and radius 0.10 m. When released, the 0.25-kg block is 0.21 m off the ground. What speed does this block have when it hits the ground?
Answer:
The answer is "0.2711 m/s".
Explanation:
Potential energy = Kinetic energy + Potential energy
[tex]m_1 gh =\frac{1}{2} m_1v^2 +\frac{1}{2} m_2v^2 + \frac{1}{2} I\omega^2 + m_1gh\\\\[/tex]
[tex](m_1- m_2)gh =\frac{1}{2} m_1v^2 +\frac{1}{2} m_2v^2 +\frac{1}{2} I\omega^2\\\\2(m_1 - m_2)gh = m_1v^2 + m_1v^2 + I\omega^2\\\\solid \ disk (I) = \frac{1}{2} \ \ M r^2 \\\\[/tex]
When there is no slipping, \omega =\frac{ v]{r}\\\\
[tex]2(m_1 - m_2)gh = m_1v^2 + m_2v^2 + (\frac{1}{2} Mr^2) (\frac{v}{r})^2\\\\2(m_1 -m_2)gh = m_1v^2 + m_2v^2 + \frac{1}{2} Mv^2\\\\4(m_1 -m_2)gh = 2m_1v^2 + 2m_2v^2 + Mv^2\\\\4(m_1 - m_2)gh = (2m_1 + 2m_2 + M) v^2\\\\[/tex]
[tex]v^2 = \frac{4(m_1 - m_2)gh}{(2m_1 + 2m_2 + M)}v[/tex]
[tex]v^2 = \frac{4 (0.25 \ kg - 0.20 \ kg) (9.8 \frac{m}{s^2}) (0.21 m)}{ (2 \times 0.25 kg + 2 \times 0.20 kg + 0.50 kg)}[/tex]
[tex]=\frac{0.1029}{1.4} \ \ \frac{m^2}{s^2}\\\\=0.0735\ \ \frac{m^2}{s^2}\\\\= 0.2711 \ \frac{m}{s}[/tex]
name 3 properties of solids
-A solid has a definite shape and volume.
-Solids in general have higher density.
-In solids, intermolecular forces are strong.
If you left a glass fiber-optic cable unshielded by any plastic covering, should the light still be able to travel through the cable?
1. Yes
2. No
Answer:
Yes
Explanation:
A 0.390 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 75.0 pC charge on its surface. What is the potential (in V) near its surface
Answer:
V = 346.15 Volts
Explanation:
Given that,
Diameter of the sphere, d = 0.390 cm
Radius, r = 0.195 cm
Charge, [tex]q=75\ pC =75\times 10^{12}\ C[/tex]
The electric potential near its surface is given by :
[tex]V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 75\times 10^{-12}}{0.195\times 10^{-2}}\\\\V=346.15\ V[/tex]
So, the potential near its surface is equal to 346.15 V.
Two long, straight wires are fixed parallel to one another a distance do apart. The wires carry equal constant currents 1, in the same direction. The attractive magnetic force per unit length between them if f = F/L. What is the force per unit length between the wires if their separation is 2d, and each carries current 210? 10 da lo A. f/4 B. f/2 C. 3f/2 D. 2f
Answer:
The option (B) is correct.
Explanation:
The magnetic force between the two current carrying wires is given by
[tex]F =\frac{\mu o}{4\pi}\times \frac{2 I I'}{r}[/tex]
where, I and I' be the currents in the wires and r is the distance between the two wires.
Here, we observe that the force between the wires is inversely proportional to the distance between them.
So, when the distance is doubled , let the new force is F'.
[tex]\frac{F'}{F}=\frac{r}{r'}\\\\\frac{F'}{F}=\frac{d}{2d}\\\\F'=\frac{F}{2}[/tex]
So, option (B) is correct.
Light energy from the Sun reaches an ocean beach, where people are
walking. Which transfer of thermal energy involved in this scenario is an
example of radiation?
Answer:
the correct answers is D
Explanation:
Thermal energy can be transferred by three methods: conduction, convention, and radiation.
Radiation transfer occurs when there is no movement of matter for the exchange of energy.
In this case, checking the correct answers is D
since in this case the transfer is between light and sand without matter exchange
Suppose that the electric field in the Earth's atmosphere is E = 131 N/C, pointing downward. Determine the electric charge (in C) in the Earth. (Assume the electric field is measured at the surface of the Earth. Include the sign of the value in your answer.) Hint
Answer:
the electric charge in the Earth is -5.91 × 10⁵ C
Explanation:
Given the data in the question;
Electric field E = 131 N/C
we know that; radius of the earth r = 6,371 km = 6371000 m
and Coulomb's constant k = 8.99 × 10⁹ Nm²/c²
Now, using the following formula to calculate the charge;
E = k × Q/r²
we make Q the subject of the formula
Q = Er² / k
so we substitute
Q = [ 131 N/C × ( 6371000 m )² ] / 8.99 × 10⁹ Nm²/c²
Q = [ 5.317242971 × 10¹⁵ ] / [ 8.99 × 10⁹ ]
Q = 5.91 × 10⁵ C
Since the electric field pointing downward
Q = -5.91 × 10⁵ C
Therefore, the electric charge in the Earth is -5.91 × 10⁵ C
What do we call air in the motion
Answer:
carbon d
Explanation:
Angles t and v are complementary.angles T has a mesure of (2X+10). Angle v has a measure of 48 what is the value of x
16 because complementary angles equal up to 90. 2x+58 = 90 x= 16
An object accelerates 3 m/s2 , when a force of 6 N acts on it. What is the object’s mass
Answer:
2 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{6}{3} \\ = 2[/tex]
We have the final answer as
2 kgHope this helps you
A truck is driving over a scale at a weight station. When the front wheels drive over the scale, the scale reads 5800 N. When the rear wheels drive over the scale, it reads 6500 N. The distance between the front and rear wheels is 3.20 m Determine the distance between the front wheels and the truck's center of gravity.
Answer:
[tex]x_2=1.60m[/tex]
Explanation:
From the Question We are told that
Initial Force [tex]F_1=5800N[/tex]
Final Force [tex]F_2=6500N[/tex]
Distance between the front and rear wheels \triangle x=3.20 m
Since
[tex]\triangle x=3.20 m[/tex]
Therefore
[tex]x_1+x_2=3.20[/tex]
[tex]x_1=3.20-x_2[/tex]
Generally the equation for The center of mass is at x_2 is mathematically
given by
[tex]x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}[/tex]
[tex]x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}[/tex]
[tex]2*F_1*x_2 =3.20F_1[/tex]
[tex]x_2=1.60m[/tex]
Center of gravity of a body is the sum of its moments divided by the overall weight of the object. The distance between the front wheels and the truck's center of gravity is 1.6 meters.
Given-
Scale reading value when the front wheels drive over the scale [tex]m_{1}[/tex] is 5800 N.
Scale reading value when the rear wheels drive over the scale [tex]m_{2}[/tex] is 6500 N
Distance between the front and rear wheel [tex]\bigtriangleup x[/tex] is 3.20 meters.
Let, the distance between the front wheels and the truck's center of gravity is [tex]x_{2}[/tex].
Since sum of the distance between front wheel to truck's center of gravity [tex]x_{1}[/tex], and rear wheel to truck's center of gravity [tex]x_{2}[/tex], is equal to the distance between the front and rear wheel [tex]\bigtriangleup x[/tex]. Therefore,
[tex]\bigtriangleup x=x_{1} +x_{2}[/tex]
[tex]3.20=x_{1} +x_{2}[/tex]
[tex]x_{1} =3.20-x_{2}[/tex]
For the distance between the front wheels and the truck's center of gravity is the formula of center of gravity can be written as,
[tex]x_{2} =\dfrac{m_{1}x_{1}+m_{2} x_{2} }{m_{1} +m_{2} }[/tex]
[tex]x_{2} =\dfrac{5800\times (3.20- x_{2})+6500\times x_{2} }{5800 +6500 }[/tex]
[tex]1230 x_{2} ={18560-5800 x_{2}+6500 x_{2} }[/tex]
[tex]x_{2}= 1.6[/tex]
Hence, the distance between the front wheels and the truck's center of gravity is 1.6 meters.
For more about the center of gravity, follow the link below-
https://brainly.com/question/20662119
A man is pushing a box of weight W with a forward force of magnitude F. The box
moves forward with a constant speed. What is the magnitude f of the friction force?
Answer:
The magnitude of the friction force is also F.
Explanation:
By the second Newton's law, we know that:
F = m*a
Net force is equal to mass times acceleration.
Here, we know that the box moves with constant speed, thus, the box has no acceleration, then the net force applied to the box is zero.
Also, remember that the friction force is given by:
[tex]F_f = -\mu*W[/tex]
Where mu is the coefficient of friction, and this force opposes to the direction of motion (that coincides with the direction of our forward force, that is why this has a negative sign)
The net force will be equal to the sum of our two horizontal forces (as the weight is already canceled by the normal force)
[tex]F_{total} = F + F_f[/tex]
And this is equal to zero, because we know that the box is non-accelerated.
Then we must have that:
[tex]F_f = -F[/tex]
Then we can conclude that the magnitude of the friction force is F.
If v = 4.00 meters/second and makes an angle of 60° with the positive direction of the y–axis, what is the magnitude of vx?
How large must the coefficient of static friction be between the tires and road if a car is to round a level curve of radius 125 m at a speed of 95km/h
Answer:
673km
Explanation:
Người ta đúc một khối nhóm (A) và một khối sắt ( (Fe) hình khối với cũng khói lượng 5 kg Khối lượng riêng (2) của Al và Fe là 2 rho ki =2.7*10^ 3 kg/m^ 3 ; mu 4 =7,8*10^ 3; kg/m^ 3 Hỏi, khối vật liệu nào có kích thước lớn hơn? (a) khối nhằm (b) khối sắt (c) hai khối có cùng kích cỡ
Calculate conductance of a conduit the cross-sectional area of which is 1.5 cm2 and the length of which is 9.5 cm, given that its conductivity is 0.65 ohm-1 cm-1.
0.15 ohm-1
0.10 ohm-1
1.2 ohm-1
7.5 ohm-1
Answer:
1.2
Explanation:
You instead want to make sure the battery for your string of lights will last as long as possible. A battery will last longer if it powers a circuit with low current. How could you hook up a battery and 2 light bulbs so the least amount of current flows through the battery? Use the measurement tools in the simulation to check your design.
Answer:
Put the 2 light bulbs in series.
Explanation:
The resistance will be the greatest if you hook up the light bulb in series, and since resistance and current are inversely proportional, the current will be the least as well.
An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and the drag coefficient is 0.065. The mass of the aircraft is 900 kg. Calculate the effective lift area for the aircraft and the required engine thrust and power to maintain level flight.
Answer:
- the effective lift area for the aircraft is 8.30 m²
- the required engine thrust is 1275 N
- required power is 79.7 kW
Explanation:
Given the data in the question;
Speed V = 225 km/hr = 62.5 m/s
The lift coefficient CL = 0.45
drag coefficient CD = 0.065
mass = 900 kg
g = 9.81 m/s²
a) the effective lift area for the aircraft
we know that for a steady level flight, weight = lift and thrust = drag
Using the equation for the lift force
F[tex]_L[/tex] = C[tex]_L[/tex][tex]\frac{1}{2}[/tex]ρV²A = W
we substitute
0.45 × [tex]\frac{1}{2}[/tex] × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )
1081.05 × A = 8829
A = 8829 / 1081.05
A = 8.30 m²
Therefore, the effective lift area for the aircraft is 8.30 m²
b) the required engine thrust and power to maintain level flight.
we use the expression for drag force
F[tex]_D[/tex] = T = C[tex]_D[/tex][tex]\frac{1}{2}[/tex]ρV²A
we substitute
= 0.065 × [tex]\frac{1}{2}[/tex] × 1.21 × ( 62.5 )² × 8.30
T = 1275 N
Since drag and thrust force are the same,
Therefore, the required engine thrust is 1275 N
Power required;
P = TV
p = 1275 × 62.5
p = 79687.5 W
p = ( 79687.5 / 1000 )kW
p = 79.7 kW
Therefore, required power is 79.7 kW
A firefly glows by the direct conversion of chemical energy to light. The light emitted by a firefly has peak intensity at a wavelength of 550 nm. Part A What is the minimum chemical energy, in eV, required to generate each photon
Answer:
Explanation:
The energy of a photon is given by the Planck relation
E = h f
the speed of light is related to wavelength and frequency
c = λ f
f- c /λ
we substitute
E = h c /λ
let's calculate
E = 6.63 10-34 3 10⁸ / 550 10-9
E = 3.616 10-19 J
let's reduce to eV
E = 3.616 10-19 J (1 eV / 1.6 10-19)
E = 2.26 eV
The Earth is a small, rocky planet. It is known as one of the terrestrial planets.
A planet that is similar to the Earth is
A. Uranus
B. Saturn
C. Jupiter
D. Venus
Answer:
D Venus
Explanation:
in terms of size, average, density, mass and surface gravity venus is similar to earth
A rocket moving around the earth at height "H", If the gravitational acceleration "g1" at height
His of gravitational acceleration 'g at earth surface. If Earth radius is "R", find "H"
using R
Answer:
At the earth's surface g = G M / R^2 where G is the gravitational constant
at H g1 = G M / (R + H)^2 using Gauss' theorem for enclosed mass
g1 = G M / (R^2 + 2 R H) ignoring H^2 as it is small compared to R^2
g / g1 = (R^2 + 2 R H) / R^2 = 1 + 2 R H
g = g1 + 2 R H g1
g1 - g = - 2 R H or H = (g1 - g) / 2 R