Answers
The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass. Mass is the measure of the amount of matter in a body. Usually, the relationship between mass and weight on Earth is highly proportional; objects that are a hundred times more massive than a one-liter bottle of soda almost always weigh a hundred times more-approximately 1,000 newtons, which is the weight one would expect on Earth from an object with a mass slightly greater than 100 kilograms. In common usage, the mass of an object is often referred to as its weight, though these are in fact different concepts and quantities. In scientific contexts, mass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity. In other words, an object with a mass of 1.0 kilogram weighs approximately 9.81 newtons. Weight and mass are considered to be the same quantities. But many people tend to misuse these terms in their daily conversations. The main difference between weight and mass is that weight is the force of gravity by which the earth attracts towards it whereas mass is the amount of matter in an object.
Related Questions
3. The figure below shows the motion of a car. It starts from the origin, O travels 8m
towards the east and then 12m towards the west.
D
8m.
X
X-8
12m.w
()What is the net displacement D from the origin to the final position?
(ii) What is the total distance travelled by the car?
Answers
Answer:
i. -4m
ii. 20m
Explanation:
The car travels 8m to the east, then travels 12m to the west which is the opposite of the east. Going west, the car travels 8m back to the origin point and then another 4m due west to make 12m. The displacement from the origin point is -4 (the negative sign shows the direction because displacement is a vector quantity)
Total distance = 8m going east + 8m back to origin + 4m west = 20m
3. Three blocks of masses m, 2m and 3m are suspended from the ceiling using ropes as shown in diagram. Which of the following correctly describes the tension in the three rope segments?
a. T1< T2 < T3
b. T1< T2 = T3
c. T1 = T2 = T3
d. T1> T2 > T3
please help.show how and which?
see attachment for more detail.
Answers
Option d (T₁ > T₂ > T₃) correctly describes the tension in the three rope system.
Let's evaluate each tension.
Case T₃.
[tex] T_{3} - W_{3} = 0 [/tex]
For the system to be in equilibrium, the algebraic sum of the tension force (T) and the weight (W) must be equal to zero. The minus sign of W is because it is in the opposite direction of T.
[tex] T_{3} = W_{3} [/tex]
Since W₃ = mg, where m is for mass and g is for the acceleration due to gravity, we have:
[tex] T_{3} = W_{3} = mg [/tex] (1) Case T₂.
[tex] T_{2} - (T_{3} + W_{2}) = 0 [/tex]
[tex] T_{2} = T_{3} + W_{2} [/tex] (2)
By entering W₂ = 2mg and equation (1) into eq (2) we have:
[tex] T_{2} = T_{3} + W_{2} = mg + 2mg = 3mg [/tex]
Case T₁.
[tex] T_{1} - (T_{2} + W_{1}) = 0 [/tex]
[tex] T_{1} = T_{2} + W_{1} [/tex] (3)
Knowing that W₁ = 3mg and T₂ = 3mg, eq (3) is:
[tex] T_{1} = 3mg + 3mg = 6mg [/tex]
Therefore, the correct option is d: T₁ > T₂ > T₃.
Learn more about tension and weight forces here: https://brainly.com/question/18770200?referrer=searchResults
I hope it helps you!
Correct answer: D. [tex]T_{1} > T_{2} > T_{3}[/tex]
First, we must construct the Equations of Equilibrium for each mass based on Newton's Laws of Motion, then we solve the resulting system for every Tension force:
Mass m:
[tex]\Sigma F = T_{3}-m\cdot g = 0[/tex] (1)
Mass 2m:
[tex]\Sigma F = T_{2}-2\cdot m \cdot g -T_{3} = 0[/tex] (2)
Mass 3m:
[tex]\Sigma F = T_{1}-3\cdot m\cdot g - T_{2} = 0[/tex] (3)
The solution of this system is: [tex]T_{3} = m\cdot g[/tex], [tex]T_{2} = 3\cdot m\cdot g[/tex] and [tex]T_{1} = 6\cdot m\cdot g[/tex], which means that [tex]T_{1} > T_{2} > T_{3}[/tex]. (Correct answer: D.)
Why must scientists be careful when studying
nanotechnology?
Answers
Answer:
When studying nanotechnology, scientists must be aware that their ideas may not work out. Their work could be very time consuming and cost a lot of money. Finally, scientists do not yet know all of the effects of nanotechnology on human health.
Hope it helps u:)
How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A) 0.50c
B) 0.65c
C) 0.78c
D) 0.87c
Answers
Answer:
Option (D) is correct.
Explanation:
Let the speed is v.
[tex]\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c[/tex]
Option (D) is correct.
A 1030 kg car has four 12.0 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles
Answers
Answer:
The required fraction is 0.023.
Explanation:
Given that
Mass of a car, m = 1030 kg
Mass of 4 wheels = 12 kg
We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.
The rotational kinetic energy due to four wheel is
[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]
Linear kinetic Energy of the car is:
[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]
Fraction,
[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]
So, the required fraction is 0.023.
An ideal parallel plate capacitor with a cross-sectional area of 0.4 cm2 contains a dielectric with a dielectric constant of 4 and a dielectric strength of 2 x 108 V/m. The separation between the plates of the capacitor is 5 mm. What is the maximum electric charge (in nC) that can be stored in the capacitor before dielectric breakdown
Answers
Answer: [tex]283.2\times 10^{-9}\ nC[/tex]
Explanation:
Given
Cross-sectional area [tex]A=0.4\ cm^2[/tex]
Dielectric constant [tex]k=4[/tex]
Dielectric strength [tex]E=2\times 10^8\ V/m[/tex]
Distance between capacitors [tex]d=5\ mm[/tex]
Maximum charge that can be stored before dielectric breakdown is given by
[tex]\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC[/tex]
Answer:
The maximum charge is 7.08 x 10^-8 C.
Explanation:
Area, A = 0.4 cm^2
K = 4
Electric field, E = 2 x 10^8 V/m
separation, d = 5 mm = 0.005 m
Let the capacitance is C and the charge is q.
[tex]q = CV\\\\q=\frac{\varepsilon o A}{d}\times E d\\\\q = \varepsilon o A E\\\\q = 8.85\times 10^{-12}\times0.4\times 10^{-4}\times 2\times 10^8\\\\q = 7.08\times 10^{-8}C[/tex]
Your forehead can withstand a force of about 6.0 kN before it fractures. Your cheekbone on the other hand can only handle about 1.3 kN before fracturing. If a 140 g baseball hits your head at 30.0 m/s and stops in 0.00150 s,
Required:
a. What is the magnitude of the ball's acceleration?
b. What is the magnitude of the force that stops the baseball?
c. What force does the baseball apply to your head? Explain?
d. Are you in danger of a fracture if the ball hits you in the forehead?
Answers
Answer:
Explanation:
a)
Final velocity v = 0 ; initial velocity u = 30 m/s , time t = .0015 s
v = u + a t
0 = 30 m/s + a x .0015 s
a = - 30 / .0015
= - 20000 m / s²
b )
Magnitude of force = m x a
= .140 kg x 20,000 m / s²
= 2800 N = 2.8 kN.
c )
The force applied by baseball = 2.8 kN .
d )
Since ball can withstand a force of 1.3 kN so it will break if 2.8 kN force acts on it . SO, head will fracture.
What is inertia of motion?
Answers
Explanation:
Inertia of motion
It is also known as Newton's first law of motion.
It states that,
An object remains in a state of rest or of uniform motion in a straight line unless compelled to change its state by an applied external force.
The position of the image obtained by convex lens when object is kept beyond 2F1(F: principal focus of the convex lens)
A. between F2 and 2F2
B. at 2F2
C. beyond 2F2
D. at infinity
Answers
Answer:
Between F2 and 2F2
Explanation:
Diagram attached from Teachoo.
Link to website if you need to refer
https://www.teachoo.com/10838/3118/Convex-Lens---Ray-diagram/category/Concepts/
A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy … (use g = 10 m/s 2)
A) at the moment of impact.
B) 2 seconds after the stone is released.
C) after the stone has fallen 40 m.
D) when the stone is moving at 20 m/s.
At the moment of impact both Kinetic Energy and Potential Energy should be 0, right? So it can't be A), right? Or is this wrong? Is it indeed A)? Please show work and explain it well.
Answers
Answer:
Explanation:
The answer is C because the building is 80 meters high. Before the stone is dropped, it has ONLY potential energy since kinetic energy involves velocity and a still stone has no velocity. At impact, there is no potential energy because potential energy involves the height of the stone relative to the ground and a stone ON the ground has no height; here there is ONLY kinetic.
From the First Law of Thermodynamics, we know that energy cannot be created or destroyed, it can only change form. Therefore, that means that at the halfway point of 40 meters, half of the stone's potential energy has been lost, and it has been lost to kinetic energy. Here, at 40 meters, there is an equality between PE and KE. It only last for however long the stone is AT 40 meters, which is probably a millisecond of time, but that's where they are equal.
Một ống dây điện thẳng dài có lõi sắt, tiết diện ngang của ống S = 20 cm2
, chiều dài
1 m, hệ số tự cảm L = 0,44 H. Cường độ từ trường trong ống dây là H = 0,8.103 A/m. Từ
thông gửi qua tiết diện ngang của ống bằng
3
0
1,6.10 Wb
. Cường độ dòng điện chạy
qua ống dây là
Answers
Answer:
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If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N , how hard does the wall push on you
Answers
Answer:
44 N
Explanation:
Given that,
If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N, then we need to find the force the wall push on you.
It is based on Newton's third law of motion which states that for an action there is an equal and opposite reaction. If the you push the wall with a force of 44 N, the wall push on you is 44 N also as it is based on Newton's third law of motion.
Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.
52 cm4
72 cm4
32 cm4
24 cm4
2 cm4
Answers
Answer:
Minimum Area of rectangle = 24 centimeter²
Explanation:
Given:
Length of rectangle = 6 centimeter
Width of rectangle = 4 centimeter
Find:
Minimum Area of rectangle
Computation:
Minimum Area of rectangle = Length of rectangle x Width of rectangle
Minimum Area of rectangle = 6 x 4
Minimum Area of rectangle = 24 centimeter²
1. A turtle and a rabbit are to have a race. The turtle’s average speed is 0.9 m/s. The rabbit’s average speed is 9 m/s. The distance from the starting line to the finish line is 1500 m. The rabbit decides to let the turtle run before he starts running to give the turtle a head start. If the rabbit started to run 30 minutes after the turtle started, can he win the race? Explain.
Answers
Answer:no
Explanation:because 0.9*(30*60)=0.9*1800=1620
The turtle has already won the race
Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
What will be the speed of the rabbit and the turtle?It is given
[tex]V_{t} = 0.9 \frac{m}{s}[/tex]
[tex]V_{r} = 9 \frac{m}{s}[/tex]
[tex]D=1500 m[/tex]
Time taken by turtle
[tex]T= \dfrac{D}{V_{t} }=\dfrac{1500}{0.9_{} }[/tex]
[tex]T=1666 minutes= 27 hours[/tex]
Time taken by rabbit
[tex]T= \dfrac{D}{V_{r} }=\dfrac{1500}{9_{} }[/tex]
[tex]T=166 minutes[/tex]
since rabbit started 30 minutes after turtle then
[tex]T= 136+30=196 minutes[/tex]
[tex]T= 3.2 hours[/tex]
Hence Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
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A 1,071.628 N painter needs to climb d=1.926 m up a ladder (measured along its length from the point where the ladder contacting the ground), without the ladder slipping. The uniform ladder is 12.014 m long and weighs 250 N. It rests with one end on the ground and the other end against a perfectly smooth vertical wall. The ladder rises at an angle of theta=51.96 degrees above the horizontal floor. What is friction force in unit of N that the floor must exert on the ladder? Use g = 10 m/s2 if you need to .
Answers
The frictional force in unit of N that the floor must exert on the ladder is approximately 232.216 N
The known values are;
The weight of the painter = 1,071.628 N
The height to which the painter needs to climb along the ladder = 1.926 m
The length of the ladder = 12.014 m
The weight of the ladder = 250 N
The points where one of the ladder's ends is resting = On the ground
The points where the other end of the ladder is resting = A perfectly smooth wall
The angle with which the ladder rises above the horizontal floor = 51.96°
The acceleration due to gravity, g ≈ 10 m/s²
The unknown values include;
The friction force that the floor must exert on the ladder
The strategy to be used;
At equilibrium, the sum of moments about a point is zero
Finding the moments about the point of contact where the ladder rests on the wall, P, is given as follows;
At equilibrium, the sum of clockwise, [tex]M_{CW}[/tex], moment about P = The sum of the counterclockwise, [tex]M_{CCW}[/tex]moment about P
[tex]\mathbf{M_{CCW}}[/tex] = (12.014 - 1.926) × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250
[tex]\mathbf{M_{CW}}[/tex] = 12.014 × cos(51.96°) × [tex]\mathbf{F_N}[/tex]
Where;
[tex]\mathbf{F_N}[/tex] = The normal reaction of the of the ground on the end of the ladder that rests on the floor
[tex]\mathbf{M_{CCW}}[/tex] = [tex]\mathbf{M_{CW}}[/tex]
∴ (12.014 - 1.926) × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250 = 12.014 × cos(51.96°) × [tex]F_N[/tex]
We get;
6,665.3068846 N·m = 7.40316448688 m × [tex]F_N[/tex]
[tex]\mathbf{F_N}[/tex] = 6,665.3068846 N·m/(7.40316448688 m) = 900.332135 N
The normal reaction of the floor on the ladder, [tex]\mathbf{F_N}[/tex] = 900.332135 N
Taking moment about the point the ladder rests on the floor, R, gives;
[tex]M_{CCW}[/tex] = 12.014 × sin(51.96°) × [tex]F_W[/tex]
Where;
[tex]\mathbf{F_W}[/tex] = The normal reaction at the wall
[tex]M_{CW}[/tex] = 1.926 × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250
At equilibrium, we have, [tex]M_{CCW}[/tex] = [tex]M_{CW}[/tex]
Therefore;
12.014 × sin(51.96°) × [tex]F_W[/tex] = 1.926 × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250
9.46199511627 m × [tex]F_W[/tex] = 2,197.22861125 N·m
[tex]F_W[/tex] = 2,197.22861125 N·m/(9.46199511627 m)
The reaction of the wall, [tex]\mathbf{F_W}[/tex] = 232.216206 N
We note that also at equilibrium, the sum horizontal forces = 0
The horizontal forces acting on the ladder = The normal reaction on the, [tex]F_W[/tex] wall and the friction force on the ground, [tex]\mathbf{F_f}[/tex]
∴ At equilibrium; [tex]\mathbf{F_W}[/tex] + [tex]\mathbf{F_f}[/tex] = 0
[tex]\mathbf{F_f}[/tex] = -[tex]\mathbf{F_W}[/tex]
[tex]\mathbf{F_W}[/tex] = 232.216206 N
Therefore;
The frictional force in unit of N that the floor must exert on the ladder, [tex]\mathbf{F_f}[/tex] = 232.216206 N ≈ 232.216 N.
(The coefficient of friction, μ = [tex]\mathbf{F_N}[/tex]/[tex]\mathbf{F_W}[/tex] = 900.332135/232.216206 ≈ 3.877).
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An evacuated tube uses an accelerating voltage of 55 kV to accelerate electrons to hit a copper plate and produce x rays. Non-relativistically, what would be the maximum speed of these electrons?
Answers
Answer:
v = 4.4 x 10⁷ m/s
Explanation:
The kinetic energy of the electrons will be equal to the energy supplied by the electric voltage:
Kinetic Energy = Electric Energy
[tex]\frac{1}{2}mv^2 = eV[/tex]
where,
m = mass of electron = 9.1 x 10⁻³¹ kg
v = speed of electron = ?
e = charge on electron = 1.6 x 10⁻¹⁹ C
V =Voltage = 55 kV = 55000 V
Therefore,
[tex]\frac{1}{2}(9.1\ x\ 10^{-31}\ kg)(v)^2 = (1.6\ x\ 10^{-19}\ C)(55000\ V)\\\\v^2 = \frac{(2)(8.8\ x\ 10^{-16}\ J)}{9.1\ x\ 10^{-31}\ kg}\\\\v = \sqrt{19.34\ x\ 10^{14}\ m^2/s^2}[/tex]
v = 4.4 x 10⁷ m/s
8. A boat moving initially at 6.5 km hr due southwest crosses a river that is flowing due south at 3 km hr.
What is the magnitude and direction of the boat relative to the ground? If the river is 1.5 mi wide how long
does it take the boat to cross?
Answers
Answer:
a) v = 8,878 km / h, θ’= 238.8º, b) t = 1890.9 s
Explanation:
a) In this exercise we must find the resulting speed of the boat.
Let's use trigonometry to break down the speed of the boat (v1)
cos 225 = v₁ₓ / v₁
sin 225 = v_{1y} / v₁
v₁ₓ = v₁ soc 225
v_{1y} = v₁ sin 225
v₁ₓ = 6.5 cos 225 = -4.596 km / h
v_{1y} = 6.5 sin 225 = -4.596 km / h
to find the velocity we add each component
vₓ = v₁ₓ
vₓ = - 4,596 km / h
v_y = v_{1y} + v₂
v_y = -4.596 - 3
v_y = - 7,596 km / h
Now let's compose the speed
Let's use the Pythagorean theorem for the module
v = [tex]\sqrt{v_x^2 + v_y^2 }[/tex]
v = Ra 4.596² + 7.596²
v = 8,878 km / h
Let's use trigonometry for the direction
tan θ = v_y / vₓ
θ = tan⁻¹ v_y / vₓ
θ = tan⁻¹ ( [tex]\frac{-7.596}{ -4.596}[/tex] )
θ = 58.8º
measured from the positive side of the x-axis
θ'= 180 + 58.8
θ’= 238.8º
b) Let's reduce the river width to the SI system
x = 1.5 miles (1,609 km / 1 mile) = 2,414 km
to cross the river the speed is on the x axis which is the width of the river
v = x / t
t = x / v
t = 2.414 /4.596
t = 0.525 h
let's reduce to the SI system
t = 0.525 h (3600 s / 1h)
t = 1890.9 s
1) Consider an electric power transmission line that carries a constant electric current of i = 500 A. The cylindrical copper cable used to transmit this current has a diameter o = 2.00 cm and a length L = 150 km. If there are 8.43x10^28 free electrons per cubic meter (m^3 ) in the cable, calculate how long it would take for an electron to cross the entire length of the transmitter line.
Answers
Answer:
t = 1.27 x 10⁹ s
Explanation:
First, we will find the volume of the wire:
Volume = V = AL
where,
A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²
L = Length of wire = 150 km = 150000 m
Therefore,
V = 47.12 m³
Now, we will find the number of electrons in the wire:
No. of electrons = n = (Electrons per unit Volume)(V)
n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)
n = 3.97 x 10³⁰ electrons
Now, we will use the formula of current to find out the time taken by each electron to cross the wire:
[tex]I =\frac{q}{t}[/tex]
where,
t = time = ?
I = current = 500 A
q = total charge = (n)(chareg on one electron)
q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)
q = 6.36 x 10¹¹ C
[tex]500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}[/tex]
Therefore,
t = 1.27 x 10⁹ s
Two point charges, the first with a charge of 4.47 x 10-6 C and the second with a charge of 1.86 x 10-6 C, are separated by 17.4 mm. What is the magnitude of the electrostatic force experienced by charge 2
Answers
Answer: [tex]247.12\ N[/tex]
Explanation:
Given
Magnitude of the charges
[tex]q_1=4.47\times 10^{-6}\ C[/tex]
[tex]q_2=1.86\times 10^{-6}\ C[/tex]
Distance between them [tex]d=17.4\ mm[/tex]
As both charges are of same sign, they must repel each other
Force experienced by second charge is
[tex]\Rightarrow F_{21}=\dfrac{kq_1q_2}{d^2}\\\\\Rightarrow F_{21}=\dfrac{9\times 10^9\times 4.47\times 10^{-6}\times 1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\\Rightarrow F_{21}=\dfrac{74.82\times 10^{-3}}{302.76\times 10^{-6}}\\\\\Rightarrow F_{21}=0.2471\times 10^3\\\Rightarrow F_{21}=247.12\ N[/tex]
Thus, charge 2 experience a force of [tex]247.12\ N[/tex]
Answer:
The force between the two charges is 247.15 N.
Explanation:
Charge, q = 4.47 x 10^-6 C
charge, q' = 1.86 x 10^-6 C
distance, d = 17.4 mm
Let the force is F.
The force is given by the Coulomb's law:
[tex]F = \frac{K q q'}{r^2}\\\\F =\frac{9\times 10^9\times 4.47\times 10^{-6}\times1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\F = 247.15 N[/tex]
When placed 1.18 m apart, the force each exerts on the other is 11.2 N and is repulsive. What is the charge on each
Answers
Answer:
[tex]q=41.62\ \mu C[/tex]
Explanation:
Given that,
Force between two objects, F = 11.2 N
Distance between objects, d = 1.18 m
We need to find the charge on each objects. The force between charges is as follows :
[tex]F=\dfrac{kq^2}{r^2}\\\\q=\sqrt{\dfrac{Fr^2}{k}} \\\\q=\sqrt{\dfrac{11.2\times (1.18)^2}{9\times 10^9}} \\\\q=41.62\ \mu C[/tex]
So, the charge on each sphere is [tex]41.62\ \mu C[/tex].
If the depth of water in a well is 10 m, what is the pressure exerted by it on the
bottom of the well? (Use g = 10 m/s)
[Ans: 10 N/m]
Answers
Answer:
Let d be the density of the water (1000 kg / m^3 eq to 1 gm / cm^3)
P = d g h for the pressure due to a column at the bottom of the column.
P = 1000 kg / m^3 * 10 m/s^2 * 10 m = 10^5 kg / m * s^2 = 10^5 N/m
At what angle torque is half of the max
Answers
An American traveler in China carries a transformer to convert China's standard 220 V to 120 V so that she can use some small appliances on her trip.
a. What is the ratio of turns in the primary and secondary coils of her transformer?
Np / Ns = ____________
b. What is the ratio of input to output current?
Iin /Iout = ___________
c. How could a Chinese person traveling in the United States use this same transformer to power her 220 V appliances from 120 V?
Answers
Answer:
(a) The ratio of turns in the primary and secondary coils of her transformer is 1.833
(b) The ratio of input to output current is 0.55
(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.
Explanation:
Given;
input voltage, [tex]V_p[/tex] = 220 V
output voltage, [tex]V_s[/tex] = 120 V
General transformer equation is given as;
[tex]\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}[/tex]
where;
Np is number of turns in the primary coil
Ns is number of turns in the secondary coil
Is - is the secondary current or output current
Ip - is the primary current or input current
(a) The ratio of turns in the primary and secondary coils of her transformer;
[tex]\frac{N_p}{N_s} = \frac{V_p}{V_s} \\\\\frac{N_p}{N_s} = \frac{220}{120} = 1.833[/tex]
(b) The ratio of input to output current;
[tex]\frac{I_p}{I_s} = \frac{V_s}{V_p} \\\\\frac{I_p}{I_s} = \frac{120}{220} \\\\\frac{I_p}{I_s} = 0.55[/tex]
(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.
find the exit angle relative to the horizontal in an isosceles triangle with 36 °
Answers
what
what
what
what
sorry
sorrry
sorry
A bird has a kinetic energy of 3 J and a potential energy of 25 J. What is the mechanical energy of the bird?
Answers
Answer:
28 j
Explanation:
because when you add you get 28
What is the name of the invisible line that runs
down the center of the axial region?
Answers
Answer:
An axis is an invisible line around which an object rotates, or spins. The points where an axis intersects with an object's surface are the object's North and South Poles.
Explanation:
The Earth's axis is represented by the red line. The white circle represents axial precission, the slow "wobble" of the axis.
An alternating voltage is connected in series with a resistance R and an inductance L If the potential drop across the
resistance is 200 V and across the inductance is 100V
then the applied voltage is
V 223.6
V 2006
V 300
V50
Please help me
Answers
Answer:
oh my God I got really confused right now
A stone is dropped from a bridge. It takes 4s to reach the water below. How high is the bridge above the water?
Answers
Answer:
height is 78.4m
Explanation:
h=u.t + 0.5.g.t^2
= 0 + 0.5x9.8x4^2
= 78.4m
Use the image of Potential vs. position in 1D to match each scenario with subsequent motion.
A (+) charge is placed at A and can only move in the x-direction. When it is released, what will happen?
Correct answer:
It will move to the left
A (-) charge is placed at A and can only move in the x-direction. When it is released, what will happen?
Incorrect answer:
It remains at where it was placed.
A (-) charge is placed at B and can only move in the x-direction. When it is released, what will happen?
Correct answer:
It remains at where it was placed.
A (+) charge is placed at B and pushed slightly to the right; it can only move in the x-direction. What will happen?
Correct answer:
It will move to the right.
A (-) charge is placed at B and pushed slightly to the right; it can only move in the x-direction. What will happen?
Correct answer:
It will oscillate around B
Continuing the previous exercise, determine the nature of work (for each force listed, not net force), KE and PE for:
1. A + charge moving away from a + charge, from rest, under field force only.
KE
[ Select ]
0 PE
[ Select ]
0 Work
[ Select ]
0
2. A + charge moving away from a + charge, from rest, with applied force slowing it.
Work is
[ Select ]
0
3. A - charge moving toward a + charge under field force only.
KE
[ Select ]
0 PE
[ Select ]
0 Work is
[ Select ]
0
4. A - charge moving toward a + charge with applied force slowing it.
Work is
[ Select ]
0
5. An applied force pulls a negative charge away from a positive charge.
Work is
[ Select ]
6. An applied force pushes 2 like charges together.
Work is
[ Select ]
Answers
Answer:
incorporators and it is the one you for the delay to get it for now that the new to me to the same as last week to week in my opinion of your
Electric field is always perpendicular to the equipotential surface.
a. True
b. False
Answers
Answer:
a: true.
Explanation:
We can define an equipotential surface as a surface where the potential at any point of the surface is constant.
For example, for a punctual charge, the equipotential surfaces are spheres centered at the punctual charge.
Or in the case of an infinite plane of charge, the equipotential surfaces will be planes parallel to our plane of charge.
Now we want to see if the electric field is always perpendicular to these equipotential surfaces.
You can see that in the two previous examples this is true, but let's see for a general case.
Now suppose that you have a given field, and you have a test charge in one equipotential surface.
So, now we can move the charge along the equipotential surface because the potential in the surface is constant, then the potential energy of the charge does not change. And because there is no potential change, then there is no work done by the electric field as the charge moves along the equipotential surface.
But the particle is moving and the electric field is acting on the particle, so the only way that the work can be zero is if the force (the one generated by the electric field, which is parallel to the electric field) and the direction of motion are perpendiculars.
Then we can conclude that the electric field will be always perpendicular to the equipotential surfaces.
The correct option is a.
