Answer:
a) U = 506 J, b) U = 37.11 J, c) U = 0
Explanation:
The gravitational power energy is given by the expression
U = m g (y -y₀)
In general, a reference system is set that allows the expression to be simplified, in this case let's assume the reference system at the child's lowest point, therefore y₀ = 0
Let's use trigonometry to find the child's height
h = y = L - L cos θ
we substitute
U = m g L (1 - cos θ)
a) when the chain is horizontal θ = 90 and cos 90 = 0
U = mg L
weight and mass are related
W = mg
m = W / g
U = 230 2.20
U = 506 J
b) θ = 33.0º
cos 33 = 0.83867
U = 230 (1 - 0.83867)
U = 37.11 J
c) in this case θ = 0 cos 0 = 1
U = 0
Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by strings with the same tension.a. Compare the final kinetic energies of pucks A and B. b. Compare the final speeds of pucks A and B.
Answer:
(a) 1 : 2
(b) same
Explanation:
Let the mass of puck A is m and the mass of puck B is 2 m.
initial speed for both the pucks is same as u and the distance is same for both is s.
let the tension is T for same.
The kinetic energy is given by
[tex]K = 0.5 mv^2[/tex]
(a) As the speed is same, so the kinetic energy depends on the mass.
So, kinetic energy of A : Kinetic energy of B = m : 2m = 1 : 2
(b) A the distance s same so the final velocities are also same.
(a) The kinetic energy of puck B is 2 times the kinetic energy of puck A.
(b) The final speed of both the puck A and B are same.
Let the mass of puck A is m and the mass of puck B is 2 m.
Initial speed for both the pucks is same as u and the distance is same for both is s.
Let the tension is T for same.
Then, the kinetic energy is given as,
[tex]KE = \dfrac{1}{2}mv^{2}[/tex]
(a)
As the speed is same, so the kinetic energy depends on the mass.
Then,
[tex]\dfrac{KE_{A}}{KE_{B}} = \dfrac{1/2 \times mv^{2}}{1/2 \times (2m)v^{2}}\\\\\\\dfrac{KE_{A}}{KE_{B}} =\dfrac{1}{2}[/tex]
So, kinetic energy of A : Kinetic energy of B = 1 : 2.
Thus, we can conclude that the kinetic energy of puck B is 2 times the kinetic energy of puck A.
(b)
The final speed for the puck is given as,
v = s/t
here, s is the distance covered.
Since, both pucks are pulled the same distance across frictionless ice. Then, the final speed of each puck is also same.
Thus, we can conclude that the final speed of both the puck A and B are same.
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Find the refractive index of a medium
having a velocity of 1.5 x 10^8*
Explanation:
someone to check if the answer is correct
NEED HELP ASAP- Please show work
The angular position of an object is given by θ = 4t3 +10t −40 , where θ is in radians and t is in seconds what is:
(a) (5 points) The angular velocity at t = 2 s?
(b) (5 points) The angular acceleration at t = 2 s?
Answer:
Look at work
Explanation:
Θ= 4t^3+10t-40
a) In order to find ω, we need to find displacement so plug in t=2 to find Θ.
Θ= 4*8+20-40=12
use ω=Θ/t
Plug in values
ω=6 rad/s
b) In order to find α we use ω/t.
Plug in values
α=6/2= 3 rad/s^2
A
Fluids in which the shear stress must reach
certain minimum value(yield stress)
before flow commences are called
Answer:
Plastic
Explanation:
Shear Modulus can be defined as the ratio of shear stress to shear strain with respect to a physical object.
This ultimately implies that, Shear Modulus arises as a result of the application of a shear force on an object or body which eventually leads to its deformation. Thus, this phenomenon is simply used by scientists to measure or determine the rigidity of an object or body.
Fluids in which the shear stress must reach certain minimum value (yield stress) before flow commences are called plastic. Thus, a plastic would only begin to flow when its shear stress attain a certain minimum value (yield stress). The unit of measurement of yield stress is usually mega pascal (MPa).
suppose a car of 1200kg is moving with a velocity of 40km/hr therefore its kinetic energy is not zero. 1. explain briefly what happens to its kinetic energy when the driver applies the breaks and the car stops
Answer:
Explanation:
For starters begin with a warning not to touch the brake drums. All of the KE is transferred to the brake drums. The result is a large rise in temperature. Heat. If you press hard on the brakes, rubber is left on the road and there is heat involved in that too.
Answer:
KInetic energy reduces.
Explanation:
Application of breaks reduces velocity. Reduction of velocity constitutes velocity reduction.
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You are driving in such a way that the car is accelerating at a constant rate in the positive direction. When you pass the first sign, you are traveling at 4 m/s. When you pass the second sign 50 m down the road, you note that the seconds indicator of your clock reads 45 seconds. You also note that your velocity is now 9 m/s.
Required:
a. What is your acceleration?
b. What was the clock’s seconds indicator reading when you passed the first sign?
Answer:
Explanation:
a)
v² = u² + 2 a s
v = 9 m/s
u = 4 m/s
s = 50 m
9² = 4² + 2 x a x 50
a = 0.65 m /s²
Acceleration is 0.65 m /s²
b )
time elapsed before velocity changed from 4 m/s to 9 m/s with acceleration of .65 m /s ²
(v - u ) / t = a
(v - u ) / a = t
(9 - 4 ) / .65 = t
t = 7.7
time when passing the first sign will be 7.7 s earlier .
Reading of time indicator = 45 - 7.7
= 37.3 seconds.
Answer:
(a) 0.45 m/s^2
(b) 33.9 s
Explanation:
initial velocity, u = 4 m/s
final velocity, v = 9 m/s
distance, s = 50 m
(a) Let the acceleration is a.
Use third equation of motion
[tex]v^2 = u^2 + 2 as \\\\9^2 = 4^2 + 2\times a\times 50\\\\a = 0.45 m/s^2[/tex]
(b) Let the time is t.
Use first equation of motion
v = u + at
9 = 4 + 0.45 x t
t = 11.1 s
So, the initial time, t' = 45 - 11.1 = 33.9 s
A mass-spring system oscillates with an amplitude of 4.20 cm. If the spring constant is 262 N/m and the mass is 560 g, determine the mechanical energy of the system.
Answer:
[tex]M.E=41J[/tex]
Explanation:
From the question we are told that:
Amplitude [tex]a=4.20cm[/tex]
Spring Constant [tex]K=262N/m[/tex]
Mass [tex]m=560g[/tex]
Generally the equation for mechanical energy is mathematically given by
[tex]M.E=\frac{1}{2}km^2[/tex]
[tex]M.E=0.5*262*0.56^2[/tex]
[tex]M.E=41J[/tex]
David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0 m/s2 at the instant when David passes. How far does Tina drive before passing David?
Which simple machine is shown in the diagram?
a wedge
a screw
an inclined plane
a wheel and axle
Answer:
Wheel and axle
Explanation:
Which simple machine is shown in the diagram?
a wheel and axle
From the given diagram, the machine shown is actually a wheel and axle
Description of wheel and axle
The wheel and axle is a machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.
Answer:
Wheel and axle
Explanation:
Part C – RC Circuits in AC Mode 1. Derive Equation 5-6 from Equation 5-5. 2. Using the τ’s you calculated and your measured resistance: a. Calculate the capacitances of the capacitors. b. Compare your calculated and measured values via percent error.
Answer: hi your question is incomplete attached below is the complete question
1) attached below
2) a) 31 Ω, 302.9 Ω
b) 17.3 Ω , 26.4 Ω
Explanation:
1) Deriving Eqn 5-6 from Eqn 5-5
attached below
2) using τ’s calculated and measured resistance
use given data
Tm1 = Rm1 * Cm1
= 329.3 * 333 * 10^-9 = 109.65 μs
Tm2 = Rm2 * Cm2
= 329.3 * 200 * 10^-9 = 658.6 μs
a) Capacitance of capacitors
For Cm1
t₁₂ = 72 μs = R₁' Cm1 log²
∴ R₁' = ( 72 * 10^-6 ) / ( 333 * 10^-9 log² )
= 31 Ω
For Cm2
t₁₂ = 40 μs , ∴ R₂' = 302.9 Ω
b) comparing calculated and measured values via percent error
errors
Rm1 - R₁' = 17.3 Ω
Rm2 - R₂' = 26.4 Ω
Light energy is part of a larger form of energy known as __________.
Light energy is part of a larger form of energy known as electromagnetic energy. Details about electromagnetic energy can be found below.
What is electromagnetic radiation?Electromagnetic spectrum is the entire range of wavelengths of all known electromagnetic radiations extending from gamma rays through visible light, infrared, and radio waves, to X-rays.
Visible light is the part of the electromagnetic spectrum, between infrared and ultraviolet, that is visible to the human eye.
Therefore, Light energy is part of a larger form of energy known as electromagnetic energy.
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A 10.0 L tank contains 0.329 kg of helium at 28.0 ∘C. The molar mass of helium is 4.00 g/mol . Part A How many moles of helium are in the tank? Express your answer in moles.
Answer:
82.25 moles of He
Explanation:
From the question given above, the following data were obtained:
Volume (V) = 10 L
Mass of He = 0.329 Kg
Temperature (T) = 28.0 °C
Molar mass of He = 4 g/mol
Mole of He =?
Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:
1 Kg = 1000 g
Therefore,
0.329 Kg = 0.329 Kg × 1000 g / 1 Kg
0.329 Kg = 329 g
Thus, 0.329 Kg is equivalent to 329 g.
Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:
Mass of He = 329 g
Molar mass of He = 4 g/mol
Mole of He =?
Mole = mass / molar mass
Mole of He = 329 / 4
Mole of He = 82.25 moles
Therefore, there are 82.25 moles of He in the tank.
Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the coefficient of kinetic friction between each of the blocks and the surface is 0.30, determine the magnitude of the force exerted on the 2.0-kg block by the 3.0-kg block.
I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)
Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.
The 2.0-kg block feels
• the downward pull of its own weight, (2.0 kg) g
• the upward normal force of the surface, magnitude n₁
• kinetic friction, mag. f₁ = 0.30n₁, pointing in the negative horizontal direction
• the contact force of the larger block, mag. c₁, also pointing in the negative horizontal direction
• the applied force, mag. F, pointing in the positive horizontal direction
Meanwhile the 3.0-kg block feels
• its own weight, (3.0 kg) g, pointing downward
• normal force, mag. n₂, pointing upward
• kinetic friction, mag. f₂ = 0.30n₂, pointing in the negative horizontal direction
• contact force from the smaller block, mag. c₂, pointing in the positive horizontal direction (this is the force that is causing the larger block to move)
Notice the contact forces form an action-reaction pair, so that c₁ = c₂, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:
• net vertical force:
n₂ - (3.0 kg) g = 0 ==> n₂ = (3.0 kg) g ==> f₂ = 0.30 (3.0 kg) g
• net horizontal force:
c₂ - f₂ = 0 ==> c₂ = 0.30 (3.0 kg) g ≈ 8.8 N
what will be the gravitational force between two heavenly bodies if the masses of both are tripled keeping the distance between them constant
Answer:
If the mass of one of the objects is tripled, then the force of gravity between them is tripled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces
A cannon and a supply of cannonballs are inside a sealed railroad car of length L, as in Fig. 7-33. The cannon fires to the right; the car recoils to the left. The cannonballs remain in the car after hitting the far wall. (a) After all the cannonballs have been fired, what is the greatest distance the car can have moved from its original position
Answer:
Initially let n cannonballs with a total mass of m be to the left of the center of mass at L /2 and the mass of the car at L/2
x1 = [-m / (m + M)] * L / 2 is the original position of the CM
x2 = (m (x + L/2) + M x) / (m + M) * L/2 final position of CM with all cannon balls to the right
[-m x - m L / 2 + m x - M x] / (M + m) * L/2
= - ( m L / 2 + M x) / (m + M) * L/2 = Xcm
Check the math, but maximum distance occurs when the cannonballs of mass m move from -L/2 to L/2 and the car of mass M moves from zero to -x
The gravitational field strength due to its planet is 5N/kg What does it mean?
Answer:
The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.
Weight can be calculated using the equation:
weight = mass × gravitational field strength
This is when:
weight (W) is measured in newtons (N)
mass (m) is measured in kilograms (kg)
gravitational field strength (g) is measured in newtons per kilogram (N/kg)
An electron has an initial speed of 8.06 x10^6 m/s in a uniform 5.60 x 10^5 N/C strength electic field.The field accelerates the electron in the direction opposite to its initial velocity.
(a) What is the direction of the electric field?
i. opposite
ii. direction to the electron's initial velocity
iii. same direction as the electron's initial velocity
iv. not enough information to decide
(b) How far does the electron travel before coming to rest? m
(c) How long does it take the electron to come to rest? s
(d) What is the electron's speed when it returns to its starting point?
Answer:
Explanation:
a)
The force on electron acts opposite to the velocity , and direction of force on electron is always opposite to direction of electric field .
Hence direction of electric field must be in the same in which electrons travels.
Hence option iii is correct.
b )
deceleration a = force / mass
= qE / m
= 1.6 x 10⁻¹⁶ x 5.6 x 10⁵ / 9.1 x 10⁻³¹
= .98 x 10²⁰ m /s²
v² = u² - 2 a s
0 = (8.06 x 10⁶ )² - 2 x .98 x 10²⁰ s
s = 64.96 x 10¹² / 1.96 x 10²⁰
= 33.14 x 10⁻⁸ m
c ) time required
= 8.06 x 10⁶ / .98 x 10²⁰
= 8.22 x 10⁻¹² s .
d ) Its speed will be same as that in the beginning ie 8.06 x 10⁶ m/s .
Answer:
(a) Option (i)
(b) 6.6 x 10^-4 m
(c) 8.2 x 10^-11 s
Explanation:
initial velocity, u = 8 .06 x 10^6 m/s
Electric field, E = 5.6 x 10^5 N/C
(a) The direction of field is opposite.
Option (i).
(b) Let the distance is s.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times \frac{qE}{m}\times s\\\\8.06\times 10^6\times 8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} s\\\\s = 6.6\times 10^{-4} m[/tex]
(c) Let the time is t.
Use first equation of motion.
[tex]v = u + a t \\\\0 = u - \times \frac{qE}{m}\times t\\\\8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} t\\\\t = 8.2\times 10^{-11} s[/tex]
3
Select the correct answer.
What is a substance?
Answer:
physical material from which something is made or which has discrete existence
Explanation:
The earth's radius is about 4000 miles. Kampala, the capital of Uganda, and Singapore are both nearly on the equator. The distance between them is 5000 miles as measured along the earth's surface.
a. Through what angle do you turn, relative to the earth, if you fly from Kampala to Singapore? Give your answer in both radians and degrees.
b. The flight from Kampala to Singapore take 9 hours. What is the plane's angular speed relative to the earth?
Answer:
a) the required angle in both radian and degree is 1.25 rad and 71.6°
b) the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec
Explanation:
Given the data in the question;
a)
we know that The expression for the angle subtended by an arc of circle at the center of the circle is,
θ = Length / radius
given that Length is 5000 miles and radius is 4000 miles
we substitute
θ = 5000 miles / 4000 miles
θ = 1.25 rad
Radian to Degree
θ = 1.25 rad × ( 180° / π rad )
θ = 71.6°
Therefore, required angle in both radian and degree is 1.25 rad and 71.6°
b)
The flight from Kampala to Singapore take 9 hours.
the plane's angular speed relative to the earth = ?
we know that, the relation between angular velocity and angular displacement is;
ω = θ / t
given that θ is 1.25 rads and time t is 9 hours or ( 9 × 3600 sec ) = 32400 sec
we substitute
ω = 1.25 rad / 32400 sec
ω = 3.86 × 10⁻⁵ rad/sec
Therefore, the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec
A uniform steel rod of length 0.9 m and mass 3.8 kg has two point masses of 2.3 kg each at the two ends. Calculate the moment of inertia of the system about an axis perpendicular to the rod, and passing through its center.
Answer: [tex]2.4705\ kg.m^2[/tex]
Explanation:
Given
length of the rod is L=0.9 m
Mass of the rod m=3.8 kg
Point masses has mass of m=2.3 kg
Moment of Inertia of the rod about the center is
[tex]\Rightarrow I_o=\dfrac{1}{12}ML^2[/tex]
Moment of inertia of combined system is the sum of rod and two point masses.
[tex]\Rightarrow I=I_o+2mr^2[/tex]
[tex]\Rightarrow I=\dfrac{1}{12}3.8\times 0.9^2+2\times 2.3\times \left(\dfrac{0.9}{2}\right)^2\\\\\Rightarrow I=1.539+0.9315\\\Rightarrow I=2.4705\ kg-m^2[/tex]
Electrical resistance is a measure of resistance to the flow of _?____
Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.
Hope this helps!!!!
Answer:
electric current
Explanation:
The answer is electric current
1. A block of mass m = 10.0 kg is released with a speed v from a frictionless incline at height 7.00 m. The
block reaches the horizontal ground and then slides up another frictionless incline as shown in Fig. 1.1. If the
horizontal surface is also frictionless and the maximum height that the block can slide up to is 26.0 m, (a) what
is the speed v of the block equal to when it is released and (b) what is the speed of the block when it reaches
the horizontal ground? If a portion of length 1 2.00 m on the horizontal surface is frictional with coefficient
of kinetic friction uk = 0.500 (Fig. 1.2) and the block is released at the same height 7.00 m with the same
speed v determined in (a), (c) what is the maximum height that the block can reach, (d) what is the speed of the
block at half of the maximum height, and (e) how many times will the block cross the frictional region before
it stops completely?
1 = 2.00 m (frictional region)
Let A be the position of the block at the top of the first incline; B its position at the bottom of the first incline; C its position at the bottom of the second incline; and D its position at the top of the second incline. I'll denote the energy of the block at a given point by E (point).
At point A, the block has total energy
E (A) = (10.0 kg) (9.80 m/s²) (7.00 m) + 1/2 (10.0 kg) v₀²
E (A) = 686 J + 1/2 (10.0 kg) v₀²
At point B, the block's potential energy is converted into kinetic energy, so that its total energy is
E (B) = 1/2 (10.0 kg) v₁²
The block then slides over the horizontal surface with constant speed v₁ until it reaches point C and slides up a maximum height of 26.0 m to point D. Its total energy at D is purely potential energy,
E (D) = (10.0 kg) (9.80 m/s²) (26.0 m) = 2548 J
Throughout this whole process, energy is conserved, so
E (A) = E (B) = E (C) = E (D)
(a) Solve for v₀ :
686 J + 1/2 (10.0 kg) v₀² = 2548 J
==> v₀ ≈ 19.3 m/s
(b) Solve for v₁ :
1/2 (10.0 kg) v₁² = 2548 J
==> v₁ ≈ 22.6 m/s
Now if the horizontal surface is not frictionless, kinetic friction will contribute some negative work to slow down the block between points C and D. Check the net forces acting on the block over this region:
• net horizontal force:
∑ F = -f = ma
• net vertical force:
∑ F = n - mg = 0
where f is the magnitude of kinetic friction, a is the block's acceleration, n is the mag. of the normal force, and mg is the block's weight. Solve for a :
n = mg = (10.0 kg) (9.80 m/s²) = 98.0 N
f = µn = 0.500 (98.0 N) = 49.0 N
==> - (49.0 N) = (10.0 kg) a
==> a = - 4.90 m/s²
The block decelerates uniformly over a distance 2.00 m and slows down to a speed v₂ such that
v₂² - v₁² = 2 (-4.90 m/s²) (2.00 m)
==> v₂² = 490 m²/s²
and thus the block has total/kinetic energy
E (C) = 1/2 (10.0 kg) v₂² = 2450 J
(c) The block then slides a height h up the frictionless incline to D, where its kinetic energy is again converted to potential energy. With no friction, E (C) = E (D), so
2450 J = (10.0 kg) (9.80 m/s²) h
==> h = 25.0 m
(d) At half the maximum height, the block has speed v₃ such that
2450 J = (10.0 kg) (9.80 m/s²) (h/2) + 1/2 (10.0 kg) v₃²
==> v₃ ≈ 15.7 m/s
The block loses speed and thus energy as it moves between B and C, but its energy is conserved elsewhere. If we ignore the inclines and pretend that the block is sliding over a long horizontal surface, then its velocity v at time t is given by
v = v₁ + at = 22.6 m/s - (4.90 m/s²) t
The block comes to a rest when v = 0 :
0 = 22.6 m/s - (4.90 m/s²) t
==> t ≈ 4.61 s
It covers a distance x after time t of
x = v₁t + 1/2 at ²
so when it comes to a complete stop, it will have moved a distance of
x = (22.6 m/s) (4.61 s) + 1/2 (-4.90 m/s²) (4.61 s)² = 52.0 m
(e) The block crosses the rough region
(52.0 m) / (2.00 m) = 26 times
need help pleaseee,question is in the pic
Explanation:
For engine 1,
Energy removed = 239 J
Energy added = 567 J
[tex]\eta_1=\dfrac{239}{567}\cdot100=42.15\%[/tex]
For engine 2,
Energy removed = 457 J
Energy added = 789 J
[tex]\eta_2=\dfrac{457}{789}\cdot100=57.92\%[/tex]
For engine 3,
Energy removed = 422 J
Energy added = 1038 J
[tex]\eta_3=\dfrac{422}{1038}\cdot100=40.65\%[/tex]
So, the engine 2 has the highest thermal efficiency.
You place an 8 kg ball on the top of your 2 cm^2 finger tip. Calculate the
PRESSURE. Show MATH, answer and unit.
Answer:
the pressure exerted by the object is 392,000 N/m²
Explanation:
Given;
mass of the object, m = 8 kg
area of your finger, A = 2 cm² = 2.0 x 10⁻⁴ m²
acceleration due to gravity, g = 9.8 m/s²
The pressure exerted by the object is calculated as;
[tex]Pressure = \frac{F}{A} = \frac{mg}{A} = \frac{8 \times 9.8}{2\times 10^{-4}} = 392,000 \ N/m^2[/tex]
Therefore, the pressure exerted by the object is 392,000 N/m²
A proton enters a region of constant magnetic field, perpendicular to the field and after being accelerated from rest by an electric field through an electrical potential difference of 330 V. Determine the magnitude of the magnetic field, if the proton travels in a circular path with a radius of 23 cm.
Answer:
B = 1.1413 10⁻² T
Explanation:
We use energy concepts to calculate the proton velocity
starting point. When entering the electric field
Em₀ = U = q V
final point. Right out of the electric field
em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
q V = ½ m v²
v = [tex]\sqrt{2qV/m}[/tex]
we calculate
v = [tex]\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }[/tex]
v = [tex]\sqrt{632.3353 \ 10^8}[/tex]
v = 25.15 10⁴ m / s
now enters the region with magnetic field, so it is subjected to a magnetic force
F = m a
the force is
F = q v x B
as the velocity is perpendicular to the magnetic field
F = q v B
acceleration is centripetal
a = v² / r
we substitute
qvB =1/2 m v² / r
B = v[tex]\frac{m v}{2 q r}[/tex]
we calculate
B = [tex]\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}[/tex]
B = 1.1413 10⁻² T
a) Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at 5.00 x10^7 m/s in a circular path 2.00m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge.b) Is this field strength obtainable with today's technology or is it a futuristic possibility?
Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 2 m/s, exactly how fast (in m2/s) is the area of the spill increasing when the radius is 39 m?
Explanation:
The area of a circle of radius r is given by
[tex]A = \pi r^2[/tex]
Taking the derivative of A with respect to time t, we get
[tex]\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}[/tex]
We also know that
[tex]\dfrac{dr}{dt} = 2\:\text{m/s}\:\text{at}\:r = 39\:\text{m}[/tex]
[tex]\dfrac{dA}{dt} = 2\pi (39\:\text{m})(2\:\text{m/s})= 490\:\text{m}^2\text{/s}[/tex]
Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
Take the coefficient of expansion to be 25 x 10-6 (oC)-1 . Write the answer in meters with three significant figures
Answer:
ΔL = 3.82 10⁻⁴ m
Explanation:
This is a thermal expansion exercise
ΔL = α L₀ ΔT
ΔT = T_f - T₀
where ΔL is the change in length and ΔT is the change in temperature
Let's reduce the length to SI units
L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m
let's calculate
ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))
ΔL = 3.8236 10⁻⁴ m
using the criterion of three significant figures
ΔL = 3.82 10⁻⁴ m
Which of the following categories of motion is mutually exclusive with each of the others? A. Translational motion B. Rectilinear motion C. Rotational motion D. Curvilinear motion
Answer:
C. Rotational motion
Explanation:
The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It only describes motion—it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion: v = v+at (constant a).
Rotational motion is mutually exclusive with each of the others. Hence, option (C) is correct.
What is Rotational motion?"The motion of an object around a circular route, in a fixed orbit, is referred to as rotational motion."
Rotational motion dynamics are identical to linear or translational dynamics in every way. The motion equations for linear motion share many similarities with the equations for the mechanics of rotating objects. Rotational motion only takes stiff bodies into account. A massed object that maintains a rigid shape is referred to as a rigid body.
What is Curvilinear motion?Curvilinear motion is the movement of an object along a curved route. Example: A stone hurled at an angle into the air.
The motion of a moving particle that follows a predetermined or known curve is referred to as curvilinear motion. Two coordinate systems—one for planar motion and the other for cylindrical motion—are used to examine this type of motion.
Learn more about motion here:
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When you hammer a nail into wood, the nail heats up. 30 Joules of energy was absorbed by a 5-g nail as it was hammered into place. How much does the nail's temperature increase (in °C) during this process? (The specific heat capacity of the nail is 450 J/kg-°C, and round to 3 significant digits.
Answer:
13.33 K
Explanation:
Given that,
Heat absorbed, Q = 30 J
Mass of nail, m = 5 g = 0.005 kg
The specific heat capacity of the nail is 450 J/kg-°C.
We need to find the increase in the temperature during the process. The heat absorbed in a process is as follows:
[tex]Q=mc\Delta T\\\\\Delta T=\dfrac{Q}{mc}\\\\\Delta T=\dfrac{30}{0.005\times 450}\\\\=13.33\ K[/tex]
So, the increase in temperature is 13.33 K.
