Answer:
2.21 N
Explanation:
The force in this case is the total mass multiplied by the acceleration due to gravity. You are not asked for the solution to be in terms of the torque which is the usual way to solve these problems. That's why you are not given where the fulcrum is.
The fulcrum feels F1 + F2 + 34 * 980
F2 = 141.7 * 980 = 138866
F1 = 50.3 * 980 = 49294
Ruler = 34 * 980= 33320
Total Force = 221480 The units here are dynes
I just saw in the middle of the question that g = 9.80
So the answer becomes 221480 / 1000 = 221.48 because we needed kg
And that answer becomes 221.48/100 2.21 because the force of gravity should be 9.8 not 980
The total force exerted on the fulcrum is
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
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.
what is the average velocity if the initial velocity is at rest and the final velocity is 16 m/s
Answer:
8m/s
Explanation:
Vavg= 16-0/2=8m/s
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]
Find the starting pressure of CCl4 at this temperature that produces a total pressure of 1.1 atm at equilibrium. Express the pressure in atmospheres to three significant figures.
The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.
Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.
Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
Explanation:
The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Initial: a 0 0
Equilibrium: (a - x) 0 2x
Total pressure = (a - x) + 2x = a + x
As it is given that the total pressure is 1.1 atm.
So, a + x = 1.1
a = 1.1 - x
Now, expression for equilibrium constant for this equation is as follows.
[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]
Hence, the value of 'a' is calculated as follows.
a + x = 1.1 atm
a = 1.1 atm - x
= 1.1 atm - 0.31 atm
= 0.79 atm
Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
A body of mass 4kg is moving with a velocity of 108km/h . find the kenetic energy of the body.
Answer:
KE = 2800 J
Explanation:
Usually a velocity is expressed as m/s. Then the energy units are joules.
[tex]\frac{108 km}{hr} * \frac{1000m}{1 km} * \frac{1 hour}{3600 seconds} =\frac{108*1000 m}{3600sec}[/tex]
v = 30 m / sec
KE = 1/2 * 4 * (30)^2
KE =2800 kg m^2/sec^2
KE = 2800 Joules
PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
Answer:
multiply mp and c^2
Explanation:
e=mc^2
A system is acted on by its surroundings in such a way that it receives 50 J of heat while simultaneously doing 20 J of work. What is its net change in internal energy
Answer:
30J
Explanation:
Given data
The total quantity of heat recieved= 50J
Quantity of heat used to do work= 20J
Hence the net change is
ΔU= Total Heat - Net work
ΔU= 50-20
ΔU= 30J
Hence the change in the internal energy is 30J
Two sinusoidal waves have the same frequency and wavelength. The wavelength is 20 cm. The two waves travel from their respective sources and reach the same point in space at the same time, resulting in interference. One wave travels a larger distance than the other. For each of the possible values of that extra distance listed below, identify whether the extra distance results in maximum constructive interference, maximum destructive interference, or something in-between.
a. 10 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
b. 15 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
c. 20 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
d. 30 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
e. 35 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
f. 40 cm - (A) in-between (2) maximum destructive (3) maximum constructive.
Answer:
Explanation:
When the path difference is equal to wave length or its integral multiple, constructive interference occurs . If it is odd multiple of half wave length , then destructive interference occurs.
For constructive interference , path diff = n λ
For destructive interference path diff = ( 2n+ 1 ) λ /2
where λ is wave length of wave , n is an integer.
a )
path diff = 10 cm which is half the wavelength , so maximum destructive interference will occur.
b )
path diff = 15 cm which is neither half the wavelength nor full wavelength , so in between is the right option.
c )
path diff = 20 cm which is equal to the wavelength , so maximum constructive interference will occur.
d)
path diff = 30 cm which is 3 times half the wavelength , so maximum destructive interference will occur.
e)
path diff = 35 cm which is neither integral multiple of half the wavelength , nor integral multiple of wavelength so in between is th eright answer.
f )
path diff = 40 cm which is 2 times the wavelength , so maximum constructive interference will occur
A 55 g soapstone cube--a whisky stone--is used to chill a glass of whisky. Soapstone has a density of 3000 kg/m3, whisky a density of 940 kg/m3. What is the approximate normal force of the bottom of the glass on a single stone?
Answer:
[tex]N=0.37N[/tex]
Explanation:
Mass [tex]m=55g=>0.055kg[/tex]
Soapstone Density [tex]\rho_s=3000kg/m^2[/tex]
Whisky Density [tex]\rho_w=940kg/m^2[/tex]
Generally the equation for Force is mathematically given by
[tex]F=U+N[/tex]
Therefore
[tex]N=m*g-(\frac{m}{\rho_s})*\rho_w*g[/tex]
[tex]N=0.055*9.81 - {(\frac{0.055}{3000})*940*9.81}[/tex]
[tex]N=0.37N[/tex]
A ball is thrown from ground level with an initial speed of 24.5 m/s at an angle of 35.5 degrees above the horizontal. The ball hits a wall that is 25.8 meters horizontally from where it started. How high (meters) does the ball hit on the wall?
6.07 m
Explanation:
Given:
[tex]v_0=24.5\:\text{m/s}[/tex]
[tex]\theta_0 = 35.5°[/tex]
First, we need to find the amount of time it takes to travel a horizontal distance of 25.8 m. We know that
[tex]x = v_{0x}t \Rightarrow t = \dfrac{x}{v_0 \cos \theta_0}[/tex]
or
[tex]t = 1.29\:\text{s}[/tex]
To find the vertical height where the ball hit the wall, we use
[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]
[tex]\:\:\:\:=(24.5\:\text{m/s})\sin 35.5(1.29\:\text{s}) \\ - \frac{1}{2}(9.8\:\text{m/s}^2)(1.29\:\text{s})^2[/tex]
[tex]\:\:\:\:=6.07\:\text{m}[/tex]
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)
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 vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?
Answer:
Explanation:
Speed= distance/time
Or time = distance/speed
According to your question
Speed=15m/s
and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s
D= 1.2km = 1.2×1000m =1200meter
Time = distance/ speed
1200/15 =80second
Or. 1min and 20 sec will be your answer.
The north pole of magnet A will __?____ the south pole of magnet B
Answer:
A will attract
B will repare
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
A student measure the length of a laboratory bench with a meter ruler. Which of the following values is the most approbriate way to record the result ? a.4.022m b.4.02m c.4.0m d.4m
Answer:
Well a meter stick has increments of a centimeter, and since 1 cm=0.01m he should record it as 4.02m(b)
Explanation:
What are stepdown transformers used for
Answer:
Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.
Explanation:
pls mark me as brainlist
Thanks a lot
g How much buoyancy force, in N, a person with a mass of 70 kg experiences by just standing in air
Answer:
686.7N
Explanation:
Given data
Mass= 70kg
We know that the buoyant force experienced by the person is equal to the weight of the person
Hence the weight is
Weight = mass* Acceleration
Weight= 70*9.81
Weight= 686.7N
Therefore the weight is 686.7N
A 64-ka base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
Required:
a. How much mechanical energy is tout due to friction acting on the runner?
b, How far does he slide?
Answer:
Explanation:
From the given information:
mass = 64 kg
speed = 3.2 m/s
coefficient of friction [tex]\mu =[/tex] 0.70
The mechanical energy touted relates to the loss of energy in the system as a result of friction and this can be computed as:
[tex]W = \Delta K.E[/tex]
[tex]\implies \dfrac{1}{2}m(v^2 -u^2)[/tex]
[tex]= \dfrac{1}{2}(64.0 \kg) (0 - (3.2 \ m/s^2))[/tex]
Thus, the mechanical energy touted = 327.68 J
According to the formula used in calculating the frictional force
[tex]F_r = \mu mg[/tex]
= 0.70 × 64 kg× 9.8 m/s²
= 439.04 N
The distance covered now can be determined as follows:
d = W/F
d = 327.68 J/ 439.04 N
d = 0.746 m
What is significant about the primary colors of pigments?
They can be mixed together to make almost any other color.
Any two primary colors of pigments combine to make white pigment.
Each primary color of pigment absorbs all other colors.
Any two primary colors of pigments combine to make black pigment.
Answer:
They can be mixed together to make almost any other color.
Explanation:
All the three primary colors can mix to form white color.
Blue and red mix to form a black color.
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
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.
Learn more about electromagnetic spectrum at: https://brainly.com/question/23727978
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A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140 N. Fp is parallel to the displacement of the block. The final speed of the block is 2.35 m/s.
a) How much work was converted to thermal energy? What work did friction do on the box?
b) What is the coefficient of friction?
Answer:
The answer is "151.25 J and -547.64 J".
Explanation:
[tex]u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\[/tex]
Using formula:
[tex]v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\[/tex]
[tex]= 0.55 \ \frac{m}{sec^2}\\\\[/tex]
[tex]F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\[/tex]
Calculating the Work by net force
[tex]W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\[/tex]
The above work is converted into thermal energy.
Now,
[tex]F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J[/tex]
g A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 3.00 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do
The desk is in equilbrium, so Newton's second law gives
∑ F (horizontal) = p - f = 0
∑ F (vertical) = n - mg = 0
==> n = mg
==> p = f = µn = µmg = 0.400 (80.0 kg) g = 313.6 N
The student pushes the desk 3.00 m, so she performs
W = (313.6 N) (3.00 m) = 940.8 Nm ≈ 941 J
of work.
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
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.
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
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?