Answer:
35.047 m
Explanation:
The time it takes the lead ball to reach the surface of the water is
s = ut+gt²/2............. Equation 1
Where t = time it takes the lead ball to reach the surface of water, u = initial velocity of the lead ball, g = acceleration due to gravity, s = heigth.
From the question,
Given: s = 5.20 m, u = 0 m/s (dropped from a height)
Constant: g = 9.8 m/s²
5.2 = 0+9.8t²/2
t² = (5.2×2)/9.8
t² = 10.4/9.8
t² = 1.06
t = √(1.06)
t = 1.03 s
Hence, time taken for the lead ball to reach the bottom of the lake is
t' = 4.5-1.03
t' = 3.47 seconds
v² = u²+2gs............... Equation 2
Where v = final velocity of the lead ball
Substitute into equation 2
v² = 0+2(9.8)(5.2)
v² = 101.92
v = √(101.92)
v = 10.1 m/s
Therefore, depth of the lake is
D = vt'
D = 10.1(3.47)
D = 35.047 m
A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force of 9.14 × 10-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.
Answer:
a) The proton's speed is 5.75x10⁵ m/s.
b) The kinetic energy of the proton is 1723 eV.
Explanation:
a) The proton's speed can be calculated with the Lorentz force equation:
[tex] F = qv \times B = qvBsin(\theta) [/tex] (1)
Where:
F: is the force = 9.14x10⁻¹⁷ N
q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C
v: is the proton's speed =?
B: is the magnetic field = 3.28 mT
θ: is the angle between the proton's speed and the magnetic field = 17.6°
By solving equation (1) for v we have:
[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]
Hence, the proton's speed is 5.75x10⁵ m/s.
b) Its kinetic energy (K) is given by:
[tex] K = \frac{1}{2}mv^{2} [/tex]
Where:
m: is the mass of the proton = 1.67x10⁻²⁷ kg
[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]
Therefore, the kinetic energy of the proton is 1723 eV.
I hope it helps you!
Given that the temperature of a body is 527K determine the value in degree C
Answer:
253.85°C
Explanation:
Here is the formula for converting K to °C
527K − 273.15 = 253.85°C
Convert 385k to temperature of
Answer:
233.33°F
Explanation:
(385K - 273.15) * 9/5 + 32 = 233.33°F
Suppose you exert a force of 314 N tangential to a grindstone (a solid disk) with a radius of 0.281 m and a mass of 84.2 kg What is the resulting angular acceleration of the grindstone assuming negligible opposing friction
Answer:
The angular acceleration is 26.6 rad/s^2.
Explanation:
Force, F = 314 N
radius, r = 0.281 m
mass, m = 84.2 kg
The grindstone is a disc.
The torque is given by
torque = force x radius
Torque = 314 x 0.281 = 88.234 Nm
The torque is given by
Torque = Moment of inertia x angular acceleration
[tex]88.234 = 0.5 mr^2 \alpha \\\\88.234 = 0.5\times 84.2\times 0.281\times 0.281\times \alpha \\\\\alpha = 26.6 rad/s^2[/tex]
Thorium-232 goes through multiple types of decay in order to reach a stable isotope. What isotope is created after the first two decays if it first goes through an alpha decay and then a beta decay?
A)uranium-236
B)protactinium-232
C)radon-224
D)Astinium-228
Answer:
The answer would be D), if the decay is beta negative.
Explanation:
Thorium-232 goes through alpha decay:
Thorium-232 --> Helium-4 + Radium-228
Radium-228 then can undergo beta positive or beta negative decay:
Beta positive = Radium-228 --> Electron + Francium-228
Beta negative = Radium-228 --> Positron + Actinium-228
Therefore, the isotope that is created is Actinium-228
A singly charged 7Li ion has a mass of 1.16 10-26 kg. It is accelerated through a potential difference of 523 V and subsequently enters a uniform magnetic field of magnitude 0.370 T perpendicular to the ion's velocity. Find the radius of its path.
Answer:
[tex]R=0.023m[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.16*10^{-26}[/tex]
Potential difference [tex]V=523V[/tex]
Magnitude [tex]m=0.370 T[/tex]
Generally the equation for Velocity is mathematically given by
[tex]\frac{1}{2}mv^2=ev[/tex]
[tex]v=\frac{2ev}{m}[/tex]
[tex]v=\frac{2*1.6*10^{-19}*542}{1.16*10^{-26}}[/tex]
[tex]v=12.22*10^4m/s[/tex]
Generally the equation for Force is mathematically given by
[tex]F=qvBsin \theta[/tex]
Where
[tex]qVB=m\frac{v^2}{R}[/tex]
[tex]F=m\frac{v^2}{R}sin\theta[/tex]
Therefore
[tex]R=\frac{mv}{qB sin \theta}[/tex]
[tex]R=\frac{1.6*10^{-26}*12.2*10^{4}}{1.60*10^{-19}*0.394 sin 90}[/tex]
[tex]R=0.023m[/tex]
There are 5640 lines per centimeter in a grating that is used with light whose wavelegth is 455 nm. A flat observation screen is located 0.661 m from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen
The minimum width of the screen is 34 cm.
For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/5640 lines per cm = 1/5640 cm per line = 1/5640 × 10⁻² m per line, θ = angle between principal maximum and the center axis of the grating, m = order of maxima = 1 (since we require the position of the principal maximum) and λ = wavelength = 455 nm = 455 × 10⁻⁹ m
So, sinθ = mλ/d
Also tanθ = L/D where θ = angle between principal maximum and the center axis of the grating, L = distance between central maximum and principal maximum and D = distance between grating and screen = 0.661 m.
For small angles sinθ ≈ tanθ
So, mλ/d = L/D
making L subject of the formula, we have
L = mλD/d
L = 1 × 455 × 10⁻⁹ m × 0.661 m ÷ 1/5640 × 10⁻² m per line
L = 1 × 455 × 10⁻⁹ m × 0.661 m × 5640 × 10² line per m
L = 1696258.2 × 10⁻⁷ m
L = 0.16963 m
L ≅ 0.17 m
So, for centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is w = 2L.
So, w = 2 × 0.17 m
w = 0.34 m
w = 34 cm
So for the centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is 34 cm.
Learn more about diffraction grating here:
https://brainly.com/question/15712101
Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.250 A when R1 is removed, leaving R2 connected across the battery.
(a) Find R1.
Ω
(b) Find R2.
Ω
Answer:
a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω
Explanation:
For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances
all resistors connected
V = i (R₁ + R₂)
with R₁ connected
V = (i + 0.5) R₁
with R₂ connected
V = (i + 0.25) R₂
We have a system of three equations with three unknowns for which we can solve it
We substitute the last two equations in the first
V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )
1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )
1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) = [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]
i² + 0.75 i + 0.125 = 2i² + 0.75 i
i² - 0.125 = 0
i = √0.125
i = 0.35355 A
with the second equation we look for R1
R₁ = [tex]\frac{V}{i+0.5}[/tex]
R₁ = 12 /( 0.35355 +0.5)
R₁ = 14.1 Ω
with the third equation we look for R2
R₂ = [tex]\frac{V}{i+0.25}[/tex]
R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]
R₂ = 19.9 Ω
Phân biệt các đặc điểm khác nhau giữa chất rắn, chất lỏng
Answer:
şen çal kapimi turkish drama
If a wave has to travel 600m and it’s wavelength is 0.4 m , with a frequency of 500Hz. How much time will it take for the wave to travel 600m ?
A student of mass 50kg takes 15seconds to run up a flight of 50 steps. If each step is 20cm, calculate the potential energy of the student at the maximum height
Answer:
the answer is 49000 joules at the maximum height
Explanation:
we know the mass (50kg)
we know the acceleration due to gravity(9.8m/s²)
we know the height too(maximum height meaning the 50th step so we multiply 50 with 20cm as each step is 20 cm and we get 1000 cm, convert to m it is 100 m
the formula is potential energy=mgh
m for mass
g for acceleration due to gravity
h for height
multiply them
50x9.8x100
we get 49000
the unit of potential energy is joules so the answer is
49000 joules
Answer:
49000 joules
Explanation:
hope it helpss
What improvements were made in measuring system with the introduction of standard units?
Answer:
Standard units are commonly used units of measurement, which help us measure length, height, weight, temperature, mass and more. These units are standardised, which means that everyone gets the same understanding of the size, weight and other properties of objects and things.
Explanation:
list at least types of motion
Answer:
These four are rotary, oscillating, linear and reciprocating. Each one moves in a slightly different way and each type of achieved using different mechanical means that help us understand linear motion and motion control.
(I got this off the web so credits to the rightful owner and I hope you have good day :)
Which of these rotational quantities is analogous to mass in a linear system?
a.
Angle in radians
b.
Angular acceleration
c.
Torque
d.
Rotational inertia
Answer:
d
Explanation:
i think it is rotational inertia
because analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in transnational motion.
hope it's right & helps !!!!!!!!!
Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.0 cm and a current of 12 A. The bigger loop has a current of 20 A. The magnetic field at the center of the loops is found to be zero.
Required:
What is the radius of the bigger loop?
Answer:
the radius of the bigger loop is 5 cm.
Explanation:
Given;
current in the smaller loop, I₁ = 12 A
current in the larger loop, I₂ = 20 A
radius of the smaller loop, r₁ = 3 cm
let the radius of the larger loop, = r₂
Apply Biot-Savart's law to determine the magnetic field at the center of the circular loops.
[tex]B= \frac{\mu_0 I}{2r}[/tex]
The magnetic field at the center of the smaller loop;
[tex]B_1 = \frac{\mu_0 I_1}{2 r_1}[/tex]
The magnetic field at the center of the bigger loop;
[tex]B_2 = \frac{\mu_0 I_2}{2 r_2}[/tex]
If the magnetic field at the center is zero, then B₁ = B₂
[tex]B_1 = B_2 = \frac{\mu_0 I_1}{2 r_1} = \frac{\mu_0 I_2}{2 r_2} \\\\\frac{I_1}{ r_1} = \frac{ I_2}{r_2} \\\\r_2 = \frac{I_2 r_1}{ I_1} = \frac{(20 \ A) \times (3.0 \ cm)}{12 \ A} = 5 \ cm[/tex]
Therefore, the radius of the bigger loop is 5 cm.
the air pressure at the base of the mountain is 75.0cm of mercury while at the top is 60cm of mercury. Given that theaverage density of air is 1.25kg/m³ and the density of mercury is 13 600 kg/m³and gravity-10N/kg, calculate the height height of the mountain
Answer:
질문?
Explanation:
평균 공기 밀도가 1.25kg/m³이고 수은 밀도가 13600kg/m³이고 g=10N/kg인 경우 산 기슭의 기압은 수은의 75.0cm이고 정상의 수은은 60cm입니다. 산의 높이를 계산?
68.6 bags are required to fill the specified volume of 10"x20"x20".
It is 40 lbs/1.9 gallons, assuming US gallons, for the topsoil.
US gallons equal 3785 ml. 1 lb = 1/2.2 = 0.454 kg or 454 grammes since 1 kilogramme equals 2.2 lbs. 3785 x 1.9 = 7192 ml is equal to 1.9 gallons. 40 lbs = 18160 g. Therefore, the topsoil density is 2.52 g/cc (18160/7192).
The volume of the peat bag is 14x20x30 inches, which is equal to 35.6x50.8x76.2 cm (1 inch = 2/54 cm)=137806 ccs.
In other words, for the 40 lb again, 18160 grams/137806 equals 0.13 g/cc. The peat is therefore significantly lighter than topsoil.
The volume of the latter volume is 120"x240"x20" or 576,000 cubic inches, and the volume of a bag is 14"x20"x30" or 8400 cubic inches, hence 576,000/8400 = 68.6 bags are required to fill the specified volume of 10"x20"x20".
Learn more about density herebrainly.com/question/4970627
SPJ1
A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hanging vertically, the mass is pulled aside a small distance of 7.6 cm and released from rest. While the mass is swinging the cord exerts an almost-constant force on it. For this problem, assume the force is constant as the mass swings. How much work in J does the cord do to the mass as the mass swings a distance of 8.0 cm
Answer:
work done is -2.8 × 10⁻⁶ J
Explanation:
Given the data in the question;
mass of the pendulum m = 6 kg
Length of core = 1.7 m
Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad
In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.
Now, the required work done will be;
[tex]W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta[/tex]
[tex]W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta[/tex]
[tex]W = -mgl \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta[/tex]
W = [tex]-mgl[[/tex] -cosθ [tex]]^{0.047}_{0.045 }[/tex]
W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]
W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]
W = -2.8 × 10⁻⁶ J
Therefore, work done is -2.8 × 10⁻⁶ J
Uuse Lenz's law to explore what happens when an electromagnet is activated a short distance from a wire loop. You will need to use the right-hand rule to find the direction of the induced current.
Answer:
Explanation:
According to the Fleming's right hand rule, if we spread our right hand such that the thumb, fore finger and the middle finger are mutually perpendicular to each other, then the thumb indicates the direction of force, fore finger indicates the direction of magnetic field, then the middle finger indicates the direction of induced current.
According to the Lenz's law, the direction of induced emf is such that it always opposes the cause due to which it is produced.
How many types of physics?
Answer:
Two Main Branches of Physics
it is Classical Physics and Modern Physics.
Explanation:
Further sub Physics branches are Mechanics, Electromagnetism, Thermodynamics, Optics, etc. The rapid progress in science during recent years has become possible due to discoveries and inventions in the field of physics.
hope it helped
Assume that the far point of a myopic (nearsighted) eye is 5.04 m in front of the eye. A lens is used to correct the vision, such that it can focus sharply an object at infinity. What is the power of the lens (in diopters; answer sign and magnitude)?
Answer:
[tex]P=-0.2D[/tex]
Explanation:
From the question we are told that:
Far-point [tex]v=-5.04m[/tex]
Where
u=-\alpha
Generally the equation for Lens is mathematically given by
[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]
[tex]P=\frac{1}{-5.04}-\frac{1}{\alpha}[/tex]
[tex]P=-0.2D[/tex]
After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving
This question is incomplete, the complete question is;
A bat flies towards a moth at 7.1 m/s while the moth is flying towards the bat at 4.4 m/s. The bat emits a sound wave of 51.7 kHz.
After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving
Answer:
the second moth is moving at 5.062 m/s
Explanation:
Given the data in the question;
Using doppler's effect
[tex]f_{moth[/tex] = f₀( [tex]v_{s[/tex] ± [tex]v_{observer[/tex] / [tex]v_{s[/tex] ± [tex]v_{source[/tex] )
f₁ = f₀( ([tex]v_{s[/tex] + v₂) / ( [tex]v_{s[/tex] - v₁ ) )
frequency reflected from the moth,
Now, moth is the source and the bat is the receiver
f₂ = f₁( ([tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) )
hence, f = f₀[ ( ( [tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) ) ( ( [tex]v_{s[/tex] + u₂ ) / ( [tex]v_{s[/tex] - u₁ ) )
we know that, the velocity of sound [tex]v_{s[/tex] = 343 m/s.
given that v₁ and v₂ { velocity of bat } = 7.1 m/s, f₀ = 51.7 kHz and f = 55.5 kHz.
we substitute
55.5 = 51.7[ ( ( 343 + 7.1 ) / ( 343 - 7.1 ) ) ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 51.7[ ( 350.1 / 335.9 ) ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 51.7[ 1.04227 ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 53.885359 ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 / 53.885359 = ( 343 + u ) / ( 343 - u )
1.02996 = ( 343 + u₂ ) / ( 343 - u )
( 343 + u₂ ) = 1.02996( 343 - u )
343 + u = 353.27628 - 1.02996u
u + 1.02996u = 353.27628 - 343
2.02996u = 10.27628
u = 10.27628 / 2.02996
u = 5.062 m/s
Therefore, the second moth is moving at 5.062 m/s
A bus moving on a straight road increases its speed uniformly from rest to 20m's over a time period of 1 min. The distance travelled during the time is (a) 150 m (b) 300 m (c) 600 m (d) 900 m
Explanation:
Given that,
Initial velocity (u) = 0 m/sFinal velocity (v) = 20 m/sTime taken (t) = 1 minute = 60 secondsIn order to find the distance travelled, firstly we need calculate the acceleration.
→ v = u + at
→ 20 = 0 + 60a
→ 20 = 60a
→ 20 ÷ 60 = a
→ ⅓ m/s² = a
Now, by using the 2nd equation of motion :
→ s = ut + ½at²
→ s = 0(60) + ½ × ⅓ × (60)²
→ s = ⅙ × 3600
→ s = 1 × 600
→ s = 600 m
Hence, the distance travelled is 600 m.
In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0E0E_0 and B0B0B_0 are the __________ of the electric and magnetic fields. Choose the best answer to fill in the blank.
Two charged objects attract each other with a force 1.0 N. What happens to the force between them if one charge is increased by a factor of 2, the other charge is increased by a factor of 4, and the separation distance between their centers is reduced to 1/4 its original value
Answer:
F' = 128 N
Explanation:
The electrostatic force of attraction between two charges is given by Colomb's Law, as follows:
[tex]F = \frac{kq_1q_2}{r^2}\\\\[/tex]
where,
F = Force of attraction = 1 N
G = universal gravitational constant
q₁ = magnitude of the first charge
q₂ = magnitude of the second charge
r = distance between charges
Therefore,
[tex]1\ N = \frac{kq_1q_2}{r^2}[/tex] --------------------- eq(1)
Now, we apply the changes given in the question:
[tex]F' = \frac{k(2q_1)(4q_2)}{(\frac{1}{4}r)^2}\\\\F' = 128(\frac{kq_1q_2}{r^2})[/tex]
using eq (1):
F' = 128(1 N)
F' = 128 N
An 1800-W toaster, a 1400-W electric frying pan, and a 55-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. (The three devices are in parallel when plugged into the same socket.)
a. Will this combination blow the 15-A fuse?
b. What current is drawn by each device?
Being in parallel each device will have an equal voltage drop of 120 V
A. Yes the combination will blow the fuse. See part B for the total current.
B. Toaster = 1800W / 120V = 15A
Frying Pan = 1400W / 120V = 11.67A
Lamp = 55W / 120V = 0.458A
Total amps = 15 + 11.67 + 0.458 = 27.128 Amps
27.128A is greater than 15A so the fuse will blow.
A simple pendulum consists of a ball of mass 3 kg hanging from a uniform string of mass 0.05 kg and length L. If the period of oscillation of the pendulum is 2 s, determine the speed of a transverse wave in the string when the pendulum hangs vertically.
Answer:
v = 3.12 m/s
Explanation:
First, we will find the length of the string by using the formula of the time period:
[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\l = \frac{T^2g}{4\pi^2}\\\\[/tex]
where,
l = length of string = ?
T = time period = 2 s
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]l = \frac{(2\ s)^2(9.81\ m/s^2)}{4\pi^2}\\\\l = 0.99\ m[/tex]
Now, we will find tension in the string in the vertical position through the weight of the ball:
T = W = mg = (3 kg)(9.81 m/s²)
T = 29.43 N
Now, the speed of the transverse wave is given as follows:
[tex]v=\sqrt{\frac{Tl}{m}}\\\\v=\sqrt{\frac{(29.43\ N)(0.99\ m)}{3\ kg}}\\\\[/tex]
v = 3.12 m/s
An astronaut on the moon drops a rock from rest. The rock falls 0.8m in one second of falling time. If the dropped rock fell for a total of two seconds of time instead of 1 second, then the distance traveled would be:____.
A) The same.
B) Doubled.
C) Tripled.
D) Quadruple.
E) None of the above.
Answer:
D) Quadruple.
Explanation:
We will use the second equation of motion to solve this problem:
[tex]s = v_it + \frac{1}{2}gt^2[/tex]
where,
s = distance travelled by the rock
vi = initial speed of rock = 0 m/s
t = time taken
g = acceleration due to gravity on the surface of the moon
Therefore,
[tex]s = (0\ m/s)t+\frac{1}{2}gt^2\\\\s =\frac{1}{2}gt^2[/tex]----------- equation (1)
Now, we double the time:
[tex]s' = \frac{1}{2}g(2t)^2\\\\s' = 4(\frac{1}{2}gt^2)[/tex]
using equation (1):
s' = 4s
Hence, the correct option is:
D) Quadruple.
how can you convert galvanometer into ammeter?
Answer:
A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.
Explanation:
This low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of the ammeter depends on the values of the shunt resistance.
The resistance of the light bulb changed as the voltage (and current) changed. Why does this resistance change occur?
When tightening a bolt, you push perpendicularly on a wrench a with force of 165 N at a distance of 0.140m from the center of the bolt. (A) How much torque are you exerting in newton x meters (relative to the center of the bolt)
Answer:
Part a)
23.1 Nm
..........
