Answer:
ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
Explanation:
Given the data in the question;
T[tex]_i[/tex] = 300 K, T[tex]_f[/tex] = 412 K, n = 0.7 mol
since helium is monoatomic;
Cv = (3/2)R, Cp = (5/2)R
W₁ = 0 J [ at constant volume or ΔV = 0]
Now for the first cylinder; from the first law of thermodynamics;
Q₁ = ΔE[tex]_{int[/tex],₁ + W₁
Q₁ = ΔE[tex]_{int[/tex],₁ = n × Cv × ΔT
we substitute
Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × ( 3/2 )8.314 × ( 412 - 300 )
Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × 12.471 × 112
Q₁ = ΔE[tex]_{int[/tex],₁ = 977.7 J
Therefore, ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
Light with a wavelength of 5.0 · 10-7 m strikes a surface that requires 2.0 ev to eject an electron. Calculate the energy, in joules, of one incident photon at this frequency. _____ joules 4.0 x 10 -19 4.0 x 10 -49 9.9 x 10 -32 1.1 x 10 -48
Answer:
pretty sure its 6.2 x 10^-13
Explanation:
I looked it up I'm not a bigbrain but want to help
An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration
Answer:
a= -80.357 m/s
Explanation:
use the formula
vf^2=vi^2+2a(xf-xi)
Plug in givens
0=(7.50)^2+2a(0.350m)
solve for acceleration
a= -80.357 m/s
When a golfer tees off, the head of her golf club which has a mass of 158 g is traveling 48.2 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.
Answer:
v₂ = 53.23 m/s
Explanation:
Given that,
The mass of a golf club, m₁ = 158 g = 0.158 kg
The initial speed of a golf club, u₁ = 48.2 m/s
The mass of a golf ball, m₂ = 46 g = 0.046 kg
It was at rest, u₂ = 0
Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s
We use the conservation of energy to find the speed of the golf ball just after impact as follows :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]
So, the speed of the golf ball just after the impact is equal to 53.23 m/s.
A student claimed that thermometers are useless because a
thermometer always registers its own temperature. How would you
respond?
[
An inductor of inductance 0.02H and capacitor of capatance 2uF are connected in series to an a.c. source of frequency 200 Hz- Calculate the Impedance in the circuit . TC
Explanation:
Given:
L = 0.02 H
C = [tex]2\:\mu \text{F}[/tex]
f = 200 Hz
The general form of the impedance Z is given by
[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]
Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to
[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]
[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]
A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside the tire is 45oC, what fraction of the original volume of air must be removed to maintain the pressure at 240 kPa?
Answer:
7.8% of the original volume.
Explanation:
From the given information:
Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C
Pressure [tex]P_1[/tex] = 240 kPa
Temperature [tex]T_2[/tex] = 45° C
At initial temperature and pressure:
Using the ideal gas equation:
[tex]P_1V_1 =nRT_1[/tex]
making V_1 (initial volume) the subject:
[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]
[tex]V_1 = \dfrac{nR*295}{240}[/tex]
Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:
the final volume [tex]V_2[/tex] can be computed as:
[tex]V_2 = \dfrac{nR*318}{240}[/tex]
Now, the change in the volume ΔV = V₂ - V₁
[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]
[tex]\Delta V = \dfrac{23nR}{240}[/tex]
∴
The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:
[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]
[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]
[tex]= 0.078[/tex]
= 7.8% of the original volume.
As you move farther away from a source emitting a pure tone, the ___________ of the sound you hear decreases.
Answer:
frequency
Explanation:
The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.
So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is
[tex]f' = \frac{v + v_o}{v+ v_s} f[/tex]
where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.
Question 3 of 10
Which statement describes the law of conservation of energy?
A. Air resistance has no effect on the energy of a system.
B. Energy cannot be created or destroyed.
C. The total energy in a system can only increase.
D. Energy cannot change forms.
هما
SUBMIT
Answer:
B . energy cannot be created or destroyed
A TV satellite dish is designed to receive radio waves of wavelength
0.0644 meters. What is the frequency of the waves it receives? _______GHz
Give your answer in gigahertz (GHz). 1 GHz = 10^9 Hz.
Give your answer to the nearest tenth of a GHz (one place after the decimal). Just enter the number; do NOT use scientific notation.
Answer:
4.7 GHz
Explanation:
Applying,
v = λf................. Equation 1
Where v = velocity of the radio wave, λ = wavelength, f = frequency
make f the subject of the equation
f = v/λ.............. Equation 2
Note: A radio wave is an electromagnetic wave, as such it moves with a velocity of 3.00 x 10⁸ m/s
From the question,
Given: λ = 0.0644 meters
Constant: v = 3.00 x 10⁸ m/s
Substitute these values into equation 2
f = (3.00 x 10⁸)/0.0644
f = 4.66×10⁹ Hz
f = 4.7 GHz
A typical incandescent light bulb consumes 75 W of power and has a mass of 20 g. You want to save electrical energy by dropping the bulb from a height great enough so that the kinetic energy of the bulb when it reaches the floor will be the same as the energy it took to keep the bulb on for 2.0 hours. From what height should you drop the bulb, assuming no air resistance and constant g?
Answer:
h = 2755102 m = 2755.102 km
Explanation:
According to the given condition:
Potential Energy = Energy Consumed by Bulb
[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]
where,
h = height = ?
P = Power of bulb = 75 W
t = time = (2 h)(3600 s/1 h) = 7200 s
m = mass of bulb = 20 g = 0.02 kg
g = acceleration due to gravity = 9.8 m/s²
Therefore,
[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]
h = 2755102 m = 2755.102 km
One hazard of space travel is the debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00×10^3 m/s, given the collision lasts 6.00×10^8s.
Answer:
F = 6666.7 N
Explanation:
Given that,
Mass of a chip, m = 0.1 mg
Initial speed, u = 0
Final speed,[tex]v=4\times 10^{3}\ m/s[/tex]
Time of collision,[tex]t=6\times 10^{-8}\ s[/tex]
We know that,
Force, F = ma
Put all the values,
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.1\times 10^{-6}\times (4\times 10^3-0)}{6\times 10^{-8}}\\\\F=6666.7\ N[/tex]
So, the required force is 6666.7 N.
Cold air rises because it is denser than water, is this true?
Answer:
true
Explanation:
im not sure please dont attack me
Question: A NEO distance from the Sun is 1.17 AU. What is the speed of the NEO (round your answer to 2 decimal places)
Answer:
v = 2.75 10⁴ m / s
Explanation:
For this exercise we must use Kepler's third law which is an application of Newton's second law to the solar system
F = ma
where force is the force of gravity
F = [tex]G \frac{m M}{r^2}[/tex]
acceleration is centripetal
a = [tex]\frac{v^2}{r}[/tex]
we substitute
G m M / r² = m v² / r
[tex]\frac{GM}{r}[/tex] = v²
v = [tex]\sqrt{GM/r}[/tex]
indicate that the radius of the orbit is r = 1.17 AU, let's reduce to the SI system
r = 1.17 AU (1.496 10¹¹ m / 1 AI) = 1.76 10¹¹ m
let's calculate
v = [tex]\sqrt{\frac{6.67 \ 10^{-11} 1.991 \ 10^{30} }{ 1.76 \ 10^{11}} }[/tex]Ra (6.67 10-11 1.991 10 30 / 1.76 10 11
v = [tex]\sqrt{7.5454 \ 10^8 }[/tex]ra 7.5454 10 8
v = 2.75 10⁴ m / s
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
[tex]A=2.01×10^{16}\:\text{nuclei}[/tex]
Explanation:
Given:
[tex]\lambda = 4.96×10^3 s[/tex]
[tex]A_0 = 3.21x10^{17}[/tex] nuclei
t = 1.98×10^4 s
[tex]A=A_02^{-\frac{t}{\lambda}}[/tex]
[tex]A=(3.21×10^{17}\:\text{nuclei}) \left(2^{-\frac{1.98×10^4}{4.96×10^3}} \right)[/tex]
[tex]\:\:\:\:\:\:\:=2.01×10^{16}\:\text{nuclei}[/tex]
how much amount of heat energy is required to convert 5 kg of ice at - 5° c into 100°c steam?
Assuming no heat lost to the surrounding,
-5⁰C ice → 0⁰C ice
Specific heat capacity of ice = 2.0 x 10³ J/kg/⁰C
Q = mc∆θ
Q = 5(2.0 x 10³) x (0-(-5))
Q = 50000J
0⁰C ice → 0⁰C water
Specific latent heat of fusion of ice = 3.34 x 10⁵J/kg
Q = mLf
Q = 5(3.34 x 10⁵)
Q = 1670000J
0⁰C water → 100⁰C water
Specific heat capacity of water = 4.2 x 10³ J/kg/⁰C
Q = mc∆θ
Q = 5(4.2 x 10³) x (100-0)
Q = 2100000J
100⁰C water → 100⁰C steam
Specific latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Q = mLv
Q = 5(2.26 x 10⁶)
Q = 11300000J
Total amount of heat required
= 50000 + 1670000 + 2100000 + 11300000
= 15120000J
The velocity-time graph of a body is given. What quantities are represented by (a) slope of the graph and (b) area under the graph?
Answer:
a) acceleration
b) displacement
Explanation:
The velocity-time graph is a graph of velocity versus time. The velocity (m/s) would be on the Y-axis while time (s) would be on the X-axis.
a) The slope of a graph is given by: change in Y-axis/change in X-axis = ΔY/ΔX
In a velocity-time graph, ΔY = change in velocity and ΔX = change in time.
Hence, the slope of a velocity-time graph becomes: change in velocity/change in time.
Also, acceleration = change in velocity/change in time.
Hence, the slope of a velocity-time graph = acceleration.
b) Assuming that the area under a velocity-time graph is a rectangle, the area is given as:
Area of a rectangle = length x breadth
= velocity x time (m/s x s)
Also, displacement = velocity x time (m)
Hence, the area under a velocity-time graph of a body would give the displacement of the body.
g As they reach higher temperatures, most semiconductors... Selected Answer: have an increased resistance. Answers: have a constant resistance. have an increased resistance. have a decreased resistance.
Answer:
have an increased resistance
The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be?
Answer:
7.9 [tex]\frac{m}{s^{2} }[/tex]
Explanation:
Take the fact that mass is inversely proportional to accelertation:
m ∝ a
Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:
[tex]\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\[/tex]
Using algebra, we can rearrange our equation to find the final acceleration, [tex]a_{2}[/tex]:
[tex]a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\[/tex]
Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:
0.85g * 9.8 [tex]\frac{m }{s^{2} }[/tex] = 8.33 [tex]\frac{m }{s^{2} }[/tex]
Plug everything in:
7.9 [tex]\frac{m }{s^{2} }[/tex] = [tex]\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}[/tex]
(1590kg the initial weight plus the weight of the added passenger)
A car is moving at a speed of 60 mi/hr (88 ft/sec) on a straight road when the driver steps on the brake pedal and begins decelerating at a constant rate of 10ft/s2 for 3 seconds. How far did the car go during this 3 second interval?
Answer:
219 ft
Explanation:
Here we can define the value t = 0s as the moment when the car starts decelerating.
At this point, the acceleration of the car is given by the equation:
A(t) = -10 ft/s^2
Where the negative sign is because the car is decelerating.
To get the velocity equation of the car, we integrate over time, to get:
V(t) = (-10 ft/s^2)*t + V0
Where V0 is the initial velocity of the car, we know that this is 88 ft/s
Then the velocity equation is:
V(t) = (-10 ft/s^2)*t + 88ft/s
To get the position equation we need to integrate again, this time we get:
P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0
Where P0 is the initial position of the car, we do not know this, but it does not matter for now.
We want to find the total distance that the car traveled in a 3 seconds interval.
This will be equal to the difference in the position at t = 3s and the position at t = 0s
distance = P(3s) - P(0s)
= ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)
= ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)
= (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft
The car advanced a distance of 219 ft in the 3 seconds interval.
a vechile having a mass of 500kg is moving with a speed of 10m/s.Sand is dropped into it at the rate of 10kg/min.What force is needed to keep the vechile moving with uniform speed
Answer:
1.67 N
Explanation:
Applying,
F = u(dm/dt)+m(du/dt)................ Equation 1
Where F = force, m = mass of the vehicle, u = speed.
Since u is constant,
Therefore, du/dt = 0
F = u(dm/dt)............... Equation 2
From the question,
Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s
Substitute these values into equation 2
F = 10(10/60)
F = 100/60
F = 1.67 N
Hi can someon help me how to answer this?
Btw I'm from Philippines
Answer:
Test 1
1.True
2.True
3.True
4.False
5.True
6.True
7.False
8.True
9.True
10.True
yung iba nasa pic
A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal frictionless surface. The mass was released from rest at
t = 0
from the position
x = 0.0480 m.
Determine the location of the mass at
t = 5.85 s?
Answer:
[tex]X=0.0389m[/tex]
Explanation:
From the question we are told that:
Period of spring [tex]T_s=2.25s[/tex]
Initial Position of Mass [tex]x=0.0480m[/tex]
Final Mass period [tex]T_f=5.85s[/tex]
Generally the equation for the Mass location is mathematically given by
[tex]X=xcos*\frac{2\pi T_s}{T_f}[/tex]
[tex]X=0.048*cos*\frac{2\pi 5.85}{2.25}[/tex]
[tex]X=0.0389m[/tex]
When a rigid body rotates about a fixed axis, all the points in the body have the same Group of answer choices linear displacement. angular acceleration. centripetal acceleration. tangential speed. tangential acceleration.
Answer:
angular acceleration.
Explanation:
Newton's law of universal gravitation states that the force of attraction (gravity) acting between the Earth and all physical objects is directly proportional to the Earth's mass, directly proportional to the physical object's mass and inversely proportional to the square of the distance separating the Earth's center and that physical object.
Generally, when a rigid body is made to rotate about a fixed axis, all the points in the body would typically have the same angular acceleration, angular displacement, and angular speed.
suppose the tank is open to the atmosphere instead of being closed. how does the pressure vary along
Answer:
Pressure is more in the open container than the closed one.
Explanation:
The pressure due to the fluid at a depth is given by
Pressure = depth x density of fluid x gravity
So, when the container is open, the atmospheric pressure is also add up but when the container is closed only the pressure due to the fluid is there.
So, when the container is open, the pressure is atmospheric pressure + pressure due to the fluid.
hen the container is closed only the pressure due to the fluid is there.
Topic: Chapter 10: Projectory or trajectile?
Projectile range analysis:
A projectile is launched from the ground at 10 m/s, at
an angle of 15° above the horizontal and lands 5.1 m away.
What other angle could the projectile be launched at, with the same velocity,
and land 5.1 m away?
90°
75°
45
50°
30°
Answer:
The other angle is 75⁰
Explanation:
Given;
velocity of the projectile, v = 10 m/s
range of the projectile, R = 5.1 m
angle of projection, 15⁰
The range of a projectile is given as;
[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]
To find another angle of projection to give the same range;
[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]
Check:
sin(2θ) = sin(2 x 75) = sin(150) = 0.5
sin(2θ) = sin(2 x 15) = sin(30) = 0.5
Can you think of reasons why the charge on each ball decreases over time and where the charges might go
Answer:
By the principle of corona discharge.
Explanation:
The charge on each ball will decreases over time due to the electrical discharge in air.
According to the principle of corona discharge, when the curvature is small, the discharge of the charge takes placed form the pointed ends.
A submarine has a "crush depth" (that is, the depth at which
water pressure will crush the submarine) of 400 m. What is
the approximate pressure (water plus atmospheric) at this
depth? (Recall that the density of seawater is 1025 kg/m3, g=
9.81 m/s2, and 1 kg/(m-s2) = 1 Pa = 9.8692 x 10-6 atm.)
Answer:
P =40.69 atm
Explanation:
We need to find the approximate pressure at a depth of 400 m.
It can be calculated as follows :
P = Patm + ρgh
Put all the values,
[tex]P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm[/tex]
So, the approximate pressure is equal to 40.69 atm.
Part AFind the x- and y-components of the vector d⃗ = (4.0 km , 29 ∘ left of +y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.d⃗ = km Part BFind the x- and y-components of the vector v⃗ = (2.0 cm/s , −x-direction).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.v⃗ = cm/s Part CFind the x- and y-components of the vector a⃗ = (13 m/s2 , 36 ∘ left of −y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.a⃗ x = m/s2
Solution :
Part A .
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, d = [tex]\text{4 km 29}[/tex] degree left of [tex]y[/tex]-axis.
So the [tex]x[/tex] component is = -4 x sin (29°) = -1.939 km
[tex]y[/tex] component is = 4 x cos (29°) = 3.498 km
Part B
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{v = 2 cm/s}[/tex] , [tex]\text{-x direction}[/tex]
So the [tex]x[/tex] component is = -2 cm/s
[tex]y[/tex] component is = 0
Part C
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{a = 13 m/s, 36 degree}[/tex] left of [tex]y[/tex]-axis.
So the [tex]x[/tex] component is = -13 x sin (36°) = -7.6412 [tex]m/S^2[/tex]
[tex]y[/tex] component is = -13 x cos (36°) = -10.517 [tex]m/S^2[/tex]
The x- and y-components of the vectors is mathematically given as as follows for each Part respectively
x= -1.939 km, y= 3.498 km
x= -2 cm/s, 0
y=, x= -7.6412m/s^2, -10.517m/s^2
What are the x- and y-components of the vectors?
Question Parameters:
Generally, we follow a basic principle where
x component= Fsin\theta
y component= Fcos\theta
Therefore
For A
x component is
x= -4 x sin (29°)
x= -1.939 km
y component is
y= 4 x cos (29°)
y= 3.498 km
For B
x component is
x= -2 cm/s
y component is
y= 0
For C
x component is
x= -13 x sin (36°)
x= -7.6412m/s^2
y component is
y= -13 x cos (36°)
y= -10.517m/s^2
Read more about Cartession co ordinate
https://brainly.com/question/9410676
An object is 2.0 cm from a double convex lens with a focal length of 1.5 cm. Calculate the image distance
Answer:
0.857 cm
Explanation:
We are given that:
The focal length for a convex lens to be (f) = 1.5cm
The object distance (u) = - 2.0 cm
We are to determine the image distance (v) = ??? cm
By applying the lens formula:
[tex]\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}[/tex]
By rearrangement and making (v) the subject of the above formula:
[tex]v = \dfrac{uf}{u-f}[/tex]
replacing the given values:
[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]
[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]
v = 0.857 cm
A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press
5.0 m from the right support column (Figure slide 8). Calculate the force
on each of the vertical support columns.
Answer:
[tex]\mathbf{F_1=4.41*10^4\ N}[/tex]
[tex]\mathbf{F_2 = 1.176*10^5 \ N}[/tex]
Explanation:
The missing image of the figure slide is attached in below.
However, from the model, it is obvious that it is in equilibrium.
As a result, the relation of the force and the torque is said to be zero.
i.e.
[tex]\sum F = 0[/tex] and [tex]\sum \tau = 0[/tex]
From the image, expressing the forces through the y-axis, we have:
[tex]F_1+F_2 = W_B + W_P \\ \\ \implies 9.8(1500+15000) \\ \\ \implies \mathtt{1.617\times 10^5 \ N}[/tex]
Also, let the force [tex]F_1[/tex] be the pivot and computing the torque to determine [tex]F_2[/tex]:
Then:
[tex]F_1(0)+F_2(20.0) = 10.0W_B + 15.0W_P[/tex]
[tex]F_2 = \dfrac{((10*1500)+(15*15000))*9.8}{20.0}[/tex]
[tex]F_2 = 117600 \ N[/tex]
[tex]\mathbf{F_2 = 1.176*10^5 \ N}[/tex]
For the force equation:
[tex]F_1+F_2=1.617*10^5 \ N;[/tex]
where:
[tex]F_2 = 1.176*10^5 \ N[/tex]
Then:
[tex]F_1+1.176*10^5 \ N=1.617*10^5 \ N[/tex]
[tex]F_1=1.617*10^5 \ N-1.176*10^5 \ N[/tex]
[tex]F_1=44100\ N[/tex]
[tex]\mathbf{F_1=4.41*10^4\ N}[/tex]
