Answer:
Explanation:
2as=vf^2-Vi^2
vf=30 m/s
vi= 0 m/s
a=g=9.8 m/s^2
s=vf^2-Vi^2/2a
s=(30)²-(0)²/2*9.8
s=900-0/19.6
s=45.9=46 m
The support condition of the truss shown as Figure 1 below are such that the truss is pinned at
A and supported by pin rollers at E. The truss is subjected to loading as shown on the Figure.
a) Calculate the reactions at supports A and E.
b) Calculate the force in each member of the truss. Reproduce and fill in the given table
by placing your answers in the columns designated (Force in the member) and
(Nature of the Force i.e. strut or tie)
80KN
с
90KN
3m
D
B
60kN
4m
А
E
F
40 KN
3m
3
3
Зm
A cement truck when filled weighs 42 000 N. Calculate its mass
Answer:
Mass = 4281.35 kg
Explanation:
From Newton's law of motion, the weight of an object is given by the formula;
W = mg
Where;
W is weight
m is mass
g is acceleration due to gravity which has a constant value of 9.81 m/s².
We are given weight W = 42000 N.
Thus;
42000 = m × 9.81
m = 42000/9.81
m ≈ 4281.35 kg
Determine the mass of an object if it's moving at 5 m/s and has a momentum of 50 kg*m/s
What constant acceleration (in ft/s2) is required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds
Answer:
6.746 ft/s^2
Explanation:
v(t)=50
v(0)=27
t=5/3600 = 1/720 hours
v(t)-v(0)= a(t-0)
50-27= a(1/720)
a= 23*720= 16560 mi/h^2
16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2
The acceleration required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds is 6.75 ft/s².
Acceleration is given by:
[tex] a = \frac{\Delta v}{\Delta t} = \frac{v_{f} - v_{i}}{t_{f} - t_{i}} [/tex]
Where:
[tex] v_{f} [/tex]: is the final speed of the car = 50 mi/h
[tex] v_{i}[/tex]: is the initial speed of the car = 27 mi/h
[tex] \Delta t[/tex]: is the change in time = 5 seconds
The constant acceleration required is:
[tex] a = \frac{v_{f} - v_{i}}{\Delta t} [/tex]
[tex] a = \frac{50 \frac{mi}{h}*\frac{1 h}{3600 s}*\frac{5280 ft}{1 mi} - 27 \frac{mi}{h}*\frac{1 h}{3600 s}*\frac{5280 ft}{1 mi}}{5 s} [/tex]
[tex] a = 6.75 ft/s^{2} [/tex]
Therefore, the constant acceleration required is 6.75 ft/s².
Find more here:
https://brainly.com/question/12134554?referrer=searchResults
I hope it helps you!
A balloon is filled with 73 L of air at 1.3 atm pressure. What pressure is needed to change the volume to 43L?
Answer:
roughly 2.2 atm
Explanation:
using Boyle's law (P1V1=P2V2),
73×1.3=43×x
x= 2.21 or 2.2 atm
Use the drop-down menus to complete the statements. v ion is formed. OS When electrons are lost, a ____ When electrons are gained, a ____ ion is formed.
Answer:
Positive
negative
this is on edgy
Explanation:
I got it right when I did this.
Answer:
1. Positive
2. Negative
Explanation: edge 2021
Hope this helps! :)
Which organ completes the majority of chemical digestion?
Answer:
The majority of chemical digestion occurs in the small intestine. Digested chyme from the stomach passes through the pylorus and into the duodenum.
A hockey puck attached to a horizontal spring oscillates on a frictionless, horizontal surface. The spring has force constant 4.50 N/m and the oscillation period is 1.20 s. What is the mass of the puck
Answer:
m = 0.164 kg
Explanation:
T (period)
k (force/spring constant)
m (mass)
T = 2*Pi*sqrt(m/k)
T/(2*Pi) = sqrt(m)/sqrt(k)
(T/(2*Pi))*sqrt(k) = sqrt(m)
m = ((T/(2*Pi))*sqrt(k))^2
m = 4.5*((1.2/(2*Pi)))^2
m = 0.1641403175
1. Newton's third law states that any action will have a(n) and reaction O A. Equal and opposite O B. Greater and opposite C. Equal and similar D. Equal and different
Answer: Equal and Opposite
Explanation:
William wanted to create a report on a geographical location with the greatest species diversity. Which ecosystem can consider for his report?
A.
the ocean
B.
a desert
C.
a tropical rain forest
D.
a coral reef
Two protons are on either side of an electron as shown below:
Diagram
The electron is 30 µm away from the proton on its left and 10 µm away from the proton on its right. What is the magnitude and direction of the net electric force acting on the electron? Note that
P=1.6×10(-19) c=-e
Select one:
a. 2.0×10−18N to the right
b. 7.0×10−24N to the right
c. 2.0×10−18N to the left
d. −9.59×10−24N to the right
e. 7.0×10−24N to the
Answer:
Choice a. approximately [tex]2 \times 10^{-18}\; \rm N[/tex] to the right.
Explanation:
Look up the Coulomb constant, [tex]k[/tex]:
[tex]k \approx 8.987552 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex].
Coulomb's Law may be used to find the magnitude of the electric force between the electron and each proton.
Let [tex]q_1[/tex] and [tex]q_2[/tex] denote the magnitude of electric charge on two point charges. Let [tex]r[/tex] denote the distance between these two charges.
By Coulomb's Law, the magnitude of the force between these two point charges would be:
[tex]\displaystyle F = \frac{k\, q_1\, q_2}{r^2}[/tex].
The unit of distance in the Coulomb's constant here[tex]k \approx 8.987552 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex] is meters. However, the distance between the electron and each proton is given in micrometers, [tex]\rm \mu m[/tex]. Note that [tex]1\; \rm \mu m = 10^{-6}\; \rm m[/tex].
Convert these units to meters:
[tex]30 \; \rm \mu m = 30 \times 10^{-6}\; \rm m[/tex].
[tex]10 \; \rm \mu m = 10 \times 10^{-6}\; \rm m[/tex].
The question states that the magnitude of electric charge on each proton and each electron is [tex]1.6\times 10^{-19}\; \rm C[/tex].
Using these information, find the magnitude of the electric force between the electron and each proton:
Between the electron and the proton on the left-hand side of the electron: [tex]\begin{aligned}F &= \frac{k\, q_1\, q_2}{r^2}\\ &\approx \frac{\left(8.987552\times 10^{9}\; \rm N \cdot m^2\cdot C^{-2}\right) \times \left(1.6 \times 10^{-19}\; \rm C\right) \times \left(1.6 \times 10^{-19}\; \rm C\right)}{\left(30 \times 10^{-6}\; \rm m\right)} \\ &\approx 0.26 \times 10^{-18}\; \rm N\end{aligned}[/tex].Between the electron and the proton on the right-hand side of the electron:[tex]\begin{aligned}F &= \frac{k\, q_1\, q_2}{r^2}\\ &\approx \frac{\left(8.987552\times 10^{9}\; \rm N \cdot m^2\cdot C^{-2}\right) \times \left(1.6 \times 10^{-19}\; \rm C\right) \times \left(1.6 \times 10^{-19}\; \rm C\right)}{\left(10 \times 10^{-6}\; \rm m\right)} \\ &\approx 2.56 \times 10^{-18}\; \rm N\end{aligned}[/tex].The two protons in this question will both attract the electron. Therefore, the force between the electron and the proton on the left-hand side of the electron would point to the left. Similarly, the force between the electron the proton on the right-hand side of the electron would point to the right.
The magnitude of the net electric force on the electron would be:
[tex]2.56 \times 10^{-18}\; \rm N - 0.26\times 10^{-18}\;\rm N \approx 2\times 10^{-18}\; \rm N[/tex].
These two forces act along the same line in opposite direction. Therefore, the resultant force of these two force would be in the direction of the larger force of the two.
In this question, the electric force between the electron and the proton on the right-hand side of the electron is larger. Hence, the net electric force of the protons on the electron should point to the right (towards the proton that is closer to the electron.)
The average swimming speed of a salmon is about 5 ft/s. If a salmon swims at this speed for a full minute, how far will the salmon travel?
300 feet because 5x60=300 and there is 60 seconds in a min
Which of the following units are used to describe acceleration?
m
b
m/s2
С
m/s
d
S
Answer:
m/s2 m/s
Explanation:
A girl swings a 0.250 kg rock attached to a taut string in a circle around her head. Her hand holds the end of the string above her head, and the string angles down slightly (11.9° below the horizontal.). The string is massless and 0.75m long.
A coordinate system lies with its origin a the location where the string comes out of the girl's hand. The positive z-axis points vertically. the projection onto the horizontal plane is such that the string makes an angle of 34.6° with the x-axis.
At a certain instant, the rock makes 2.50 revolutions per second (rev/s) in a counterclockwise direction as seen from above.
What is the tangential velocity vector at this instant?
Complete Question
The diagram of with this question is shown on the first uploaded image
Answer:
The value is [tex]v = -6.543 \^ i + 9.47 \^ j + 0 \^ k[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m = 0.250 \ kg[/tex]
The length of the string is [tex]L = 0.75 \ m[/tex]
The angle the string makes horizontal is [tex]\theta = 11.9^o[/tex]
The angle which the projection of the string onto the xy -plane makes with the positive x-axis is [tex]\phi = 34.6^o[/tex]
The angular velocity of the rock is [tex]w = 2.50 rev/s = 2.50 * 2\pi =15.7 \ rad/s[/tex]
Generally the radius of the circle made by the length of the string is mathematically represented as
[tex]r = L cos(\theta )[/tex]
=> [tex]r = 0.75 cos(11.9 )[/tex]
=> [tex]r = 0.734 \ m[/tex]
Generally the resultant tangential velocity is mathematically represented as
[tex]v__{R}} = w * r[/tex]
=> [tex]v__{R}} = 15.7 *0.734[/tex]
=> [tex]v__{R}} = 11.5 \ m/s[/tex]
Generally the tangential velocity along the x-axis is
[tex]v_x = -v__{R}} * sin(\phi)[/tex]
=> [tex]v_x =- 11.5 * sin(34.6)[/tex]
=> [tex]v_x = -6.543 \ m/s[/tex]
The negative sign show that the velocity is directed toward the negative x-axis
Generally the tangential velocity along the y-axis is
[tex]v_y = v__{R}} * cos(\phi)[/tex]
=> [tex]v_y = 11.5 * cos(34.6)[/tex]
=> [tex]v_y = 9.47 \ m/s[/tex]
Generally the tangential velocity along the y-axis is
[tex]v_z = v__{R}} * cos(90)[/tex]
=> [tex]v_z = 0 \ m/s[/tex]
Generally the tangential velocity at that instant is mathematically represented as
[tex]v = -6.543 \^ i + 9.47 \^ j + 0 \^ k[/tex]
5
b. What is the molecular shape of the molecule? (3 points)
Answer:
tetrahedral shape.
Explanation:
Answer:
Molecular Geometries. The VSEPR theory describes five main shapes of simple molecules: linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. Apply the VSEPR model to determine the geometry of molecules where the central atom contains one or more lone pairs of electrons.
Debbie touches the metal doorknob to her bedroom at 7 a.m. and feels a static electric shock. In the afternoon, Debbie comes back and touches the doorknob again, but she experiences no shock.
Answer:
she had socks one and was shuffeling so she had static
Explanation:
5) A train travels 120 km in 90 minutes (min). What is its average speed?
Answer:
1.33 km per minute
Explanation:
120 km divided by 90 is 1.33 km
The two blocks are attached to each other by a massless string that is wrapped
around a frictionless pulley as shown in figure-2. When the bottom 4.00 kg block is
pulled to the left by the constant force P~ = 45 N, the top 2.00 kg block slides across
it to the right. Assume that the coefficient of kinetic friction between all surfaces
is μk = 0.400
Answer:
a) The complete free body diagram of the system is presented in the image attached below according to the Newton's Law.
b) The acceleration of the top block is 0.962 meters per square second in the -x direction and the acceleration of the bottom block is 0.962 meters per square second in the -x direction.
c) The tension on the rope is 9.770 newtons.
Explanation:
a) The complete free body diagram of the system is presented in the image attached below according to the Newton's Law.
b) At first we have to construct corresponding equations of equilibrium for each mass:
Mass A
[tex]\Sigma F_{x} = T - \mu_{k}\cdot N_{A} = m_{A}\cdot a_{A}[/tex] (1)
[tex]\Sigma F_{y} = N_{A}-m_{A}\cdot g = 0[/tex] (2)
Mass B
[tex]\Sigma F_{x} = -P+\mu_{k}\cdot N_{A}+\mu_{k}\cdot N_{B}+T= -m_{B}\cdot a_{A}[/tex] (3)
[tex]\Sigma F_{y} = N_{B}-N_{A}-m_{B}\cdot g= 0[/tex] (4)
Where:
[tex]P[/tex] - Force exerted on the bottom block, measured in newtons.
[tex]N_{A}[/tex], [tex]N_{B}[/tex] - Normal forces, measured in newtons.
[tex]m_{A}[/tex], [tex]m_{B}[/tex] - Masses of the top and bottom blocks, measured in kilograms.
[tex]\mu_{k}[/tex] - Kinetic coeffcient of friction, dimensionless.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]a_{A}[/tex], [tex]a_{B}[/tex] - Net accelerations of the top and bottom blocks, measured in meters per square second.
From (2):
[tex]N_{A} = m_{A}\cdot g[/tex] (5)
(2) in (4):
[tex]N_{B}=(m_{A}+m_{B})\cdot g[/tex] (6)
(5) in (1):
[tex]T = m_{A}\cdot a_{A}+\mu_{k}\cdot m_{A}\cdot g[/tex] (7)
(6) in (3):
[tex]T = -m_{B}\cdot a_{A}+P-\mu_{k}\cdot m_{A}\cdot g-\mu_{k}\cdot (m_{B}+m_{A})\cdot g[/tex]
[tex]T = -m_{B}\cdot a_{A}+P -\mu_{k}\cdot (2\cdot m_{A}+m_{B})\cdot g[/tex] (8)
And we obtain an expression for the magnitude of the acceleration by eliminating tension:
[tex]m_{A}\cdot a_{A} +\mu_{k}\cdot m_{A}\cdot g = -m_{B}\cdot a_{A}+P-\mu_{k}\cdot (2\cdot m_{A}+m_{B})\cdot g[/tex]
[tex](m_{A}+m_{B})\cdot a_{A} = P-\mu_{k}\cdot (3\cdot m_{A}+m_{B})\cdot g[/tex]
[tex]a_{A} = \frac{P-\mu_{k}\cdot (3\cdot m_{A}+m_{B})\cdot g}{m_{A}+m_{B}}[/tex]
If we know that [tex]P = 45\,N[/tex], [tex]m_{A} = 2\,kg[/tex], [tex]m_{B} = 4\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\mu_{k} = 0.4[/tex], then the magnitude of the acceleration is:
[tex]a_{A} = \frac{45\,N-(0.4)\cdot [3\cdot (2\,kg)+4\,kg]\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{2\,kg+4\,kg}[/tex]
[tex]a_{A} = 0.962\,\frac{m}{s^{2}}[/tex]
The acceleration of the top block is 0.962 meters per square second in the -x direction and the acceleration of the bottom block is 0.962 meters per square second in the -x direction.
c) By means of (7) and knowing that [tex]m_{A} = 2\,kg[/tex], [tex]a_{A} = 0.962\,\frac{m}{s^{2}}[/tex], [tex]\mu_{k} = 0.4[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], we find that the tension on the rope is:
[tex]T = (2\,kg)\cdot \left(0.962\,\frac{m}{s^{2}} \right)+(0.4)\cdot (2\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]T = 9.770\,N[/tex]
The tension on the rope is 9.770 newtons.
Which best describes the way a sound wave is sent through the radio
Answer:
sound wave-electrical wave-radio wave
Answer:
D
Explanation:
on edg
What is a difference between an object's speed and velocity?
Answer:
.
Explanation:
Speed, being a scalar quantity, is the rate at which an object covers distance. ... On the other hand, velocity is a vector quantity; it is direction-aware. Velocity is the rate at which the position changes. The average velocity is the displacement or position change (a vector quantity) per time ratio.
Answer:
velocity is the magnitude and the direction and speed is the distance something traveled in a certain amount of time
SOMETHING:
hope this helps
A mnemonic device in which phrases or poems use the first letter of each word to help a person remember the information is an example of a(n) __________.
A.
acronym
B.
rhyme
C.
list
D.
acrostic
Answer:
D. Acrostic
Explanation:
Answer:
D.
acrostic
Explanation:
Determine the velocity (in m/s) of the object during the last six seconds. Include a numerical answer accurate to the second decimal place. If negative, include the - sign
Answer:
0.33 m/s
Explanation:
The graph given is that of distance against time. The slope of such a graph gives the velocity of the object while travelling.
In this graph the change in distance for the last 6 seconds is given by;
The distance is : 10-8 = 2 m
The time in seconds is: 6 s
The velocity = 2/6 = 1/3 m/s
The velocity expressed in decimal form is : 0.33 m/s
Which of the following is a scalar quantity?
Answer:
55
Explanation:
I hope this answer help u
Answer:
i think no. A is the coeerct answer
hope this will help you
what problems might it cause if a company tried to recycle materials without sorting them first? at least 2 or 3 sentences please!!!!
Answer:
They would probably not be looking good.. like if u don't sort it out they won't be shaped well and there are many types of plastic and if we combine them together they won't give the correct form..
hope this helps :)
When you use a match to light a candle _______________________________ transformations of energy
Answer:
thermal or heat energy
describe the difference between particle motion in solids and liquids.
Answer:
Particles in a: gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.
i hope this helps your answer
Answer:
liquids vibrate, move about, and slide past each other. solids vibrate but generally do not move from place to place.
Please help with this question
Answer:
I believe the answer is 5718.75. Respond if it is wrong please.
Explanation:
I used a calculator.
Question 7
A thick uniform rubber rope of density 0.7 gcm-3 and Young Modulus 5 x 106 Nm 2 has a length 6.7 m. when
hung from the ceiling of the room, calculate the increase in length due to its own weight ?
O A. 35.79 x 10-2 m
O B. 43.81 x 10-2 m
OC.52.34 x 10-2 m
O D.61.65 x 10-2m
Answer:
Increase in length, e = 61.59 * 10⁻²m
The closest value from the options is option D. 61.65 x 10-2m
Explanation:
Young Modulus = stress/strain = (F/A)(/e/L) = F * L/A * e
where F = force; L = original length; A = area; e = increase in length
making e subject of formula; [tex]e = F * L/Y * A[/tex]
Also force, F = mass * acceleration due to gravity;
mass = density * volume = density * area * length = ρ * A * L
Therefore, force, F = ρ * A * L * g
e = ρ *A * L* L * g/Y * A = ρgL²/Y
density = 0.7 gcm⁻³ = 0.7 * 10³ kgm⁻³; g = 9.8 ms⁻²
e = {0.7 * 10³ kgm⁻³ * (6.7 m)² * 9.8 ms⁻²}/5 * 10⁶ Nm²
e = 61.59 * 10⁻² m
What would happen to moon if gravity no longer affected it?
Answer: The moon would continue to move on its orbital path.
Explanation:
How do the forces on a skydiver affect the velocity of the skydiver as they fall?
Hellp
Answer:
Explanation:
As a skydiver falls, he accelerates downwards, gaining speed with each second. The increase in speed is accompanied by an increase in air resistance. This force of air resistance counters the force of gravity.
When a skydiver falls, he constantly accelerates. Therby, falling faster and faster. But also he is slowed down a bit due to the force of air resistance.