Answer:
Answer d) Lab B's reading is less precise than Lab A's
Explanation:
Recall that accuracy is a measure of how close to the actual physical quantity the experiment leads to. In this case, we don't know what the true age is,so we cannot tell which reading is more accurate.
But we can tell which reading is more precise given the value of their reported errors. Since Lab A shows a smaller error in years (+/- 20 years versus 30 years), as well as a smaller percent error (3.9% versus 5.1%), we can say that Lab's A reading is more precise than Lab B's reading.
This agrees with answer labeled d).
Determine the maximum velocity and maximum acceleration of a particle which moves in simple harmonic motion with an amplitude of 0.2 in and a period of 0.1 s.
Answer:
The maximum velocity and maximum acceleration of the particle are approximately 12.566 inches per second and 789.568 inches per square second.
Explanation:
The equation of motion for the position of a particle experimentating a simple harmonic motion ([tex]x(t)[/tex]), measured in inches, is described by the following expression:
[tex]x(t) = A\cdot \cos \left(\frac{2\pi\cdot t}{T} +\phi\right)[/tex] (1)
Where:
[tex]A[/tex] - Amplitude, measured in inches.
[tex]t[/tex] - Time, measured in seconds.
[tex]T[/tex] - Period, measured in seconds.
[tex]\phi[/tex] - Phase, measured in radians.
Then, we obtain the formulas for the velocity and acceleration of the particle by differentiating (1):
[tex]v(t) = -\frac{2\pi\cdot A}{T}\cdot \sin \left(\frac{2\pi\cdot t}{T}+\phi \right)[/tex] (2)
[tex]a(t) = -\left(\frac{2\pi}{T} \right)^{2}\cdot A\cdot \cos \left(\frac{2\pi\cdot t}{T}+\phi \right)[/tex] (3)
From (2) and (3) we find that maximum velocity ([tex]v_{max}[/tex]), measured in inches per second, and maximum acceleration ([tex]a_{max}[/tex]), measured in inches per square second, are defined by the following formulas:
[tex]v_{max} = \frac{2\pi\cdot A}{T}[/tex] (4)
[tex]a_{max} = \left(\frac{2\pi}{T} \right)^{2}\cdot A[/tex] (5)
If we know that [tex]A = 0.2\,in[/tex] and [tex]T = 0.1\,s[/tex], then the maximum velocity and maximum acceleration of the particle are, respectively:
[tex]v_{max} = \frac{2\pi\cdot (0.2\,in)}{0.1\,s}[/tex]
[tex]v_{max} \approx 12.566\,\frac{in}{s}[/tex]
[tex]a_{max} = \left(\frac{2\pi}{0.1\,s} \right)^{2}\cdot (0.2\,in)[/tex]
[tex]a_{max} \approx 789.568\,\frac{in}{s^{2}}[/tex]
The maximum velocity and maximum acceleration of the particle are approximately 12.566 inches per second and 789.568 inches per square second.
When an object executes a to and fro motion is said to be a simple harmonic motion. The maximum velocity and maximum acceleration of the particle will be 12.566 in/s and 789.568 in/s².
What is simple harmonic motion?When an object executes a to and fro motion in the definite plane when it is tied with the string. The type of motion will be the simple harmonic motion.
Simple harmonic motion is a form of periodic motion in mechanics and physics in which the restoring force on the moving item is directly proportional to the size of the object.
The value of maximum velocity is given by the formula in the SHM motion is
[tex]\rm V_{max = \frac{2\pi A}{T}[/tex]
[tex]\rm V_{max} = \frac{2\times 3.14 \times 0.2 }{0.1}[/tex]
[tex]\rm V_{max = 12.566 \;in/sec[/tex]
The value of maximum acceleration is given by the formula in the SHM motion is
[tex]\rm a_{max}=(\frac{2\pi}{T} )^2.A\\\\ \rm a_{max}=(\frac{2\times 3.14 }{0.1} )^2 \times 0.2 \\\\ \rm a_{max}=789.568\; in/sec^2[/tex]
Hence the maximum velocity and maximum acceleration of the particle will be 12.566 in/s and 789.568 in/s²
To learn more about the simple harmonic motion refer to thr link;
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What is calibration of thermometer?
Answer:
HACCP based food safety programs require accurate record keeping to be successful. Temperature is often the parameter of interest when monitoring a critical control point (CCP). ... To be considered accurate, a thermometer must be calibrated to measure within +/- 2° F (1.1° C) of the actual temperature.
Explanation:
An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4
sec. What is the average speed of the
electron during this time?
Answer:
Average speed = 10,000 m/s
Explanation:
Given the following data;
Distance = 2m
Time = 0.0002secs
To find the average speed;
Average speed = distance/time
Average speed = 2/0.0002
Average speed = 10,000 m/s
Therefore, the average speed of the
electron is 10,000 meters per seconds.
2- A car on a straight highway goes in the positive direction for 8 km and then backs up for 3.6 km. What are the distance and displacement covered by the train?
Answer:
11.6km
4.4km in the negative direction
Explanation:
Distance is the total length of path covered and traveled by a body.
So, for this car on a straight line;
Total distance = 8km + 3.6km = 11.6km
Displacement is the distance traveled along a path and the direction it takes.
It is a vector quantity with magnitude and directional attributes.
For this journey;
Displacement = 8km - 3.6km = 4.4km in the negative direction.
A low rider truck accelerates from 0 m/s to 30.0 m/s with a displacement of 50m. What is the vehicle's acceleration?
a. 2.00m/s
b. 5.00m/s
c. 9.0m/s
d. 10.0m/s
Answer:
a = 9 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}^{2}=v_{o}^{2} +2*a*x[/tex]
where:
Vf = final velocity = 30 [m/s]
Vo = initial velocity = 0
a = acceleration [m/s²]
x = displacement = 50 [m]
Now replacing:
[tex]30^{2}=0 +2*a*50\\100*a=900\\a=9[m/s^{2} ][/tex]
Mass is:
A.how much an object weighs
B.everything around you
C.how much matter is in an object
D.the state of matter such a as solid liquid or gas
Answer:
C-How much matter is in an object
Explanation:
Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are tripled and their separation is also tripled, what happens to the magnitude of the electrostatic force that each charge exerts on the other
Answer:
The force remains the same.
Explanation:
Let the magnitude of the forces in each case be F1 and F2 respectively.
The charges are Q1 and Q2.
The distance of separation between them is R.
Hence, for F1;
F1 = KQ1Q2/R^2
For F2:
F2 = K * 3Q1 * 3Q2/(3R)^2
F2 = 9KQ1Q2/9R^2
F2 =KQ1Q2/R^2
Hence F1=F2
The force is the same in both cases!
A 105 kg football player runs at 8.5 m/s and plows into the back of an 85 kg referee running at 3.5 m/s on the field causing the referee to fly forward at 6.0 m/s.
What must the momentum of the football player be after the collision?
Answer:
680 Kg.m/s
Explanation:
Mass of player; m_p = 105 kg.
Speed of player before Collision; v_pi = 8.5 m/s
Mass of referee; m_r = 85 kg
Speed of referee before collision; v_ri = 3.5 m/s
Speed of referee after collision; v_rf = 6 m/s
From conservation of momentum,
Initial momentum = final momentum
Thus;
(m_p × v_pi) + (m_r × v_ri) = (m_p × v_pf) + (m_r × v_rf)
Where (m_p × v_pf) is the momentum of the player after collision.
Thus, Plugging in the relevant values, we have;
(105 × 8.5) + (85 × 3.5) = (m_p × v_pf) + (85 × 6)
(m_p × v_pf) = (105 × 8.5) + (85 × 3.5) - (85 × 6)
(m_p × v_pf) = 680 Kg.m/s
Convert 41.3 kilocalories into joules.
Explanation:
41.3 kilocalories = 172 799.2 joules
[tex]hope \: it \: will \: help[/tex]
Sam is driving along the highway towards Savannah. He travels 150 m in 3 seconds. What is his speed
Answer:
50m/s
Explanation:
Speed = distance/ time
Distance = 150m
Time = 3 secs
Therefore Speed =150/3
Speed = 50 m/s
An electric heater is 90% efficient. How much useful energy is produced from 8000 J?
Answer:
7200J
Explanation:
you have to find 90 percent of 8000 J to find total efficiency
Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is released
Answer:
171.5m
Explanation:
The velocity of sound in water = 343m/s
Time taken = 1.00secs
using the formula to calculate the distance
2x = vt
x is the distance
v is the speed of sound
t is the time
x = vt/2
x = 343(1)/2
x = 171.5m
hence their separation 1.00 s after the second object is released is 171.5m
Please help! Will give a lot of points
g Observer A, who is at rest in the laboratory, is studying a particle that is moving through the laboratory at a speed of 0.624c and determines its lifetime to be 159 ns. (a) Observer A places markers in the laboratory at the locations where the particle is produced and where it decays. How far apart are those markers in the laboratory
Answer:
[tex]D=29.7648m[/tex]
Explanation:
From the question we are told that
Speed of the particle[tex]V=0.624c[/tex]
Lifetime [tex]t=159ns[/tex]
Generally the equation for distance is mathematically given by
[tex]D=Vt[/tex]
[tex]D=0.624c*159*10^-^9[/tex]
[tex]D=0.624*3*10^8*159*10^-^9[/tex]
[tex]D=29.7648m[/tex]
Therefore the markers are [tex]D=29.7648m[/tex] apart
(Are these statements about the spring constant true or
false?
a. The spring constant is a measure of the stiffness of
the spring.
b. The spring constant tells you how many newtons of
force it takes to stretch the spring one meter.
c. If a spring stretches easily, it will have a high spring
constant.
d. The spring constant of a spring varies with the
amount of stretch or compression of the spring.
Answer:
true
true
false
true
pls follow me and thank you
what kind of mathematical relationship do you predict wll exist between velocity and number of filters
Answer:
[tex]PK_T = constant[/tex]
Explanation:
[tex]PK_T[/tex] = constant
terminal velocity is the maximum const. velocity to achieved by the partical when particle is free fall with the effect of gravity
so that here
eq of motion is
Fnet = mg - kv ................1
here a = 0
v = terminal velocity
0 = mg - kV
v = [tex]\frac{mg}{k}[/tex] m/s
Which term is the product of force and distance? power work net force acceleration please help
Answer: WORK
Explanation:
Answer:
work
Explanation:
got it right on edge:p
What does the law of conservation of energy state? *
1. Total energy before transfer is equal to total energy after transfer
2. Total energy before transfer is less than total energy after transfer
3. Total energy before transfer is more than total energy after transfer
Answer:
total energy before transfer is equal to total energy after transfer
Answer:
1 is the correct one Answer
What is the distance of separation between objects of masses 5.6 x 10 5 kg and 8.8 x 10 6 kg if the force of gravity between them is 440 N
Answer:
2.87m
Explanation:
Using the law of gravitation to solve this question
F = GMm/r²
G is the gravitational constant
M and m are the masses
r is the distance between the masses
Substitute the given values
G = 6.67×10^-11 m³/kgs²
M =8.8 x 10^6 kg
m = 5.6 x 10^5 kg
F =440N
400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²
400r² = 328.698×10
400r² = 3286.98
r² = 3286.98/400
r² = 8.21745
r = √8.21745
r = 2.87m
Hence the distance of separation is 2.87m
The distance of separation will be "0.86 m".
Given:
Mass,
[tex]m_1 = 5.6\times 10^5 \ kg[/tex][tex]m_2 = 8.8\times 10^6 \ kg[/tex]Force,
[tex]F = 440 \ N[/tex]As we know,
→ [tex]F = \frac{Gm_1 m_2}{r^2}[/tex]
By putting the values, we get
→ [tex]440=\frac{6.67\times 10^{-11}\times 5.6\times 10^5\times 8.8\times 10^6}{r^2}[/tex]
→ [tex]r^2 = \frac{6.67\times 5.6\times 8.8}{440}[/tex]
→ [tex]r^2 = 0.74704[/tex]
→ [tex]r = \sqrt{0.74704}[/tex]
→ [tex]= 0.86 \ m[/tex]
Thus above approach is correct.
Learn more:
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A speedboat moving at 29.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.4 m/s2 by reducing the throttle. (a) How long does it take the boat to reach the buoy
Answer:
[tex]t =4.8sec[/tex]
Explanation:
From the question we are told that
Velocity of speed boat [tex]V=29.0m/s[/tex]
Distance to Marker [tex]d=100[/tex]
Acceleration of [tex]a=-3.4m/s^2[/tex]
Generally the Newtons 3rd motion equation is given as
[tex]v^2 = u^2 + 2 * a* s[/tex]
[tex]v^2 = 29^2 + 2 * -3.4* 100[/tex]
[tex]v = \sqrt{161}[/tex]
[tex]v=12.68m/s[/tex]
Generally the Newton's first equation of motion is given as
[tex]v = u + a*t[/tex]
[tex]12.68 = 29 -3.4*t[/tex]
[tex]12.68-29 = -3.4t[/tex]
[tex]-16.32 = -3.4t[/tex]
[tex]t =\frac{-16.32}{-3.4}[/tex]
[tex]t =4.8sec[/tex] .
prove principle of conversation of linear momentum in accordance to principle of classical relativity
Answer:
Cuz im batman hahahhahah
the amount of power exerted by a machine that applies a force of 550 N over a distance of 10 m in a time of 5 seconds
Answer:
The meaning of work is a force that acts on an object and the object moves in respond and the meaning of power is how much work is done in a given amount of time. The two terms are related by is that power is the rate of doing work.
Explanation:
During an automobile crash test, the average force exerted by a solid wall on a 1,700 kg car that hits the wall is measured to be 170,000 N over a 0.14 s time interval. What was the initial speed of the car (in m/s) prior to the collision, assuming the car is at rest at the end of the time interval
Answer:
14 m/s
Explanation:
Step One:
Given data:
mass of automobile m= 1700kg
Force F= 170,000 N
time t= 0.14s
v= ??
Required
The velocity of the car
Step two:
From the expression given below, we can find the velocity
Ft= mv
make v subject of the formula
v= Ft/m
v= 170000*0.14/1700
v=23800/1700
v=14 m/s
A 1250-kg car moves at 20.0 m/s. How much work must be done on the car to increase its speed to 30.0 m/s.
Answer:
312,497.5Joules
Explanation:
Work done = force × distance
W = FS
Get the force
F = ma
F = 1250×9.8
F = 12250N
Get the distances using the equation of motion
v² = u² +2gS
30² =20²+2(9.8)S
900 =400+19.6S
900-400 =19.6S
500 = 19.6S
S = 500/19.6
S = 25.51m
Get the work done
Work done = 12250×25.51
Workdone = 312,497.5Joules
Answer:
3.13 x 10^5 J
Explanation:
The work done increases the car's kinetic energy so that its new speed is 30 m/s. So, we set t, the net work required to increase the car's speed equal to the change in the car's kinetic energy.
Wnet = △K = Kf - Ki = 1/2mvf^2 - 1/2mvi^2 = 1/2m(vf^2-vi^2) = 1/2(1250kg)((30m/s)^2 - (20m/s)^2) = 3.13x10^5 J
In a chemical reaction, the mass of the reactants is 13,3 grams. Which of the
following could represent the mass of the products?
Answer:
my assuytyyhyyyyyy gftdrrdtiifyb tvhyvth rv b yy
A helicopter flies with an air speed of 175 km/h, heading south. The wind is blowing at 85 km/h to the east relative to the ground. Calculate the speed and direction of the helicopter.
Answer:
154° at 195 km/h
Explanation:
The helicopter is moving south at 175 km/h, relative to the wind.
But the wind is moving east at 85 km/h, relative to the ground.
This means that the helicopter is moving south east relative to the ground.
Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.
This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.
Refer to the triangle b1.
The distance traveled by the helicopter in 1 hour is denoted by d.
d is the hypotenuse of the right triangle.
Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)
Hence the helicopter is traveling at 195 km/h relative to the ground.
To calculate the direction we use,
tan (x) = opposite/adjacent = 85/175
So the angle x is,
[tex]x = arctan (\frac{85}{175} )[/tex] = 25.9°
Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)
If force c is 30n and force d us 20 N what is the size of the horizontal force
A constant force of 20. newtons applied to a box causes it to move at a constant speed of 4.0 meters per second. How much work is done in the boxin 6.0 seconds?
1) 240 joules
2) 120 joules
3) 480 joules
4) 80. joules
The work done is 480 Joules.
Option 3 is correct.
The work done is computed as multiplication of force and displacement.
Given that, Force = 20 N
Displacement = 4 meter per second
For 6 second,
Displacement [tex]=4*6=24m[/tex]
[tex]Workdone=Force*displacement\\\\Workdone=20*24=480J[/tex]
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NEED Answer asap plz!!
Answer:
I think a is the right not clear but
A 10 kg weight is suspended in the air by a strong cable. How much work is done, per unit time, in suspending the weight
Answer:
There is no work done.
Explanation:
Given the following data;
Mass = 10 kg
To find the work done?
In Physics, work done can be defined as the amount of energy transfered when an object or body is moved over a distance due to the action of an external force.
Mathematically, work done is given by the formula;
Work done = force * distance
[tex] W = F * d[/tex]
Where,
W is the work doneF represents the force acting on a body. d represents the distance covered by the body.However, the weight suspended in the air by a strong cable does not move or experience any form of displacement. Therefore, there is no work done.