Answer: B
Explanation: The forces are equal and opposite each other.
When two objects are interacting but stay stationary, then the forces are equal and opposite each other.
What do you mean by Force?Force may be defined as the process of pushing and pulling an object with an actual mass that stimulates its velocity to be changed. It is a type of vector quantity because it has both magnitude and direction.
It is the simple and fundamental concept of physics that when two or more objects are interacting with one another but do stimulate any change in their position, the forces among them are definitely equal and opposite to one another. It is the most plausible explanation of Newton's third law of motion.
Therefore, when two objects are interacting but stay stationary, then the forces are equal and opposite each other.
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Your question seems incomplete. The most probable complete question is as follows:
The forces are equal and opposite each other. The forces are not equal and opposite to each other. The forces are equal but not opposite to each other.The forces are not equal but opposite to each other.Two pieces of steel wire with identical cross sections have lengths of L and 2L. The wires are each fixed at both ends and stretched so that the tension in the longer wire is four times greater than in the shorter wire. If the fundamental frequency in the shorter wire is 60 Hz, what is the frequency of the second harmonic in the longer wire?
Answer:
Explanation:
Expression for fundamental frequency of tone produced in a wire under tension of T and length L is given as follows
[tex]f=\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]
m is mass per unit length .
We shall apply this formula for given wires .
For shorter wire
[tex]60 =\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]
For longer wire for second harmonic
length of wire is 2L , tension is 4T ,
[tex]f =\frac{2}{4L} \times \sqrt{\frac{4T}{ m} }[/tex]
[tex]f =\frac{2\times 2}{4L} \times \sqrt{\frac{T}{ m} }[/tex]
f = 2 x 60 = 120 Hz .
true or false A person's speed around the Earth is faster at the poles than it is at the equator.
Answer:False
Explanation:The Earth rotates faster at the equator than at the poles.
A 2 kg toy car moves at a speed of 5 m/s. If a child applies a 5N force for 2 m in the same direction the car is already moving, what is the change in kinetic energy of the car?
Answer:
[tex]10\: \mathrm{J}[/tex]
Explanation:
The kinetic energy of an object is [tex]KE=\frac{1}{2}mv^2[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.
The toy car's initial kinetic energy is [tex]KE_{i}=\frac{1}{2}\cdot 2\cdot 5^2=25\: \mathrm{J}[/tex].
After the child applies a 5N force on it in the same direction, its velocity will increase but its mass will stay the same.
To find the final velocity of the toy car, we can use kinematic equation [tex]v_f^2=v_i^2+2a\Delta x, \\ v_f=\sqrt{v_i^2+2a\Delta x}[/tex]
We are given [tex]v_i=5\: \mathrm{m/s}[/tex] and [tex]\Delta x = 2\: \mathrm{m}[/tex].
To find acceleration:
[tex]F=ma, a=\frac{F}{m}=\frac{5}{2}=2.5\: \mathrm{m/s^2}[/tex].
Now substitute [tex]v_i=5\: \mathrm{m/s}, \: a=2\: \mathrm{m/s^2}, \: \Delta x = 2\: \mathrm{m}[/tex] into [tex]v_f=\sqrt{v_i^2+2a\Delta x}[/tex] to get [tex]v_f\approx 5.92\: \mathrm{m/s}[/tex].
Using this, we can find the final kinetic energy of the toy car is [tex]KE_f=\frac{1}{2}\cdot 2\cdot 5.92^2[/tex].
Thus, the change in kinetic energy is [tex]KE_f-KE_i=\frac{1}{2}\cdot2\cdot 5.92^2-\frac{1}{2}\cdot 2\cdot 5^2=\fbox{$10\: \mathrm{J}$}[/tex] (one significant figure).
The change in the kinetic energy of the car is 10 J.
The given parameters;
mass of the car, m = 2 kginitial velocity of the car, u = 5 m/sforce applied by the child, F = 5 Ndistance traveled, s = 2 mThe acceleration of the car is calculated as follows;
[tex]F = ma\\\\a = \frac{F}{m} \\\\a = \frac{5}{2} \\\\a = 2.5 \ m/s^2[/tex]
The final velocity of the car is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v = \sqrt{u^2 + 2as} \\\\v = \sqrt{5^2 \ + \ 2(2.5)(2)} \\\\v = 5.92 \ m/s[/tex]
The change in the kinetic energy of the car is calculated as follows;
[tex]\Delta K.E = \frac{1}{2} m(v^2 - u^2)\\\\\Delta K.E = \frac{1}{2} \times 2 \times (5.92^2\ - \ 5^2)\\\\\Delta K.E = 10 \ J[/tex]
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How long ago did most Middle Eastern countries gain their independence?
A.
10-20 years ago
B.
50-100 years ago
C.
200-300 years ago
D.
400-500 years ago
Please select the best answer from the choices provided
A
B
C
D
Answer:
the correct answer is B
Explanation:
Answer:
.B
Explanation:
This is the answer on edge 2021
Have a good day!
Solve each of the following problems to 3 sig figs and correct Sl units, showing all work.
1. A cart with a mass of 45.0 kg is being pulled to the right with a force of 250 N giving it an
acceleration of 1.30 m/s2. The wheels of the cart are locked and the cart must be dragged.
a) Draw a free body diagram of the cart.
b) Calculate the net force acting on the cart.
c) Create a force table and fill it in.
d) Find the coefficient of kinetic friction.
Answer:
every number to 3 sf = 1) 45.0 2) 250 3) 1.30
Explanation:
your welcome :)
Various amplifier and load combinations are measured as listed below using rms values. For each, find the voltage, current, and power gains ( A v , Ai , and Ap, respectively) both as ratios and in dB:
(a) vI= 100 mV, iI = 100 μA, vO = 10 V, RL = 100Ω
(b) vI = 10 μV, iI = 100 nA, vO =1 V, RL= 10 kΩ
(c) vI =1 V, iI = 1 mA, vO =5 V, RL = 10Ω
Answer:
The solution to this question can be defined as follows:
Explanation:
In point (a):
[tex]v_i= 100 \ mV\\\\ i_I = 100 \mu \ A\\\\ v_O = 10 \ V\\\\ R_L = 100 \ \Omega \\\\i_L = \frac{V_0}{R_L} = \frac{10}{100} = 100 \ MA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{100 \times 10^{-3}} =100 \\\\A_v(db) = 20 \lag (100) =40 \ db \\\\ A_i= \frac{i_L}{i_i} = \frac{100 \times 10^{-3}}{100 \times 10^{-6}} =1000 \\\\A_i(db) = 20 \lag (100) =60 \ db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_L}{v_i i_i} = \frac{ 10(100 \times 10^{-3})}{100 \times 10^{-6} \times 100 \times 10^{-3}} =100000\\\\ A_p(db) =10 \log (100000) =50 \ db \\\\[/tex]
In point (b):
[tex]v_i = 10 \mu V\\\\ i_i = 100 \ nA \\\\ v_O =1 \ V \\\\ R_L= 10 \ k \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{1}{10 \ K} = 100 \ \muA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{10 \times 10^{-6}} =100000 \\\\A_v(db) = 20 \lag (100000) =100 \ db \\\\ A_i= \frac{i_0}{i_i} = \frac{100 \times 10^{-6}}{100 \times 10^{-9}} =1000 \\\\A_i(db) = 20 \lag (1000) =60 \ db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 1 \times 100 \times 10^{-6})}{10 \times 10^{-6} \times 100 \times 10^{-9}} =100000000\\\\A_p(db) =10 \log (100000000) =80 \ db \\\\[/tex]
In point (C):
[tex]v_i =1\ V\\\\ i_I = 1 \ mA\\\\ v_O =5\ V \\\\ R_L = 10 \ \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{5}{10 } = 0.5 \ A \\\\A_v = \frac{V_0}{V_i} = \frac{5}{1} =5 \\\\A_v(db) = 20 \log 5 =13.97 \ db = 14 \db \\\\ A_i= \frac{i_0}{i_i} = \frac{0.5}{1\times 10^{-3}} =500 \\\\A_i(db) = 20 \log (500) =53.97 \ db = 54 \db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 5 \times 0.5 }{1 \times 1 \times 10^{-3}} =2500\\\\A_p(db) =10 \log (2500) = 33.97 \ db = 34 \db\\\\[/tex]
If an atom of oxygen has an atomic number of eight that means...…
E. there are 8 protons
F. there are 8 neutrons
G. it weighs 8 amu
H. it is in group 8
I will give brainly
Defend Democritus' work on the atom and its contribution to the modern atomic model.
John runs 3 km north then walks 2 km south. What is his total distance traveled and displacement?
Answer:
the total distance is 5km and the displacement is 1km
Explanation:
The total distance would be the addition of John running both ways so 3 km, 2 km.
However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.
Think about 2 km as a positive value for the first part of the question and a negative value for the second part.
In which part of a lab report would be the following sentence most likely occur? “Since the data showed that the
Answer:
most likely be included in the analysis section of a lab report
Explanation:
help me help me help me
Sultan walks for 15 km at 35° south of east.
Which of the following journeys would result in the same displacement?
Answer:
☝
Explanation:
What relationship must exist between an applied force and the velocity of a moving object if uniform circular motion is to result?
Answer:
See explanation
Explanation:
Centripetal force is defined as the inward force required to keep an object moving with a constant speed in a circular path.
The magnitude of this force depends on the mass of the object, radius of the object and the velocity of the body.
So we can write;
F = mv^2/r
Particle A with charge q and mass ma and particle B with charge 2q and mass
mb, are accelerated from rest by a potential difference AV and subsequently
deflected by a uniform magnetic field into semicircular paths. The radii of the
trajectories by particle A and B are R and 3R, respectively. The direction of
the magnetic field is perpendicular to the velocity of the particle. Determine
their mass ratio?
I need this done by tonight!! Can anyone help me please? Answer these 4 questions
Answer:
1. 14 g of chocolate mixture.
2. 24 fl oz of chocolate milk
3. 10 cups of chocolate milk.
4. 12½ cups.
Explanation:
From the question given above, the following data were obtained:
1 TBSP = 7 g
1 Cup = 8 fl oz
2 Table spoons (TBSP) for 1 cup (8 fl oz) of milk.
1. Determination of the mass of chocolate mixture in 1 cup of chocolate milk.
From the question given above,
1 Cup required 2 Table spoons (TBSP)
But
1 TBSP = 7 g
Therefore,
2 TBSP = 2 × 7 = 14 g
Thus, 1 Cup required 14 g of chocolate mixture.
2. Determination of the number fl oz of chocolate milk in 3 cups
1 Cup = 8 fl oz
Therefore,
3 Cups = 3 × 8
3 Cups = 24 fl oz
Thus, 24 fl oz of chocolate milk are in 3 cups.
3. Determination of the number of cups of chocolate milk produce from 20 TBSP.
2 TBSP is required to produce 1 cup.
Therefore,
20 TBSP will produce = 20/2 = 10 Cups.
Thus, 10 cups of chocolate milk produce from 20 TBSP.
4. Determination of the number of cups obtained from 100 fl oz chocolate milk.
8 fl oz is required to produce 1 cup.
Therefore,
100 fl oz will produce = 100 / 8 = 12½ cups.
Thus, 12½ cups is obtained from 100 fl oz chocolate milk.
A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
was the runner's average speed?
Answer:
7.01yard/sec
Explanation:
Given parameters:
Initial position = 50yard
Final position = 12yard
Time = 5.42s
Unknown:
Average speed of runner = ?
Solution:
To solve this problem;
Speed = [tex]\frac{distance}{time}[/tex]
Distance covered = Initial position - final position = 50 - 12 = 38yards
So;
Speed = [tex]\frac{38}{5.42}[/tex] = 7.01yard/sec
You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the standard error of your results to ONE significant digit.
Answer:
0.01
Explanation:
Given the data:
10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90
True value = 9.81
Mean value :
Σx / n
Sample size, n = 9
(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9
= 88.69 / 9
= 9.854
Standard deviation (σ) :
Sqrt (Σ(X - m)² / n)
[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9
Sqrt(0.113824 / 9)
Sqrt(0.0126471)
σ = 0.1124593
Standard Error = σ / sqrt(n)
Standard Error = 0.1124593 / 9
Standard Error = 0.0124954
Standard Error = 0.01 ( 1 significant digit)
Can a single atom be considered a molecule?
A:only if the atom is found in water
B:no, it takes two or more atoms bonded to create a molecule
C:only if it is an oxygen atom floating in the air
D:yes, all atoms are made up of many different molecules
What is the function
of second plate in
parallel plate capacitor?
a student lift a 25kg mass at vertical distance of 1.6m in a time of 2.0 seconds. a. Find the force needed to lift the mass (in N ). b. Find the work done by the student (in J). c. Find the power exerted by the student (in W)
Answer:
a. F = 245 Newton.
b. Workdone = 392 Joules.
c. Power = 196 Watts
Explanation:
Given the following data;
Mass = 25kg
Distance = 1.6m
Time = 2secs
a. To find the force needed to lift the mass (in N );
Force = mass * acceleration
We know that acceleration due to gravity is equal to 9.8
F = 25*9.8
F = 245N
b. To find the work done by the student (in J);
Workdone = force * distance
Workdone = 245 * 1.6
Workdone = 392 Joules.
c. To find the power exerted by the student (in W);
Power = workdone/time
Power = 392/2
Power = 196 Watts.
1. An engine absorbs 600 J of heat while doing 650 J of work. What is the change
in internal energy of the enginge? *
1250 J
-50 J
-1250 J
50 J
The change in internal energy of the engine is -50 joule. Hence, option (B) is correct.
What is law of conservation of energy?Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.
The absorb energy: Q = 600 Joule
Work done: W = 650 Joule.
Let, the change in internal energy of the engine= dU.
According to conservation of energy:
The absorb energy = change in internal energy + Work done
Q = dU + W
dU = Q - W
= 600 joule - 650 joule
= - 50 joule.
Hence, the change in internal energy of the engine is -50 joule.
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the diameter of the wheels on your car ( including the tires) is 25 inches. you are going to drive 250 miles today. each of your wheels is goingnto turn by an angle of
What is the instantaneous velocity of a freely falling object 11 s after it is released from a position of rest
Answer:
v= -107.8 m/s
Explanation:
Since the object is in free fall, this means that is moving at an accceleration equal to the one due to gravity.Since it starts at rest, we can apply the definition of acceleration, rearranging terms as follows:[tex]v_{f} = v_{o} + a*t = a*t = -g*t = 9.8m/s2*11s = -107.8 m/s (1)[/tex]
(Assuming as positive the upward direction)An object is released from rest at a height H near and above the surface of Earth. As the object falls toward the surface, Earth’s atmosphere exerts a resistive force on the object such that it reaches a terminal velocity before it reaches the ground. Which of the following claims is true? Select two answers.
The system consisting of only the object is an open system.
The system consisting of only the object is an open system.
A
Earth’s atmosphere does negative work on the object as it falls toward the surface.
Earth’s atmosphere does negative work on the object as it falls toward the surface.
B
The change in the object’s kinetic energy from the instant it is released from rest, to the instant it reaches terminal velocity, is zero.
The change in the object’s kinetic energy from the instant it is released from rest, to the instant it reaches terminal velocity, is zero.
C
The total mechanical energy of the object-Earth system remains constant at all times in which the object is in motion.
Answer:
Second and Last Option Are Correct
Explanation:
Suppose it takes a constant force a time of 6.0 seconds to slow a 2500 kg truck
from 26.0 m/sec to 18.0 m/sec. What is the magnitude of the force? Give
your answer in scientific notation rounded correctly.
Answer:
[tex]3.3\cdot 10^3\:\mathrm{N}[/tex]
Explanation:
Impulse on an object is given by [tex]\mathrm{[impulse]}=F\Delta t[/tex].
However, it's also given as change in momentum (impulse-momentum theorem).
Therefore, we can set the change in momentum equal to the former formula for impulse:
[tex]\Delta p=F\Delta t[/tex].
Momentum is given by [tex]p=mv[/tex]. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:
[tex]p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}[/tex].
Now we plug in our values and solve:
[tex]\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}[/tex](two significant figures).
A ball is thrown vertically upward from the top of a 100 foot tower, with an initial velocity of 10 ft/sec. Its position function is s(t)=−16t2+10t+100.
a. What is its velocity in ft/sec when t = 2 seconds? (Solve by using instantaneous rate.)
b. Determine the equation of a line, in slope-intercept form, that passes through the points (5, 6) and (10, 2).
Answer;
-54ft/s
y = -4/5 x + 10
Explanation
Given the position of an object expressed by the function
s(t)=−16t²+10t+100
Velocity is the change in position with respect to time
v(t) = ds(t)/dt
v(t) = -32t + 10
When t = 2
v(2) = -32(2)+10
v(2) = -64+10
v(2) = -54
Hence the velocity of the object is -54ft/s
b) The standard equation of a line in point slope form is expressed as;
y = MX+c
M is the slope
c is the y-intercept
Given the coordinate (5, 6) and (10, 2)
M = 2-6/10-5
M = -4/5
Get the y-intercept
Substitute m = -4/5 and any point say (5,6) into the expression y = mx+c
6 = -4/5 (5) + c
6 = -4+c
c = 6+4
c = 10
Get the required equation
Recall that: y = mx+c
y = -4/5 x + 10
Hence the equation of a line, in slope-intercept form is y = -4/5 x + 10
describe measurement in our daily life
Which image shows the difference between the speed of molecules in hot and cold water? Explain your answer choice.
HELP ME,EVERYONE!!!!!!!! :(
Answer:
the answer is B
Explanation:
I think its B because on the top it shows the molecule speed and A looks like the water is cold, C shows that the hot
water is cooler, and D shows that both are cold
Why is damage from sound waves is an issue on the launchpad but not in the air
(I would have done more points for answering but I'm almost out sry. just pls answer and help.)
The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.
Closeness to the Sound Source: When a rocket is fired on the launchpad, it creates a tremendous amount of noise in proximity to the nearby equipment and structures.
When a rocket is launched, concentrated sound waves are created that can seriously harm neighboring structures, especially if such structures are not built to handle such strong vibrations. In contrast, once a rocket is in the air, the sound waves spread out and become less forceful as they travel through the atmosphere, decreasing the possibility that they may cause harm.
Reflection and Amplification: The launchpad environment can serve as an echo chamber for sound waves because of its huge, solid structures.
Hence, The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.
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A safety plug is designed to melt when the pressure inside a metal tank becomes too high. A gas
at 51.0 atm and a temperature of 23.0°C is contained in the tank, but the plug melts when the
pressure reaches 75.0 atm. What temperature did the gas reach?
