Answer:
D. A large force produces a large change in the object’s momentum only if the force is applied over a very short time interval.
Explanation:
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = mass * velocity [/tex]
Also, the impulse of an object is given by the formula;
[tex] Impulse = force * time [/tex]
In accordance with the impulse-momentum theorem, the statement which is true about the effect of a force exerted upon an object is that a large force produces a large change in the object’s momentum only if the force is applied over a very short time interval.
We intend to measure the open-loop gain (LaTeX: A_{open}A o p e n ) of an actual operational amplifier. The magnitude of LaTeX: A_{open}A o p e n is in the range of 106 V/V. However, the signal generator in measurement setup can supply minimal voltage of 1 mV, and the oscilloscope used at amplifier output can measure maximal voltage level of 10 V. Can you design a simple measurement setup using this signal generator and oscilloscope, and accurately measure the LaTeX: A_{open}A o p e n
Answer:
voltage divider, R₂ = 1000 R₁
measuring the output in the resistance R₁
Explanation:
Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V
in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.
To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.
If we use two resistors whose relationship is
R₂ / R₁ = 10³
R₂ = 1000 R₁
When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator
a race car goes around a circular track of radius 150 m at speed of 10.0 m/s. How long does it take to complete one lap?
Answer:
94.25 seconds
Explanation:
Solve for period (T) using: v=(2*pi*r)/T
rearrange: vT=2*pi*r
rearrange: T=(2*pi*r)/v
Plug in values.
T=(2*pi*150)/10
T=94.25 seconds
If a race car goes around a circular track of a radius of 150 m at speed of 10.0 m/s ,then the time taken to complete the one lap would be 94.25 seconds.
What is speed?The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.
As given in the problem a race car goes around a circular track of radius 150 m at speed of 10.0 m/s.
vT = 2 × π × r
T = (2 × π × r)/ v
T = (2 × π× 150)/10
T = 94.25 seconds
Thus, the time taken to complete the one lap would be 94.25 seconds.
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An 5kg object is released from rest near the surface of a planet. The vertical position of the object as a function of time is shown in the graph. All frictional forces are considered to be negligible. What is the closest approximation of the weight of the object.
a) 300N
b) 30N
c) 5N
d) 150N
Answer:
The correct option is b: 30 N.
Explanation:
First, we need to find the acceleration due to gravity (a):
[tex] y_{f} - y_{0} = v_{o}t - \frac{1}{2}a(\Delta t)^{2} [/tex] (1)
Where:
[tex]y_{f}[/tex]: is the final vertical position (obtained from the graph)
[tex]y_{0}[/tex]: is the initial vertical position (obtained from the graph)
v₀: is the initial speed = 0 (it is released from rest)
Δt: is the variation of time (from the graph)
From the graph, we can take the following values of height and time:
t₀ = 0 s → [tex]t_{f}[/tex] = 5 s
y₀ = 300 m → [tex]y_{f}[/tex] = 225 m
Now, by entering the above values into equation (1) and solving for "a" we have:
[tex] a = 2\frac{y_{0} - y_{f}}{(t_{f} - t_{0})^{2}} = 2\frac{300 m - 225 m}{(5 s - 0)^{2}} = 6 m/s^{2} [/tex]
Finally, the weight of the object is:
[tex] W = ma = 5 kg*6 m/s^{2} = 30 N [/tex]
Therefore, the correct option is b: 30 N.
I hope it helps you!
An inclined plane consists of a 25 m length that raises an object 5 m above the ground. When pushing a 4500 N crate to the top of the ramp you exert 1000 N. What is the ideal Mechanical Advantage of this machine?
The IDEAL mechanical advantage of this ramp is (25m / 5m) = 5 .
But it's only giving you a real MA of (4500N/1000N) = 4.5 .
The friction between the crate and the surface of the ramp is robbing some of the work you do as you slide the crate up the ramp, which degrades the mechanical advantage.
A physics student sits in a chair. The chair pushes up on the student's body. Identify the other force of the interaction force pair.
Answer:
gravity pulls down on student the chair pushes up on the student's body with the same force gravity is pulling down on the student
What is the chemical formula for magnesium sulfide?
Answer:
MgS
Explanation:
Which statement about oceans is incorrect?
A Evaporation occurs when water is warmed by the sun.
B Most evaporation and precipitation occur over the ocean.
C 97 percent of Earth's water is fresh water from the ocean.
D Water leaves the ocean by the process of evaporation.
Which of the following is NOT a reason
why gravity is important?
A It holds the planets in
orbit around the sun
B. It causes the ocean tides
C. It guides the growth of plants
D. None of the above
Answer:
I'm gonna say d
Explanation:
bc they all seem very important
hope this helped
A wave with a frequency of 5Hz travels a distance of 40mm in 2 seconds.What is the speed of the wave
Answer:
20mm per second
Explanation:
4. When you are holding a book, energy is stored between the book and the Earth.
This type of energy is called
potential energy.
A. Elastic potential energy
B. Chemical potential energy
C. Gravitational potential energy
D. Kinetic energy
Answer:
gravitational potential energy
PLEASE CLICK ON THIS IMAGE I NEED HELP
Answer:
Second option
Explanation:
"Uniform" pretty much means the same thing happens.
why type of volcano is built almost entirely from ejected lava fragments
Answer:
Shield volcanoes
Explanation:
Which image best illustrates diffraction?
Answer: A
Explanation:
what dose current equal?
Planets don't collide into
the sun because they
A. Are moving
B. Have too much mass
C. Have their own gravity
D. Are more attracted to each other
HELP me please cause I don't understand it.
Answer:
Force = 0.49 N (Approx)
Explanation:
Given:
Mass = 50 grams = 0.05 Kg
Acceleration = 9.81 m/s²
Find:
Force
Computation:
Force = Mass x Acceleration
Force = 0.05 x 9.81
Force = 0.4905
Force = 0.49 N (Approx)
Which of the following is an example of kinetic mechanical energy?
Immersive Reader
(2 Points)
A. A bike rolling down a hill
B. An elevated wrecking ball
C. A compressed spring
D. A loaded gun
E. A set mouse trap
Answer:
A
Explanation:
Kinetic energy must be moving. Potential energy has the ability to move but is not doing so at the moment.
A is likely the answer. But there's lots involved in that kind of motion.
B If the ball is elevated, it implies it is not moving yet. It has potential energy.
C Again, the spring is compressed. It will push something when it moves, but it is not moving yet.
D The load gun's bullet is not moving. It's still potential energy.
E. The mouse trap is set, but it is not moving. When the mouse eats the bait then it's potential energy will transform into kinetic energy.
Which word best completes the sentence?
Select the word from the drop-down menu
He is quite
Choose...
despite never having left his smalL TOWEN
Answer:
it’s cosmopolitan
Explanation:
k12
how many electrons can occupy each sublevel?
in a class where the number of girls is 36% of the total number,there are 48 boys.how many students are there in the class?
Answer:
There are 75 people in the class. The number of boys is 48 and the number of girls is 27. The percentage of girls is 36% of 75.
Explanation:
Waves are ?
that can travel through matter.
Answer:
A wave can be thought of as a disturbance or oscillation that travels through space-time, accompanied by a transfer of energy. The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.
If a 15 N box is lifted a distance of 3 m, how much work is done?
0 J
45 J
5 J
5 N
Answer:
W=45J
Explanation:
W=Fd
W=15(3)=45
W=45J
What is the average kinetic energy of particles in a gas at a temperature of 245 Kelvins?
Answer:
2)
Explanation:
A ball of mass m=10g, carrying a charge q =-20μe is suspended from a string of length L= 0.8m above a horizontal uniformly charged infinite plane sheet of charge density σ = 4μe/m^2. The ball is displaced from the vertical by an angle and allowed to swing from rest.
Required:
a. Obtain the equations of motion of the charged ball based on Newtonian laws of motion.
b. Assume the displaced angle θ is small and simplify the results obtained in part (a) to obtain the frequency of oscillations of the charged ball.
Answer:
a) [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] = [tex]\frac{d^2 \theta}{d t^2}[/tex]
b) f = 2π [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]
Explanation:
a) To have the equations of motion, let's use Newton's second law.
Let's set a reference system where the x-axis is parallel to the path and the y-axis is in the direction of tension of the rope.
For this reference system the tension is in the direction of the y axis, we must decompose the weight and the electrical force.
Let's use trigonometry for the weight that is in the vertical direction down
sin θ = Wₓ / W
cos θ = W_y / w
Wₓ = W sin θ
W_y = W cos θ
we repeat for the electric force that is vertical upwards
F_{ex} = F_e sin θ
F_{ey} = F_e cos θ
the electric force is
F_e = q E
where the field created by an infinite plate is
E = [tex]\frac{ \sigma}{2 \epsilon_o}[/tex]
let's write Newton's second law
Y axis
T - W_y = 0
T = W cos θ
X axis
F_{ex} - Wₓ = m a (1)
we use that the acceleration is related to the position
a = dv / dt
v = dx / dt
where x is the displacement in the arc of the curve
substituting
a = d² x /dt²
we substitute in 1
q E sin θ - mg sin θ = m [tex]\frac{d^2 x}{dt^2}[/tex]
we have angular (tea) and linear (x) variables, if we remember that angles must be measured in radians
θ = x / R
x = R θ
we substitute
sin θ (q E - mg) = m \frac{d^2 R \ theta}{dt^2}
[tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] = [tex]\frac{d^2 \theta}{d t^2}[/tex]
this is the equation of motion of the system
b) for small oscillations
sin θ = θ
therefore the solution is simple harmonic
θ = θ₀ cos (wt + Ф)
if derived twice, we substitute
- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{\theta}{R } θ₀ cos (wt + Ф) = -w² θ₀ cos (wt + Ф)
w² = [tex]\frac{g}{R}[/tex] - [tex]\frac{q}{m} \frac{ \sigma }{2 \epsilon_o} \frac{1}{R}[/tex]
angular velocity is related to frequency
w = 2π f
f = 2π / w
f = 2π/w
f = 2π [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]
One of the smallest planes ever flown was the Bumble Bee II, which had a mass of 180 kg. If the pilot’s mass was 70 kg, what was the velocity of both plane and pilot if their momentum was 20,800 kg∙m/s to the west?
Answer:
83.2 m/s to the West
Explanation:
From the question given above, the following data were obtained:
Mass of plane = 180 Kg
Mass of pilot = 70 Kg
Momentum = 20800 Kg∙m/s West
Velocity =?
Next, we shall determine the total mass. This can be obtained as follow:
Mass of plane = 180 Kg
Mass of pilot = 70 Kg
Total mass =?
Total mass = Mass of plane + Mass of pilot
Total mass = 180 + 70
Total mass = 250 Kg
Finally, we shall determine the velocity. This can be obtained as follow:
Total mass = 250 Kg
Momentum = 20800 Kg∙m/s West
Velocity =?
Momentum = mass × Velocity
20800 = 250 × Velocity
Divide both side by 250
Velocity = 20800 / 250
Velocity = 83.2 m/s West
Thus, the velocity of both plane and pilot is 83.2 m/s to the West
A 50 kg mass is sitting on a frictionless surface. An unknown constant force called force A pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s. If the 50 kg mass is now pushed by an unknown force B and reaches the velocity of 3 m/s in 4 seconds, compare the impulse delivered to the mass when acted upon by force A with the impulse delivered to the mass when acted on by force B? *
A) The impulse delivered to the mass when acted upon by force A is greater
B) The impulse delivered to the mass when acted upon by force B is greater
C) The impulse is the same in each case
D) We need to know the value of force A and force B in order to determine this
Answer:
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1. What is matter?
2. What are the three phases of matter?
3. Describe how gas particles move.
4. What is temperature?
5. The slower the particles, the ______________ the temperature.
6. A change in temperature causes what?
7. What is the difference between boiling and evaporation?
8. What is sublimation?
9. Name the three ways thermal energy is transferred.
10. Sunburn is an example of what?
11. Give an example of convection.
12. What is conduction?
13. What is the difference between conductors and insulators?
Answer:
1.matter is any substance that has mass and takes up space by having volume.
2.The three fundamental phases of matter are solid, liquid, and gas (vapour),
3.In gases the particles move rapidly in all directions, frequently colliding with each other and the side of the container. With an increase in temperature, the particles gain kinetic energy and move faster. ... In liquids, particles are quite close together and move with random motion throughout the container.
3.In gases the particles move rapidly in all directions, frequently colliding with each other and the side of the container. With an increase in temperature, the particles gain kinetic energy and move faster. ... In liquids, particles are quite close together and move with random motion throughout the container.
Each vertical line on the graph is 1 millisecond (0.001 s) of time. What is the period and
frequency of the sound waves?
Explanation:
Given that,
Each vertical line on the graph is 1 millisecond (0.001 s) of time.
We need to find the period and the frequency of the sound wave. The period of a wave is equal to the each vertical line on graph i.e. 0.001 s.
Let f be the frequency of the sound wave. So,
f = 1/T
i.e.
[tex]f=\dfrac{1}{0.001 }\\\\f=1000\ Hz[/tex]
So, the period and the frequency of the sound waves is 1 milliseond and 1000 Hz respectively.
A physics student sits in a chair. The chair pushes up on the student's body. Identify the other force of the interaction force pair.
Answer:
The other force is the weight of the student.
Explanation:
With respect to Newton's third law of motion, for the student to sit and balance on the chair, there must be two equal and opposite forces involved. The student applies his/ her weight on the chair which acts downwards, while the chair applies an equal but opposite force to the weight of the student.
The force applied by the chair on the student's body is counter balanced by the student's weight. Note that, if the weight of the student is greater than the opposing force from the chair, the chair would collapse.
The local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bumper car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.
Calculate the time required for bumper car A to travel the 10 m to collide with bumper car B.
Calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s.
Is the direction of motion for bumper car A following the collision with bumper car B to the right, to the left, or is bumper car A at rest?
Is the collision elastic? Justify your answer.
Answer:
a. 20 s
b. 0 m/s
c. right
d.no its inelastic because when the car b was at rest and a was coming in at it, since b had no force what so ever car a swept it away with it moving to the right
Explanation:
im not sure though
By applying conservation of linear momentum, the answers are:
1. Time = 2 s
2. 3 m/s
3. same direction
4. Inelastic collision
COLLISIONThere are for types of collision. They are;
Elastic CollisionPerfectly elastic collisionInelastic collisionPerfectly Inelastic collisionGiven that a local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bumper car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.
1. The time required for bumper car A to travel the 10 m to collide with bumper car B can be calculated by using first equation of linear motion.
V = U + at
Where
V = 4 m/s
U = 2 m/s
a = 1 m/[tex]s^{2}[/tex]
Substitute all the parameters into the formula
4 = 2 + t
t = 4 - 2
t = 2s
2. To calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s, we will apply conservation of linear momentum
[tex]m_{1}u_{1}[/tex] = [tex]m_{1}v_{1}[/tex] + [tex]m_{2}v_{2}[/tex]
300 x 4 = 300V + 200 x 3
1200 = 300V + 300
300V = 1200 - 300
300V = 900
V = 900/300
V = 3 m/s
3. Since the final velocity of car A is positive, the direction of motion for bumper car A follows the collision with bumper car B to the right.
4. Since the both move at the same velocity, the collision inelastic.
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