Answer:
Fluid mechanics is considered one of the toughest subdisciplines within mechanical and aerospace engineering. It is unique from almost any other field an undergraduate engineer will encounter. It requires viewing physics in a new light, and that's not always an easy jump to make.
Which is true regarding a child standing up for their own rights?
Answer:
hey mate......looks like the question is incomplete
A 41.6-kg person, running horizontally with a velocity of +4.21 m/s, jumps onto a 14.6-kg sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?
Answer:
a) v = 3.116 m / s, b) μ = 1.65 10⁻²
Explanation:
a) to find the velocity of the set, let's define a system formed by the person and the sled, so that the forces during the collision are internal and the moment is conserved
initial instant. Before the crash
p₀ = M v₀
final instant. After the crash
p_f = (M + m) v
the moment is preserved
M v₀ = (M + m) v
v = [tex]\frac{M}{M+m} \ v_o[/tex]
let's calculate
v = [tex]\frac{41.6}{41.6 + 14.6} \ 4.21[/tex]
v = 3.116 m / s
b) for this part let's use the relationship between work and kinetic energy
W = ΔK
as the body has its final kinetic energy is zero
the work of the friction forces is
W = - fr x
the negative sign is because the friction forces always oppose the movement
let's write Newton's second law
Y axis
N - W_sled -W_person = 0
N = mg + M g
N = (m + M) g
X axis
fr = ma
the friction force has the expression
fr = μ N
fr = μ g (m + M)
we substitute
- μg (m + M) x = 0- ½ (m + M) v²
μ = [tex]\frac{1}{2} \ \frac{v^2 }{g \ x }[/tex]
let's calculate
μ = [tex]\frac{1}{2} \ \frac{3.116^2}{9.8 \ 30.0}[/tex]
μ = 0.0165
μ = 1.65 10⁻²
PLEASE HELP! I'LL GIVE BRAINLEST
Answer:
Weight = 8.162 Newton.
Explanation:
Given the following data;
Mass = 2.2 kg
Acceleration due to gravity = 3.71 N/kg
To find the weight of the textbook;
Weight = mass * acceleration due to gravity
Weight = 2.2 * 3.71
Weight = 8.162 N
Therefore, the weight of the science textbook in mars is 8.162 Newton.
Disk A, with a mass of 2.0 kg and a radius of 70 cm , rotates clockwise about a frictionless vertical axle at 50 rev/s . Disk B, also 2.0 kg but with a radius of 50 cm , rotates counterclockwise about that same axle, but at a greater height than disk A, at 50 rev/s . Disk B slides down the axle until it lands on top of disk A, after which they rotate together. After the collision, what is their common angular speed (in rev/s) and in which direction do they rotate?
Answer:
w = - 197.5 rad / s
The negative sign indicates that the rotations are clockwise
Explanation:
To solve this exercise, let's use the concept of conservation of the angular number.
We create a system formed by the two discs, in this case the forces last the shock are internal
initial instant .. just before shock
L₀ = I₀ w₀ + I₁ w₁
instnte final. Right after crash
L_f = (I₀ + I1) w
angular momentum is conserved
I₀ w₀ + I₁ w₁ = (I₀ + I₁) w
w = I₀ w₀ + I₁ w₁ / Io + I1
The moment of inertia of a disk with an axis passing through its thermometric center
I₀ = ½ m² r₀²
I₁ = ½ m₁ r₁²
we substitute
I₀ = ½ 2.0 0.70²
I₀ = 0.49 kg m
I₁ = ½ 2.0 0.5²
I₁ = 0.25
₁
let's reduce the magnitudes the SI system
w₀ = -50 rev / (2pi rad / 1rev) = -314.15 rad / s
w₁ = 70 rev (2pi rad / 1rev) = 439.82 rad / s
we will assume that the counterclockwise turns are positive
w = -0.49 314.15 + 0.25 439.82 / (0.49 + 0.25)
w = (- 4.696 + 1.0995) 102) / 0.74
w = -197.75 + 0.25
w = - 197.5 rad / s
The negative sign indicates that the rotations are clockwise
what is the formula for Mass when you have been provided force, time and distance
Answer:
The mass formula is also given as m = F / a If acceleration itself is the gravity, then M = F / g
please mark me as brainleist
An object A with mass 200 kg and an another object B with mass 1000 kg are moving with same speed. The ratio of kinetic energy of object A to B is
Answer:
Ratio of kinetic energy of object A to B = 1:5
Explanation:
Given:
Mass of object A = 200 kg
Mass of object B = 1,000 kg
Find:
Ratio of kinetic energy of object A to B
Computation:
Kinetic energy = (1/2)(m)(v²)
Kinetic energy of object A = (1/2)(200)(v²)
Kinetic energy of object A = (100)(v²)
Kinetic energy of object B = (1/2)(1,000)(v²)
Kinetic energy of object B = (500)(v²)
Ratio of kinetic energy of object A to B = Kinetic energy of object A / Kinetic energy of object B
Ratio of kinetic energy of object A to B = (100)(v²) / (500)(v²)
Ratio of kinetic energy of object A to B = 100 / 500
Ratio of kinetic energy of object A to B = 1/5
Ratio of kinetic energy of object A to B = 1:5
(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg buddy Bubba has been riding at the front of your 60 kg, 4 m long boat. You come to a stop in the water and then switch places. A) What is the center of mass before you switch places
Answer:
Explanation:
From the given information:
Let the first weight be [tex]m_ 1[/tex] = 80 kg
The weight of the buddy be [tex]m_2[/tex] = 120 kg
The weight of Bubba be [tex]m_3[/tex] = 60 kg
Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:
[tex]x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2[/tex]
The length of the boat be [tex]x_2[/tex] = 4 m
∴
We can find the center of mass of the system by using the formula:
[tex]X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}[/tex]
Which of the following is true of the deep
water layer of the ocean?
A. warmest and least dense of the ocean layers
B. experiences a rapid decrease in temperature
C. is warm in the summer and cold in the winter
D. cold all year round
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m the floor is frictionless, and for the next 10.5 m the coefficient of friction is 0.17.
What is the final speed of the crate after being pulled these 20.5 meters?
Answer:
The final speed of the crate is 12.07 m/s.
Explanation:
For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:
[tex] F = ma [/tex]
[tex] a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2} [/tex]
Now, we can calculate the final speed of the crate at the end of 10.0 m:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad_{1} [/tex]
[tex] v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s [/tex]
For the next 10.5 meters we have frictional force:
[tex] F - F_{\mu} = ma [/tex]
[tex] F - \mu mg = ma [/tex]
So, the acceleration is:
[tex] a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2} [/tex]
The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad_{2} [/tex]
[tex] v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s [/tex]
Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.
I hope it helps you!
Sparks occur when the electric field in air exceeds 3 x 106 N/C. This is because free electrons normally present in air are accelerated to such high speeds that their kinetic energy will overcome the potential energy holding other electrons to atoms. When those electrons rearrange themselves after such a collision, a flash of light is emitted. Let us suppose that the work done on an electron must give it an energy of 3 x 10-19 J to cause this ionization. How far does an electron involved in making in a spark travel through the air before it collides with an atom
Answer:
h = 5.38 10¹⁶ m
Explanation:
Let's start this exercise by assuming that all the potential energy of the electron is converted into kinetic energy, let's use the conservation of energy
starting point. Just before ionization
Em₀ = U = qE
final point. Right after ionization
Em_f = K = ½ m v²
Energy is conserved
Em₀ = Em_f
q E = ½ m v²
v² = 2qE / m
Now we can use the relationship between net work and kinetic energy
W_net = ΔK
net work is the work done by the electron minus the binding energy with the atom, called the work function, Ф = 3 10-19 J
W - Ф = K_f - K₀
we assume that the electron converts all its initial initial kinetic energy to be zero
W -Ф = ½ m v² - 0
W = ½ m v² +Ф
we substitute
W = 1/2 m 2qE/m + E
W = qE +Ф
W = 1.6 10⁻¹⁹ 3 10⁶ + 3 10⁻¹⁹
W = 4.8 10⁻¹³ + 3 10⁻¹⁹
W = 4.8 10⁻¹³ J
When the electron is in air, its kinetic energy can be transformed into gravitational potential energy
As the electron is in the air, all work is transformed into scientific energy
W = K
starting point Em₀ = K = W
end point Em_F = U = m g h
energy conservation Em₀ = Em_f
W = m g h
h = [tex]\frac{W}{mg}[/tex]
let's calculate
h = [tex]\frac{4.8 \ 10^{-13} x}{9.1 \ 10^{-31} \ 9.8 }[/tex]
h = 5.38 10¹⁶ m
Electron involved in making in spark travel through the air before it collides with an atom will be at the distance of 5.38 10¹⁶ m.
What is an electric field?An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.
Let's begin this exercise by assuming that all of the electron's potential energy is turned into kinetic energy, and then we'll apply the law of conservation of energy.
Energy before ionization;
[tex]\rm Em_0 = U = qE[/tex]
Energy after ionization;
[tex]Em_f = K = \frac{1}{2} mv^2[/tex]
From the law of conservation of energy principle;
[tex]Em_0 = Em_f \\\\ q E =\frac{1}{2} m v^2\\\\ v^2 = \frac{2qE }{m}[/tex]
The relationship between net work and kinetic energy;
[tex]W_{net} = \triangle K[/tex]
The work function is defined as net work, which is the work done by the electron minus the binding energy with the atom.
[tex]W - \phi = K_f - K_0[/tex]
[tex]W = K_f+ \phi[/tex]
[tex]W = \frac{1}{2} m \times \frac{2qE}{m} + E\\ \\W = qE + \phi \\\\ \rm W = 1.6 \times 10^{-19}\times 3 \tims 10^6 3 10⁶ +3 \times 10^{-19} \\\\ W = 4.8 \times 10^{-13}+ 3 \times 10^{-19}\\\\ W = 4.8 \times 10^{-13} J[/tex]
EMF at starting point;
[tex]\rm Em_0 = K = W[/tex]
EMF at the endpoint;
[tex]\rm Em_F = U = m g h[/tex]
From the law of conservation of energy principle;
[tex]Em_0 = Em_f \\\\ W = m g \\\\ h = \frac{W}{mg}\\\\\ h = \frac{4.8 \timjes 10^{-13}}{9.1 \times 10^{-31} \times 9.81 }\\\\ \rm h= 5.38 \times 10^{16}[/tex]
Hence electron involved in making in spark travel through the air before it collides with an atom will be at a distance of 5.38 10¹⁶ m.
To learn more about the electric field refer to the link;
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conduction happens because of
a)touching
b)liquid or gas
c)space
A 1-kg ball is 12 m above the ground, with an initial velocity = 0 m/s.
Use the following formulas
[ KE = 1/2 xm x V2]
[P.E = mxgxh] . g =9.8 m/s2
[ Mechanical energy = K.E + P.E]
Answer:
M = 117.6 J
Explanation:
Given that,
The mass of a ball, m = 1kg
The height of the ball, h = 12 m
At point A, its initial velocity, v = 0
The mechanical energy is the sum of kinetic and potential energy such that,
[tex]M=\dfrac{1}{2}mv^2+mgh\\\\M=0+mgh\\\\M=1\times 9.8\times 12\\\\M=117.6\ J[/tex]
So, the mechanical energy is equal to 117.6 J.
What is the difference between a positively and negatively charged object?
Answer:
Positively charged objects have electrons; they simply possess more protons than electrons. Negatively charged objects have protons; it's just their number of electrons is greater than their number of protons.
The difference between a positively charged object and a negatively charged object is the number of protons and electrons. The imbalance in charge results into formation of charged objects.
What are Charged objects?
Charged objects have an imbalance of charge that is either more negative electrons than the positive protons or more positive protons than the negative electrons in the object. The neutral objects are those species which have a balance of charge with equal number of protons and electrons.
A positively charged object is formed when an atom has more protons than electrons. And, a negatively charged object is formed when an atom has more electrons than protons. As, electrons have a negative charge and protons have a positive charge.
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A copper wire has a mass of 29.33 mg/cm and has a length of 2.5 cm.
Find the weight of the copper wire.
What does a step-up transformer do?
A. It steps up the energy.
B. It steps up the power.
C. It steps up the voltage.
D. It steps up the current.
40 POINTS!!! PLEASE HELPP!!!
Please select the word from the list that best fits the definition
One form of energy changes to one other form
(A.) single transformation
(B.) multiple transformation
Answer:
A
Explanation:
if it was B it would say from one to another to another
ʕ•ᴥ•ʔhi how are you ？
Answer:
I’m not okay . Me and my bf are bickering once again
Look at this attached photo
Answer:
C) Kinetic energy
Explanation:
Since the skateboarder is ALREADY going down the hill, he is using kinetic energy.
Kinetic energy is when you're moving and in motion
How much mechanical work is required to catch a 14.715N ball traveling at a velocity of 37.5m/s?
To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.
What is work?Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.
In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that
Work is the product of the component of the force acting in the displacement's direction and its magnitude.
Weight of the ball = 14.715 N.
Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.
Velocity of the ball = 37.5 m/s
Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.
Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.
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question is included in the picture!!! PUT REAL ANSWERS OR I WILL REPORT YOU
Answer:
Explanation:
this is like rubbing a balloon on your head to make your hair stand up. Do that to the can. The balloon is filled , ofc, and then just rub the balloon on the can. This will charge the can with static electricity. :P
A roller coaster works by being hoisted to the top of the highest peak on the track and then released. As the roller coaster car moves along the track the energy gets converted back and forth between kinetic and gravitational potential energy. The entire time the roller coaster car is in motion air friction and friction in the wheels act to oppose the motion of the car. At which point during the ride is the total mechanical energy of the roller coaster car the greatest?
a. At the bottom of the first dip in the track when the car is moving the fastest.
b. When the car is exactly at half its original height half way through the ride.
c. At the top of the highest peak as the very start of the ride.
d. At the end of the ride as the car nears the station.
Answer:
a. At the bottom of the first dip in the track when the car is moving the fastest.
Explanation:
P.S - The exact question is -
The correct option is - a. At the bottom of the first dip in the track when the car is moving the fastest.
Reason -
From the diagram , we can see that the total mechanical energy of the roller coaster car is greatest At the bottom of the first dip in the track when the car is moving the fastest.
35 POINTSS!!! PLSSSS HELLPPP!!!
Work is the transfer of power from one object to another.
Please select the best answer from the choices provided
T
F
Answer:
T
beacuse:
Energy can be transferred from one object to another by doing work. ... When work is done, energy is transferred from the agent to the object, which results in a change in the object's motion (more specifically, a change in the object's kinetic energy).
1. Although mercury is a metal, it is a liquid at room temperature. Mercury melts at about -39°C. If
the temperature of a block of mercury starts at -54°C and increases by 22°C, does the mercury melt?
Explain your answer.
Answer:
I think so because if it starts at a low temperature for that material, it should melt when you bring it up to that temperature.
Answer: Mercury
Explanation: Mercury is kinda like the opposite of regular semi-liquid. For ice to melt you need Fahrenheit to melt ice into water while you need Celsius to melt mercury.
A moving man is lifting a box up an inclined plane. Halfway up he sets the box down and rests. Which of the following explains why the box does not slide back down the inclined plane?
a. The force of friction balances the force of gravity.
b. The force of gravity does not affect inclined planes.
c. The force of friction does not affect inclined planes.
d. The force of friction is less than the force of gravity.
Answer: A or B
Explanation: I’m guessing that they even each other out depending on the incline, gravity will help keeping it in place
Plz answer asap I need all of the answer
Answer:
Turned up side down
Explanation:
A rod that is 96.0 cm long is made of glass that has an index of refraction equal to 1.60. The rod has its ends ground to convex spherical surfaces that have radii equal to 8.00 cm and 19.1 cm. An object is in air on the long axis of the rod 19.9 cm from the end that has the 19.1-cm radius.
(a) Find the image distance due to refraction at the 19.1-cm radius surface.
(b) Find the position of the final image due to refraction at both surfaces.
(c) Is the final image real or virtual?
A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force P. Cylinders A and B weigh, respectively, 5 and 20 lb. The shaft of cylinder A is free to slide in a vertical slot and the coefficients of friction between the belt and each of the cylinders are µs = 0.50 and µk = 0.40. For P = 3.6 lb,
Determine:
(a) Whether slipping occurs between the belt and either cylinder,
(b) The angular acceleration of each cylinder.
Slipping doesn't occur between the belt and cylinder B because the force of static friction is greater than force exerted on cylinder B.
Given the following data:
Mass A = 5 lb to kg = 2.27 kg.
Mass B = 20 lb to kg = 9.02 kg.
Force = 3.6 lb to N = 16.02 Newton.
How to calculate angular acceleration.In order to calculate the angular acceleration of each cylinder, we would take moment about the two cylinders.
For cylinder A:
[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_A\alpha _A = F_A (0.1)\\\\(\frac{m_Ar^2}{2}) \alpha _A = F_A (0.1)\\\\(\frac{2.27 \times 0.1^2}{2}) \alpha _A = F_A (0.1)\\\\0.1F_A=0.01135\alpha _A\\\\F_A=\frac{0.01135\alpha _A}{0.1} \\\\F_A= 0.1135\alpha _A[/tex]
For cylinder B:
[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_B\alpha _B = F_B (0.2)\\\\(\frac{m_Br^2}{2}) \frac{\alpha _A}{2} = F_B (0.2)\\\\(\frac{9.02 \times 0.1^2}{4}) \alpha _A = F_B (0.2)\\\\0.1F_B=0.02255\alpha _A\\\\F_B=\frac{0.02255\alpha _A}{0.2} \\\\F_B= 0.1128\alpha _A[/tex]
For the belt, we have
[tex]\sum F_A =0\\\\P-F_B-F_A=0\\\\16.02-0.1128\alpha _A-0.1135\alpha _A=0\\\\16.02=0.2263\alpha _A\\\\\alpha _A=\frac{16.02}{0.2263} \\\\\alpha _A=70.79 \;rad/s^2[/tex]
Also, we would determine the angular acceleration of cylinder B:
[tex]0.1\alpha _A=0.2\alpha _B\\\\0.1 \times 70.79 = 0.2\alpha _B\\\\7.079= 0.2\alpha _B\\\\\alpha _B=\frac{7.079}{0.2} \\\\\alpha _B=35.40\;rad/s^2[/tex]
Next, we would calculate the forces acting on the cylinders:
[tex]F_A = 0.1135\alpha _A\\\\F_A = 0.1135 \times 70.79\\\\F_A = 8.04 \;Newton[/tex]
[tex]F_B = P-F_A\\\\F_B = 16.02 - 8.04\\\\F_B = 7.98\;Newton[/tex]
Next, we would determine the force of static friction:
[tex]F_s = \mu_s N = \mu_s m_B g\\\\F_s = 0.50 \times 9.02 \times 9.8\\\\F_s=44.198\;Newton[/tex]
From the above calculation, we can deduce that the force of static friction is greater than force exerted on cylinder B. Therefore, slipping doesn't occur between the belt and cylinder B.
Read more on angular acceleration here: https://brainly.com/question/6860269
assuming weightless pulleys and 100% efficiency, what is the minimum input force required to lift a 120 N weight using a single fixed pulley?
A. 21 N
B. 61 N
C. 121 N
D. 241 N
A wave has a frequency of 2 Hz. Find its period
Explain why synovial joints have greater flexibility than fixed and cartilaginous joint.
Answer:Unlike fixed joints or cartilaginous joints, where the bones are connected by either connective tissue or cartilage, the bones in synovial joints are not directly joined by anything, which allows for a much greater range of motion.
Explanation: I got a 100% on my test