Answer:
7 Newton
Explanation:
Dado
Longitud de la cuerda = 50 m
El cable se dobla en 0,058 m.
Masa de roedor = 350 gramos = 0,35 kg
T = m * a + m × v2 / r
Sustituyendo los valores dados obtenemos
T = 0,35 (10 + 10)
T = 0,35 * 20
T = 7 Newton
Equilibrium of forces
Answer:
If the size and direction of the forces acting on an object are exactly balanced, then there is no net force acting on the object and the object is said to be in equilibrium. Because the net force is equal to zero, the forces in Example 1 are acting in equilibrium.
Equilibrium of forces means that the net force is 0. It can either be when there is no force acting on the object or when the force acting on the object are balanced.
c) You wish to put a 1000-kg satellite into a circular orbit 300 km above the earth's surface. (a)
What speed, period, and radial acceleration will it have? (b) How much work must be done to the
satellite to put it in orbit? (c) How much additional work would have to be done to make the
Answer:
Scalar
Explanation:
No direction
A 2120 kg car traveling at 13.4 m/s collides with a 2810 kg car that is initally at rest at a stoplight. The cars stick together and move 1.97 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.
Answer:
The coefficient of friction between the cars and the road is 0.859.
Explanation:
The two cars collide each other inelastically, then we can determine the resulting velocity by the Principle of Momentum Conservation:
[tex]m_{A}\cdot v_{A} + m_{B}\cdot v_{B} = (m_{A} + m_{B})\cdot v[/tex] (1)
Where:
[tex]m_{A}[/tex], [tex]m_{B}[/tex] - Masses of the cars, in kilograms.
[tex]v_{A}[/tex], [tex]v_{B}[/tex] - Initial velocities of the cars, in meters per second.
[tex]v[/tex] - Velocity of the resulting system, in meters per second.
If we know that [tex]m_{A} = 2120\,kg[/tex], [tex]v_{A} = 13.4\,\frac{m}{s }[/tex], [tex]m_{B} = 2810\,kg[/tex] and [tex]v_{B} = 0\,\frac{m}{s}[/tex], then the velocity of the resulting system:
[tex]v = \frac{m_{A}\cdot v_{A}+m_{B}\cdot v_{B}}{m_{A}+m_{B}}[/tex]
[tex]v = \frac{(2120\,kg)\cdot \left(13.4\,\frac{m}{s} \right)+(2810\,kg)\cdot \left(0\,\frac{m}{s} \right)}{2120\,kg + 2810\,kg}[/tex]
[tex]v = 5.762\,\frac{m}{s}[/tex]
By Principle of Energy Conservation and Work-Energy Theorem, we understand that the initial translational kinetic energy ([tex]K[/tex]), in joules, is dissipated due to work done by friction ([tex]W_{f}[/tex]), in joules, that is to say:
[tex]K = W_{f}[/tex] (2)
[tex]\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2} = \mu\cdot (m_{A}+m_{B})\cdot g \cdot s[/tex]
[tex]\frac{1}{2}\cdot v^{2} = \mu \cdot g\cdot s[/tex] (2b)
Where:
[tex]\mu[/tex] - Coefficient of friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex]- Travelled distance, in meters.
If we know that [tex]v = 5.762\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]s = 1.97\,m[/tex], then the coefficient of friction is:
[tex]\mu = \frac{v^{2}}{2\cdot g\cdot s}[/tex]
[tex]\mu = \frac{\left(5.762\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.97\,m)}[/tex]
[tex]\mu = 0.859[/tex]
The coefficient of friction between the cars and the road is 0.859.
How is the kinetic energy of the particles that make up a substance different
after it changes state?
Answer:
the kinetic energy is less after a substance changes from liquid to gas
Explanation:
the kinetic energy is greater after a substance changes from a solid to liquid
Answer:
The kinetic energy is less after a substance changes from a gas to a liquid.
Explanation:
A flywheel with radius of 0.400 mm starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2rad/s2. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Throwing a discus. Part A For a point on the rim of the flywheel, what is the magnitude of the tangential acceleration after 2.00 ss of acceleration
Answer: [tex]0.00024\ m/s^2[/tex]
Explanation:
Given
Radius of flywheel is [tex]r=0.4\ mm[/tex]
Angular acceleration [tex]\alpha=0.6\ rad/s^2[/tex]
For no change in radius, tangential acceleration is given as
[tex]\Rightarrow a_t=a\lpha \times r[/tex]
Insert the values
[tex]\Rightarrow a_t=0.6\times 0.4\times 10^{-3}\ m/s^2\\\Rightarrow a_t=2.4\times 10^{-4}\ m/s^2\ \text{or}\ 0.00024\ m/s^2[/tex]
How many times will the temperature of oxygen with a mass of 1 kg increase if its volume is increased by 4 times, and the pressure is decreased by 2 times?
Round off the answer to the nearest whole number.
Answer:
9.2 Relating Pressure, Volume,
Figure 1. In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.
Explanation:
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Its volume is 20 cm3, and its mass is 100 grams. What is the sample’s density?
Which of the following describes the relationship between the weight of fluid
displaced by an object and the buoyant force exerted on the object?
A. Archimedes' principle
B. Flow rate equation
C. Pascal's principle
D. Bernoulli's principle
¿cual es la presión que se aplica sobre un líquido encerrado en un tanque, por medio de un pistón que tiene un aria de 0.02 metros cuadrados ya aplica una fuerza con una magnitud de 100 newtons?
Answer:
Podemos decir que la presión que se aplica sobre un liquido encerrado en un tanque es de 5000 Pa.
Explanation:
Answer:
Podemos decir que la presión que se aplica sobre un liquido encerrado en un tanque es de 5000 Pa.
Explanation:
Transfer of thermal energy between air molecules in closed room is an example of
conduction
convection
radiation
Answer and I will give you brainiliest
Answer: Conduction
Explanation: Conduction is the process by which heat energy is transmitted through collisions between neighboring atoms or molecules. Conduction occurs more readily in solids and liquids, where the particles are closer to together, than in gases, where particles are further apart.
Pete is investigating the solubility of salt (NaCl) in water. He begins to add 50 grams of salt to 100 grams of
room temperature tap water in a beaker. After adding all of the salt and stirring for several minutes, Pete
notices a solid substance in the bottom of the beaker. Which statement best explains why there is a solid
substance in the bottom of the beaker?
A. The salt he is using is not soluble in water.
B. The salt is changing into a new substance that is not soluble in water,
C. The dissolving salt is causing impurities in the water to precipitate to the bottom
D. The water is saturated and the remaining salt precipitates to the bottom
Answer:
would the answer be c
Explanation: that what i think in my opian
Answer:
A
Explanation:
The wave functions for states of the hydrogen atom with orbital quantum number l=0 are much simpler than for most other states, because the angular part of the wave.
a. True
b. False
By what amount does the 52-cmcm-long femur of an 85 kgkg runner compress at this moment? The cross-section area of the bone of the femur can be taken as 5.2×10−4m25.2×10−4m2 and its Young's modulus is 1.6×1010N/m2.1.6×1010N/m2.
Answer:
0.156 mm
Explanation:
Here is the complete question
The normal force of the ground on the floor can reach three times a runner's body weight when the foot strikes the pavement. By what amount does the 52-cm-long femur of an 85 kg runner compress at this moment? The cross-section area of the bone of the femur can be taken as 5.2 × 10⁻⁴ m² and its Young's modulus is 1.6 × 10¹⁰ N/m²
The Young's modulus of the bone Y = stress/strain = σ/ε = F/A ÷ ΔL/L = FL/AΔL where F = force on bone = 3mg(since it is 3 times his weight) where m = mass of runner = 85 kg and g = acceleration due to gravity = 9.8 m/s². L = length of femur = 52 cm = 0.52 m, A = cross-sectional area of femur = 5.2 × 10⁻⁴ m² and ΔL = compression of femur.
Making ΔL subject of the formula,
ΔL = FL/AY
ΔL = 3mgL/AY
Substituting the values of the variables into the equation, we have
ΔL = 3mgL/AY
ΔL = 3 × 85 kg × 9.8 m/s² × 0.52 m/(5.2 × 10⁻⁴ m² × 1.6 × 10¹⁰ N/m²)
ΔL = 1299.48 kgm²/s² ÷ 8.32 × 10⁻⁶ N
ΔL = 156.1875 × 10⁻⁶ m
ΔL = 0.1561875 × 10⁻³ m
ΔL = 0.1561875 mm
ΔL ≅ 0.156 mm
The amount does the 52-cm long femur of 85 kg is 0.156 mm.
Calculation of the amount:Since
The Young's modulus of the bone Y should be
= stress/strain
= σ/ε
So,
= F/A ÷ ΔL/L
here F = force on bone = 3mg
m = mass of runner = 85 kg
and g = acceleration due to gravity = 9.8 m/s²
L = length of femur = 52 cm = 0.52 m,
A = cross-sectional area of femur = 5.2 × 10⁻⁴ m²
and ΔL = compression of femur.
Now
ΔL = FL/AY
ΔL = 3mgL/AY
Now
ΔL = 3mgL/AY
= 3 × 85 kg × 9.8 m/s² × 0.52 m/(5.2 × 10⁻⁴ m² × 1.6 × 10¹⁰ N/m²)
= 1299.48 kgm²/s² ÷ 8.32 × 10⁻⁶ N
= 156.1875 × 10⁻⁶ m
= 0.1561875 × 10⁻³ m
= 0.1561875 mm
= 0.156 mm
Learn more about moment here: https://brainly.com/question/24717686
Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 3.3 m long, its mass is 0.52 kg, and the force exerted on it by the children is 47 N. (a) What is the linear mass density of the rope (in kg/m)
Answer:
The linear mass density of rope is 0.16 kg/m.
Explanation:
mass, m = 0.52 kg
force, F = 47 N
length, L = 3.3 m
(a) The linear mass density of the rope is defined as the mass of the rope per unit length.
Linear mass density = m/L = 0.52/3.3 = 0.16 kg/m
why material selection is important to design and manufacturing?
Answer:
. You want your product to be as strong and as long lasting as possible. There are also the safety implications to consider. You see, dangerous failures arising from poor material selection are still an all too common occurrence in many industries. yep that the answer have a Great day
Explanation:
(◕ᴗ◕✿)
A cement block accidentally falls from rest from the ledge of a 53.4-m-high building. When the block is 19.4 m above the ground, a man, 2.00 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way
Answer:
The time required by the man to get out of the way is 0.6 s.
Explanation:
height of building, H = 53.4 m
height of block, h = 19.4 m
height of man, h' = 2 m
Let the velocity of the block at 19.4 m is v.
use third equation of motion
[tex]v^2 = u^2 + 2 gh\\\\v^2 = 0 + 2 \times 9.8 \times (53.4 - 19.4)\\\\v = 25.8 m/s[/tex]
Now let the time is t.
Use second equation of motion
[tex]h = u t + 0.5 gt^2\\\\19.4 - 2 = 25.8 t + 4.9 t^2\\\\4.9 t^2 + 25.8 t - 17.4= 0 \\\\t = \frac{-25.8\pm\sqrt{665.64 + 341.04}}{9.8}\\\\t = \frac{-25.8\pm31.7}{9.8}\\\\t = 0.6 s, - 5.9 s[/tex]
Time cannot be negative so time t = 0.6 s.
please help me and i will mark u as brainlist
Answer:
a) 8 secs I think
b)2m/s^2
Copy the diagram. add a voltmeter to show how you would measure the voltage of the cell
Answer: the answer is 23voltage
Explanation: because the voltage and time put together is 23
Ramesh announced in class: ''Yesterday I had fever and my body temperature was 100 degrees.'' Ravi said: ''We learnt in the last period that water boils at a temperature of 100 degrees'' Sonal said: "So Ramesh’s temperature yesterday was close to the boiling point of water.'' What can we say about that conversation?Single choice.
(1 Point)
(a) All are correct: human body temperature during fever is close to the boiling point of water.
(b) Ramesh is making some mistake - he is not remembering his temperature correctly.
(c) Ravi is incorrectly recalling the boiling point of water he learnt about in class.
(d) Ramesh and Ravi are correct, but they are using different measurement scales.
Answer:
D. Ramesh and Ravi are correct, but they are using different measurement scales.
\Huge{\underline{\textrm{Explanation}}}Explanation
Here, Ravi says that his body temperature is 100 degrees, but does not mention that whether it is 100 degrees Celsius or 100 degrees Fahrenheit. When the temperature of a human body is more than 100.4 degree Fahrenheit (38°C), or near to it, the person is considered to have fever.
The boiling point of water is 100 degrees Celsius and not 100 degrees Fahrenheit.
Thus, they both are using different measurement scales.
An amusement park ride whisks you vertically upward. You travel at a constant speed of 15 m/s during the entire ascent. You drop your phone 4.0 s after you (and your phone) begin your ascent from ground level.
a. How high above the ground is your phone when you drop it?
b. Find the maximum height above the ground reached by your phone.
Answer:
a. 60 m
b. 71.48 m
Explanation:
Below are the calculations:
a. The phone's height above the ground = Speed x Time
The phone's height above the ground = 15 x 4 = 60 m
b. Speed when phone drops, u = 15 m/s
At maximum height, v = 0
Use below formula:
v² = u² -2gh
0 = 15² + 2 × 9.8 × h
h = 11.48 m
Total height = 60 + 11.48 = 71.48 m
1. a. What is the pressure on a surface when a force of 500 N acts on an area of 2 m2
250 pascal
Explanation:
Pressure is defined as the force me unit area
Mathematically:
Pressure = Force/area
i.e = P=F/A
Define Kinetic Energy:
Begin by defining kinetic energy in your own words within one concise eight word sentence
Answer:
kinetic energy is the energy an object has beause of it's motion
Answer:
See explanation below
Explanation:
Kinetic energy is energy stored in a moving object.
Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate would the car have to accelerate so that a quarter placed on the back wall would remain in place?
Answer:
25.59 m/s²
Explanation:
Using the formula for the force of static friction:
[tex]f_s = \mu_s N[/tex] --- (1)
where;
[tex]f_s =[/tex] static friction force
[tex]\mu_s =[/tex] coefficient of static friction
N = normal force
Also, recall that:
F = mass × acceleration
Similarly, N = mg
here, due to min. acceleration of the car;
[tex]N = ma_{min}[/tex]
From equation (1)
[tex]f_s = \mu_s ma_{min}[/tex]
However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.
Thus,
[tex]F = f_s[/tex]
[tex]mg = \mu_s ma_{min}[/tex]
[tex]a_{min} = \dfrac{mg }{ \mu_s m}[/tex]
[tex]a_{min} = \dfrac{g }{ \mu_s }[/tex]
where;
[tex]\mu_s = 0.383[/tex] and g = 9.8 m/s²
[tex]a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }[/tex]
[tex]\mathbf{a_{min}= 25.59 \ m/s^2}[/tex]
5. Tests performed on a 16.0 cm strip of the donated aorta reveal that it stretches 3.37 cm when a 1.80 N pull is exerted on it. (a) What is the force constant of this strip of aortal material
Answer:
53.41 N/m
Explanation:
From Hooke's law,
Applying,
F = ke............. Equation 1
Where F = Force, e = extension, k = force constant of the aortal material
Make k the subject of the equation
k = F/e............. Equation 2
From the question,
Given: F = 1.8 N, e = 3.37 cm = 0.0337 m
Substitute these values into equation 2
k = 1.8/(0.0337)
k = 53.41 N/m
Hence the force constant of the aortal material is 53.41 N/m
b) When the muscles connected to the crystalline lens contract fully, its focal length is 16.5000 cm. With this focal length, how far away must an object be to form sharply focused images on the retina? (Note: this distance is called the far point of vision.)
c) When the muscles connected to the crystalline lens relax, the focal length is 9.0000 cm. With this focal length, how close must an object be to form sharply focused images on the retina? (Note: this distance is called the near point of vision.)
d) As people age, the crystalline lens hardens (a condition called presbyopia or “old-age” eyes) and can only vary in focal length from 12 to 15.60 cm. Calculate range of vision (the new near point and far point) for this older eye.
e) Based on part d) why might an older person hold the newspaper at arm’s length to read it?
Answer:
I have to go to work and figure it out
If you drive first at 40 km/h west and later at 60 km/h west, your average velocity is 50 km/h west.
and what else? is that all?
One way families influence healthy technology use is when siblings explain the use of media to each other. Which of these outfits would you expect if this guideline was followed?
Answer:
The answer would be C.
Explanation:
This is what I would expect when you show someone else how to do something then is also known as teaching.
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When two bodies at different temperatures are placed in thermal contact with each other, heat flows from the body at higher temperature to the body at lower temperature until them both acquire the same temperature. Assuming that there is no loss of heat to the surroundings, the heatSingle choice.
(1 Point)
(a) gained by the hotter body will be equal to the heat lost by the colder body
(b) the heat gained by the hotter body will be less than the heat lost by the colder body
(c) the heat gained by the hotter body will be greater than the heat lost by the colder body
(d) the heat lost by the hotter body will be equal to the heat gained by the colder body.
Answer:
Part d is correct.
help asap PLEASE I will give u max everything all that
steps if possible
Explanation:
2. [tex]R_T = R_1 + R_2 + R_3 = 625\:Ω + 330\:Ω + 1500\:Ω[/tex]
[tex]\:\:\:\:\:\:\:= 2455\:Ω = 2.455\:kΩ[/tex]
3. Resistors in series only need to be added together so
[tex]R_T = 8(140\:Ω) = 1120\:Ω = 1.12\:kΩ[/tex]
A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is 48.0 kg, air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.50 m/s. Determine the work done (in J) by the child as the tricycle travels down the incline.
Answer:
The work done by the child as the tricycle travels down the incline is 416.96 J
Explanation:
Given;
initial velocity of the child, [tex]v_i[/tex] = 1.4 m/s
final velocity of the child, [tex]v_f[/tex] = 6.5 m/s
initial height of the inclined plane, h = 2.25 m
length of the inclined plane, L = 12.4 m
total mass, m = 48 kg
frictional force, [tex]f_k[/tex] = 41 N
The work done by the child is calculated as;
[tex]\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech} + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96 \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J[/tex]
Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J
