Answer:
right now I see some of you have a great day
Answer: (mI^2)/3
Explanation:
The parallel axis theorem for the calculation of inertia is: I = I CM + Md^2
So, I is the apathy from an axis that is at distance d from the center of mass and LCM the apathy when the axis passes through the center of mass. Do to this, the axis passes through the end of the rod. In analysis, d=l/2
So, we have the equation:
I = mI^2/12 + m (1/2)^2 = mI^2/12 + mI^2/4 = mI^2/12 + 3mI^2/12 = mI^2/3
This relents us the terminal result: (ml^2)/3.
¿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:
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.
please help me and i will mark u as brainlist
Answer:
a) 8 secs I think
b)2m/s^2
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:
hope its help :)
nicsfrom #philippines
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:
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
A(n) 28.3 g bullet is shot into a(n) 5004 g wooden block standing on a frictionless surface. The block, with the bullet in it, acquires a speed of 1.7 m/s. Calculate the speed of the bullet before striking the block. Answer in units of m/s
Answer:
the speed of the bullet before striking the block is 302.3 m/s.
Explanation:
Given;
mass of the bullet, m₁ = 28.3 g = 0.0283 kg
mass of the wooden block, m₂ = 5004 g = 5.004 kg
initial velocity of the block, u₂ = 0
final velocity of the bullet-wood system, v = 1.7 m/s
let the initial velocity of the bullet before striking the block = u₁
Apply the principle of conservation of linear momentum to determine the initial velocity of the bullet.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
0.0283u₁ + 5.004 x 0 = 1.7(0.0283 + 5.004)
0.0283u₁ = 8.5549
u₁ = 8.5549 / 0.0283
u₁ = 302.3 m/s
Therefore, the speed of the bullet before striking the block is 302.3 m/s.
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
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
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?
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
Four identical masses m are evenly spaced on a frictionless 1D track. The first mass is sent at speed v toward the other three. As the masses collide they stick together. Mass 1 sticks to 2, then 1 2 sticks to 3, then 1 2 3 sticks to 4. When the combined 1 2 3 mass strikes mass 4, by what percentage does the speed decrease in %
Answer:
The speed decreases 75%.
Explanation:
Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.For the first collission, only mass 1 is moving before it, so we can write the following equation:[tex]p_{i} = p_{f} = m*v_{o} (1)[/tex]
Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:[tex]p_{f1} = 2*m*v_{1} (2)[/tex]
From (1) and (2) we get:v₁ = v₀/2 (3)Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:[tex]p_{1} = 2*m*v_{1} = 2*m*\frac{v_{o} }{2} = m*v (4)[/tex]
Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:[tex]p_{2} = 3*m*v_{2} (5)[/tex]
From (4) and (5) we get:v₂ = v₀/3 (6)Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:[tex]p_{2} = 3*m*v_{2} = 3*m*\frac{v_{o} }{3} = m*v (7)[/tex]
Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:[tex]p_{3} = 4*m*v_{3} (8)[/tex]
From (7) and (8) we get:v₃ = v₀/4This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.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]
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
Its volume is 20 cm3, and its mass is 100 grams. What is the sample’s density?
What type of lens is this?
Tip: A convex lens is thicker at the centre than at the edges.
When light goes through a concave lens, the light that goes through that lens spreads out more due to the structure of the surface of the lens. A concave lens does not form a focal point at the opposing side of the lens. Instead, it spreads out, doing the exact opposite of what a convex lens does. In the picture, we see a lens where the light goes through and unifies at the opposing side, which is what a convex lens does. So, the lens in that image is a convex lens.
3. What type of lens is this?
Answer: Convex
Wrong Answer: Concave
The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-second time interval.
During this same time interval, the velocity of the object changes its direction by 90°. What is the
magnitude of the average total force acting on the object during this time interval?
a. 30 N
b. 20 N
c. 15 N
d. 40 N
e. 10 N
Which is the correct answer?
Answer:
F = 2 * 30 / 5 = 12 N to stop forward motion
F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees
(12^2 + 16^2)^1.2 = 20 N average force applied
The magnitude of the average total force acting on the object during this time interval is 20 N.
The given parameters:
Mass of the object, m = 2.0 kgInitial velocity, u = 30 m/sFinal velocity, v = 40 m/sTime of motion, t = 5.0 sThe magnitude of the average total force acting on the object during this time interval is calculated as follows;
[tex]F = \frac{mv }{t} \\\\F_1 = \frac{2(40)}{5} \\\\F_1 = 16\ N\\\\F_2= \frac{2(30)}{5} \\\\F_2 = 12 \ N\\\\F = \sqrt{F_1^2 + F_2^2} \\\\F = \sqrt{16^2 + 12^2} \\\\F = 20 \ N[/tex]
Learn more about resultant force here: https://brainly.com/question/25239010
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.
Please help I need this done
Are you aware of human rights violation happening in the community and explain
Answer:
Individuals who commit serious violations of international human rights or humanitarian law, including crimes against humanity and war crimes, may be prosecuted by their own country or by other countries exercising what is known as “universal jurisdiction.”
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
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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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Hope this Helps
1 airplane
travel due north at 300 km while another airplane travels Due South and 300 km are there speed the same or their velocities the same
Answer:
Explanation:
Speed is scalar and velocity is vector. Vector values imply direction as well as magnitude. Therefore, speed and velocity are not the same. The speeds of these 2 planes are the same at 300km/hr, but the velocity of the plane traveling north is +300km/hr while the velocity of the plane traveling south is -300km/hr if we define north as positive and south as negative.
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
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.
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1177 g, and its velocity is +0.681 m/s after the bullet passes through it. The mass of the second block is 1626 g.
(a) What is the velocity of the second block after the bullet imbeds itself?
(b) Find the ratio of the total kinetic energy after the collision to that before the collision.
Answer:
(a)0.531m/s
(b)0.00169
Explanation:
We are given that
Mass of bullet, m=4.67 g=[tex]4.67\times 10^{-3} kg[/tex]
1 kg =1000 g
Speed of bullet, v=357m/s
Mass of block 1,[tex]m_1=1177g=1.177kg[/tex]
Mass of block 2,[tex]m_2=1626 g=1.626 kg[/tex]
Velocity of block 1,[tex]v_1=0.681m/s[/tex]
(a)
Let velocity of the second block after the bullet imbeds itself=v2
Using conservation of momentum
Initial momentum=Final momentum
[tex]mv=m_1v_1+(m+m_2)v_2[/tex]
[tex]4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2[/tex]
[tex]1.66719=0.801537+1.63067v_2[/tex]
[tex]1.66719-0.801537=1.63067v_2[/tex]
[tex]0.865653=1.63067v_2[/tex]
[tex]v_2=\frac{0.865653}{1.63067}[/tex]
[tex]v_2=0.531m/s[/tex]
Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s
(b)Initial kinetic energy before collision
[tex]K_i=\frac{1}{2}mv^2[/tex]
[tex]k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)[/tex]
[tex]k_i=297.59 J[/tex]
Final kinetic energy after collision
[tex]K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2[/tex]
[tex]K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2[/tex]
[tex]K_f=0.5028 J[/tex]
Now, he ratio of the total kinetic energy after the collision to that before the collision
=[tex]\frac{k_f}{k_i}=\frac{0.5028}{297.59}[/tex]
=0.00169
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]
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
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 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.
