Answer:
from which perspective the story is being told to the reader
Explanation:
i.e. first person on through third person or narrated.
Answer:
Deciding on the narrative point of views mean that an author must choose who is telling the story.
Explanation:
A narrative point of view is when the narrator becomes the character of the story who also knows the emotions or feelings and thoughts of the character. The narrator can be the first person or can be the third person depending on the characters included in the story.
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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]
Its volume is 20 cm3, and its mass is 100 grams. What is the sample’s density?
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
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
¿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:
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]
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
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 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.
how can u tell matched forces act on objects?
Answer:If an object's speed changes, or if it changes the direction it's moving in,
then there must be forces acting on it. There is no other way for any of
these things to happen.
Once in a while, there may be a group of forces (two or more) acting on
an object, and the group of forces may turn out to be "balanced". When
that happens, the object's speed will remain constant, and ... if the speed
is not zero ... it will continue moving in a straight line. In that case, it's not
possible to tell by looking at it whether there are any forces acting on it
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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A large dump truck can move 1,170 tons/h of gravel from one point to another on a work site. What is this rate in lb/s
Answer:
The rate of the dump truck is 650 [tex]\frac{lb}{s}[/tex]
Explanation:
A large dump truck can move 1,170 tons/h of gravel from one point to another on a work site.
To convert the units from tons/h to lb/s, you should know that:
1 ton= 2000 lb1 h= 3600 s (1 h= 60 minutes and 1 minute= 60 seconds)To carry out the unit conversion you must perform the following steps:
[tex]1170 \frac{ton}{h}*\frac{2000 lb}{1 ton} *\frac{1 h}{3600 s}[/tex]
Solving:
[tex]1170 \frac{ton}{h}*=650 \frac{lb}{s}[/tex]
So, the rate of the dump truck is 650 [tex]\frac{lb}{s}[/tex]
Give the missing ammeter reading a and b. suggest why more current flow through some bulbs than through others Grade 10 question and Answer
Answer:
becaude of electricity
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.
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.
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.
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
A uniform copper wire has a resistance of 100 ohms. If the wire is cut into 10 equal lengths, what will be the resistance of each piece
Answer:
Since resistance is proportional to length R = K L / A
the resistance of each piece would be R / 10 = 10 ohms
A copper wire is the conductor wire. If the wire is cut into 10 equal lengths, the resistance of each piece will be 10 ohms.
What is resistance?Resistance is the opposition to the flow of electrons in the conductor wire.
Given is the total resistance of the wire is 100 ohms.
Resistance is directly proportional to the length of the wire. When the wire is cut into 10 pieces, the new resistance will be
[tex]\dfrac{R}{10} =\dfrac{100}{10} =10 \:\rm ohms[/tex]
Hence, the resistance of each piece will be 10 ohms.
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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
If you are driving a car with a velocity of -25 m/s and you have an acceleration of -2 m/s^2, are you speeding up or slowing down? Why?
Answer:
Hmmm...
This is a bit tricky
Ok...
Negative Velocity means you're Moving in the Opposite direction....
Negative Acceleration (deceleration) means you're slowing down.
Deceleration would mean slowing down if you were Moving with a Positive velocity.
But In this case...
You're Moving with negative velocity and Negative acceleration...
This simply means that the acceleration and velocity vector are in the same direction....
Its means that...
"YOU'RE SPEEDING UP"
Just that you're doing it in the opposite direction.
Hope this helps.
what is the formula to calculate the speed
Answer:
speed=to distance ÷time , thank me later
A rock is pulled back in a slingshot as shown in the diagram below. The elastic on the slingshot is displaced 0.2 meters from its initial position. The rock is pulled back with a force of 10 newtons.
When the rock is released, what is its kinetic energy?
Answer:
id
Explanation:
i don't know
The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.
What will be the speed of the rock?Initial speed of the rock, u = 40m/s
Final position of the rock s = 0m taking the release point as reference. The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.
Nuclear energy is a useful source of power but has disadvantages. The disadvantage of nuclear energy is it produces dangerous waste.
Initial speed of the rock, u = 40m/s
Final position of the rock s = 0m taking the release point as reference
From the second equation of motion:
solving above we get:
t = 0s or t = 8.16s, t =0 seconds is neglected since it represents the initial position which is the same as the final position at t = 8.16s
So, the rock takes 8.16 seconds to return to the release point.
Therefore, The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.
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The heat capacity of sodium metal is 1500 JK-1, if the mass of the sodium metal is 75 kg, the specific
heat capacity would be
Explanation:
the answer is in the image above
steps btw if possible
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Answer:
(4) 50 ohms (5) 11.76 ohms
Explanation:
In the parallel combination, the equivalent resistance is given by :
[tex]\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....[/tex]
4. When three 150 ohms resistors are connected in parallel, the equivalent is given by :
[tex]\dfrac{1}{R}=\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}\\\\R=50\ \Omega[/tex]
5. Three resistors of 20 ohms, 40 ohms and 100 ohms are connected in parallel, So,
[tex]\dfrac{1}{R}=\dfrac{1}{20}+\dfrac{1}{40}+\dfrac{1}{100}\\\\=11.76\ \Omega[/tex]
Hence, this is the required solution.
One of the most common diseases a pet gets intestate with, is a worm infection in the stomach. 'Tape worm' is one such worm which lives on the nutrients from the stomach of an animal. Tape worms are, therefore, (a) Saprophytes
(b) Autotrophs
(c) Heterotrophs
(d) Parasites
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◕✿)
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
2.
A rock, mass 4 kg, and a 1 kg stone fall freely from rest from a height of
100 m. After they fall for 2 seconds, the ratio of the rock's speed to the
stone's speed is
Answer:
44.72m/s
Explanation:
use th formula:vf²=vi²at
and then substitute the values
remember the units
If there is no air resistance, the speed of the free falling object from the same height will be the same. Therefore, the ratio of speed of the stone to that of rock for 2 seconds falling will be 1 : 1.
What is free falling ?An object freely falling under the force of gravitation is called the free falling body. Here, the acceleration of the object is the acceleration due to gravity.
For a freely falling body, the velocity v = g t
where g is the acceleration due to gravity and for earth g is 9.8 m/s².
Everybody falls with the same rate freely under the acceleration due to gravity if there is no air resistance at all.
Here, the time t = 2 s
then v = 9.8 m/s² × 2 s = 19.6 m/s.
the speed of both the rock and stone will be 19.6 m/s in the absence of air resistance. Hence, the ratio of their speed is 1 : 1.
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nariz (am
miria amy
0 = 0 +260 + (0)
U= 29 mb
6= ut +1 (04)
Car I was sitting at rest when it nous hit from
the rear by car 2 of identical mass. Both cant had
their heaks on and they stidled together Guy
in the original directioned of motion. If the stopping
force is notx (Combined weight of the cars), die
u=0 to find the approximate speed of car a just
before the collision took place on
Answer:
33 mph
Explanation:
My best guess
plz answer the question
Answer:
Ray A = Incidence ray
Ray B = Reflected ray
Explanation:
From the law of reflection,
Normal: This is the line that makes an angle of 90° with the reflecting surface.
Ray A is the incidence ray: This is the ray that srikes the surface of a reflecting surface. The angle formed between the normal and the incidence ray is called the incidence angle
Ray B is the reflected ray: This is the ray leaves the surface of a reflecting surface. The angle formed between the reflected ray and the normal is called reflected angle
