Answer:
Chemical Bonding
Explanation:
Name the physical quantity and write its SI unit which gives the slope Of I-V graph?
Slope of I-V graph is Resistance.
Its S.I. unit is Ohm.
The symbol of ohm is Ω.
Una grúa eleva un tubo de concreto de
400 kg, con Movimiento rectilíneo uniforme,
con el
cable ABC. Determine la tensión que pueden
soportar los cables AB, BC y BD, sabiendo
que los cables AB=BC y la tensión que
soporta el cable AB es de 150 N.
Explanation:
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13.Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attains the speed u again before striking the ground. It implies that the magnitude of initial and final momentums of the ball are same. Yet, it is not an example of conservation of momentum. Explain why?
Answer:
Law of conservation of momentum is applicable to isolated system (no external force is applied).
In this case, the change in velocity is due to the gravitational force of earth.When the ball is thrown up, negative force of gravity is applied and so the speed becomes zero.After that due to gravitational force the ball falls down with same initial velocity.
9. In a __________ collision, 100% of both vehicles' speed is directed towards the point of impact. A. head-on B. rear-end C. side-impact
Answer: A
Explanation:
:)
Liquid to solid with explanation on the basis of kinetic model and freezing point.
Answer:
As a liquid is cooled its molecules lose kinetic energy and their motion slows. When they've slowed to where intermolecular attractive forces exceed the collisional forces from random motion, then a phase transition from liquid to solid state takes place and the material freezes.
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Answer:
01.
Explanation:
Half the acceleration. Its heavier and moves slower. If it moved the same acceleration, the forces would also have to be doubled since the mass was.
The reservoir stores 6 500 000 m3 of water. The density of the water is 998 kg/m3. Calculate the mass of water in the reservoir. Give your answer in standard form.
Answer:
Mass = 64,870,000,000 kilograms
Explanation:
Given the following data;
Density = 998 kg/m³
Volume = 6,500,000 m³
To find the mass of water in the reservoir;
Density can be defined as mass all over the volume of an object.
Simply stated, density is mass per unit volume of an object.
Mathematically, density is given by the equation;
Density = mass/volume
Making mass the subject of formula, we have;
Mass = density * volume
Mass = 998 * 6,500,000
Mass = 64,870,000,000 kilograms
The electric motor in the car is powered by a battery.
To charge the battery, the car is plugged into the mains supply at 230 V
The power used to charge the battery is 6.9 kW
Calculate the current used to charge the battery.
Answer:
I = 30 A.
Explanation:
Given that,
The voltage of the battery, V = 230 V
Power used to charge the battery, P = 6.9 kW
We need to find the current used to charge the battery. The formula for the power is given by :
P = VI
Where
I is current
So,
So, the required current is 30 A.
A hot-air balloon is rising upward with a constant speed of 2.85 m/s. When the balloon is 2.50 m above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground
Answer:
Explanation:
You have to declare which way is plus -- up or down. I will use down.
vi = - 2.85 The balloon is going up. That is the minus direction.
a = 9.81
d = 2.50 meters distance in this case is from the object to the ground.
d = vi*t + 1/2 a t^2
-2.50 = -2.85*t + 1/2 * 9.81 * t^2
-2.50 = -2.85*t + 4.905 * t^2 transfer the left to the right.
-4.905 t^2 + 2.85*t + 2.50 = 0
Use the quadratic formula to solve for t.
It turns out that t = 1.06
A ball of mass 0.175 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.805 m. What impulse was given to the ball by the floor
Answer:
The impulse received by the ball is - 1.561 kg.m/s
Explanation:
Given;
mass of the ball, m = 0.175 kg
initial displacement of the ball, h₁ = 1.25 m
final displacement of the ball, h₂ = 0.805 m
Assumptions:
let the downward direction of the ball be positive
let the upward direction of the ball be negative
The following equation of motion will be used to determine the final velocity of the ball at each displacement.
v² = u² ± 2gh
The final velocity of the ball when it is dropped downwards to 1.25 m;
v² = u² + 2gh
v² = 0 + 2gh
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 1.25)
v = 4.95 m/s
The final velocity of the ball when it rebounds from the floor to 0.805 m;
vf² = u² - 2gh
vf² = 0² - 2gh
vf² = -2gh
vf = -√2gh
vf = - √(2 x 9.8 x 0.805)
vf = -3.97 m/s
The impulse received by the ball is calculated as;
J = ΔP = mΔv = m(vf - v)
= 0.175(-3.97 - 4.95)
= - 1.561 kg.m/s
The negative sign indicates upward direction of the impulse.
Find the sum. Express the answer in scientific notation. (1.54 x 10^6)+(6.15 x 10^6)
Answer:
[tex] { \tt{(1.54 \times {10}^{6}) + (6.15 \times {10}^{6}) }} \\ = { \tt{(1.54 + 6.15) \times {10}^{6} }} \\ = { \tt{7.69 \times {10}^{6} }}[/tex]
Students are completing a lab in which they let a lab cart roll down a ramp. The students record the mass of the cart, the height of the ramp, and the velocity at the bottom of the ramp. The students then calculate the momentum of the cart at the bottom of the ramp.
A 4 column table with 3 rows. The first column is labeled Trial with entries 1, 2, 3, 4. The second column is labeled Mass of Cart in kilograms with entries 200, 220, 240, 260. The third column is labeled Height of ramp in meters with entries 2.0, 2.1, 1.5, 1.2. The fourth column is labeled Velocity at Bottom in meters per second with entries 6.5, 5.0, 6.4, 4.8.
Which trial’s cart has the greatest momentum at the bottom of the ramp?
Answer:
second column
Explanation:
Answer:
Trial 3 is the answer.
Explanation:
A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other, each with same KE, E 0 . The energy of explosion will be Aa
Answer:
6Eo
Explanation:
Si la rapidez del sonido en el agua es de 1450 m/s. ¿Cuánto tiempo tardará en recorrer 500 m en el fondo del lago?
Answer:
2.9
Explanation:
es 2 minutos con 9 segundos seria algo así (1450)÷(500)
Imagine you see Mars rising in the east at 6:30 pm. Six hours later what direction would you face (look) to see Mars when it is highest in the sky
Answer:
The Mars appears in the direction of South.
Explanation:
Mars is rising in the east at 6: 30 PM. The period of rotation of earth is 24 hours.
So, 6 hours is the one fourth of the period of rotation of earth. Earth rotates counter clockwise on its axis, so after 6 hours, we see the Mars in the direction of South.
3.00 m^3 of water is at 20.0°C.
If you raise its temperature to
60.0°C, by how much will its
volume expand?
Water
ß = 207•10-6C-1
(Unit = m^3)
Answer:
9m^3
Explanation:
Given data
volume v1= 3m^3
volume v2= ???
Temperature T1= 20.0°C.
Temperature T2= 60.0°C.
Applying the relation for temperature and volume
V1/T1= V2/T2
substitute
3/20= V2/60
3*60= V2*20
180= 20*V2
180/20= V2
V2= 9m^3
Hence the final volume is 9m^3
scientists are seen very busy in designing the solar power equipments, why?
Answer:
it is because solar energy is the perpetual source of energy and by using it non renewable sources of energy can be conserved for the future
Answer the question plz
Answer:
b and d
Explanation:
What factors could the skater change to apply the same amount of torque but increase the rate of his spin
Answer:
The moment of inertia should be decreased.
Explanation:
The torque is given by
Torque = Moment of inertia x angular velocity
To keep the torque constant, the spin rate be increased when the moment of inertia decreases.
The moment of inertia of the body are the efforts to put the object in rotation.
What happen to the frequency of transverse vibration of a stretched string if its tension is halved and the area of cross section of the string is doubled?
Answer:
The fundamental frequency of the stretched string is:
[tex]f=[/tex] [tex]\frac{1}{2} \sqrt{\frac{T}{L} }[/tex] [ T = Tension and μ = mass per unit length]
Here,
μ = [tex]\frac{m}{L} = \frac{Vp}{L} = Ap[/tex]
[tex]f= \frac{1}{2} \sqrt{\frac{T}{Ap} }[/tex]
If T is halved and A is doubled,
[tex]f= \frac{1}{2} \sqrt{\frac{T'}{A'p} } = \sqrt{\frac{1}{2* 2* A* p} } = \frac{1}{2} (\frac{1}{2} \sqrt{\frac{T}{Ap} } = \frac{1}{2} f[/tex]
Thus, the frequency is reduced to half if its tension is halved and the area of cross-section of the string is doubled.
When the tension is halved and the area of cross section is doubled, the frequency increases by a factor of 1.414.
The frequency of transverse vibration of a stretched string is calculated as follows;
[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]
where;
T is the tension on the stringμ is the mass per unit length[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }\\\\f = \frac{1}{2l} \sqrt{\frac{Tl}{m} } \\\\f = \frac{l^2}{2l} \sqrt{\frac{T}{m} } \\\\f = \frac{A}{2l} \sqrt{\frac{T}{m}}[/tex]
when the tension is halved and the area of cross section is doubled, the frequency is calculated as;
[tex]\frac{f_1}{A_1 \sqrt{T_1} } = \frac{f_2}{A_2\sqrt{T_2} } \\\\f_2 = \frac{f_1 A_2\sqrt{T_2}}{A_1 \sqrt{T_1} }\\\\f_2 = \frac{f_1 \times 2A_1\sqrt{0.5T_1} }{\sqrt{T_1} }\\\\f_2 = \frac{f_1 \times 2A_1 \times 0.7071\sqrt{T_1} }{\sqrt{T_1} }\\\\f_2 = 1.414 f_1[/tex]
Thus, when the tension is halved and the area of cross section is doubled, the frequency increases by a factor of 1.414.
Learn more about tension in a string here: https://brainly.com/question/25743940
Atoms contain both positive nuclei and a negative electron cloud. Which of the following would cause an attraction between two atoms?
A. The nucleus of one atom and the electron cloud another atom
B.The nucleus of one atom and the nucleus of another atom
C. The electron cloud of one atom and the electron cloud of another atom
D.None of these would cause an attraction between atoms
Answer:
Option (D)
Explanation:
A. The nucleus of one atom and the electron cloud of the another atom will not the cause of attraction.
B. The nucleus of one atom and the other atom will not cause the attraction.
C. The electron cloud of one atom and the electron cloud of the another will not cause attraction.
D. So, this option is correct.
5- Clasifica los siguientes cambios de la materia, anotando delante de cada uno cambio físico (F) o cambio químico (Q): • Disolver azúcar en agua • Freir una chuleta • Arrugar un papel • El proceso de la digestión • Secar la ropa al sol • Congelar una paleta de agua • Hacer un avión de papel • Oxidación del cobre • Romper un lápiz • Prender fuegos artificiales • Excavar un hoyo • Quemar basura
Answer:
1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.
2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.
3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.
4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.
5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.
6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.
7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.
8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.
9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.
10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.
11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.
12) Quemar basura - Cambio químico - Reacción de combustión.
Explanation:
A continuación, veremos que representa cada caso:
1) Disolver azúcar en agua - Cambio químico - Es un caso de una solución en donde el solvente es el azúcar y el soluto es el agua.
2) Freir una chuleta - Cambio físico - Es un proceso de cocinado por un transferencia de calor y transferencia de masa.
3) Arrugar un papel - Cambio físico - Se aplica fuerzas externas para deformar el papel.
4) El proceso de la digestión - Cambio químico - Reducción de los alimentos a desechos y la absorción de nutrientes por el contacto con jugos gástricos o ambientes intestinales.
5) Secar la ropa al sol - Cambio físico - El secado es un fenómeno de transferencia de masa.
6) Congelar una paleta de agua - Cambio físico - Cambio del estado líquido al estado sólido por transferencia de calor.
7) Hacer un avión de papel - Cambio físico - Aplicación de fuerzas externas para plegar y doblar la hoja de papel.
8) Oxidación del cobre - Cambio químico - Proceso de corrosión por contacto con iones que se transportan en el ambiente.
9) Romper un lápiz - Cambio físico - Proceso de ruptura por esfuerzo normal a causa de un momento como consecuencia de una fuerza externa aplicada sobre el lápiz.
10) Prender fuegos artificiales - Cambio químico - Reacción química de reducción-oxidación.
11) Excavar un hoyo - Cambio físico - Remoción de tierra por trabajo físico.
12) Quemar basura - Cambio químico - Reacción de combustión.
what is chemical change ??
Answer:
the combination,decomposition or replacement that occurs in the molecules of matter during chemical change called chemical reaction
Answer:
A chemical change happens when one chemical substance is transformed into one or more different substances, such as when iron becomes rust. ... A chemical change is different from a physical change, which doesn't rearrange atoms or molecules and produce a completely new substance.
A stone is dripped into the well 44 meter dip the spalash in heard in 3.12 s find speed of sound in air.
Answer:
The speed of sound is 366.67 m/s.
Explanation:
height, h = 44 m
total time, T = 3.12 s
Let the time taken by the stone to hit the water is t.
use second equation of motion
[tex]h = u t + 0.5 gt^2\\\\44 = 0.5\times 9.8 t^2\\\\t = 3 s[/tex]
Time taken by sound to g up
t'=T - t = 3.12- 3 = 0.12 s
The speed of sound is
[tex]v = \frac{h}{t'}\\\\v = \frac{44}{0.12}\\\\v = 366.7 m/s[/tex]
A 5.0 kg block of ice is at rest at the top of a smooth inclined plane. The block is released and slides 2.0 m down the plane. Assuming there is no friction between the block and the surface, calculate
a) the gravitational potential energy at the top of the plane
b) the component of the weight parallel to the plane
c) the acceleration of the block
d) the velocity of the block at the bottom of the plane
e) the kinetic energy at the bottom of the plane.
Answer:
a) 98.1 Joules
b) 49.05 N × sin(θ)
c) 9.81 × sin(θ)
d) The velocity of the block at the bottom of the plane, v is approximately 6.264 m/s
e) 98.1 Joules
Explanation:
The given parameters of the block are;
The mass of the block, m = 5.0 kg
The distance down the plane the block slides, h = 2.0 m
The friction between the block and the surface = 0
Let θ represent the angle of inclination oof the plane
a) The gravitational potential energy, P.E. = m·g·h
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
∴ P.E. ≈ 5.0 kg × 9.81 m/s² × 2.0 m = 98.1 Joules
The gravitational potential energy, P.E. ≈ 98.1 Joules
b) The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex], is given as follows;
[tex]w_{\parallel}[/tex] = w × sin(θ) = m·g·sin(θ)
∴ [tex]w_{\parallel}[/tex] ≈ 5.0 kg × 9.81 m/s² × sin(θ) = 49.05 × sin(θ) N
The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex] ≈ 49.05 N × sin(θ)
c) The component of the weight along the inclined plane = The force with which the block moves along the inclined plane, therefore;
[tex]w_{\parallel}[/tex] = m·g·sin(θ) = m·a
Where a represents the acceleration of the block along the plane
Therefore, by comparison, we have;
g·sin(θ) = a
∴ a ≈ 9.81 × sin(θ)
d) Given that the motion of the block is 2.0 m downwards, we have;
The velocity of the block at the bottom of the plane, v² = 2·g·h
Therefore, v² ≈ 2 × 9.81 m/s²× 2.0 m = 39.24 m²/s²
v = √(39.24 m²/s²) ≈ 6.264 m/s
e) The kinetic energy at the bottom of the plane, K.E. = (1/2)·m·v²
∴ K.E. = (1/2) × 5.0 kg × 39.24 m²/s² = 98.1 J
A body of mass 8000 moving with a velocity of 2.5m/s collides with a stationery twice the mass of the first body. After impact the mass moves to 1.0m/s Find the speed of the first body.
Answer:
the final speed of the first ball is 0.5 m/s
Explanation:
Given;
mass of the first body, m₁ = 8000 kg
mass of the second body, m₂ = 16,000 kg
initial velocity of the first body, u₁ = 2.5 m/s
initial velocity of the second body, u₂ = 0
final velocity of the stationary mass, v₂ = 1 m/s
let the final velocity of the first body = v₁
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
8000(2.5) + (16,000 x 0) = 8000v₁ + 16,000 (1)
20,000 + 0 = 8000v₁ + 16,000
8000v₁ = 4000
v₁ = 4000/8000
v₁ = 0.5 m/s
Therefore, the final speed of the first ball is 0.5 m/s
What is the relation between acceleration due to gravity and radius of the earth?
Answer:
As the earth is an oblate spheroid, its radius near the equator is more than its radius near poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.
Formula: g = GM/r2
Dimensional Formula: M0L1T-2
Values of g in SI: 9.806 ms-2
Explanation:
Please Mark me brainliest
Answer:
♡
Explanation:
As the earth is an oblate spheroid, its radius near the equator is more than its radius near poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.
[tex]what \: is \: reflection \: of \: light \: \: \: \: {?}[/tex]
A penny is dropped into a well. It takes 5 seconds to fall. Calculate the depth of the well in feet.
Answer:
d=1/2 (a)(t^2)
D = distance
A = acceleration
T = time
acceleration due to gravity is 32 ft/second
so, d=1/2 (32)(5^2)
d=16(25)
d=400
Explanation:
Answer:
400 ft.
Explanation:
D= 1/2 gt^2
=1/2(-32 ft/sec^2)(5 sec^2)
= -(1/2)(32)(25) ft
D= -400 ft, down
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How does the construction of dams positively affect natural resources?
by providing water for irrigation and restoring trees to areas where forests once existed
by creating reservoirs, preventing flooding, and renewing destroyed ecosystems
by preventing flooding, creating reservoirs, and providing water for irrigation
by renewing destroyed ecosystems and restoring trees to areas where forest once ersted
Answer:
by preventing flooding, creating reservoirs, and providing water for irrigation
Dams and waterways store and provide water for irrigation so farmers can use the water for growing crops. This idea goes way back into history. Irrigation is an important part of using water. In areas where water and rain are not abundant (like the desert), irrigation canals from rivers and dams are used to carry water.
Dams help in preventing floods. They catch extra water so that it doesn’t run wild downstream. Dam operators can let water out through the dam when needed.
