Answer:
(a) The aspect of the upper string supporting the weight and the applied force
Student 1 is correct because the upper string is the source of support of the large weight and the force applied to the short string reacts at the support of the long string
(b) The aspect of Student's (2) reasoning that is correct is that the shorter piece of string will always break first, however, the statement is only true for sudden pull due to the increased force experienced by the shorter string from a more rapid change in momentum
(c) The aspect of Student 1's statement that is incorrect is the that the upper string will always break first
The aspect of Student 2's statement that is incorrect is the that the shorter piece of string will always break first
(d) A string will break when subject to a force equivalent to its breaking force. The force experienced by the string increases as the rate of pull (suddenness) increases and the suddenness increases inversely with the length of the string, as such the shorter lower string will break first from a sudden pull before the force of the pull is completely transmitted to the upper string. Whereby the lower string is slowly pulled, the force is evenly transmitted to the upper string which is then taking up the load of the weight and the applied force together and is likely to break first
Explanation:
describe where the information and communication technology can be used it different section?
guys help me i will make you brainliest <3
Answer:
See below
Explanation:
ICT " Information and the communication technology" can be used in sectors like Telecommunications. ADB support for telecommunications and ICT is changing lives across the region. ...
ICT in Education. ICT can improve the efficiency and quality of education at all levels. ...
ICT in Public Sector Management. ...
ICT in Health. ...
ICT in Agriculture and Food Security.
( hope it helped <3 )
Please help I will give brainliest
Answer:
ask questions that the findings bring up
Explanation:
hope this helps :)
Answer:
ask questions that the findings bring up
A car of mass 1167 kg accelerates on a flat highway from 10 m/s to 28.0 m/s. How much work does the car's engine do on the car?
Answer:
Workdone = 465766038 Joules.
Explanation:
Given the following data;
Mass = 1167
Initial velocity = 10m/s
Final velocity =28m/s
To find the workdone;
We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.
Mathematically, it is given by the equation;
W = Kf - Ki
W = ½MVf² - ½MVi²
Substituting into the equation
W = ½(1167)*28² - ½(1167)*10²
W = ½ * 1361889* 784 - ½ * 1361889 * 100
W = 533860488 - 68094450
Workdone = 465766038 Joules.
A student determines the density ρ of steel by taking measurements from a steel wire
Mass- 6.2 +-0.1g
Length- 25.0 +-0.1m
Diameter- 2.00 +-0.01mm
He uses the equation ρ= 4m/πd^2l
What is the percentage uncertainty in his calculated value of density ?
Answer:
The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].
Explanation:
We can estimate the absolute uncertainty by the definition of total differential. That is:
[tex]\Delta \rho \approx \frac{\partial \rho}{\partial m}\cdot \Delta m + \frac{\partial \rho}{\partial d}\cdot \Delta d + \frac{\partial \rho}{\partial l}\cdot \Delta l[/tex] (1)
Where:
[tex]\frac{\partial \rho}{\partial m}[/tex] - Partial derivative of the density with respect to mass, measured in [tex]\frac{1}{mm^{3}}[/tex].
[tex]\frac{\partial \rho}{\partial d}[/tex] - Partial derivative of the density with respect to diameter, measured in grams per cubic milimeter.
[tex]\frac{\partial \rho}{\partial l}[/tex] - Partial derivative of the density with respect to length, measured in grams per cubic milimeter.
[tex]\Delta m[/tex] - Mass uncertainty, measured in grams.
[tex]\Delta d[/tex] - Diameter uncertainty, measured in milimeters.
[tex]\Delta l[/tex] - Length uncertainty, measured in milimeters.
[tex]\Delta \rho[/tex] - Density uncertainty, measured in grams per cubic milimeters.
Partial derivatives are, respectively:
[tex]\frac{\partial \rho}{\partial m} = \frac{4}{\pi\cdot d^{2}\cdot l}[/tex] (2)
[tex]\frac{\partial \rho}{\partial d} = -\frac{8\cdot m}{\pi\cdot d^{3}\cdot l}[/tex] (3)
[tex]\frac{\partial \rho}{\partial l} = - \frac{4\cdot m}{\pi\cdot d^{2}\cdot l^{2}}[/tex] (4)
And we expand (1) as follows:
[tex]\Delta \rho \approx \frac{4\cdot \Delta m}{\pi\cdot d^{2}\cdot l} - \frac{8\cdot m\cdot \Delta d}{\pi\cdot d^{3}\cdot l}-\frac{4\cdot m\cdot \Delta l}{\pi\cdot d^{2}\cdot l^{2}}[/tex]
[tex]\Delta \rho \approx \left(\frac{4}{\pi\cdot d^{2}\cdot l}\right)\cdot \left(\Delta m -\frac{m\cdot \Delta d}{d}-\frac{m \cdot \Delta l}{l} \right)[/tex] (5)
If we know that [tex]d = 2\,mm[/tex], [tex]l = 25\,mm[/tex], [tex]m = 6.2\,g[/tex], [tex]\Delta m = \pm 0.1\,g[/tex], [tex]\Delta d = \pm 0.01\,mm[/tex] and [tex]\Delta l = \pm 0.1\,mm[/tex], then the absolute uncertainty is:
[tex]\Delta \rho \approx \pm\left[\frac{4}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)} \right]\cdot \left[(0.1\,g)-\frac{(6.2\,g)\cdot (0.01\,mm)}{2\,mm} -\frac{(6.2\,g)\cdot (0.1\,mm)}{25\,mm} \right][/tex]
[tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex]
And the expected density is:
[tex]\rho = \frac{4\cdot m}{\pi\cdot d^{2}\cdot l}[/tex] (6)
[tex]\rho = \frac{4\cdot (6.2\,g)}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)}[/tex]
[tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex]
The percentage uncertainty in his calculated value of density is:
[tex]\%e = \frac{\Delta \rho}{\rho}\times 100\,\%[/tex] (7)
If we know that [tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex] and [tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex], then the percentage uncertainty is:
[tex]\%e = \frac{\pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}} }{78.941\times 10^{-3}\,\frac{g}{mm^{3}} }\times 100\,\%[/tex]
[tex]\%e = \pm 0.713\,\%[/tex]
The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].
A 7.40-kg object initially has 347 J of gravitational potential energy. Then an elevator lifts the object a distance of 20.6 m above its previous position. How much work did the elevator perform on the object?
Answer:
W = 1493.9 J = 1.49 KJ
Explanation:
The work done by the elevator on the object will be equal to the gain in is potential energy:
W = ΔP.E
W = mgΔh
where,
W = Work = ?
m = mass of object = 7.4 kg
g = 9.8 m/s²
Δh = gain in height = 20.6 m
Therefore,
W = (7.4 kg)(9.8 m/s²)(20.6 m)
W = 1493.9 J = 1.49 KJ
Grade 8 Science Admin. May 2018 Released
37 A physics teacher performed a demonstration for a science class by pulling a crate across the
floor and measuring the force with a spring scale. While she pulled, a student measured the
acceleration of the crate with a handheld electronic device. The results of three trials are
shown below
Motion
F = 900 N
Actual Acceleration
of Crate
Trial
1
140 kg
Acceleration
(m/s)
0.36
0.34
0.38
0.36
2
3
Average
The teacher asked the class to calculate the acceleration of the crate based on the crate's
mass and the force she applied. What conclusion can be made about the difference between
the calculated acceleration and the actual acceleration that occurred in the trials?
A Another force in the direction of the motion produced a lower acceleration than calculated.
B An opposing force caused by friction produced a lower acceleration than calculated.
C Another force in the direction of the motion produced a higher acceleration than
calculated.
D An opposing force caused by friction produced a higher acceleration than calculated,
Answer:
B. An opposing force caused by friction produced a lower acceleration than calculated
Explanation:
Answer:
An opposing force caused by friction produced a lower acceleration than calculated.
Explanation:
An archer stands on the ground and fires an arrow at a target. A second archer stands at the top of a building and holds an arrow in his hand. Which arrow has more potential energy?
In the same scenario described in question 7, which arrow has more kinetic energy? Explain.
Answer:
The arrow of the second archer standing on a high building will have more potential energy.
The arrow of the first archer standing on ground will have more kinetic energy.
Explanation:
POTENTIAL ENERGY:
The potential energy of an object depends upon its height, as given in the formula:
P.E = mgh
Hence, the arrow with the greater height will have more potential energy.
Therefore, the arrow of the second archer standing on a high building will have more potential energy.
KINETIC ENERGY:
The kinetic energy of an object depends upon its speed, as given in the formula:
K.E = (1/2)mv²
Hence, the arrow with the greater speed will have more kinetic energy energy. Since, the second arrow is stationary, it will have zero kinetic energy. But, the first arrow will have some K.E due to its speed.
Therefore, the arrow of the first archer standing on ground will have more kinetic energy.
(a) The second archer standing at the top of the building will have more potential energy.
(b) The first archer on the ground level will have the more kinetic energy.
The potential energy of an object is the energy possessed by the object by virtue its position above above the ground.
P.E = mgh
where;
h is the height above the groundThus, the second archer standing at the top of the building will have more potential energy.
(b) The kinetic energy of an object is the energy possessed by the object due to its motion.
K.E = ¹/₂mv²
where;
v is the speed of the objectAn object in motion, has zero velocity at maximum height but maximum velocity on the ground level.
Thus, the first archer on the ground level will have the more kinetic energy.
Learn more here:https://brainly.com/question/21866017
how we will solve this question?
36.45cL=______=μL
Answer:
0.364
I believe... Good luck!
Newton's first law of motion states than an object's motion will not change
unless?
Answer:
it is hit by an external force
Explanation:
Convert 318 meters per second to kilometers per hour
Answer:
1144.8
Explanation:
1 meter per second is equal to 3.6 kilometers per hour
what do humming birds eat? and how do they get their food?
A box sitting still on the ground by itself has a Normal Force of 700N, what is the mass? (gravity’s acceleration is 9.80 m/s²)
Answer:
Mass, m = 71.42 kg
Explanation:
Given that,
Normal force acting on a box, F = 700 N
We need to find the mass of a box. Let it is m. Normal force acting on an object is balanced by its weight such that,
F = mg
Where m is the mass of the box
[tex]m=\dfrac{F}{g}\\\\m=\dfrac{700\ N}{9.8\ m/s^2}\\\\m=71.42\ kg[/tex]
So, the mass of the box is 71.42 kg.
Part 2: Identify the independent, dependent, and constant variables
Experiment 1: A soap manufacturer runs an experiment to compare the foaming action of different dish detergents. Equal
amounts of each brand of detergent are placed in identical containers half-filled with water. The water and dish detergent are at
a temperature of 20°Celsius. Each container is agitated for 30 seconds, and then the height of the foam is measured.
a. independent variable:
b. dependent variable:
C. constant variable(s):
A stone is thrown vertically upwards. It takes 2 seconds to reach it highest point the acceleration due to gravity is 10m/s. With what velocity must it be thrown ?
Answer:
The velocity the stone must be thrown with is 20 m/s.
Explanation:
Given;
time taken for the stone to reach it highest point, t = 2 s
acceleration due to gravity, g = 10 m/s²
The maximum height reached by the stone at the given time is calculated as;
[tex]h = \frac{1}{2}gt^2\\\\h = \frac{1}{2}*10*2^2\\\\h = 20 \ m[/tex]
Determine the initial of the stone;
v² = u² + 2gh
where;
v is the final velocity of the stone at maximum height = 0
u is the initial velocity of the stone = ?
0 = u² - 2(10 x 20)
0 = u² - 400
u² = 400
u = √400
u = 20 m/s
Therefore, the velocity the stone must be thrown with is 20 m/s.
A dart is thrown horizontally at a target's center that is 5.00\,\text m5.00m5, point, 00, start text, m, end text away. The dart hits the target 0.150\,\text m0.150m0, point, 150, start text, m, end text below the target's center.What was the initial horizontal velocity of the dart?
Answer:
Explanation:
Given
the range = 5.00m (distance moved in horizontal direction)
Height = 0.150m
Required
Initial velocity of the dart
Using the formula for calculating range of a projectile;
R = U√2H/g
5 = U√2(0.15)/9.8
5 = U√0.0306
5 = 0.1749U
U = 5/0.1749
U = 28.59m/s
Hence the initial horizontal velocity of the dart is 28.59m/s
Answer:
28.59
Explanation:
khan academy
A vector is expressed in polar notation as A⃗ = ( 35.0 N, 37 ∘) Calculate the components of the vector, and enter them (separated by a comma) in the answer box to express the vector in rectangular notation.
Answer:
A = (27.95 N, 21 N)
Explanation:
The polar co-ordinates are given as:
(r,θ) = (35 N, 37°)
Now, to convert this into polar co-ordinates (x, y), we will use following relations:
r² = x² + y²
(35)² = x² + y²
1225 = x² + y² ----------- equation (1)
and
tan θ = y/x
tan 37° = y/x
y = 0.753 x ------------------- equation (2)
Substituting this value in equation (1):
1225 = x² + (0.753 x)²
1225 = 1.567 x²
x² = 1225/1.567
x = √781.32
x = 27.95 N
using this value in equation (2)
y = (0.753)(27.95 N)
y = 21 N
Therefore, the vector can be represented in polar co-ordinates as:
A = (27.95 N, 21 N)
For a substance to change phases, the amount of Internal energy must change. Water exists in three phases: liquid, solid (ce), and gas (water vapor). Which of the following lists the
phases in order of increasing total energy?
O
gas, liquid, solid
solid, gas, liquid
solid, liquid, gas
oliquid, gas, solld
Which car is accelerating?
A) a car that is unmoving
B) a car that rounds a curve at a constant speed
C) a car travels in a straight line at a constant speed
D) a car is set to a constant speed of 60 miles per hour
Answer:
The car that is accelerating is B a car that rounds a curve at a constant speed
Explanation:
Although all of the cars are at a constant speed or not moving acceleration is the change in speed or the change of directions therefore making the only car changing directions your answer.
Answer:
its B (:
Explanation:
i took the test :)
41. A statue weighs 1,000N and exerts a pressure of 20,000 Pa. How big is
the base of the statue in square meters?
please help
Answer:
The answer is 0.05 m²Explanation:
The area of the base of the statue can be found by using the formula
[tex]a = \frac{f}{p} \\ [/tex]
f is the force
p is the pressure
From the question we have
[tex]a = \frac{1000}{20000} = \frac{1}{20} \\ [/tex]
We have the final answer as
0.05 m²Hope this helps you
!!please help !!!!! I’ll give brainliest
Answer:
45 degrees
Explanation:
a projectile travels the farthest when it is launches at the angle of 45 degree.
The
of an object shows how fast the object is moving at a specific
Answer:
Speed is a description of how fast an object moves; velocity is how fast and in what direction it moves. In physics, velocity is speed in a given direction. When we say a car travels at 60 km/h, we are specifying its speed.
40 POINTS!
A 2,200 kg SUV is traveling at 25 m/s. What is the magnitude of its momentum?
A. 55,000 kg·m/s
B. 550 kg·m/s
C. 2,200 kg·m/s
D. 88 kg·m/s
Answer:55,000 kg•m/s
Explanation:
Help on this pls. !!!!!
Answer:
C
Explanation:
El cloro (Z igual a 17) es un elemento utilizado generalmente en la desinfección de piscinas en la naturaleza hay 2 isótopos del cloro: el Cl-35, con una abundancia del 75% y el Cl-37 con una abudancia del 25%. Determina : a)los protones y los neutrones que hay en el nucleo de cada uno de estos isotopos b) la masa atomica media del cloro expresada en u c) el numero de atomos de cloro que hay en 15g de cloro
Answer:
a) El número de neutrones de Cl-35 es 18 y del Cl-37 es 20.
b) La masa atómica media del cloro es 35.5 uma.
c) En 15 g de cloro hay 2.55x10²³ átomos.
Explanation:
a) Para calcular los protones y neutrones de los isótopos Cl-35 y Cl-37 debemos usar la siguiente fórmula:
[tex] A = Z + N [/tex]
En donde:
Z: es el número de protones = 17
N: es el número de neutrones
A: es la masa atómica
Los isótopos de un elemento dado tienen el mismo número de protones (número atómico) y en ellos sólo varia la la masa atómica (y por ende el número de neutrones); por lo que el Cl-35 y Cl-37 tienen igual Z (17).
Para el Cl-35 tenemos:
[tex] N = A - Z = 35 - 17 = 18 [/tex]
Y para el Cl-37 tenemos:
[tex]N = 37 - 17 = 20[/tex]
Por lo tanto, el número de neutrones de Cl-35 es 18 y del Cl-37 es 20.
b) La masa atómica media del cloro está dada por:
[tex]M_{Cl} = A_{Cl-35}*\%_{Cl-35} + A_{Cl-37}*\%_{Cl-37} = 35 u*\frac{75}{100} + 37 u*\frac{25}{100} = 35.5 u[/tex]
Entonces, la masa atómica media del cloro es 35.5 uma.
c) Para encontrar el número de átomos de Cl que hay en 15 g debemos usar el número de Avogadro (6.022x10²³ átomos/mol):
[tex] N = \frac{1 mol}{35.5 g}*15 g*\frac{6.022 \cdot 10^{23} atomos}{1 mol} = 2.55 \cdot 10^{23} atomos [/tex]
Finalmente, en 15 g de cloro hay 2.55x10²³ átomos.
Espero que te sea de utilidad!
Lucia kicks a ball on a level playing field with an initial velocity of 11.3 m/s at an angle of 35° above the horizontal. Find: Time of flight? Horizontal distance traveled? Maximum height?
Explanation:
Given that,
Initial velocity, u = 11.3 m/s
Angle above the horizontal, [tex]\theta=35^{\circ}[/tex]
Time of flight :
[tex]t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s[/tex]
Horizontal distance traveled is given by :
x = ut
x = 11.3 m/s × 1.32 s
x = 14.916 m
Maximum height is given by :
[tex]H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m[/tex]
Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.
The planes LA534 and LA639 are coming in for a landing on the same runway which means they are each lowering their altitude. Note that LA534 is traveling at twice the speed of LA 639. What must they do to keep from crashing into each other on the runway
Answer:
plane LA534 will have to decrease it speed by twice that of plane LA639.
Explanation:
Since both planes want to land there must be a decrease in speed.
We are told that plane LA534 velocity is twice that of plane LA639.
Now for the two planes to land without a crash. the opposite or reverse will be the case in terms of deceleration, that is
Plane LA639 will have to decrease its speed by twice the decrease in speed of plane LA534
You throw an orange out a window at a height of 12.00 meters
upwards at an angle of 32° to the horizontal at a velocity of 3.5
m/s. What is the maximum height that the orange will reach?
Answer:
12.18
Explanation:
The maximum height that the orange will reach is 12.18 m.
What is projectile?When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion.
You throw an orange out a window at a height of 12.00 meters upwards at an angle of 32° to the horizontal at a velocity of 3.5 m/s.
If u is the initial speed, the maximum height of the ball from the point of throwing is
H = u² sin²θ / 2g
Put the values, we get
H = 3.5² sin²32° / (2x9.81)
H = 0.1753 m
So, the maximum height calculated from the ground is
12 + 0.18 = 12.18 m
Thus ,the maximum height that the orange will reach is 12.18 m.
Learn more about projectile.
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Which two staments explain how a cell's parts help it get nutrients
Answer:
I think it's A and D.
The force that causes an object to follow a
circular path is
tion:
at o
1. an inertiali force.
2. a centripetal force.
3. a gravitational force.
CI
nents
stant
4. a fluid force.
Answer:
2.a centripetal force.
in a free body diagram the weight is
Answer:
Explanation:
a. WEIGHT: As we have seen, weight is the gravitational force exerted on an object by the Earth (or any other celestial body). If an object is near the Earth's surface and has mass, then the object has a weight. The magnitude of its weight is w = mg and its direction is toward the center of the Earth.