Answer:
he string has POTENTIAL ELASTIC energy
Explanation:
he catapult is a very good example of energy transformations.
Initially when the string is taut it has an elastic potential energy, when releasing the lever it is transformed into cinetic energy in the ball until it shoots out.
Consequently the string has POTENTIAL ELASTIC energy
Answer:
The answer is elastic
TRUST ME
Explanation:
What is the characteristic of the columns in the periodic table?
Answer: The periodic table also has a special name for its vertical columns. Each column is called a group. The elements in each group have the same number of electrons in the outer orbital.
Explanation: :)
Can someone please explain this to me?
Answer:
the anser should be d
Explanation:
A baseball travels at a velocity of 45.3 m/s towards second base for 17 s. Calculate the displacement of the ball.
Answer:
770.1 m
Explanation:
From the question given above, the following data were obtained:
Velocity (v) = 45.3 m/s
Time (t) = 17 s
Displacement (d) =?
Velocity is defined according to the following formula:
Velocity = Displacement /Time
With the above formula, we can obtain the displacement of the ball as follow:
Velocity = Displacement /Time
45.3 = Displacement / 17
Cross multiply
Displacement = 45.3 × 17
Displacement = 770.1 m
Therefore the displacement of the ball is 770.1 m
Answer: 770.1 gravity multiplied by time
A plastic rod is rubbed with a cloth and becomes positively charged. After charging, the rod is held close to the suspended table-tennis ball shown in Fig. 9.1. The table-tennis ball is covered with metal paint and is uncharged.
Describe what happens to the charges on the metal-painted table-tennis ball as the positively-charged rod is brought close to the ball.
PLZZZZ HELPPPPPP MEEEEEE!!!!!! ASAP!!! ILL GIVE 20 POINTS AND BRAINLIEST!!!
:'(
I am soooooooooooo desperate.
Answer:
the answers the correct one is C
Explanation:
In radioactive decay processes, a very important concept is the half-life time, which is the time it takes for the parent nuclei to decompose into child nuclei.
expressed this mathematically is
N = N₀ [tex]e^{- \lambda t}[/tex]
for the average life time
N₀ / 2 = N₀ [tex]e^{- \lambda T_{1/2}[/tex]
T =[tex]\frac{ln 2x}{ \lambda}[/tex]
Analyzing this expression, the nuclei that decay are the so-called parents and the new nuclei called children appear.
when checking the answers the correct one is C
PLA help if you understand
Answer:
b - it is not based on experimental data collected by scientists
how much does the angle of refraction change from 380nm to 700nm when the incident angle is 80?
Answer:
35 degrees
Explanation:
The incident angle, an 80-degree angle, and as the angle of incidence increases, the amount of refraction also increases. So, we can expect a significant change in the angle of refraction from blue to red light at an 80-degree angle of incidence.
What is refraction?Refraction refers to the bending of light when it passes from different mediums.The amount of refraction that occurs at an interface between two materials depends on the wavelength of light, the angle of incidence, and the properties of the materials. The relationship between these factors is described by Snell's law:
n1 sinθ1 = n2 sinθ2
where n1 and n2 are the refractive indices of the two materials, θ1 is the angle of incidence, and θ2 is the angle of refraction.
For a given angle of incidence, the amount of refraction that occurs is greater for shorter wavelengths (i.e., blue light) than for longer wavelengths (i.e., red light). This effect is known as dispersion, and it is why we see rainbows when sunlight is refracted through water droplets.
In the absence of any information about the properties of the materials involved, we can assume that the amount of refraction will increase as the wavelength decreases.
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A spring has an unstretched length of
12.0 cm. Its stiffness k is 8 N/cm. What load is
needed to stretch the spring to a length of 15.0 cm?
Answer:
24 N
Explanation:
Given,
Unstretched length or original length , x1 = 12cm
Stiffness constant, k = 8 N/cm
Load required to take spring to a length , x2 of 15cm
Recall the relation :
F = Ke
Where, e = extension
e = x2 - x1
e = (15 - 12) = 3
F = ke
F = 8 N/cm * 3cm
F = 24 N/cm*cm
F = 24 N
Hence, required load = 24 N
Define centre of mass.
Answer:
The center of mass is a position defined relative to an object or system of object
If you go twice as fast, your kinetic energy becomes.
A.2 times smaller
B.4 times smaller
C.2 times bigger
D.4 times bigger
PLZ HELP ME ASAP FOR 50 POINTS AND BRAINLIEST!!!
Answer:
1/32
Explanation:
It takes 1600 years for it to "die", when it dies it decomposes by half its original amount, then that half becomes the new total so it goes from 1 to 1/2 to 1/4 to 1/8 to 1/16 to 1/32
A graph with resistance in ohms on the x-axis from 0 to 200 and current in amps on the y-axis from 0 to 5. The graph is labeled as current as a function of resistance at 50 V. Points are plotted at (10, 5), (20, 2.5), (100, .5) and (200, .25). There is an inverse curve through the points. Use the graph to answer the questions. Based on the graph, what happens to current as resistance increases? This type of graph shows which type of relationship between the two variables? What would be a reasonable estimate for current at a resistance of 50 Ω?
Answer:
1. decreases
2. inverse proportionality
3. 1 A
Explanation
Just did it
Answer:
1. decreases
2. inverse proportionality
3. 1 A
Explanation
Just did it
A spherical conducting shell has charge -5 C. A particle with charge -8 C is placed at the center of the cavity. The charges (in C) on the inner surface and on the outer surface of the shell, respectively, are:
Answer:
Explanation:
The charge of - 5C will be spread on the outer surface of the shell . The presence of - 8C charge at the Centre will create + 8C charge on the inner surface and - 8C charge on the outer surface . The result is that the total charge on the outer surface of the shell will be - 5C + - 8C = - 13C and the inner surface will have charge of + 8C . + 8C charge will be locked to centrally situated - 8C charge and will not be neutralized by negative charge of outer surface .
How much heat does it take to raise 0.0438 kg of aluminum from room temperature (20.0°C) to its melting point (660°C)?
HELP
Answer:
2.5x10^4 J (scientific notation)
25,000 (Acellus)
Equation:
Q=mcΔT
where:
Q=heat
m = mass
c = specific heat
ΔT=change in temperature
Any rearranging?
Let's see.
No, we need heat.
So no rearranging.
Plug in the numbers and solve
It takes approximately 28,224 joules of heat to raise 0.0438 kg of aluminum from room temperature to its melting point.
To calculate the heat required to raise the temperature of aluminum from room temperature to its melting point, we need to use the specific heat capacity formula: Q = m × c × ΔT
where:
Q = heat (in joules)
m = mass of aluminum (in kilograms)
c = specific heat capacity of aluminum (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
Given:
Mass of aluminum (m) = 0.0438 kg
Specific heat capacity of aluminum (c) = 900 J/kg°C (approximate value for aluminum)
Change in temperature (ΔT) = melting point (660°C) - room temperature (20°C) = 640°C
Now, let's calculate the heat (Q):
Q = 0.0438 kg × 900 J/kg°C × 640°C
Q = 28,224 J (joules)
So, it takes approximately 28,224 joules of heat to raise 0.0438 kg of aluminum from room temperature to its melting point.
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A wave carries _____ from one place to another
Mechanical waves carry energy through ______
Answer:
Energy
Matter
Explanation:
Bcuz I just did it on the warm up for edge
Answer:
the answer is energy and matter trust me
Explanation:
6CO2 + 6H20 → C6H12O6 + 602
What type of model is shown?
a
b
computer model
conceptual model
physical model
C
mathematical model
Answer:
C. physical model
Explanation:
Sana nakatulong
The Circulatory System is made up of ?
A Blood and Lungs
B Heart and Blood Veins
C arteries, veins, and heart
How much acceleration does a car need if it is to go a distance of 620 ft. In a time of 20 sec while it is speeding up
A toy train moves round a circular track of radius 0.50 m.
How many radians has the trained turned through in moving 1.4 m along a track?
Answer:
2.8 radians
Explanation:
[tex]s = r \beta \\ 1.4 = 0.50 \beta \\ \beta = 2.8 \: rad[/tex]
A force of 20 N acts upon a 5000g block. Calculate the acceleration of the object
Answer:
a = 4 m/s²
Explanation:
Given: 5000 g, 20N
To find: Acceleration (a)
Solution: Firstly, convert 5000g to kilograms
1000g = 1kg
∴ 5000g = 5kg if divided by 1000
A = F ÷ m
= 20 ÷ 5
= 4 m/s²
Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.
7. What is the climate like on Mars?
Answer:
Temperatures on Mars average about -81 degrees F. However, temperatures range from around -220 degrees F. in the wintertime at the poles, to +70 degrees F.
Explanation:
Convert 86400 seconds into days.
Answer:
1 day
Explanation:
86400s = 1440min
1440min = 24h
24h = 1 day
Answer:
1 Day
86400 seconds is equal to a day
work done in taking charge from one point of a conductor to is another point is called
Answer:
⁸
Explanation:
electric potential
I think so
Tarzan, whose mass is 100 kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets go, his center of mass is at a height 2.9 m above the ground and the bottom of his dangling feet are at a height 2.1 above the ground. When he first hits the ground he has dropped a distance 2.1, so his center of mass is (2.9-2.1) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height 0.5 above the ground. Consider the point particle system. What is the speed v at the instant just before Tarzan's feet touch the ground?
Answer:
[tex]6.42\ \text{m/s}[/tex]
Explanation:
v = Final velocity
u = Initial velocity = 0
s = Displacement of Tarzan's feet = 2.1 m
a = g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
From the kinematic equations we get
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 2.1+0}\\\Rightarrow v=6.42\ \text{m/s}[/tex]
At the instant just before Tarzan's feet touch the ground the velocity of his feet are [tex]6.42\ \text{m/s}[/tex].
The speed at an instant just before Tarzen's feet touch the ground will be
[tex]V=6.42\ \dfrac{m}{s}[/tex]
What is speed?The speed of any body is defined as the movement of the body with respect to the time.
It is given in the question that
Mass of the Tarzen m=100 kg
The distance of the Tarzen's feet from the ground s=2.1 m
The acceleration due to gravity [tex]a=9.81\ \frac{m}{s^2}[/tex]
The speed of the Tarzen will be calculated by applying the equation of motion.
[tex]v^2=u^2+2as[/tex]
Since initial velocity is zero u=0
[tex]v^2=2as[/tex]
[tex]v=\sqrt{2as}[/tex]
[tex]v=\sqrt{2\times 9.81\times 2.1} =6.42 \ \frac{m}{s}[/tex]
Thus the speed at an instant just before Tarzen's feet touch the ground will be
[tex]V=6.42\ \dfrac{m}{s}[/tex]
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Need help pls. It is for acellus
Answer:
The height of the water column = 1.62405[tex]\overline{30}[/tex] × 10⁻¹ m
Explanation:
The air cavity in the Coke bottle = 0.220 m deep
The fundamental (frequency) it plays when water is added to shorten the column and it is blown across the top, f = 528 Hz
The given speed of sound in air, v = 343 m/s
We note that the air cavity in the coke bottle is equivalent to a tube closed at one end
The fundamental frequency for a tube closed at one end, 'f', is given as follows;
f = v/(4·L) = v/λ
Where;
L = The height of the water column
λ = The wavelength of the wave
∴ 4·L = v/f = (343 m/s)/(528 Hz) = 0.6496[tex]\overline{21}[/tex] m
∴ L = 0.6496[tex]\overline{21}[/tex] m/4 = 0.162405[tex]\overline{30}[/tex] m
The height of the water column = 1.62405[tex]\overline{30}[/tex] × 10⁻¹ m.
Answer:
0.058
Explanation:
got it right on acellus
someone pls answer! i will give u brainliest
Answer:
Liquid
Explanation:
Before anything evaporates, it is in a liquid state.
Mark Brainiest if right plzz. Noah wants observe what happens when zinc is placed in a solution of
copper sulfate, as shown in the photo. But when he tries it, nothing happens.
He knows that the reaction might be happening too slowly to see results in a
few minutes. Which action should Noah take to speed up the reaction?
A. Replace the piece of zinc with a larger piece.
B. Decrease the concentration of the copper sulfate solution
C. Decrease the temperature of the solution.
D. Increase the concentration of the copper sulfate solution.
Answer:
A. the place that piece of with a larger piece
Explanation:
Sana nakatulong
Answer:
D. Increase the concentration of the copper sulphate.
Work done depends on
A. Force
B. Displacement
C. Both force and displacement
D. Neither force nor displacement
Answer:
C. Both force and displacement
Explanation:
Hope this helps
A horizontal spring is lying on a frictionless surface. One end of the spring is attaches to a wall while the other end is connected to a movable object. The spring and object are compressed by 0.065 m, released from rest, and subsequently oscillate back and forth with an angular frequency of 14.7 rad/s. What is the speed of the object at the instant when the spring is stretched by 0.032 m relative to its unstrained length
Answer:
0.832 m/s
Explanation:
The work done by the spring W equals the kinetic energy of the object K
The work done by the spring W = 1/2k(x₀² - x₁²) where k = spring constant, x₀ = initial compression = 0.065 m and x₁ = final compression = 0.032 m
The kinetic energy of the object, K = 1/2mv² where m = mass of object and v = speed of object
Since W = K,
1/2k(x₀² - x₁²) = 1/2mv²
k(x₀² - x₁²) = mv²
mv² = k(x₀² - x₁²)
v² = [(k/m)(x₀² - x₁²)]
taking square root of both sides, we have
v = √[(k/m)(x₀² - x₁²)] since ω = angular frequency = √(k/m),
v = √[(k/m)√(x₀² - x₁²)]
v = ω√(x₀² - x₁²)]
Since ω = 14.7 rad/s, we substitute the other variables into the equation, so we have
v = 14.7 rad/s × √((0.065 m)² - (0.032 m)²)]
v = 14.7 rad/s × √(0.004225 m² - 0.001024 m²)]
v = 14.7 rad/s × √(0.003201 m²)
v = 14.7 rad/s × 0.056577
v = 0.832 m/s
An eighth-grade class is studying the motion of objects. They have learned the following concepts about motion.
Motion occurs when there is a change in the position of an object with respect to a reference point.
While the speed of the object may change at various times during a trip the average speed for the trip is found by dividing the total distance by the total time.
The motion of the object can be changed by varying the amount of net force acting on the object or by varying the mass of the object.
The teacher asked students to design investigations to determine how changing the forces acting on an object or the mass of the object would change its motion.
The materials for the investigation included:
1 - toy cart
3 - 10N weights
1 - pulley
2 - meters of string
One group of students decided to design an investigation to determine how different amounts of mass on the toy cart would affect the speed of the toy cart. They set up their investigation as shown below:
The students performed several trials using different masses in the toy cart (M1). They hung a mass shown as (M2) to pull the toy cart as it dropped.
Below is a table they made to record their results.
Which of the following would be a way to test the effect changing the amount of force acting on the object has on the average speed of the object?
Move the cart at various speeds and determine the force at M1
Add varying numbers of weights to M1 and measure the speed of the car.
Add varying numbers of weights to M2 and measure the speed of the cart.
Move the cart at various speeds and determine the force exerted by M2
Add varying numbers of weights to M1 and measure the speed of the car.
Add varying numbers of weights to M2 and measure the speed of the cart.
How to test the effect of weight on the speed of object?In order to test the changing amount of force that is acting on the object has effect on the average speed of the object we have to change the amount of weight on the object and allows it to move to a certain point in this way we can check the effect of changing amount of weight on the speed of the object.
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