multiply energy of these cannot be done by a machine.
What kind of machine is it?The inclined plane, the wedge, the screw, the lever, the wheel and axle, and the pulley are six examples of primitive machines. Although several may function in a similar manner, these six have distinctive qualities and perform distinctive tasks.
A machine's role may be as simple as modifying and transmitting forces and motions, or it may include converting chemical, thermal, electrical, or nuclear energy into mechanical energy or vice versa. Every machine has an input, an output, as well as a device for altering, changing, and conveying information.
Any tool that simplifies work by adjusting a force is a machine. Machines may broaden the area over which a force is delivered or enhance the force's power.
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What is the first step in scientific method?
A) Gather information
B) state the problem
C) form a hypothesis
D) perform the experiment
The first step in scientific method is to state the problem. That is option B.
What are scientific methods?The scientific methods are those methods that are used by scientists and researchers to discover answers to specific questions that are being asked and it involves the use of scientific steps and guidelines.
These steps include the following:
State the problem: This is the first step where the problem to be investigated is being defined.Form a hypothesis: This is the step that involves the making of predictions based on the research.Gather information: The data that is related to the problem are being collected.Perform the experiment: Analysis are being carried out on the collected data.Learn more about scientific methods here:
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Dylan and Sophia are walking along Bluebird Lake on a perfectly calm day. Dylan, determined to impress Sophia by his ability to skip rocks, picks up the flattest rock he can find and gives it a sidearm launch from the edge of the water. The rock acquires a completely horizontal speed of 27.2 m/s from a height of 0.54 meters above the water surface. How far does it go horizontally before it hits the water?
Question 17 options:
0.3 meters
9.1 meters
0.1 meters
14.7 meters
The horizontal distance travelled by the rock before it hits the water is 9.0 m.
What is the time of motion of the rock?The time of motion of the rock is the time taken for the rock to fall from the given height.
h = vt + ¹/₂gt²
where;
h is the height of fallv is the vertical velocity = 0t is the time of motionh = ¹/₂gt²
t = √(2h/g)
t = √(2 x 0.54 / 9.8)
t = 0.33 s
The horizontal distance travelled by the rock before it hits the water is calculated as follows;
X = vt
where;
v is horizontal velocityt is time of motionX = 27.2 m/s x 0.33 s
X = 9.0 m
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A 3.5-inch floppy disk in an old computer rotates with a period of 2.00 x 10^-1 s. Calculate the angular speed of the disk, the linear speed (in inches/sec) of a point on the rim of the disk, and the linear speed (in inches/sec) of a point 0.750 inches from the center of the disk. (Hint: a 3.5 inch floppy disk has a 3.50 inch diameter)
We have the next information
T=2.00 x 10^-1 s=0.2s
r=3.5 inch
For the angular speed
[tex]\omega=\frac{2\pi}{T}[/tex]where omega is the angular speed, T is the period
We substitute
[tex]\omega=\frac{2\pi}{0.2}=31.41\text{ rad/s}[/tex]For the linear speed on the rim of the disc, we will use the next formula
[tex]v=\omega\cdot r[/tex]in this case r= 3.5/2=1.75 inch
[tex]v=31.41(1.75)=54.97\text{ }\frac{inches}{\text{sec}}[/tex]Then for the linear speed on the point at 0.750 inches from the center of the disk.
[tex]v=31.41(0.750)=23.56\frac{inches}{\text{sec}}[/tex]ANSWER
ω=31.41 rad/sec
v on the rim= 54.97 inches/sec
v on the point=23.56 inches/sec
Conversion of PE into KE
1. A ball is thrown up with a velocity of 10 m/s. How high will it go? (Solve the problem
without using the equation v2=v0−2gΔy. This equation is true, but here we are trying
use PE and KE). Never mind that you are not told the mass of the ball – it will cancel
out.)
If a ball is thrown up with a velocity of 10 meters per second then it will go up to a height of 5.10 meters.
What is mechanical energy?Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total potential energy stored energy in the system which is represented by total potential energy.
As given in the problem if a ball is thrown up with a velocity of 10 meters per second,
The kinetic energy of the ball is getting converted into the potential energy
mg×h = 1/2×m×v²
v² = (2×g×h)
10² = 2×9.8×h
h = 100 / 19.6
h = 5.10 meters
Thus, If a ball is thrown up with a velocity of 10 meters per second then it will go up to a height of 5.10 meters.
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A person starts at a point on a square which has a side of 1.5 metres in the clockwise direction and reaches the same point after time of 4 s . The value of distance is
To find the distance traveled, we just have to add all the side lengths of the square, in other words, the perimeter.
[tex]\begin{gathered} P=1.5m+1.5m+1.5m+1.5m \\ P=6m \end{gathered}[/tex]Therefore, the value of the distance is 6 meters.Energy stored in the kinetic energy container is fundamentally different from energy stored in the elastic energy container.True or false
The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]Where:
m = mass of the object
v = Speed of the object
The elastic potential energy is given by:
[tex]U=\frac{1}{2}k\Delta x[/tex]Where:
k = Constant of the spring
Δx = Displacement
As we can see, both energy depends on different factors, therefore we can conclude the statement is true
Answer:
True
A 131 kg electric motorcycle moving at 30 mph accelerates by 0.03 m/s^2 for 60 s in the same direction that it’s moving. What is the average velocity during this period? Answer in km/hr.
We would calculate the distance travelled by the motor by applying one of Newton's formula for calculating motion which is expressed as
s = ut + 1/2at^2
where
s is the distance travelled
u is the initial velocity
t is the time
a is the acceleration
From the information given,
u = 30 mph
We would convert it to m/s
Recall,
1 mile = 1609.344 m
1 hour = 3600 s
Thus,
30mph = (30 x 1609.344)/3600
u = 13.4112 m/s
t = 60 s
a = 0.03m/s^2
Thus,
s = 13.4112 x 60 + 1/2 x 0.03 x 60^2
s = 804.672 + 54
s = 858.672 m
We would convert 858.672 m to km and 60s to hours. It becomes
858.672 m = 858.672/1000 = 0.858672 km
60s = 60/3600 = 1/60 hr
Average velocity = 0.858672 km/(1/60 hr)
Average velocity = 51.5 km/hr
Which is the best statement of the Law of Conservation of Energy? (1 point)
The amount of energy put out by a simple machine is always less than the amount of energy put into the machine
The amount of energy put into a simple machine should be conserved Item
The amount of energy put out by a simple machine depends on its efficiency
The amount of energy put out by a simple machine must equal the amount of energy put into the machine
The best statement of the Law of Conservation of Energy is option D. The amount of energy put out by a simple machine must equal the amount of energy put into the machine.
The regulation of conservation of energy states that the entire electricity of an isolated system remains steady; it's far stated to be conserved over time.
The law of conservation of power is a physical law that states strength cannot be created or destroyed but can be modified from one form to some other. any other manner of pointing out this regulation of chemistry is to mention the full electricity of a remoted device stays regular or is conserved within a given body of reference.
The law of conservation of energy states that power can neither be created nor destroyed - best converted from one form of energy to some other. this means that a device constantly has an equal quantity of power unless it is added from the out of doors.
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The velocity-time table represents the motion of a leftward moving car.
What is the magnitude and direction of the acceleration?
Given velocity-time table represents the motion of a leftward moving car.
Magnitude of acceleration = -4 m/s^2
Direction is Left.
When any object is slowing down then the acceleration is in opposite direction. Then the object will have a negative acceleration.
Deceleration is the opposite of acceleration. The deceleration is calculated by dividing the difference final and initial velocity by the amount of time taken for the drop in velocity. The formula for acceleration with a negative sign is used to identify the deceleration value.
Deceleration = (Final Velocity–Initial Velocity)/Time taken
It is written' –a,' where a is acceleration.
When initial velocity, final velocity and time are given, then Deceleration Formula is given by,
Given in the graph, u= 20m/s and v= 4m/s
t= 4sec
a = (v−u)/t
= (4 -20)/ 4
=- 16/4
a= -4 m/s^2
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A car accelerates from rest at -1.60 m/s2. (Include the sign in your answers.)(a) What is the velocity at the end of 5.6 s? m/s(b) What is the displacement after 5.6 s?
Given data
*The given initial velocity of the car is u = 0 m/s
*The given acceleration of the car is
[tex]undefined[/tex]A football is kicked with a velocity of 32.0m/s and at an angle of 42 degrees it takes 4.40s. How high does it travel?
23.4 m
Explanation
to solve this we need to use the expression:
[tex]h=v\sin (\theta)t-\frac{gt^2}{2}[/tex]where
[tex]\begin{gathered} h\text{ is the heigth} \\ v\text{ is initial velocity} \\ \theta\text{ is the angle} \\ t\text{ is the time} \end{gathered}[/tex]then, let
[tex]\begin{gathered} v=32\text{ m/s} \\ \theta=42\text{ \degree} \\ t=4.4\text{ s} \\ g=9.8\text{ }\frac{m}{s^2} \end{gathered}[/tex]now, replace
[tex]\begin{gathered} h=v\sin (\theta)t-\frac{gt^2}{2} \\ h=32\sin (42)(4.40)-\frac{(9.8)(4.4^2)}{2} \\ h=94.21-94.864 \\ h=-0.654 \end{gathered}[/tex]let's check the time of flight
[tex]\begin{gathered} t=\frac{2v\sin\theta}{g} \\ t=\frac{2\cdot32\cdot\sin(42)}{9.8} \\ t=4.36 \end{gathered}[/tex]it means after 4.36 the ball is on the ground.
Step 2
so, to find the maximum heigth we need to use the expression
[tex]\begin{gathered} y_{\max }=\frac{v^2\sin ^2\theta}{2g} \\ \text{replace} \\ y_{\max }=\frac{(32)^2\sin ^2(42)}{2(9.8)} \\ y_{\max }=\frac{1024\cdot0.4477}{2(9.8)} \\ y_{\max }=\frac{458.48}{19.6} \\ y_{\max }=23.39 \end{gathered}[/tex]therefore, the answer is
23.4 m
I hope this helps you
An ocean fishing boat is drifting just above a school oftuna on a foggy day. Without warning, an engine backfireoccurs on another boat at a distance of d = 1.55 km(Figure 1). The speed of sound in air is 343 m/s, and insea water is 1560 m/s.How much time elapsed before the backfire is heard by the fishermen? Assume that the fishermen hears the sound that travels only through air.
For the fisher man use speed of sound in air as v = 343
d= vt
t= d/v
where:
t= time
d= distance = 1.55 km = 1550 m
v= speed= 343 m/s
t= 1550m / 343 m/s = 4.52 s
a lot of this content is conceptual but I'm having a hard time understanding this
If momentum is denoted by p and wavelength by
[tex]\lambda[/tex]Then, the wavelength and momentum are related as
[tex]p\propto\frac{1}{\lambda}[/tex]Thus if momentum is doubled, then the wavelength is halved.
An automobile with an initial speed of 4.30 m/s accelerates uniformly at the rate of 3.0 m/s/s. Find the final speed and the displacement after 5.0s
The final speed of the automobile with an initial speed of 4.30 m/s that rates uniformly at the rate of 3.0 m/s/s for 5 s is 19.3 m / s and its displacement will be 59 m
v = u + at
v = Final velocity
u = Initial velocity
a = Acceleration
t = Time
u = 4.3 m / s
a = 3 m / s / s
t = 5 s
v = 4.3 + ( 3 * 5 )
v = 19.3 m / s
s = ut + 1 / 2 at²
s = Displacement
s = ( 4.3 * 5 ) + ( 1 / 2 *3 * 5 * 5 )
s = 21.5 + 37.5
s = 59 m
Therefore,
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Biggs and Smalls play tug-of-war with a 12-meter rope in P.E. class. Both are in their socks on a waxed gym floor, which is nearly frictionless. After a brief time, they meet. If Biggs has twice the mass of Smalls, how far has Biggs moved?
After a brief time, Biggs and smalls would meet at 4m from Biggs' side if the rope is 12 meter long and the floor is frictionless.
[tex]X_{cm}[/tex] = ( m1 x1 + m2 x2 ) / ( m1 + m2 )
[tex]X_{cm}[/tex] = Center of mass
After a brief time, Biggs and smalls would meet at the point of centre of mass. Let Biggs be 1 and smalls be 2 and the distances be measured from Biggs' side assuming that Biggs is standing on the left side.
m1 = 2 * m2
x1 = 0 m
x2 = 12 m
[tex]X_{cm}[/tex] = [ ( 2 * m2 * 0 ) + ( m2 * 12 ) ] / [ ( 2 * m2 ) + m2 ]
[tex]X_{cm}[/tex] = 12 m2 / 3 m2
[tex]X_{cm}[/tex] = 4 m
Center of mass is the average position of all the objects on a system, weighed according to the mass of individual objects. It is a position relative to an object.
Therefore, after a brief time, Biggs and smalls would meet at 4 m from Biggs' side.
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Change 72km/h into m/s
Answer:
20 m/s
Explanation:
72 km/hr * 1 hr/ 3600sec * 1000m/km = 20 m/s
If Mercury's average orbital distance from the sun is 0.312 AU and k for our sun is 1.00 AU3/yr2, then what is Mercury's orbital period?
Take into account that by the Kepler's Third Law, you can use the following formula:
[tex]k^{}=\frac{a^3}{T^2}[/tex]where,
T: period of Mercury
a: average distance from Mercuty to Sun = 0.312AU
k: Kepler's constant = 1.00 AU^3/yr^2
Then, by replacing the previous values of the parameters, you obtain for T:
[tex]T=\sqrt[]{\frac{a^3}{k}}=\sqrt[]{\frac{(0.312AU)^3}{1.00\frac{AU^3}{yr^2}}}\approx0.56yr[/tex]Hence, the Mercury's orbital period is approximately 0.56 yr
Two cars have a ‘rear end’ collision. A 120 kg Honda moving at 20 m/s strikes a 100 kg Ford moving at 15 m/s. Their bumpers become locked and they continue to move as one mass. What is their final velocity?
Given
Mass of the Honda car, m=120 kg
Velocity of the Honda car, v=20 m/s
Mass of the Ford, m'=100 kg
Velocity of the Ford , v'=15 m/s
To find
The final velocity.
Explanation
The combined velocity is V
By conservation of momentum,
[tex]\begin{gathered} mv+m^{\prime}v^{\prime}=(m+m^{\prime})V \\ \Rightarrow120\times20+100\times15=(120+100)V \\ \Rightarrow V=17.72\text{ m/s} \end{gathered}[/tex]Conclusion
The combined velocity is 17.72 m/s
Exactly 3.3 s after a projectile is fired into the air from the ground, it is observed to have a velocity v⃗ = (8.6 i^ + 4.7 j^)m/s, where the x axis is horizontal and the y axis is positive upward. aDetermine the horizontal range of the projectile.bDetermine its maximum height above the ground.cDetermine the speed of motion just before the projectile strikes the ground.dDetermine the angle of motion just before the projectile strikes the ground.= ∘ below the horizontal
a. Horizontal range is given by,
[tex]Range=\frac{u^2sin2\theta}{g}[/tex]where u is the initial speed, and theta is the angle of projection,
Now velocity vector as a function of time is given by,
[tex]\vec{v}(t)=(ucos\theta)\hat{i}+(usin\theta-gt)\hat{j}[/tex]Given,
[tex]\begin{gathered} \vec{v}(3.3s)=8.6\text{ }\hat{i}+4.7\text{ }\hat{j} \\ \Rightarrow ucos\theta=8.6,usin\theta-(10)(3.3)=4.7 \\ \Rightarrow ucos\theta=8.6,usin\theta=37.7 \\ Dividing, \\ \Rightarrow tan\theta=\frac{37.7}{8.6}=4.3837 \\ \Rightarrow\theta=77.15\degree \\ Hence \\ ucos77.15\degree=8.6 \\ \Rightarrow u=\frac{8.6}{0.2224}=38.67\text{ m/s} \end{gathered}[/tex]And so, the horizontal range will be,
[tex]Range=\frac{(38.67)^2sin(154.3\degree)}{(10)}=64.85\text{ m}[/tex]b. Maximum height reached by a projectile is given by,
[tex]\begin{gathered} H_{max}=\frac{u^2sin^2\theta}{2g} \\ \Rightarrow H_{max}=\frac{(38.67)^2sin^2(77.15\degree)}{(2)(10)}=71.07\text{ m} \end{gathered}[/tex]c. Now time of flight of the projectile is given by,
[tex]t_{flight}=\frac{2usin\theta}{g}=\frac{(2)(38.67)sin(77.15\degree)}{(10)}=7.54\text{ s}[/tex]this is the time at which the projectile strikes the ground, and so its velocity then would be,
[tex]\begin{gathered} \vec{v}(7.54s)=8.6\text{ }\hat{i}+(37.7-10\times7.54)\hat{j} \\ \Rightarrow\vec{v}(7.54s)=8.6\text{ }\hat{i}-37.7\text{ }\hat{j} \end{gathered}[/tex]As expected the vertical component of velocity has just been flipped, and therefore its speed will be same as its initial speed i.e.
[tex]v_{strikes\text{ }ground}=u=38.67\text{ m/s}[/tex]d. Below the horizontal, let the angle be φ, then
[tex]tan\varphi=-\frac{v_y}{v_x}[/tex]extra -ve sign since its below the horizontal, when the projectile strikes the ground,
[tex]\begin{gathered} v_y=-37.7\text{ m/s} \\ v_x=8.6\text{ m/s} \\ \Rightarrow tan\varphi=-\frac{(-37.7)}{8.6}=\frac{37.7}{8.6}=tan\theta \\ \Rightarrow\varphi=\theta \\ \Rightarrow\varphi=77.15\degree \end{gathered}[/tex]Result: a. 64.85 m, b. 71.05 m, c. 38.67 m/s, d. 77.15°.The inside of the Earth is hot due to heat left over from when the Earth formed and due to heat generated by the radioactive decay of elements in the Earth's mantle and crust. Heat from these sources flows to the surface at a total rate of of 47 terrawatts. If the Earth were instantly teleported into deep space so that absorption of any outside radiation stopped, what would the equilibrium surface temperature of the Earth be, using e=0.6 for the emissivity, and R = 6400km for the radius? Write the temperature in Kelvin, just the number without units.
The temperature at the earth's surface would be
40.48K
This is further explained below.
What is Stefan Boltzmann's law:?Generally, In accordance with the Stefan-Boltzmann equation, the total radiant heat output that is radiated from a surface is proportional to the fourth power of that surface's absolute temperature.
According to the Stefan-Boltzmann law, the total energy radiated per unit surface area of a black body across all wavelengths per unit of time is directly proportional to the fourth power of the black body's thermodynamic temperature T: j=.
To be more specific, the Stefan-Boltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit of time is directly proportional to the black
[tex]P=\sigma e AT^4[/tex]
Where P stands for power, e stands for emissivity, A stands for the surface area of the earth, and T stands for the temperature at the surface.
[tex]&P=47 * 10^{12} W, \sigma \\\\=5.67 * 10^{-8} W / m^2 . K^4, e ,[/tex]
=0.6
[tex]\\\\&A=4 \pi R^2 \\\\=4 \pi *\left(6400 * 10^3\right)^2 \\\\[/tex]
=5.147 * 10^{14}m^2
[tex]T=\left(\frac{P}{\sigma e A}\right)^{1 / 4} \\\\=\left(\frac{47 * 10^{12}}{5.67 * 10^{-8} * 0.6 * 5.147 * 10^{14}}\right)^{1 / 4}[/tex]
=40.48K
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Kelley is playing pool and hits the cue into the nine ball, which is directly in the center of the pool table. The nine ball travels a total of 7 meters before stopping, 1 meter to the left of the initial starting point of the nine ball. Which sentence reflects the correct statement regarding distance and displacement when considering the motion of the nine ball?(1 point)
I NEED THE ANSWER NOW. . .I really don't understand displacement
The displacement of the nine balls is 1 m while the distance travelled by the nine balls is 7 m.
What is the displacement of he ball?The displacement of the ball is the change in the position of the ball. Displacement describes the shortest distance between the initial and final position of an object.
Mathematically, the displacement of the nine balls is given as;
Δx = xf - xi
where;
xf is the final position of the objectxi is the initial position of the objectIf the balls travels 1 meter to the left of the initial starting point, then the displacement of the balls is 1 m. The 1 meter is the shift from the initial position.
The distance of the balls is the total path covered by the balls.
distance = 7 m
Thus, the displacement of the nine balls is the change in the position of the balls while the distance of the nine balls is the total path covered.
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Can you please explain to me how a steam engine is a example of the 2nd law of thermodynamics
To find:
How a steam engine is an example of the second law of thermodynamics?
Explanation:
The second law of thermodynamics has more than one statement.
Kelvin-Plank statement:
No process is possible whose sole result is to convert the heat absorbed completely into useful work.
Clausius statement:
No process is possible whose sole result is the transfer of heat from a colder object to a hotter object.
Both of the statements are equivalent.
A steam engine takes the heat from a hot reservoir and converts some parts of it into useful work and the rest of the heat absorbed will be sent to a cold reservoir. Thus the steam engine is an example of the 2nd law of thermodynamics.
A soccer ball kicked with a force of 13.5 N accelerates at 6.5 m/s^2. What is the mass of the ball?
ANSWER:
2.08 kg
STEP-BY-STEP EXPLANATION:
The force is given by the multiplication of the mass and the acceleration, like this:
[tex]\begin{gathered} f=m\cdot a \\ f=13.5\text{ N} \\ a=6.5\frac{m}{s^2} \end{gathered}[/tex]We replace and calculate for m:
[tex]\begin{gathered} 13.5=m\cdot6.5 \\ m=\frac{13.5}{6.5} \\ m=2.08\text{ kg} \end{gathered}[/tex]The mass is 2.08 kg
Look at the diagram below.
Encoding technique - Decoding technique
"Written language
Pictures
Billboards
Advertisements
E-mails
Reading
Viewing
Interpreting
The communication process, which involves understanding parts and meanings, consists of eight key elements: Source Message Channel Receiver Feedback Environment Context, and Interference
Common examples of character encoding systems are Morse Code, Bodo Code, American Standard Code ASCII for Information Interchange, and Unicode. The purpose of encoding is to transform data so that it can be properly and safely processed by another type of binary system data transmitted via e-mail or special characters that appear on web pages.
The goal is not to keep information secret, but to make it available for proper consumption. Coding is the process of transforming data into the formats required for various information processing needs, such as compiling and running programs. Coding uses different patterns of voltage or current levels to represent the 1s and 0s of a digital signal on a transmission line. Common types of line coding are unipolar, polar, bipolar, and Manchester.
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A 20.0 N block slides over a horizontal table. If it takes 5.0 N to move the block at constant velocity, what is the coefficient of friction?
A. 0.50
B. 4.0
C. 0.10
D. 0.25
A 20.0 N block slides over a horizontal table. If it takes 5.0 N to move the block at a constant velocity, the coefficient of friction is D. 0.25.
Given,
Frictional force (F) is 5.0 N
Normal force (N) is 20.0 N
The coefficient of friction (µ), a numerical value, is obtained by dividing the resistive force of friction (F) by the normal or perpendicular force (N) pushing the objects together.
i.e., µ = F/N
Calculating the coefficient of friction :
The frictional force is equal to and opposing the applied force since the box is traveling at a constant speed.
µ = F / N
µ = 5.0 N / 20.0 N
µ = 1/4 N
µ = 0.25 N
Thus, the coefficient of friction is D. 0.25.
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What space telescope is used to look deep into the universe?
O A.
OB.
O C.
Hubble
Galileo
Penzias
OD. Edwin
The gravitational force between two masses is 36 N. What is the gravitational force if the distance between them is tripled? (G = 6.673 x10 (Power of-11)
If the distance between them is tripled, the gravitational force is F2= 36/9=4N. The force of attraction between any two bodies is proportional to the product of their masses and inversely proportional to the square of their distance.
WHAT IS GRAVITATIONAL FORCE?
It is the force that connects all masses in the universe, particularly the pull of the earth's mass on objects near its surface. It follows the inverse square law. The force of attraction exerted on a body by the earth is known as gravitational force. For example, the leaves and fruits of a tree fall to the ground, water in a river flows down streams, and a ball thrown up travels to a height before returning to the ground are all examples of motion caused by gravitational force.The formula for Gravitational force:
F=G m1*m2/r² ,
G: gravitational constant = 6.67x10⁻¹¹ N*m²/kg²
m1: is the mass of the first object
m2: is the mass of the second object
r: is the distance between the center of the masses of the objects
F∝1/r²
so F1/F2=( r2/r1)².......(1)
Given F1= 36N F2=?
let r1=r, r2= 3r
putting all this value in equation (1)
36/F2= (3r/r)²
⇒F2= 36/9=4N
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Find the answer to the problem
The box is moving at a 2.06 m/s rate.
What causes acceleration?Acceleration is the rate at which velocity changes over time in both the direction and the speed. Anything is said to have been accelerated when it goes faster or slower in a straight line. Even travel on a circle accelerates steadily since the direction is continuously changing.
What metrics are used to quantify acceleration?Acceleration is measured in meters per second per second (m/s2).
Formulas for
F NET are F NET = m and F NET = F1 + F2.
3 + 3 + 7 Equals 13 when F NET is used.
F NET = m * a then,
where a = F NET/m and
a = 13N / 6.3 kg and 2.06m/s respectively.
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List three ways that you can use water in each of its states.
Identify the time interval during which the velocity of shuttle bus is zero.
Explanation:
during D and E, so between 13 and 16 (whatever time unit this is - hours, minutes, ... ?).
Answer:
Explanation:
The time interval 13 to 16 seconds
