Answer:
am not sure about the answer
Explanation:
you need to find out the amount of force it's going in for example 10n or 100n then you need to times it the distance then devide by the time
6.
ribbon
AA
SON
120 N
Two teams of students are competing in a tug-o-war contest, as shown in the
picture above. How does the ribbon move?
Answer:
The ribbon will move to the right.
Explanation:
To know the the correct answer to the question, we shall determine the net force and direction. This can be obtained as follow:
Force to the right (Fᵣ) = 120 N
Force to the left (Fₗ) = 80 N
Net force (Fₙ) =?
Fₙ = Fᵣ – Fₗ
Fₙ = 120 – 80
Fₙ = 40 N to the right.
From the calculation made above, the net force is 40 N to the right. Thus, the ribbon will move to the right.
Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.
How much kinetic energy does the rake gain?
Answer:
10 J.
Explanation:
Given that,
Net force acting on the rake, F = 2 N
Distance moved by the rake, d = 5 m
We need to find the kinetic energy gained by the rake. We know that,
Kinetic energy = work done
So,
K = F×d
K = 2 N × 5 m
K = 10 J
So, 10 J of kinetic energy is gained by the rake.
Violet pulls a rake horizontally on a frictionless driveway with a net force of 2.0 N for 5.0 m.
How much kinetic energy does the rake gain?
Answer: 10 J
Rank the four fundamental forces from strongest to weakest. Use 1 to indicate the strongest force and 4 to indicate the weakest force. The gravitational force: The electromagnetic force: The strong nuclear force: The weak nuclear force:
Answer:
4
2
1
3
Explanation:
Be safe, lovelies <3
Which elements have one valence electron?
A. Sodium
B. Carbon
C. Fluorine
D. Magnesium
The answer is A
Answer:
Well, you said the answer is A, so it’s A!
A spacecraft and a staellite are at diametrically opposite position in the same circular orbit of altitude 500 km above the earth. As it passes through point A, the spacecraft fires its engine for a short interval of time to increase its speed and enter an elliptical orbit. Knowing that the spacecraft returns to A at the same time the satellite reaches A after completing one and a half orbits, determine (a) the increase in speed required, (b) the periodic time for the elliptic orbit
Answer:
Hello the diagram related to your question is attached below
answer: a) 851 m/s
b) 8506.1 secs
Explanation:
calculate the periodic time of the satellite using the equation below
t = [tex]\frac{2\pi }{R} \sqrt{\frac{(R+h)^{3} }{g} }[/tex] -- ( 1 )
where ; R = 6370 km
h = 500 km
g = 9.81 m/s^2
input given values into equation 1
t = 5670.75 secs
next calculate the periodic time taken by the space craft
a) determine the increase in speed
V = v - [tex]\sqrt{\frac{gR^2}{R + h} }[/tex]
where ; v = 8463 m/s , R = 6370 km, h = 500 km
V = 851 m/s
b) Determine the periodic time for the elliptic orbit
τ = [tex]\frac{3t}{2}[/tex]
= [tex]\frac{3*5670.76}{2}[/tex] = 8506.1 secs
attached below is the remaining part of the detailed solution
Potential energy is defined as
A. energy of motion
B. moving another object
C. stored energy
Answer:
c
Explanation:
it is stored energy because it is built up in said object
the water behind hoover dam in nevada is 206 m higher than the colorado river below it. at what rate must water pass through the hydraulic turbines of this dam to produce 100 mw of power if the turbines are 100 percent efficient
Answer:
the required mass flow rate is 49484.37 kg/s
Explanation:
Given the data in the question;
we first determine the relation for mass flow rate of water that passes through the turbine;
so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;
[tex]W_{net}[/tex] = m ( Δ P.E )
so we substitute (gh) for ( Δ P.E );
[tex]W_{net}[/tex] = m (gh)
m = [tex]W_{net}[/tex] / gh
so we substitute our given values into the equation
m = 100 MW / ( 9.81 m/s²) × 206 m
m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m
m = 10 × 10⁷ / 2020.86
m = 49484.37 kg/s
Therefore, the required mass flow rate is 49484.37 kg/s
According to some nineteenth-century geo-
logical theories (now largely discredited), the
Earth has been shrinking as it gradually cools.
If so, how would g have changed over geo-
logical time?
1. It would increase; g is inversely proportional to the square of the radius of the Earth
2. It would decrease; the Earth’s radius is decreasing
3. It would not change; the mass of the Earth remained the same.
I really need this answer NOW. i’m taking a timed test. Will mark brainliest answer.
Answer:
What was it
Explanation:
It would increase; g is inversely proportional to the square of the radius of the Earth. The correct option is A.
What is geological theory?A current idea in geology that describes how the earth's crust is made up of a few big, hard plates that move independently of one another, causing deformation, volcanism, and seismic activity along their boundaries.
Because it explains how mountain ranges, earthquakes, volcanoes, shorelines, and other features often emerge where the moving plates contact along their boundaries, plate tectonics provides "the overall picture" of geology.
The Earth has been shrinking as it gradually cools, according to some geological hypotheses from the nineteenth century that have now been completely debunked.
If that were the case, it would rise since g is inversely proportional to the square of the Earth's radius.
Thus, the correct option is A.
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Micro-bats use a form of radar called echolocation to navigate and find their prey such as flying insects. They locate the surrounding objects by bouncing sound wave pulses off these objects and detecting the time delay between the emitted pulses and the reflected pulses. Determine the time delay between the pulse emitted by the micro-bat and the detected pulse reflected from an insect located 10 m away from the micro-bat. Assume the approximate speed of sound waves to be 340 m-s-1
Answer:
t = 5.88 10⁻² s
Explanation:
The speed of the sound wave after it is emitted by the bat is constant, so we can use the uniform motion relationships
v = [tex]\frac{x}{t}[/tex]
t = [tex]\frac{x}{v}[/tex]
in this case the distance is that of the sound in going from the bat to the insect and back
x = 2d
x = 2 10
x = 20 m
let's calculate
t = 20/340
t = 5.88 10⁻² s
We can see that the time is very short, so the distance traveled by the two animals has little influence on the result.
(a) The electric potential due to a point charge is given by V = kq⁄r where q is the charge, r is the distance from q and k = 8.99 × 109 ????m2 C 2 ⁄ . Show, in detail, that the SI unit of electric potential is a volt. ( b ) What are the equipotential lines? (c) How are equipotential lines used to obtain the electric field lines? (
Answer:
a) [volts] = [N m / C],
b) The lines or surface that has the same potential are called equipotential
c) the equipotential lines must also be perpendicular to the electric field lines
Explanation:
a) find the units of the volt
the electric potential energy is
V = k q / r
V = [N m² / C²] C / m
V = [N m / C]
The electric potential is defined as
V = E .s
V = [N / C] [m]
V = [N m / C] = [volt]
we see that in the two expressions the same result is obtained therefore the volt is
[volts] = [N m / C]
b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.
c) The electric potential is defined as the gradient of the electric field
v = [tex]- \frac{dE}{dx} i^[/tex]
therefore the equipotential lines must also be perpendicular to the electric field lines
(a) The unit of electric potential is volts.
(b) Equipotential lines are lines along which the electric potential is constant.
(c) Equipotential lines are obtained by drawing a line perpendicular to electric field lines.
SI unit of electric potentialElectric potential is the work done in moving a unit positive from infinite to a point in the electric filed.
Electric potential = EQ
Electric potential = (V/C) x (C)
Electric potential = Volts.
Equipotential linesThis is a line along which the electric potential is constant.
How to obtain equipotential linesEquipotential lines are obtained by drawing a line perpendicular to electric field lines.
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If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the initial displacement is 2.25 m, and the change in momentum is 0.58 kg*m/s, how far will it coast back up the ramp before changing directions
Answer:
l = 0.548 m
Explanation:
For this exercise we compensate by finding the speed of the car
p = m v
v = p / m
v = 0.58 / 0.2
v = 2.9 m / s
this is how fast you get to the ramp, let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ m v²
final point. Point where it stops on the ramp
[tex]Em_{f}[/tex] = U = m g h
mechanical energy is conserved
Em₀ = Em_{f}
½ m v² = m g h
h = [tex]\frac{m v^2}{2 g}[/tex]
let's calculate
h = [tex]\frac{0.2 \ 2.9^2}{2 \ 9.8}[/tex]
h = 0.0858 m
to find the distance that e travels on the ramp let's use trigonometry, we look for the angle
tan θ = y / x
tan θ = 12/75 = 0.16
θ = tan⁻¹ 0.16
θ = 9º
therefore
sin 9 = h / l
l = h / sin 9
l = 0.0858 / sin 9
l = 0.548 m
NEED HELP WITH THIS PLEASE
Answer:
A
Explanation:
8) A train enters a curved horizontal section of the track at a speed of 100 km/h and slows down with constant deceleration to 50 km/h in 12 seconds. If the total horizontal acceleration of the train is 2 m/s2 when the train is 6 seconds into the curve, calculate the radius of curvature of the track for this instant.
Answer:
the radius of curvature of the track for this instant is 266 m
Explanation:
Given that;
The Initial Velocity u = 100 km/h = 100 × [tex]\frac{5}{18}[/tex] = 27.77 m/s
velocity of the train at t=12 s is;
[tex]V_{t=12}[/tex] = 50 km/h = 50 × [tex]\frac{5}{18}[/tex] = 13.89 m/s
now, we calculate the deceleration of the train
[tex]V_{t=12}[/tex] = u + at
13.89 = 27.77 + [tex]a_{t}[/tex]12
[tex]a_{t}[/tex] = (13.89 - 27.77) / 12
[tex]a_{t}[/tex] = -13.88 / 12
[tex]a_{t}[/tex] = - 1.1566 m/s²
Now, the velocity of the train at 6 seconds is;
[tex]V_{t=6}[/tex] = u + at
[tex]V_{t=6}[/tex] = 27.77 + ( - 1.1566 m/s²)6
[tex]V_{t=6}[/tex] = 27.77 - 6.9396
[tex]V_{t=6}[/tex] = 20.83 m/s
The acceleration at t=6 s is;
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]
we substitute
2m/s² = √[ (- 1.15 )² + ([tex]a_{n}[/tex])²]
4 = (- 1.1566 )² + ([tex]a_{n}[/tex])²
4 = 1.3377 + ([tex]a_{n}[/tex])²
([tex]a_{n}[/tex])² = 4 - 1.3377
([tex]a_{n}[/tex])² = 2.6623
[tex]a_{n}[/tex] = √2.6623
[tex]a_{n}[/tex] = 1.6316 m/s²
Now the radius of curve is;
a = V² / p
[tex]p_{t=6}[/tex] = ( [tex]V_{t=6}[/tex] )² / [tex]a_{n}[/tex]
[tex]p_{t=6}[/tex] = ( 20.83 m/s )² / 1.6316 m/s²
[tex]p_{t=6}[/tex] = 433.8889 / 1.6316
[tex]p_{t=6}[/tex] = 265.9 m ≈ 266 m
Therefore; the radius of curvature of the track for this instant is 266 m
a point charge q1 = 2.40 uC is held stationary at the origin. A second point charge q2 = -4.30uC moves from the point x= .150 m, y= 0.0 m, to the point x = .250 m, y= 0.0m
a) what is the charge in potential energy of the pair of charges?
b) How much work is done by the electric force on q2
Answer:150M
Explanation:
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.0988 N when their center-to-center separation is 44.5 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0276 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other
Answer:
(a) The negative charge on one of the charges is -8.79630245 × 10⁻⁷C
(b) The positive charge on one of the other charges is 8.79630245 × 10⁻⁷C
Explanation:
The given parameters are;
The force of attraction between the two spheres = 0.0988 N
The distance between their centers = 44.5 cm = 0.445 m
Therefore, we have;
[tex]F = \dfrac{k \cdot q_1 \cdot q_2}{d^2}[/tex]
Therefore, we have;
[tex]0.0988 \ N = -\dfrac{8.99 \times 10^9 N\cdot m^2 \cdot C^{-2}\cdot q_1 \cdot q_2}{(0.445 \ m)^2}[/tex]
Therefore, we have;
q₁·q₂ = -0.0988 N × (0.445 m)²/(8.99 × 10⁹ N·m²·C⁻²) = -2.17629255 × 10⁻¹² C²
q₁·q₂ = -2.17629255 × 10⁻¹² C²...(1)
When the two charges are connected, we get;
[tex]F_2 = \dfrac{k \cdot \left (\dfrac{q_1 + q_2}{2} \right) ^2}{d^2}[/tex]
Therefore, we have;
[tex]q_1 + q_2 = \sqrt{\dfrac{4 \cdot F_2 \cdot d^2}{k} }[/tex]
[tex]q_1 + q_2 = \sqrt{\dfrac{4 \times 0.0276 \ N \times(0.445 \ m)^2}{8.99 \times 10^9 N\cdot m^2 \cdot C^{-2}} } = 1.59446902743 \times 10^{-6} \ C[/tex]
q₁ + q₂ = 1.59446902743 × 10⁻⁶ C...(2)
From, equation (2), we have;
q₁ = 1.59446902743 × 10⁻⁶ C - q₂
Plugging in the value of q₁ in equation (1) givens;
q₁·q₂ = -2.17629255 × 10⁻¹²
Therefore, we have;
(1.59446902743 × 10⁻⁶ - q₂) × q₂ = -2.17629255 × 10⁻¹²
Which gives;
-q₂² + 1.59446902743 × 10⁻⁶·q₂+2.17629255 × 10⁻¹² = 0
Solving, with a graphing calculator, we get;
q₂ = 2.4741×10⁻⁶ C, or -8.79630245 × 10⁻⁷C
q₁ = 8.79630245 × 10⁻⁷C or -2.4741×10⁻⁶ C
Therefore, we have;
(a) The negative charge on one of the charges = -8.79630245 × 10⁻⁷C
(b) The positive charge on one of the other charges = 8.79630245 × 10⁻⁷C
The force of friction occurs primarily because:
A) two surfaces in contact have magnetic forces of attraction.
B) on the microscopic level, two surfaces in contact are rough even if they appear smooth to the touch.
C) two surfaces in contact have a gravitational attraction to one another.
D) both A and B.
Answer:
B
Explanation:
Friction is a force that opposes motion between any surfaces that are touching. Friction occurs because no surface is perfectly smooth Friction produces heat because it causes the molecules on rubbing surfaces to move faster and have more energy.
What is the displacement of the particle in the time interval 7 seconds to 8 seconds?
Answer:
it 1.5 meters
Explanation:
if u could put the option number it will be cool and hope it help and if it doesnt am really sorry ;)
true or false solubility can be used to identify an unknown substance
The head of a rattlesnake can accelerate at 50 m/s2 in striking a victim. If a car could do as well, how long would it take to reach a speed of 100 km/h from rest
Answer:
the time for the car to reach the final velocity is 0.56 s.
Explanation:
Given;
acceleration of the car, a = 50 m/s²
final velocity of the car, v = 100 km/h = 27.778 m/s
the initial velocity of the car, u = 0
The time for the car to reach the final velocity is calculated as;
v = u + at
27.778 = 0 + 50t
27.778 = 50t
t = 27.778 / 50
t = 0.56 s
Therefore, the time for the car to reach the final velocity is 0.56 s.
Could I get help on this question please
Answer:
124.51 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 49.4 m/s
Final velocity (v) = 0 m/s (at maximum height)
Maximum height (h) =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
The maximum height to which the cannon ball attained before falling back can be obtained as illustrated below:
v² = u² – 2gh ( since the ball is going against gravity)
0² = 49.4² – (2 × 9.8 × h)
0 = 2440.36 – 19.6h
Collect like terms
0 – 2440.36 = –19.6h
–2440.36 = –19.6h
Divide both side by –19.6
h = –2440.36 / –19.6
h = 124.51 m
Therefore, maximum height to which the cannon ball attained before falling back is 124.51 m
you describe a friend’s position by including distance, direction, and what other term?
Answer choices:
A. Acceleration
B.displacement
C.Average speed
D. Reference point
PLEASE HELP I NEED THIS IN AN HOUR
Answer:
Acceleration
Explanation:
Answer: Acceleration
Explanation:
A building inspector standing on the top floor of a building wishes to determine the depth of the elevator shaft. They drop a rock from rest and hear it hit bottom after 2.56 as. (a) How far (in m) is it from where they drop the rock to the bottom of the shaft
Answer:
d = 29.89 m
Explanation:
To solve this, we need to separate this problem in two parts.
One part would be the the time taken by the rock to actually hit the bottom, and the other part would be the time taken by the sound to reach the inspector.
Joining these two times we have:
t = t₁ + t₂ (1)
This time is 2.56 s.
Now, as we are asked to determine the distance from the top floor to the bottom, and we have two times taken in different ways, one by sound and the other the actual, we can say the same thing on distance, we need a distance relationed to the time taken by rock to hit the bottom, and the other distance relationet to the time taken by sound to reach the inspector.
Doing this we have that the distance traveled by the rock is:
y₁ = gt²/2
y₁ = 9.8t²/2 = 4.9t₁² (2)
Now, the distance traveled by sound would be:
y₂ = v * t₂ = 336t₂ (3)
Remember that the speed of the sound is 336 m/s
From this last expression (3), we can actually write t₂ in function of t₁, using (1):
2.56 = t₁ + t₂
t₂ = 2.56 - t₁ (4)
Replacing (4) in (3):
y₂ = 336(2.56 - t₁) (5)
Now that we have y₁ and y₂, we can equal (2) and (5), both expressions to get the value of t₁, and then, calculate the distance:
4.9t₁² = 336(2.56 - t₁)
4.9t₁² = 860.16 - 336t₁
4.9t₁² + 336t₁ - 860.16 = 0
Using the quadractic formula, we can calculate t₁:
t₁ = -336 ±√(336)² + 4*4.9*860.16 / (2*4.9)
t₁ = -336 ±√129,7555.136 / 9.8
t₁ = -336 ± 360.21 / 9.8 Using only the positive value we have:
t₁ = 2.47 s
This means that the rock hits the bottom in 2.47 s, and the remaining 0.09 s belongs to the time taken by sound. (2.47 + 0.09 = 2.56 s)
With this, we can calculate the distance of the rock using expression (2):
y₁ = 4.9 * (2.47)²
y₁ = 29.89 mHope this helps
Renee looks out a window. The window is clear, or transparent. This means most of the light that hits the window is:
А
scattered
B
reflected
с
transmitted
D
absorbed
Answer:
the answer is transmitted
Explanation:
When the material in the mantle cools off near the surface then sinks
down towards the core and get heated again and rises back towards the
surface it is called?
Condensation
О
The water cycle
O
Convection currents
О
Apples and Bananas.
PLEASE HELP THIS IS URGENT ITS FOR A TEST
Answer:
convection currents
Explanation:
A constant torque of 3 Nm is applied to an unloaded motor at rest at time t = 0. The motor reaches a speed of 1,393 rpm in 4 s. Assuming the damping to be negligible, calculate the motor inertia in Nm·s2.
Answer:
The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.
Explanation:
From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system ([tex]\tau[/tex]), measured in Newton-meters, is:
[tex]\tau = I\cdot \alpha[/tex] (1)
Where:
[tex]I[/tex] - Moment of inertia, measured in Newton-meter-square seconds.
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
If motor have an uniform acceleration, then we can calculate acceleration by this formula:
[tex]\alpha = \frac{\omega - \omega_{o}}{t}[/tex] (2)
Where:
[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.
[tex]\omega[/tex] - Final angular speed, measured in radians per second.
[tex]t[/tex] - Time, measured in seconds.
If we know that [tex]\tau = 3\,N\cdot m[/tex], [tex]\omega_{o} = 0\,\frac{rad}{s }[/tex], [tex]\omega = 145.875\,\frac{rad}{s}[/tex] and [tex]t = 4\,s[/tex], then the moment of inertia of the motor is:
[tex]\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}[/tex]
[tex]\alpha = 36.469\,\frac{rad}{s^{2}}[/tex]
[tex]I = \frac{\tau}{\alpha}[/tex]
[tex]I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }[/tex]
[tex]I = 0.0823\,N\cdot m\cdot s^{2}[/tex]
The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.
Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Falling off of a skateboard after you run into a curb A ball hits the ground and the ground pushes up on it with the same force
Answer:
A ball hits the ground and the ground pushes up on it
Explanation:
Newton's third law basically states that for every action, there's a reaction.
a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.
Hope this Helps!!! :)
A skydiver is using wind to land on a target that is 120 m away horizontally. The skydiver starts from a height of 70 m and is falling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target
Answer:
13.9 m/s.
Explanation:
Since the vertical velocity of the skydiver is constant at v = 7.0 m/s, we find the time, t it takes him to drop from a height of h = 70 m.
So, distance = velocity time
h = vt
t = h/v = 70 m/7 m/s = 10 s
This is also the time it takes him to move horizontally a distance of d = 120 m to the target.
So, his horizontal velocity is v' = distance/time = d/t = 120m/10 s = 12 m/s.
Since both vertical and horizontal velocities are perpendicular, we add them vectorially to obtain the skydivers total speed, V.
So, V = √(v² + v'²)
= √((7.0 m/s)² + (12.0 m/s)'²)
= √(49 m²/s² + 144 m²/s²)
= √(193 m²/s²)
= 13.9 m/s.
The direction of this velocity is Ф = tan⁻¹(v/v')
= tan⁻¹(7 m/s/12 m/s)
= tan⁻¹(0.5833)
= 30.3°
An 80 N rightward force is applied to a 10 kg object to accelerate it to the right.
The object encounters a friction force of 50 N.
net force = 30 N
mass = 8.16 kg
acceleration = 3.68 m/s²
Further explanationGiven
80 N force applied
mass of object = 10 kg
Friction force = 50 N
Required
Net force
mass
acceleration
Solution
net forceNet force = force applied(to the right) - friction force(to the left)
Net force = 80 - 50 = 30 N
massGravitational force(downward) : F = mg
m = F : g
m = 80 : 9.8
m = 8.16 kg
accelerationa = F net / m
a = 30 / 8.16
a = 3.68 m/s²
Concept Simulation 2.3 offers a useful review of the concepts central to this problem. An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of 17.4 m/s and measures a time of 12.4 s before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet
Answer:
1.40 m/s^2
Explanation:
Given data
Velocity= 17.4 m/s
time= 12.4 seconds
We want to find the acceleration of the rock
We know that
acceleration = velocity/time
Substitute
acceleration= 17.4/12.4
acceleration=1.40 m/s^2
Hence the acceleration is 1.40 m/s^2
How high above the ground would a 2 kg object need to be in order to have 180 J
of gravitational potential energy?
Answer:
energy= MGH
2*9.8*h=180
h=180/19.6
h=9.32 m
The height of the object above the ground would be equal to 9.18 m.
What is gravitational potential energy?When an object of mass (m) is moved from infinity to a certain point inside the gravitational influence, the amount of work done in displacing it is stored in the form of potential energy and is known as gravitational potential energy.
The mathematical equation for gravitational potential energy can be written as:
Gravitational potential energy = m⋅g⋅h
Where m is the mass, g is the gravitational acceleration and h is the height above the ground.
Given, the mass of the given object, m = 2 Kg
The gravitational potential energy = 180 J
[tex]GPE = m\times g\times h[/tex]
180 = 2 × 9.8 × h
h = 9.18 m
Therefore, the object should be at a height of 9.18 meters in order to have 180 J of gravitational potential energy.
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