If 2 joules of work is done in raising an apple 8 meters, how much force was exerted?

Final answer:

To determine how far the base of a 15-foot slanted retaining wall is from the base of a hill, we can use the cosine of the angle (75°) the wall makes with the ground. The calculation shows the base of the wall is approximately 3.88 feet from the hill's base.

Explanation:

To find out how far the base of a 15-foot slanted wall, which is supposed to shore up the side of a hill at a 75° angle with the ground, is from the base of the hill, we can use trigonometry. Specifically, we can use the cosine function which is defined in a right-angled triangle as the adjacent side (base) over the hypotenuse (the slanted wall). The angle given is 75° and the length of the slanted wall (hypotenuse) is 15 feet.

Thus, the formula to calculate the distance from the base of the hill (base of the wall) is:

Cos(75°) = Base / 15 feet

Rearranging the formula to solve for the base gives:

Base = 15 feet × Cos(75°)

Using a calculator, Cos(75°) is approximately 0.2588. Therefore:

Base ≈ 15 feet × 0.2588 ≈ 3.882 feet

This means the base of the wall is approximately 3.88 feet from the base of the hill.

Answer: Mercury is used for making thermometers, barometers, diffusion pumps, and other laboratory instruments. It can also be used to make mercury-vapor lamps, advertising signs, and mercury switches as well as pesticides, dental preparations, batteries, and catalysts.

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Answer: Charge and distance

Explanation:

There is a pattern of current flow. Based on the above scenario,  Option b is the correct answer as bulb B will be brighter than before (Image attached).

Based on the current flowing through bulb A  is said to remains the same in both scenario. It is the current through bulb B that changes when the switch  are said to be closed.

There is something that happens to the brightness of a bulb when the switch is said to be closed. Note that when the switch is closed, the light bulb often operates because of the current flows via the circuit. The bulb (B) tends to glows at its full brightness because it receives more volts.

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To solve for speed or rate use the formula for speed, s = d/t which means speed equals distance divided by time.

speed = distance/time

outer core is your answer

Answer:

c. outer core

Explanation:

The inner core of the Earth is solid and contains iron. The outer core is in molten state and contains iron and nickel in liquid state. As the Earth spins, the domains of this magnetic material aligned in single direction generating huge magnetic field. Thus, earth acts like an enormous bar magnet.

Correct option is c. outer core.

The car, running out of gas and hence momentum, will convert its initial kinetic energy into gravitational potential energy as it moves uphill. In the absence of friction, it will ascend approximately 6.27 meters up the hill before it stops.

The subject of the question is related to physics, particularly to the principle of conservation of energy. When the car runs out of gas, its initial kinetic energy is converted into potential energy as it goes uphill. The initial kinetic energy of the car is given by 1/2*m*v^2 kg (where m is the mass and v is the velocity of the car) which equals 1/2*1400*11^2 = 85,800 Joules.

The car will continue to move up the hill until all this kinetic energy is converted into gravitational potential energy, given by m*g*h = 1400*9.8*h, where g is the acceleration due to gravity and h is the height the car reaches up the hill. Setting these two equations equal allows us to determine how high the car will go: 85,800 = 1400*9.8*h, which gives h = 85,800/(1400*9.8) = 6.27 meters. Therefore, in the absence of friction, the car will move approximately 6.27 meters up the hill before coming to a stop.

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The car can climb a hill approximately 6.17 meters in height before coming to a complete stop due to its kinetic energy. This is calculated using the principles of energy conservation and the formulas for kinetic and potential energy.

This physicist question is about the concept of conservation of energy. The 1400 kg car runs out of gas, but it maintains its kinetic energy. This kinetic energy will then be converted into potential energy as the car climbs the hill. The calculation of how high the car will climb can be done using the formula for kinetic energy, which is 0.5*m*v2, and the formula for gravitational potential energy, m*g*h.

First, we determine the initial kinetic energy of the car, which comes out to be 0.5*1400 kg*(11 m/s)2 = 85,250 Joules. This will be equal to the potential energy at the peak of the hill; thus, we have the equation 1400 kg*9.8 m/s2*h = 85,250 Joules, so solving for height, we find h = 85,250/ (1400*9.8) = approximately 6.17 meters. So, the car should be able to climb a hill of approximately 6.17 meters before it comes to a stop.

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The gravitational force the asteroids exert on each other will be one fourth of their initial force.

Any two bodies will be attracted to one another by the force of gravity, also known as gravity. There is an attraction between every thing in the cosmos, but most of the time it is too faint to be noticed due to the extreme distance between the objects. Furthermore, although the influence of gravity is weaker as objects are moved away, its range is infinite.

We know that, gravitational force acting between two bodies,

F=[tex]\frac{ G m_1 m_2}{r^2}[/tex]

Where, G = universal gravitational constant

m₁ and m₂ are masses of the two bodies and r is distance between them.

Let, the masses of the two asteroids are M₁ and M₂ and initial distance between them is R.

Hence, gravitational force they exert on each other, F₁ =[tex]\frac{ G M_1 M_2}{R^2}[/tex]

Now, when the distance between two asteroids is doubled, that is 2R, the gravitational force they exert on each other will, F₂ = [tex]\frac{ G M_1 M_2}{(2R)^2}[/tex] = [tex]\frac{1}{4}[/tex] [tex]\frac{ G M_1 M_2}{R^2}[/tex] = [tex]\frac{1}{4}[/tex] F₁

Hence, the final gravitational force is one-fourth of the initial one.

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The answer in this question is scale. The scale is used to determine the actual distance between two points in a planimetric map. This map only represents the horizontal position of features. This is also called by others as the line map.  This is usually consisted of man-made and natural features. These are from aerial shots by the use of modern technology. This map’s common features are street and water centerlines, sidewalks, culverts, utility lines, building footprints and vegetation. Sometimes, there are also other useful data attached in this map like titles of the properties, owner, assessed value, etc. It is because of these, this map becomes a storage of valuable data which can be easily and quickly accessed. 

The scale on a planimetric map is used to determine the actual distance between two points by converting measured map distances to real-world distances.

On a planimetric map, the scale is used to determine the actual distance between two points. By measuring the distance between two points on the map with a ruler and then using the map scale to find the real distance on the land surface, it is possible to convert map distances to real-world distances. For example, if you measure 10 centimeters between two points on a map with a scale of 1:10,000, you multiply the measured distance by the scale factor, resulting in an actual distance of 100,000 centimeters on the Earth's surface.

It is important to note that large-scale maps, like a standard topographic map, typically have less distance distortion, making it easier to measure straight line distances. However, when measuring distances along curved features, techniques such as using multiple straight-line segments or tools like an opisometer may be employed for greater accuracy. Map scales can be complex, as they can vary with latitude due to the Earth's curvature, particularly on maps that cover large areas such as equidistant projections which are true-to-scale only along meridians.

Answer:

The car is parked.

The car is moving at a fixed speed and direction.

Explanation:

The car is either at rest, or traveling at a constant velocity.

Hope this helps. Good luck!

Answer:

130 days

Explanation:

Strontium-85 has a half-life of 65 days. how long will it take for the radiation level of strontium-85 to drop to one-fourth of its original level?

Strontium-85 is a radioactive element. so it takes 65days for one strontium of a particular mass to decrease to half of its initial size

Half life  is the time taken for a radioactive element to decrease to half of its initial size. it decays by half every 65 days.

from 1 to 1/2  = 65 days

from 1/2 to 1/4  =65 days

from 1/4 to 1/8 = 65 days

from 1/8 to 1/16 = 65 days

total time taken to decay from original amount to 1/4 is two half lives =

65+65=130 days

The time it will take for the radiation level of Strontium-85 to drop to one-fourth of its original level is 130 days.

To calculate the time it will take for Strontium-85 to drop to one-fourth of its original value, we use the formula below.

Formula:

Where:

From the question,

Given:

Assuming,

Substitute the given values and the assumed values into equation 1

Solve for a

Equating the base,

Hence, The time it will take for the radiation level of Strontium-85 to drop to one-fourth of its original level is 130 days.

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Answer:

The answer on ED is A

Explanation:

Answer: These planets are formed by rocks.

Explanation:

There are 2 types of planets in Our Solar System:

1. Inner planets: There are 4 planets which are considered as inner planets, these are Mercury, Venus, Earth and Mars. These planets are small and have rocky surface. Their time of revolution around the Sun is less than the outer ones. These planets have maximum number of moon till 2.

2. Outer planets: There are 4 planets which are considered as outer planets, these are Jupiter, Saturn, Uranus and Neptune. These planets are large and gaseous planets. Their time of revolution around the Sun is more than the inner ones. These planets have many moon.

Hence, the inner planets are believed to be formed of rocks.

Answer:

a) The time the police officer required to reach the motorist was 15 s.

b) The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) The total distance traveled by the officer was 225 m.

Explanation:

The equations for the position and velocity of an object moving in a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

a)When the officer reaches the motorist, the position of the motorist is the same as the position of the officer:

x motorist = x officer

Using the equation for the position:

x motirist = x0 + v · t (since a = 0).

x officer = x0 + v0 · t + 1/2 · a · t²

Let´s place our frame of reference at the point where the officer starts following the motorist so that x0 = 0 for both:

x motorist = x officer

x0 + v · t = x0 + v0 · t + 1/2 · a · t²      (the officer starts form rest, then, v0 = 0)

v · t = 1/2 · a · t²    

Solving for t:

2 v/a = t

t = 2 · 15.0 m/s/ 2.00 m/s² = 15 s

The time the police officer required to reach the motorist was 15 s.

b) Now, we can calculate the speed of the officer using the time calculated in a) and the  equation for velocity:

v = v0 + a · t

v = 0 m/s + 2.00 m/s² · 15 s

v = 30 m/s

The speed of the officer at the moment she overtakes the motorist is 30 m/s

c) Using the equation for the position, we can find the traveled distance in 15 s:

x = x0 + v0 · t + 1/2 · a · t²

x = 1/2 · 2.00 m/s² · (15s)² = 225 m

The baseball player hitting a baseball is the best example of Newton's Second Law of Motion, as the force exerted on the ball causes it to move in the opposite direction with the same magnitude.

The best example of Newton's Second Law of Motion is the baseball player hitting a baseball that is pitched to him. When the player hits the ball, the force exerted on the ball causes it to move in the opposite direction with the same magnitude. This is known as the law of action and reaction, which is a part of Newton's Second Law of Motion.

Another example of Newton's Second Law is a student on roller skates pushing against a railing and immediately rolling backwards. The force exerted by the student on the railing causes an equal but opposite force on the student, resulting in the backward motion.

Lastly, the example of a car making a quick right turn and the passenger sliding across the back seat can be considered an example of Newton's Second Law. The passenger tends to continue moving in a straight line due to inertia but is pushed to the side due to the force of the car turning.

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Driving your boat with its engine onto a trailer can cause uncontrollable movement, potential damage, erosion of the launch area, and dangerous propeller strikes.

There are several problems related to using your boat's engine to drive it onto a trailer. Firstly, the propellers of the boat create a turbulent water flow that will provide thrust in an uncontrolled manner, likely leading to imprecise movement and potential damage to either the boat or the trailer. Moreover, the propeller wash can cause erosion and possibly damage the launching area. Secondly, using the engine to drive onto the trailer may lead to prop strike, where the boat's propeller might come into contact with the ramp, causing significant damage to the motor.

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