For every increase in mass the gravitational force blank If the total mass increase by effective for the gravitational force

Answers

Answer 1

Answer:

For every increase in mass, the gravitational force increases. Gravitational force is directly proportional to the mass of the object.

What is gravitational force?

Gravitational force is the force by which an object attracts other objects into its center of mass. Earth attracts other objects gravitationally and that keep everyone stand to the ground.

Gravitational force directly proportional to the mass and inversely proportional to the distance between the objects. The expression relating the force and mass is written as:

g = G m/r²

Where G is the universal gravitational constant.

Therefore, as the mass of the object increase, the gravitational force exerted also increases. Similarly massive object experience more gravitation force by earth.

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Answer 2

Answer:

Answer:

Increases by the same amount.Increases by a factor of 4.

Explanation:

i took it

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Which type of diffraction occurs when the point source and the screen are at finite distances from the obstacle forming the diffraction pattern?A. fraunhofer B. fresnelC. far-fieldD. single slit

Which of the following statements are true?A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping. B. The increase in amplitude of an oscillation by a driving force is called forced oscillation. C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced. D. An oscillation that is maintained by a driving force is called forced oscillation.

Answers

Statements that are right as regards oscillation are:

A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping.

B. The increase in amplitude of an oscillation by a driving force is called forced oscillation.

C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced.

D. An oscillation that is maintained by a driving force is called forced oscillation.

Amplitude can be regarded as magnitude of change that is been experienced by oscillating variable with each oscillation.

When there is a decrease in the amplitude of an oscillation as a result dissipative forces, then it is regarded as damping.

When there is increase in amplitude of an oscillation as a result of driving force then it is termed forced oscillation.

Therefore, the options are correct.

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Answer:

right A, B, C, D

Explanation:

They ask which statements are true

A) Right. The decrease in amplitude is due to the dissipation of energy by friction and is called damping

B) Right. In resonant processes the amplitude of the oscillation increases, being a forced oscillation

C) Right. In a system with energy loss, the amplitude must decrease, therefore energy must be supplied to compensate for the loss.

D) Right. It is a resonant process the driving force keeps the oscillation of the system

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air?
(b) What must have been the initial horizontal component of the velocity?
(c) What is the vertical component of the velocity just before the ball hits the ground?
(d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

Answers

Answer:

Explanation:

Given

height of building (h)=60 m

Range of ball=100 m

(a)time travel to cover a vertical distance of 60 m

$h=ut+(at^2)/(2)$

$60=0+(9.8* t^2)/(2)$

$t^2=12.24$

t=3.49 s

(b)To cover a range of 100 m

R=ut

$100=v_x* 3.49$

$v_x=28.57 m/s$

(c)vertical component of velocity just before it hits the ground

$v_y=u+at$

$v_y=0+9.81* 3.49=34.202 m/s$

(d) $v_(net)=√(v_y^2+v_x^2)$

$v_(net)=√(1986.02)=44.56 m/s$

Final answer:

The ball is in the air for about 3.5 seconds. The initial horizontal velocity would have been approximately 28.6 m/s. The vertical component of the velocity just before the ball hits the ground is nearly 34.3 m/s. The overall velocity of the ball just prior to impact is roughly 44.6 m/s.

Explanation:

The problem given is about projectile motion which can be approached by splitting the motion into the horizontal and vertical components. We can work out the durations for each.

For the time the ball is in the air, we know that it falls vertically under gravity. Use the equation of motion, h = 0.5gt^2 where h is the height (60.0 m) and g is the acceleration due to gravity (approx 9.81 m/s^2). Solving for t takes approximately 3.5 seconds.
The initial horizontal component of the velocity can be calculated by the distance it traveled horizontally divided by the time it spent in the air. So, 100 m / 3.5 s = 28.6 m/s.
The vertical component of velocity just before the ball hits the ground can be calculated using v = gt where g is the acceleration due to gravity and t is the time. Solving gives around 34.3 m/s.
The overall velocity can then be calculated using the Pythagorean theorem, combining the horizontal and vertical components of the velocity - sqrt((28.6 m/s)^2 + (34.3 m/s)^2). This will approximately equal 44.6 m/s.

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An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?

Answers

Answer:

11405Volt

Explanation:

To solve this problem it is necessary to use the concept related to induced voltage or electromotive force measured in volts. Through this force it is possible to maintain a potential difference between two points in an open circuit or to produce an electric current in a closed circuit.

The equation that allows the calculation of this voltage is given by,

$\epsilon = BAN \omega$

Where

B = Magnetic field

A= Area

N = Number of loops

$\omega$ = Angular velocity

Our values previously given are:

$N = 140$

$A = 71.2*10^(-2)m*22.6*10^(-2)m=0.1609m^2$

$B = 4.32 T$

$\omega = 1120 rev / min$

We need convert the angular velocity to international system, then

$\omega = 1120 rev/min$

$\omega = 1120rev/min*(2\pi)/(1rev)*(1min)/(60sec)$

$\omega = 117.2rad/s$

Applying the equation for emf, we replace the values and we will obtain the value.

$\epsilon = BAN \omega$

$\epsilon = (4.32)(0.1609)(140)*117.2$

$\epsilon = 11405Volt$

A CO2 gun shoots a 0.2 gram round pellet (bb) at 2800 ft/sec, and as the bb leaves the gun it gets charged by friction . If Earths magnetic field points South to North at an intensity of 20 uT, and the bb is shot W->E. Find the charge the bb would need to stay level by balancing out the force of gravity.

Answers

Answer:

So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.

Explanation:

Fb = Fg

qvb= mg ⇒ q = mg/vB = 0.2 *10∧-3 * 9.8/853.44 * 20 * 10∧-6

= 0.115C

note:2800ft/sec = 853.44m/s

So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.

Lab: Weather Patterns

Answers

A weather pattern is defined as a period of time when the weather remains consistent. In the lab, a lot of observation about weather is obtained

What is the definition of a weather pattern?

A weather pattern is defined as a period of time when the weatherremains consistent. Weather changes are crucial to humanexistence.

because they influence our everyday activities and provide moisture for crops.

The rain does not always end within the day, and gloomy days might last just as long as sunny days. Tornadoes and hurricanes, for example, may inflict tremendous damage.

In the lab the following observation about weather is obtained;

1. We will find the graphs and statistics that indicate signs of climate change and engage with an interactivegraphic.

2. You'll also look at and debate maps of global temperature and precipitation patterns that are changing.

3. This lab will teach you about Earth's biomes and the close relationship that exists between them and the climates that serve to define them.

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Final answer:

The question pertains to meteorology, climatology, and atmospheric science. These are disciplines that study weather and climate, respectively, and their effects on the planet. Atmospheric Science is a broad field that includes both and employs physics principles.

Explanation:

The question refers to the subjects of meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric impacts on the Earth's weather. It involves the prediction of weather in the short term based on thousands of measurements of variables such as air pressure and temperature.

Climatology, on the other hand, is the study of climate, which involves analyzing averaged weather conditions over longer time periods using atmospheric data. Unlike meteorologists, climatologists focus on patterns and effects that occur over longer timescales of decades, centuries, and millennia.

Atmospheric Science is a broad field that encompasses both meteorology and climatology, as well as other disciplines that study the atmosphere. This discipline is typically based heavily on physics and involves the study of weather and climate patterns, predictions of developments in weather and climatic events, and the analysis of the effects of these events on the planet and its inhabitants.

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A car is making a 50 mi trip. It travels the first half of the total distance 25.0 mi at 7.00 mph and the last half of the total distance 25.0 mi at 52.00 mph. (a) What is the total time in hours of the trip? Keep two decimal places. 4.05 Correct (100,0%) (b) What is the car's average speed in mph for the entire trip? Keep two decimal places. 12 35 Correct (100,0%) Submit The car travels the same distance again, but this time, in the first half of the time its speed is 7.00 mph and in the second half of the time its speed is 52.00 mph. c) What is the total time in hours of the trip? Keep two decimal places.

Answers

Answer:

a) The total time of the trip is 4.05 h.

b) The average speed of the car is 12.35 mi/h.

c) The total time of the trip is 1.69 h.

Explanation:

Hi there!

a) The equation of traveled distance for a car traveling at constant speed is the following:

x= v · t

Where:

x = traveled distance.

v = velocity.

t = time.

Solving the equation for t, we can find the time it takes to travel a given distance "x" at a velocity "v":

x/v = t

So, the time it takes the car to travel the first half of the distance will be:

t1 = 25.0 mi / 7.00 mi/h

And for the second half of the distance:

t2= 25.0 mi / 52.00 mi / h

The total time will be:

total time = t1 + t2 = 25.0 mi / 7.00 mi/h + 25.0 mi / 52.00 mi / h

total time = 4.05 h

The total time of the trip is 4.05 h.

b) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it takes to travel that distance (t). In this case, the traveled distance is 50 mi and the time is 4.05 h. Then:

a.s = d/t

a.s = 50 mi / 4.05 h

a.s = 12.35 mi/h

The average speed of the car is 12.35 mi/h

c) Let's write the equations of traveled distance for both halves of the trip:

For the first half, you traveled a distance d1 in a time t1 at 7.00 mph:

7.00 mi/h = d1/t1

Solving for d1:

7.00 mi/h · t1 = d1

For the second half, you traveled a distance d2 in a time t2 at 52.00 mph.

52.00 mi/h = d2/t2

52.00 mi/h · t2 = d2

We know that d1 + d2 = 50 mi and that t1 and t2 are equal to t/2 where t is the total time:

d1 + d2 = 50 mi

52.00 mi/h · t/2 + 7.00 mi/h · t/2 = 50 mi

Solving for t:

29.5 mi/h · t = 50 mi

t = 50 mi / 29.5 mi/h

t = 1.69 h

The total time of the trip is 1.69 h.

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