6) The magnitude of the force the Sun exerts on Uranus is 1.41 x 1021 newtons. Explain how it is possible for the Sun to exert agreater force on Uranus than Neptune exerts on Uranus.

Answer:206.3

Explanation:

Explanation:

Kinetic energy is the energy by virtue of

object's motion whereas Thermal energy is

the internal energy of an object due to the

kinetic energy of its atoms.

On Increasing temperature, they both

increases

Answer:

The new gravitational force will be of 4.5 units

Explanation:

Recall that the formula for the gravitational force between two objects of masses m1 and m2 separated by a distance d is given by:

\(F_g=G\,\frac{m1*m2}{d^2}\)

in our case, we are told that such gives 72 units of force:

\(F_g=G\,\frac{m1*m2}{d^2} =72\)

Then we change the distance between the objects to 4 times the original (4 * d), such will produce a new gravitational force Fg':

\(F_g'=G\,\frac{m1*m2}{(4*d)^2} =G\,\frac{m1*m2}{16*d^2} = \frac{1}{16} *G\,\frac{m1*m2}{d^2}=\frac{1}{16} *\,72=4.5\)

Therefore the new gravitational force would be of 4.5 units

Answer:

125J

Explanation:

\(work \: = force \: \times distance \\ = 25 \times 5 \\ = 125joules\)

The two risk factors that can affect the ability of a child to learn in school is poor parenting and malnutrition.

A risk factor can be defined as any predisposing factor that can expose an individual to harm.

A risk factor that affects a child is an aspect of the child or environment that increases the probability of poor outcomes.

The two risk factors that can affect the ability of a child to learn in school include the following:

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Hi there!

We can use the following kinematic equation:

\(v_f^2 = v_i^2 + 2ad\)

vf = final velocity (? m/s)
vi = intial velocity (0 m/s)

a = acceleration (5 m/s²)

d = displacement (8 m)

Plug in the givens and solve.

\(v_f^2 = 0 + 2(5)(8)\\\\v_f = \sqrt{80} = \boxed{8.944 \frac{m}{s}}\)

Using the division operation, the number of tubes gotten by each patient can be obtained by dividing 20 by 5. Hence, each patient will get 4 tubes of solution.

Tubes of solution received :

Total tubes received :

Number of tubes of solution gotten by each patient :

Tubes of solution per patient = (20 ÷ 5) = 4 tubes

Therefore, each patient will get 4 tubes of solution.

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Since, density of the bracelet is not equal to the density of gold, then, the bracelet is not pure gold.

To know if the bracelet is pure gold, we calculate the density of the bracelet and compare it to the density of pure gold (19.3 g/cm³).

That is, for the bracelet to pure gold,

Density of bracelet ≈ 19.3 g/cm³

Density can be defined as the ratio of the mass and the volume of a substance.

The formula of Density is give as

⇒ Where:

From the question,

⇒ Given:

⇒ Substitute these values into equation 1

Hence, since the density of the bracelet is not equal to the density of gold, then, the bracelet is not pure gold.

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Answer:

a camera employs camera lens to firm some images.

Explanation:

hope this helps.

Answer:

Short-term regulation of blood pressure is controlled by the autonomic nervous system (ANS). Changes in blood pressure are detected by baroreceptors. These are located in the arch of the aorta and the carotid sinus. Increased arterial pressure stretches the wall of the blood vessel, triggering the baroreceptors.

hope this helps and can i have brainliest please.

*★,°*:.☆(￣▽￣)/$:*.°★* 。

Answer:

Thus, the velocity at the time of strike is same as the velocity at the time of projection.

Explanation:

Let a projectile is projected vertically upwards with a speed of u and reaches to the maximum height H.

At maximum height , the speed is zero and then the projective comes back on the ground.

Use the third equation of motion

\(v^2 = u^2 + 2 g h \\\\0 = u^2 - 2 g H\\\\\u =\sqrt{2gH}\)

Now let the velocity at the time of strike is v'.

Use third equation of motion, here initial velocity is zero.  

\(v'^2 = 0 + 2 g H \\\\v = \sqrt{2gH}\)

Thus, the velocity at the time of strike is same as the velocity at the time of projection.

An object in motion tends to stay in motion in an object in rest tends to stay in rest. "many years before newton wrote this law, Galileo's the same idea as the principle of inertia.

This is further explained below.

Generally, When an item experiences inertia, it keeps moving in the same direction or at the same pace until another force changes it.

The phrase "the principle of inertia" as it is used in Newton's first rule of motion is correctly interpreted as the word "inertia."

In conclusion,  The concept of inertia was first proposed by Galileo many years before Newton codified it as "an object in motion tends to remain in motion whereas an object in rest tends to stay in rest."

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Answer:

V = a t      velocity after time t

t = 33.1 / 9.80 = 3.38 sec   (time ball had been falling)

S = 1/2 a t^2 = 55.9 m

So the ball had been falling  for 7 * 8 = 56 m  

The child was 7 floors from the top

Since he was on the eight floor the floors below him were

7 * 8 + 12 = 68 m     total floors below child

68 + 56 = 124 m      total height of building

Total floors in building = 7 + 7 + 1 = 15 floors

PE at top = KE at bottom

KE = m g h = .125 * 9.80 * 124 = 152 Joules

The 99% confidence interval for the average maximum HP for the experimental engine is (610.12, 689.88).

To calculate the confidence interval for the experimental engines' average maximum HP, we can use the following formula:

To find the z-score for α=0.01, we can refer to a standard normal distribution table or use a calculator. The z-score is approximately 2.58.

Substituting the given values into the formula, we get:

CI = 650 ± 2.58*(60/√25) CI = 650 ± 30.96

Rounding to two decimal places, the confidence interval for the experimental engines' average maximum HP is:

CI = [619.04 HP, 680.96 HP]

Therefore, we can say with 99% confidence that the true average maximum HP for the experimental engines falls between 619.04 HP and 680.96 HP. Thus, we can conclude that the experimental engines' average maximum HP is likely to be within this range. However, note that this range does not include the manufacturer's claimed maximum HP of 630 HP, which may indicate that the engines are performing below expectations.

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Ellie's experience of a gender identity that doesn't align with her body is likely linked to gender dysphoria, a condition characterized by distress caused by the incongruence between one's internal sense of gender and assigned sex at birth.

The correct answer is option B.

Gender dysphoria refers to the distress or discomfort individuals may experience when their gender identity does not match the sex they were assigned at birth. It involves a deep-rooted sense of incongruence between one's internal sense of gender and the external physical characteristics.

It is important to note that gender dysphoria is not a result of misinformation or a lack of understanding. It is a genuine psychological condition recognized by medical and mental health professionals. Individuals with gender dysphoria often experience significant distress and may seek gender-affirming interventions, such as hormone therapy or gender-affirming surgeries, to align their physical appearance with their gender identity.

Misinformation, on the other hand, refers to inaccurate or misleading information, which may not directly relate to Ellie's experience. Construed body image and positive body image are also not directly linked to Ellie's situation. Construed body image refers to the way individuals perceive their own bodies, which may be influenced by various factors, while positive body image refers to a healthy and accepting attitude toward one's own body.

In summary, Ellie's experience of a gender identity that doesn't align with her body is likely linked to gender dysphoria. Gender dysphoria involves distress or discomfort resulting from a mismatch between an individual's internal sense of gender and their assigned sex at birth.

Therefore, among the options provided the correct answer is option B.

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Pressure on the bottom when water is added to fill the pipe to its top is 180kPA.

In the first scenario, the atmospheric pressure is:

In the second scenario, h = 8m + 1 m. Therefore, the pressure at the bottom of the barrel is:

What is hydrostatic pressure meaning?

Hydrostatic pressure refers to the pressure that any fluid in a confined space exerts. If the fluid is in a container, there will be some pressure on the wall of that container.

What is hydrostatic pressure used for?

Hydrostatic pressure is one reason (along with the lack of oxygen) why it's not safe for humans to travel unprotected in space. You rely on hydrostatic pressure to keep your lungs at the right inflation and to keep the water in your body from vaporizing.

How is hydrostatic pressure calculated?

The pressure in a liquid at a given depth is called hydrostatic pressure. This can be calculated using the hydrostatic equation:

P = rho * g * d, where P is the pressure, rho is the density of the liquid, g is gravity (9.8 m/s^2) and d is the depth (or height) of the liquid.

Thus, the pressure on the bottom when water is added to fill the pipe to its top is 180kPa.

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Answer:

pictures please

Explanation:

I need a picture so I can tell you

Explanation:

a. For constant acceleration:

v_avg = ½ (v + v₀)

v_avg = ½ (60 m/s + 15 m/s)

v_avg = 37.5 m/s

b. a = (v − v₀) / t

a = (60 m/s − 15 m/s) / 20 s

a = 2.25 m/s²

c. x = v_avg t

x = (37.5 m/s) (20 s)

x = 750 m

Answer:

The bag of sand

Explanation:

I think it is the bag of sand because according to the definition of impulse, impulse is the average force acting on a particule when an external force is being acted on it.

The minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.

To maintain equilibrium, the torques acting on the meter stick must balance each other. The torque is given by the formula:

τ = r * F * sin(θ)

where τ is the torque, r is the distance from the pivot point to the point where the force is applied, F is the force applied, and θ is the angle between the force vector and the lever arm.

In this case, the meter stick is in equilibrium when the torques on both sides of the pivot point cancel each other out. The torque due to the weight of the meter stick itself is acting at the center of mass of the meter stick, which is at the 50.0 cm mark.

Let's denote the mass to be placed on the 0.00 cm mark as M. The torque due to the weight of M can be calculated as:

τ_M = r_M * F_M * sin(θ)

where r_M is the distance from the pivot point to the 0.00 cm mark (which is 30.0 cm), F_M is the weight of M, and θ is the angle between the weight vector and the lever arm.

Since the system is in equilibrium, the torques on both sides of the pivot point must be equal:

τ_M = τ_stick

r_M * F_M * sin(θ) = r_stick * F_stick * sin(θ)

Substituting the given values:

30.0 cm * F_M = 20.0 cm * (0.0400 kg * 9.8 m/s^2)

Solving for F_M:

F_M = (20.0 cm / 30.0 cm) * (0.0400 kg * 9.8 m/s^2)

F_M = 0.0264 kg * 9.8 m/s^2

F_M = 0.25872 N

Finally, we can convert the force into mass using the formula:

F = m * g

0.25872 N = M * 9.8 m/s^2

M = 0.0264 kg

Therefore, the minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.

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Answer:

if the frequency is double, the wavelength is only half as long

Explanation:

Considering the definition of wavelength, frequency and propagation speed, if the wavelength is doubled, the frequency is reduced by half.

In a periodic wave the wavelength (λ) is the physical distance between two points from which the wave repeats itself. That is, the wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).

The frequency (f) is a measure of the number of cycles or repetitions of the wave per unit of time. Its unit is s⁻¹ or hertz (Hz).

The wavelength and its frequency are related from the speed at which the wave travels. The propagation speed (v) is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave's disturbance propagates along its displacement. So, the speed expression is:

v=λ×f

This indicates that the higher the frequency, the shorter the wavelength and the lower the frequency, the longer the wavelength.

All electromagnetic waves propagate in a vacuum at a constant speed of 300,000,000 m/s, the speed of light. Then, since an inversely proportional relationship is established between the frequency and the wavelength, if the wavelength is doubled, the frequency is reduced by half.

In summary, if the wavelength is doubled, the frequency is reduced by half.

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Answer: 6067.5 N

Explanation:

Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.

Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.

The velocity of particle 2 when it is 1cm away from particle 1 is determined as 62.18 m/s.

The electrostatic force between the two charges is calculated as follows;

F = (kq1q2)/(r²)

F = (9 x 10⁹ x 4.5 x 10⁻⁶ x 3.1 x 10⁻⁶)/(0.037)²

F = 91.71 N

F = ma

a = F/m

a = (91.71 N)/(0.0069 kg)

a = 13,291.2 m/s²

v² = u² + 2as

v² = 60² + 2(13,291.2)(0.01)

v² = 3865.82

v = 62.18 m/s

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Answer:

Answer

3.0/5

3

rubisnithi

Ace

413 answers

76.3K people helped

The fauna of the eastern Himalayas is similar to that of the southern Chinese and Southeast Asian region. Many of those species are primarily found in tropical forests and are only secondarily adapted to the subtropical, mountain, and temperate conditions prevailing at higher elevations and in the drier western areas. The animal life of the western Himalayas, however, has more affinities with that of the Mediterranean, Ethiopian, and Turkmenian regions. The past presence in the region of some African animals, such as giraffes and the hippopotamuses, can be inferred from fossil remains in deposits found in the Siwalik Range. The animal life at elevations above the tree line consists almost exclusively of cold-tolerant endemic species that evolved from the wildlife of the steppes after the uplift of the Himalayas. Elephants and rhinoceroses are restricted to parts of the forested Tarai region—moist or marshy areas, now largely drained—at the base of the low hills in southern Nepal. Asiatic black bears, clouded leopards, langurs (a long-tailed Asian monkey), and Himalayan goat antelopes (e.g., the tahr) are some of the denizens of the Himalayan forests. The Indian rhinoceros was once abundant throughout the foothill zone of the Himalayas but is now endangered, as is the musk deer; both species are dwindling, and few live, other than those in a handful of reserves set up to protect them. The Kashmir stag, or hangul, is near extinction.

Himalayan tahr

Himalayan tahr

Himalayan tahr (Hemitragus jemlahicus)

Arthur W. Ambler—The National Audubon Society Collection/Photo Researchers

In remote sections of the Himalayas, at higher elevations, snow leopards, brown bears, lesser pandas, and Tibetan yaks have limited populations. The yak has been domesticated and is used as a beast of burden in Ladakh. Above the tree line the most numerous animals, however, are diverse types of insects, spiders, and mites, which are the only animal forms that can live as high up as 20,700 feet (6,300 metres).

Himalayas: yak

Himalayas: yak

A yak in the Himalayas, Nepal.

© Galyna Andrushko/Fotolia

Fish of the genus Glyptothorax live in most of the Himalayan streams, and the Himalayan water shrew inhabits stream banks. Lizards of the genus Japalura are widely distributed. Typhlops, a genus of blind snake, is common in the eastern Himalayas. The butterflies of the Himalayas are extremely varied and beautiful, especially those in the genus Troides.

Bird life in the Himalayas is equally rich but is more abundant in the east than in the west. In Nepal alone almost 800 species have been observed. Among some of the common Himalayan birds are different species of magpies (including the black-rumped, the blue, and the racket-tailed), titmice, choughs (related to the jackdaw), whistling thrushes, and redstarts. A few strong fliers, such as the lammergeier (bearded vulture), the black-eared kite, and the Himalayan griffon (an Old World vulture), also can be seen. Snow partridges and Cornish choughs are found at elevations of 18,600 feet (5,700 metrers)

Explanation:

The intrepid hero has done 2.332 x Joules of work in pushing the crate.

To ascertain the work done by the traveler, we first need to find the power he applied on the case. As per Newton's subsequent regulation, force is equivalent to mass times speed increase, so the power applied by the traveler on the container is:

Force = mass x speed increase = 220 kg x 2  = 440 N

Then, we really want to work out the distance the case was moved. The pilgrim pushed the box a distance of 5.3 km, or 5,300 m.

At long last, we can compute the work done by the pioneer utilizing the equation:

Work = force x distance = 440 N x 5,300 m = 2.332 x 10^6 Joules

Thusly, the valiant legend has done 2.332 x  Joules of work in pushing the case.

The space pilgrim takes care of business on the case by applying a power that makes it speed up. The work done is equivalent to the power duplicated by the distance over which the power is applied. Involving the recipe for force, F=ma, and the given qualities for mass and speed increase, we can ascertain the power applied. Then, at that point, involving the recipe for work, W=Fd, and the given distance, we can ascertain the work done. The work done by the adventurer is 2.332 x  J.

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Answer:

d) 15.12 N

e) 15.12 N

Explanation:

Draw a free body diagram of the each block.

Block A has three forces on it: weight force mAg pulling down, normal force N pushing up, and tension force T pulling down.

Block B has two forces on it: weight force mBg pulling down, and tension force T pulling up.

Sum of forces on A in the y direction:

∑F = ma

T + N − mAg = mAa

N = mAa + mAg − T

N = mA (a + g) − T

Sum of forces on B in the y direction:

∑F = ma

T − mBg = mBa

T = mBa + mBg

T = mB (a + g)

Plug in values:

T = (1.80 kg) (-1.60 m/s² + 10 m/s²)

T = 15.12 N

N = (3.60 kg) (-1.60 m/s² + 10 m/s²) − 15.12 N

N = 15.12 N

So the answers to (d) and (e) are both 15.12 N.

Answer:

a) interior,  b) inside,   c) minor

Explanation:

In this exercise you are asked to select the correct words so that the statements have been correct

Electric charges always repel each other when they are of the same sign, in conductors this has the consequence that charges accumulate on the surface and the interior remains without electric charges.  with this we analyze the statements

a) interior

b) inside

c) minor

therefore the phrase would be:

(a) The net charge is always zero ---INTERIOR--- the surface of an isolated conductor.

(b) The electric field is always zero ---INSIDE--- a perfect conductor.

(c) The charge density on the surface of an isolated, charged conductor is highest where the surface is ---MINOR---to)

Answer:  Therefore, the change in entropy when 536 g of gold are melted is 0.132 J/K.

Explanation:  To calculate the change in entropy when 536 g of gold are melted, we need to know the entropy of fusion of gold and the temperature at which it melts.

The entropy of fusion of gold is 2.35 J/g·K, and the melting point of gold is 1064 °C or 1337 K.

The change in entropy when gold is melted can be calculated using the formula:

ΔS = Q/T

where ΔS is the change in entropy, Q is the heat absorbed during the process, and T is the temperature at which the process occurs.

The heat absorbed when gold is melted can be calculated using the formula:

Q = m × ΔH_fus

where m is the mass of the gold and ΔH_fus is the enthalpy of fusion of gold, which is 64.9 kJ/mol.

Converting the mass of gold to moles:

536 g / 196.97 g/mol = 2.72 mol

The heat absorbed by the gold when it is melted is:

Q = 2.72 mol × 64.9 kJ/mol = 176.2 kJ

Finally, we can calculate the change in entropy:

ΔS = Q/T = 176.2 kJ / 1337 K = 0.132 J/K