Suppose you go outside and look at three stars. Star A is blue, Star B is white, and Star C is red. Which star is the hottest and which star is the coldest?

Answer:

The weight of the truck is 22072.5 N.

Explanation:

Given that,

Width of the brick =4.3 m

Length of the brick = 8.5 m

Distance = 6.15 cm

We need to calculate the volume of the water displaced by the boat with truck

Using formula of volume

[tex]V=l\times w\times d[/tex]

Put the value into the formula

[tex]V=8.5\times4.3\times6.15\times10^{-2}[/tex]

[tex]V=2.25\ m^3[/tex]

We need to calculate the mass of the water displaced by the boat with truck

Using formula of density

[tex]\rho=\dfrac{m}{V}[/tex]

[tex]m=\rho\times V[/tex]

Put the value into the formula

[tex]m=1000\times2.25[/tex]

[tex]m=2250\ kg[/tex]

We need to calculate the weight of the truck

Using formula of weight

[tex]W= mg[/tex]

Put the value into the formula

[tex]W=2250\times9.81[/tex]

[tex]W=22072.5\ N[/tex]

Hence, The weight of the truck is 22072.5 N.

Answer:

[tex]\theta=44.03^{o}[/tex]

Explanation:  

Here we have an inelastic collision problem. We can use the momentum (p = mv) conservation law in each component of the displacement.

So, [tex]p_{i}=p_{f}[/tex]

X-component:

[tex]m_{1}v_{i1x}+m_{2}v_{i2x}=m_{1}v_{f1x}+m_{2}v_{f2x}[/tex] (1)

Now,

Then, using this information we can rewrite the equation (1).

[tex]m_{2}v_{i2x}=v_{fx}(m_{1}+m_{2})[/tex]

[tex]v_{fx}=\frac{m_{2}v_{i2x}}{m_{1}+m_{2}}=\frac{2597.2*164}{1720+2597.2}[/tex]

[tex]v_{fx}=98.66 km/h[/tex]

Y-component:

[tex]m_{1}v_{i1y}+m_{2}v_{i2y}=m_{1}v_{f1y}+m_{2}v_{f2y}[/tex] (2)

We can do the same but with the next conditions:

Then, using this information we can rewrite the equation (2).

[tex]m_{1}v_{i1y}=v_{fy}(m_{1}+m_{2})[/tex]

[tex]v_{fy}=\frac{m_{1}v_{i1y}}{m_{1}+m_{2}}=\frac{1720*239.44}{1720+2597.2}[/tex]

[tex]v_{fy}=95.39 km/h[/tex]

Now, as we have both components of the final velocity, we can find the angle East of North. Using trigonometric functions, we have:

[tex]tan(\theta)=\frac{v_{y}}{v_{x}}[/tex]

[tex]\theta=arctan(\frac{v_{y}}{v_{x}})=arctan(\frac{95.39}{98.66})[/tex]

[tex]\theta=44.03^{o}[/tex]

I hope it helps you!

Answer: The balance of care is sympathetic and loving

Explanation: According to Erikson's theory, when the balance of care is sympathetic and loving the psychological conflict of the first year is resolved on the positive side.

According to Erikson's theory of psychosocial development, the psychological conflict of the first year is resolved on the positive side when the infant develops a sense of trust in their caregivers.

According to Erik Erikson's theory of psychosocial development, when an infant develops a sense of trust in their caregivers during the first year of life, the psychological conflict of that stage is resolved on the positive side. Erickson named this first stage trust versus mistrust. In this stage, caregivers who are responsive and sensitive to their infant's needs help the child to develop a sense of trust; thus, the child will see the world as a safe, predictable place. If the caregivers are unresponsive and do not meet the infant's needs, feelings of anxiety, fear, and mistrust emerge; the child may see the world as unpredictable.

Successful completion of each developmental task at various stages results in a sense of competence and a healthy personality, according to Erikson. Failure to master these tasks may lead to feelings of inadequacy and affect the formation of a positive self-concept.

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Answer:

The kinetic energy when the film vault landed is 12744000J.

Explanation:

The kinetic energy is defined as:

[tex]k_{e} = \frac{1}{2}mv^{2}[/tex]  (1)

Where m is the mass and v is the velocity.

By means of equation 1, the kinetic energy of the film vault when it landed can be determined

[tex]k_{e} = \frac{1}{2}(1770kg)(120m/s)^{2}[/tex]  

But [tex]1 J = kg.m^{2}/s^{2}[/tex]

[tex]k_{e} = 12744000J[/tex]

Hence, the kinetic energy when the film vault landed is 12744000J.

The kinetic energy of the film vault when it landed was 1,919,200 Joules.

The kinetic energy of the film vault can be calculated

The kinetic energy of the film vault can be calculated using the formula: KE = 0.5 * mass * velocity^2. Given that the mass of the film vault is 1770 kg and its velocity is 120 m/s, we can substitute these values into the formula:

KE = 0.5 * 1770 kg * (120 m/s)^2 = 0.5 * 1770 * 14400 = 1,919,200 J

Therefore, the kinetic energy of the film vault when it landed was 1,919,200 Joules. using the formula: b mass ity^2. Given that the mass of the film vault is 1770 kg and its velocity is 120 m/s, we can substitute these values into the formula:

KE = 0.5 * 1770 kg * (120 m/s)^2 = 0.5 * 1770 * 14400 = 1,919,200 J

Therefore, the kinetic energy of the film vault when it landed was 1,919,200 Joules.

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Answer:

Time period, T = 0.5 seconds

Explanation:

Given that,

One complete wave passes a putrefying pile of poached pickerel in 0.50 seconds, T = 0.5 s

The piled pickerel protrude 20 cm along the beach, A = 20 cm

We need to find the period of the wave. The time period of the wave is defined as the time taken by the wave to complete one oscillation. So, the time period of the wave will be 0.5 seconds.

Hence, this is the required solution.      

Answer:

I₁ > I₃ > I₂

Explanation:

Taking the pic shown, we have

m₁ = 10m₀

m₂ = 2m₀

m₃ = m₀

r₁ = r₀

r₂ = 2r₀

r₃ = 3r₀

We apply the formula

I = mr²

then

I₁ = m₁r₁² = (10m₀)(r₀)² = 10m₀r₀²

I₂ = m₂r₂² = (2m₀)(2r₀)² = 8m₀r₀²

I₃ = m₃r₃² = (m₀)(3r₀)² = 9m₀r₀²

finally we have

I₁ > I₃ > I₂

Final answer:

To rank the objects according to their moments of inertia, calculate I = mr² for each, where m is mass and r is the distance from the axis. Rank from largest to smallest moment of inertia value.

Explanation:

The student's question involves ranking objects according to their moments of inertia, which requires an understanding of rotational motion in physics. The moment of inertia (I) of an object is calculated as I = Σmr², where m represents the mass of the object and r the perpendicular distance from the object to the axis of rotation. This equation states that for point masses, the moment of inertia is a product of the mass and the square of its distance from the rotation axis.

The student would need to calculate the moment of inertia for each object by multiplying its given mass by the square of its specified distance from the rotation axis (I = mr² for point masses). The objects should be ranked from the highest value of the moment of inertia to the lowest.

The concepts of rotational inertia and the parallel-axis theorem may also be relevant if the objects are not point masses, but the question provided seems to simplify the scenario to treat the masses as points. Additionally, the question posed hints that integration could be required for more complex shapes, but for the simple case of point masses or uniformly distributed mass like a hoop or rod, the formulas are straightforward.

Answer:

V = 145 cm³

Explanation:

given,

mass of the solid = 2.8 Kg

                             = 2800 g

density of the gold = 19.3 g/cm³

volume of the water displaced = ?

volume of water displaced is equal to the volume of statue.

Volume of statue = ?

[tex]Volume = \dfrac{mass}{density}[/tex]

[tex]Volume = \dfrac{2.8\times 1000}{19.3}[/tex]

        V = 145 cm³

volume of the water displaced will be equal to 145 cm³

The volume of water displaced by the 2.8 kg solid gold statue to verify its purity should be 145.08 cm3, considering the density of pure gold is 19.3 g/cm3.

To determine whether a 2.8 kg solid gold statue is made of pure gold using its density, we need to calculate the volume of gold that mass would occupy and compare it with the volume of water displaced. Since density (d) is mass divided by volume, we can rearrange this formula to solve for volume (V) where V = m/d. First, we need to convert the mass of the gold statue from kilograms to grams because the density of gold is given in grams per cubic centimeter (19.3 g/cm3).

2.8 kg = 2,800 g. Using the density of gold, the volume (in cm3) the statue should displace if it is pure gold is calculated as follows:

Volume of gold (V) = Mass of gold (m) / Density of gold (d)

V = 2,800 g / 19.3 g/cm3

V = 145.08 cm3 (rounded to two decimal places)

In conclusion, to verify that the statue is pure gold, the volume of water displaced should be 145.08 cm3. If the volume of displaced water matches this calculation, it suggests that the statue is indeed made of pure gold.

Answer:

The vasa vasorum, the collection of small arteries and veins supplying blood to the smooth muscle and fibroblasts within blood vessels, is found in the tunica externa.

Explanation:

In order to understand the answer above let explain the concept of tunica externa

This in Latin means outer coat and can also be called 'tunica adventitia' which also means "additional coat in Lain",it is the outermost layer of the a blood vessel covering the tunica media(This is a Latin word  translated as middle coat in English which is the middle layer of an artery or a vein it lies between the tunica externa  and the tunica initima which is the innermost part of the artery or vein).

An Artery

We can define an artery as a blood vessel that transports blood from the heart to other parts of the body.

An Vein

We can define a vein as a blood vessel that transports blood to the heart

and we can define  blood vessel as a tube that carries blood in the circulatory system.  

Explanation:

The given statement is absolutely true.  this is because magnitude of a vector is always non negative, it can not be zero unless its a zero vector. So, in the given question, final position vector of a moving object has a smaller magnitude than the initial position vector, so, magnitude is neither zero nor negative. Hence, it has a positive magnitude.

Final answer:

The displacement of an object is the final position vector minus the initial position vector and is a vector with both magnitude and direction. The magnitude of displacement does not have to be positive if the final position vector has smaller magnitude than the initial; it depends on the direction relative to the chosen coordinate system.

Explanation:

When discussing the displacement of an object, it is important to understand that it refers to the change in the object's position and is defined as the final position vector minus the initial position vector. Displacement, being a vector quantity, has both magnitude and direction. If the magnitude of the final position vector is smaller than the magnitude of the initial position vector, this does not necessitate that the magnitude of the displacement is positive. In fact, the magnitude of the displacement vector could be either positive or negative based on the assigned coordinate system.

In a one-dimensional coordinate system, the direction of motion can be designated as positive or negative. For example, if rightward motion is deemed positive, then a leftward motion would be negative. Similarly, in vertical motion, upward movement is often taken as positive and downward as negative, though these conventions can be reversed based on the scenario and convenience. The direction of the displacement vector essentially depends on the chosen orientation of the positive direction.

Therefore, if the final position vector has a smaller magnitude than the initial position vector, and if the positive direction is assigned towards the final position, the displacement vector magnitude could indeed be negative, not necessarily positive, as it signifies a change in position opposite to the assigned positive direction.

Answer:

Height will be equal to 0.4489 m

Explanation:

We have given mass of the toy m = 0.50 kg

Initial kinetic energy K = 2.2 J

We have to fond the height of the hill over which car roast

When car will roast the hill its kinetic energy will be converted into potential energy and at maximum height all kinetic energy will be converted into potential energy

So at maximum height [tex]mgh=K[/tex]

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

So 0.50×9.8×h = 2.2

h = 0.4489 m

Answer:

(A) t_H = Vo×Sin (theta)/g

(B) t,_R = (2VoSin(theta))/g

(C) H = Vo²Sin²(theta) / 2g

(D) Range = Vo²Sin(2×theta) / g.

Explanation:

The detailed solution to the problem can be found in the attachment below

To find the time it takes for the projectile to reach its maximum height, use the equation t_H = v_0 sin(θ) / g. To find the time the projectile hits the ground, use the equation t_R = 2v_0 sin(θ) / g. The maximum height attained by the projectile can be found using the equation H = (v_0^2 sin^2(θ)) / (2g). The total distance traveled in the x-direction (range) can be found using the equation R = (v_0^2 sin(2θ)) / g.

To find the time it takes for the projectile to reach its maximum height, we can use the equation for vertical motion:

t_H = v_0 sin(\theta) / g

To find the time the projectile hits the ground, we can use the equation for vertical motion:

t_R = 2v_0 sin(\theta) / g

The maximum height attained by the projectile can be found using the equation:

H = (v_0^2 sin^2(\theta)) / (2g)

The total distance traveled in the x-direction (range) can be found using the equation:

R = (v_0^2 sin(2\theta)) / g

Answer:

0.15756 m/s

Explanation:

There are 10 blades and 10 gaps

To move through one blade or gap the windmill has to rotate

[tex]\dfrac{2\pi}{20}=0.31415\ rad[/tex]

This divided by the angular velocity is gives us the time

[tex]\dfrac{0.31415}{1.1}=0.28559\ s[/tex]

When the ball moves it does in a way that the ball must travel a distance of its own diameter which is [tex]4.5\times 10^{-2}\ m[/tex]

[tex]Speed=\dfrac{Distance}{Time}[/tex]

[tex]v=\dfrac{4.5\times 10^{-2}}{0.28559}\\\Rightarrow v=0.15756\ m/s[/tex]

The minimum linear speed is 0.15756 m/s

Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 8.5 m/s

     Final velocity, v = 0 m/s    - At maximum height

     Time, t = ?

     Acceleration , a = -9.81 m/s²

     Substituting

                      v = u + at  

                      0 = 8.5 + -9.81 x t

                      t = 0.867 s

  It takes 0.867 seconds to get to the top of its motion

Answer:

0.87 s

Explanation:

initial velocity, u = 8.5 m/s

Let it takes time t to reach to maximum height. At maximum height the velocity is zero, so, v = 0

Use first equation of motion

v = u - gt

where, g be the acceleration due to gravity

0 = 8.5 - 9.8 t

t = 0.87 s

Thus, the time taken to reach at top is 0.87 s.

Answer:

[tex]F_a=1470\ N[/tex]

Explanation:

Friction Force

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.  

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

[tex]\displaystyle F_a-F_{r1}-F_{r2}=m.a=0[/tex]

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

[tex]\displaystyle F_a=F_{r1}+F_{r2}.....[1][/tex]

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

[tex]\displaystyle F_{r2}-T=0[/tex]

The friction forces are computed by

[tex]\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g[/tex]

[tex]\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g[/tex]

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

[tex]\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g[/tex]

Simplifying

[tex]\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)[/tex]

Plugging in the values

[tex]\displaystyle F_{a}=0.25(9.8)[400+2(100)][/tex]

[tex]\boxed{F_a=1470\ N}[/tex]

Final answer:

To make the 100 kg block slip, a horizontal force of at least 245 N must be applied to the 400 kg lower block, calculated based on the maximum static friction force determined by the coefficient of friction and the weight of the 100 kg block.

Explanation:

To find the horizontal force F applied to the lower block that is necessary for the 100 kg block (upper block) to start slipping, we need to calculate the force required to overcome the static friction between the 100 kg block and the 400 kg block. The coefficient of friction (μ) is given as 0.25 for both contacts.

First, calculate the maximum static friction force that can act on the 100 kg block:

The weight of the 100 kg block (W) = 100 kg × 9.8 m/s² = 980 N.

Maximum static friction force (Ffriction) = μ × Normal force = 0.25 × 980 N = 245 N.

To initiate slipping, the horizontal force (F) applied must at least be equal to the maximum static friction force, which is 245 N. However, since this force will be applied to the combined system of both blocks, we must consider the total mass involved when finding the acceleration caused by the applied force.

The total mass of the system is 500 kg (400 kg + 100 kg). To move this mass with an acceleration that would cause the upper block to slip, we calculate:

F = ma, where m is the total mass and a is the acceleration. Since the force to overcome static friction is 245 N, this gives us the minimum force needed to initiate slipping when applied to the 400 kg block.

Answer:

Change in potential energy, [tex]U=1.2\times 10^8\ J[/tex]

Explanation:

Given that,

The potential difference between a storm cloud and the ground is 100 million V, [tex]V=100\ million V=100\times 10^6\ V=10^8\ V[/tex]

If a charge of 1.2 C flashes in a bolt from cloud to Earth, q = 1.2 C

We need to find the change of potential energy of the charge. The relation between the potential difference and the potential energy of the charge is given by :

[tex]U=qV[/tex]

U is the potential energy of the charge

[tex]U=1.2\times 10^8\ J[/tex]

So, the change of potential energy of the charge is [tex]U=1.2\times 10^8\ J[/tex]. Hence, this is the required solution.

According to the question,

                                                = 100×10⁶ V

                                                = 10⁸ V

The change in Potential energy:

→ [tex]U = qV[/tex]

By substituting the values, we get

      [tex]= 1.2\times 10^{8} \ J[/tex]

Thus the above response is right.

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Answer:

46 days

Explanation:

It is a very logical question which does not require any mathematical calculation rather requires thinking.

It is said in the question every day, the patch doubles in size. if it takes 48 days for the patch to cover the entire lake.

Let total area of the lake be 100 m^2. The at the end of  48th day, it would have covered 100 m^2. So, at the end of 47th day it would have covered only 50 m^2, since every day, the patch doubles in size. Similarly on 46th day it would have covered 25 m^2, which is quarter of 100 m^2.

So ,it would take 46 days for the patch to cover quarter the lake.

Answer:

Angle at which object is launched is 17.15°

Explanation:

We have given initial velocity at which object is launched u = 15 m/sec

It rises to a height of 1 m

So height h = 1 m

We have to find the angle of projection [tex]\Theta[/tex]

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

We know that maximum height is given by [tex]h=\frac{u^2sin^2\Theta }{2g}[/tex]

So [tex]1=\frac{15^2\times\ sin^2\Theta }{2\times 9.8}[/tex]

[tex]sin^2\Theta =0.0871[/tex]

[tex]sin\Theta =0.295[/tex]

[tex]\Theta =sin^{-1}0.295=17.15^{\circ}[/tex]

So the angle at which object is launched is 17.15°

Answer:

b. is the process in which two smaller nuclei combine to form a larger nucleus

Answer: 16N

Explanation:

According to Coulomb's law, the force existing between two charges is directly proportional to the product of their charges and inversely proportional to the distance between the charges. Mathematically,

F = kq1q2/r²

Where F is the force between them

q1 and q2 are the charges

r is the distance between them.

If the force between them is 10N, the formula becomes;

10 = kq1q2/r²... (1)

If the distance between them is now halved, the force will become

F = kq1q2/(r/2)²

F = kq1q2/(r²/4)

F = 4kq1q2/r² ... (2)

Dividing equation 1 by 2 we have;

4/F = (kq1q2/r²)÷ (4kq1q2/r²)

4/F = kq1q2/r²×r²/4k1q1q2

4/F = 1/4

Cross multiplying we have;

F = 4×4

F = 16N

Therefore the new force between the charges is 16N

Answer:

option B

Explanation:

given,

Satellite B has an orbital radius nine times that of satellite A.

R' = 9 R

now, orbital velocity of the satellite A

[tex]v_a =\sqrt{\dfrac{GM}{R}}[/tex]........(1)

now, orbital velocity of satellite B

[tex]v_b=\sqrt{\dfrac{GM}{R'}}[/tex]

[tex]v_b=\sqrt{\dfrac{GM}{9R}}[/tex]

[tex]v_b=\dfrac{1}{3}\sqrt{\dfrac{GM}{R}}[/tex]

from equation 1

[tex]v_b=\dfrac{v_a}{3}[/tex]

hence, the correct answer is option B

The speed of Satellite B is one-third of the speed of Satellite A (vA/3), based on Kepler's laws of planetary motion and centripetal force requirements for circular orbits. So the correct option is B.

The question asks for the speed of Satellite B assuming that Satellite A has speed vA and that the orbital radius of Satellite B is nine times that of Satellite A. To find the speed of Satellite B, we can use Kepler's third law and the fact that the centripetal force required for circular motion is provided by the gravitational force.

The orbital speed equation reveals a relation v ≈ r^-1/2, meaning that if the radius increases, the speed decreases. Specifically, if the radius becomes nine times larger, then the speed will be the square root of 1/9 times the initial speed, which is 1/3. So, the correct answer is that the speed of Satellite B is Satellite A's speed divided by 3, which is vA/3.

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Answer:

Moment of inertia of the flywheel, [tex]I=1.82\ kg-m^2[/tex]          

Explanation:

Given that,

The maximum energy stored on  flywheel, [tex]E=4\ MJ=4\times 10^6\ J[/tex]

Angular velocity of the flywheel, [tex]\omega=20000\ rev/s=2094.39\ rad/s[/tex]

We need to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

[tex]E=\dfrac{1}{2}I\omega^2[/tex]

I is the moment of inertia of the flywheel

On rearranging we get :

[tex]I=\dfrac{2E}{\omega^2}[/tex]

[tex]I=\dfrac{2\times 4\times 10^6}{(2094.39)^2}[/tex]

[tex]I=1.82\ kg-m^2[/tex]

So, the moment of inertia of the flywheel is [tex]I=1.82\ kg-m^2[/tex]. Hence, this is the required solution.

The moment of inertia of the flywheel [tex]I= 1.82\ kg-m^2[/tex]

It is given that,

The maximum energy stored on the flywheel is given as

E=4 MJ=[tex]4\times 10^6\ J[/tex]

Angular velocity of the flywheel is

[tex]w=20000\ \dfrac{Rev}{Sec} =2094.39 \frac{rad}{sec}[/tex]

So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

[tex]E= \dfrac{1}{2} Iw^2[/tex]

By rearranging the equation

[tex]I=\dfrac{2E}{w^2}[/tex]

[tex]I=1.82\ kg-m^2[/tex]

Thus the moment of inertia of the flywheel [tex]I= 1.82\ kg-m^2[/tex]

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Answer:

69.74 N

Explanation:

We are given that

Weight of sled=49 N

Coefficient of kinetic friction[tex],\mu_k=0.11[/tex]

Weight of person=585 N

Total weight==mg=49+585=634 N

We know that

Force needed to pull the sled across the snow at constant speed,F=Kinetic friction

[tex]F=\mu_k N[/tex]

Where N= Normal=mg

[tex]F=0.11\times 634=69.74 N[/tex]

Hence, the force is needed  to pull the sled across the snow at constant speed=69.74 N

Answer:

U₂=45 kJ

Explanation:

Given Data

Initial Energy U₁=19.5 kJ

Qin=30 kJ

Win=500 N.m

Qout= 5 kJ

To find

Final energy U₂

Solution

From The first law of thermodynamic

ΔE = Ein - Eout + ΔQ -ΔW

or

U₂=U₁+Qin+Win-Qout

where

U₂ is final energy

U₁ is initial energy

q = energy transferred as heat to a system

w = work done on a system

U₂=(19.5+30+500×10⁻³-5) kJ  

U₂=45 kJ

Final answer:

To calculate the final energy of a system involving heat transfer and work, first, find the net heat transfer by considering heat added and lost, and then include the work done. The first law of thermodynamics is applied to combine these energies with the initial energy to find the final energy, which is 45 kJ.

Explanation:

The question involves calculating the final energy of a system after heat transfer and work done on a closed pan of water on a stove. Given that 30 kJ of heat is transferred to the water, 5 kJ is lost to the surrounding air, and the paddle-wheel work on the system is 500 N·m (or 0.5 kJ since 1 N·m = 1 J), we can use the first law of thermodynamics to find the final energy of the system. The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

First, we calculate the net heat transfer (Q) to the system by subtracting the heat lost to the air from the heat supplied: Q = 30 kJ - 5 kJ = 25 kJ. Then, we add the work done on the system (which is positive because the work is done on the system, not by it): 25 kJ + 0.5 kJ = 25.5 kJ. This is the net energy input into the system.

Finally, to find the final energy of the system, we add this net input to the initial energy of the system: 19.5 kJ (initial energy) + 25.5 kJ (net input) = 45 kJ. Therefore, the final energy of the system is 45 kJ.

This observation is not consistent with a Sun-centered model, because in this model "the rise and set of all objects depends only on Earth’s rotation".

Answer: Option D

Explanation:

Copernican heliocentrism, the astronomical model's title which was created by Nicolaus Copernicus and released in 1543. This model placed the Sun in motionless position near the Universe's center, with Earth and the planets of other which orbiting it in a circular orbits, altered by epicycles, and also at constant speeds.

Out of several planets in a determined order the Earth orbits around a stationary sun having three motions like annual revolution, daily rotation, and its axis is tilted annually. The planets' retrograde motion is understood by motion from the Earth.

Answer:

λ = 502 n m

Explanation:

given,

slit separation,  d = 0.040 mm

                             = 4 x 10⁻⁵ m

Distance between the fringes, D = 4.70 m

distance between the fringes, x = 5.9 cm

                                                     = 0.059 m

wavelength of the light = ?

the separation between the fringe is given by

[tex]x = \dfrac{n\lambda\ D}{d}[/tex]

[tex]\lambda = \dfrac{x d}{n\ D}[/tex]

where as n = 1

[tex]\lambda = \dfrac{0.059\times 4\times 10^{-5}}{1\times 4.70}[/tex]

 λ = 5.02127 x 10⁻⁷ m

λ = 502 n m

hence, the wavelength of the light is equal to λ = 502 n m

Answer

given,

Force on car, F = 300 N

time, t = 0.250 s

distance of the force from the pivot, r= 0.3 m

Angular momentum = ?

Now,

torque acting is calculated by multiplying force with displacement.

    [tex]\tau = F \times r [/tex]

    [tex]\tau = 300 \times 0.3 [/tex]

    [tex]\tau = 90\ N.m[/tex]

we know,

torque is equal to change in angular momentum per unit time.

[tex]\tau = \dfrac{\Delta L}{\Delta t}[/tex]

Δ L = τ Δ t

initial angular  momentum is zero

L = 90 x 0.25

L = 22.5 kg.m²/s

Angular momentum is equal to 22.5 kg.m²/s

Answer:

Explanation:

Here is the full question:

Two neutron stars are separated by a distance of 1.0 × 1010 m. They each have a mass of 1.0 × 1030 kg and a radius of 1.0 × 105 m. They are initially at rest with respect to each other. As measured from that rest frame, how fast are they moving when (a) their separation has decreased to one-half its initial value and (b) they are about to collide?

solution:

G is the gravitational constant, the value is,

[tex]G=6.67\times 10^{-11}\frac{m^3}{kg.s^2}[/tex]

For half distance is,

[tex]U'=2U\\\\-dU=\frac{-2Gm^2}{R}\\\\dKE=-dU[/tex]

a)

If the sepperation confined to one-half its initial value,

The velocity is,

[tex]K=\frac{Gm^2}{2R}\\\\\frac{1}{2}mv^2=\frac{Gm^2}{2R}\\\\v=\sqrt{\frac{Gm}{R}}\\\\=\sqrt{\frac{(6.67\times 10^{-11}\frac{m^3}{kg.s^2})(10^{30}kg)}{(10^{10}m)}}\\\\=81,670m/s\\\\v=8.2\times 10^4m/s[/tex]

b)

[tex]dU=Gm^2(\frac{1}{R}-\frac{1}{2r})\\\\dKE=Gm^2(\frac{1}{2r}-\frac{1}{R})=mv^2\\\\\therefore dKE=-dU\\\\v=\sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}\\\\\sqrt{(6.67\times 10^{-11}\frac{m^3}{kg.s^2})(10^{30}kg))(\frac{1}{2(10^5m)}-\frac{1}{(10^{10}m)})}\\\\v=1.8\times 10^7m/s[/tex]

Answer:

The magnitude of the average force exerted on the water by the blade is 960 N.

Explanation:

Given that,

The mass of water per second that strikes the blade is, [tex]\dfrac{m}{t}=30\ kg/s[/tex]

Initial speed of the oncoming stream, u = 16 m/s

Final speed of the outgoing water stream, v = -16 m/s

We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :

[tex]F=\dfrac{\Delta P}{\Delta t}[/tex]

[tex]F=\dfrac{m(v-u)}{\Delta t}[/tex]

[tex]F=30\ kg/s\times (-16-16)\ m/s[/tex]

F = -960 N

So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.