Adding or removing protons from an atom does what to the atom

Answers

Answer 1

Answer:

Changes the element of the atom.

Explanation:

The elements are classified by the number of protons they have in their nucleus, so if the number of protons is changed (added or removed), that atom will become one of a different element.

For example, hydrogen has only 1 proton in its nucleus, and helium has 2. So if a proton is added to a hydrogen atom, it becomes a helium atom, and consequently its atomic number, wich is determided by the protons in an element, will also change.


Related Questions

A 531.7-W space heater is designed for operation in Germany, where household electrical outlets supply 230 V (rms) service. What is the power output of the heater when plugged into a 120-V (rms) electrical outlet in a house in the United States? Ignore the effects of temperature on the heater's resistance.

Answers

Answer:

P=144.74W

Explanation:

We can model the power output of a resistance by using the following formula:

[tex]P=\frac{V^{2}}{R}[/tex]

Wehre P is the power output, V is the rms voltage and R is the resistance. The resistance of the space heater will remain the same, so we can calculate it from the power output in Germany and its rms voltage. So when solving for R, we get:

[tex]R=\frac{V^{2}}{P}[/tex]

and we can now use the provided data:

[tex]R=\frac{(230V)^{2}}{531.7W}[/tex]

which yields:

R= 99.49 Ω

Once we know what the heater's resistance is, we can now go ahead and calculate the power outpor of the heater in the U.S.

[tex]P=\frac{V^{2}}{R}[/tex]

so

[tex]P=\frac{(120V)^{2}}{99.49\Omega}[/tex]

P=144.74W

According to the law of reflection, which statement must be true? The measure of angle A equals the measure of angle B. The measure of angle B equals the measure of angle C. The measure of angle A equals the measure of angle C. The measure of angle B equals the measure of angle D.

Answers

Answer:

The measure of angle B equals the measure of angle C.

Explanation:

When a ray of light falls on a smooth polished surface and the light ray bounces back, it is called the reflection of light.

There are two laws of reflection;

1) The incident ray and reflected ray and normal all lie in the same plane.

2) The angle of incidence is equal to angle of reflection.

Here angle of incidence is represented by B and angle of reflection is by C.

So according to law of reflection the measure of angle B equals the measure of angle C.

Answer:

The measure of angle B equals the measure of angle C

Explanation:

"B" is the correct answer on edge

Explain why atomic radius decreases as you move to the right across a period for main-group elements but not for transition elements.Match the words in the left column to the appropriate blanks in the sentences on the right.

Answers

Answer:

Explained.

Explanation:

Only the first question has been answered

In a period from left to right the nuclear charge increases and hence nucleus size is compressed. Thus,  atomic radius decreases.

In transition elements, electrons in ns^2 orbital remain same which is the outer most orbital having 2 electrons and the electrons are added to (n-1) d orbital. So, outer orbital electron experience almost same nuclear attraction and thus size remains constant.

Final answer:

The atomic radius of main-group elements decreases as you move to the right across a period due to increased positive charge, while the atomic radius of transition elements remains relatively constant.

Explanation:

The atomic radius of main-group elements decreases as you move to the right across a period because the number of protons in the nucleus increases. This increased positive charge pulls the electrons closer to the nucleus, reducing the size of the atom. In contrast, the atomic radius of transition elements remains relatively constant as you move across a period because their outermost electrons are in different energy levels or subshells. The addition of protons does not significantly affect the size of the atom.

Moving across a period, the number of protons in the nucleus increases, leading to a greater positive charge. This increased positive charge exerts a stronger pull on the electrons, pulling them closer to the nucleus and resulting in a smaller atomic radius.

However, electron shielding, or the repulsion between electrons in different energy levels, also plays a role. As you move across a period, the number of electrons in the same energy level (shell) remains constant, providing consistent shielding effects. This partial counteraction to the increased positive charge contributes to the overall trend of decreasing atomic radius.

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A 3,000-kg truck traveling 8 m/s collides with a 500-kg car that is at rest. After the collision, the car is traveling at 10 m/s. How fast will the truck be moving?

Answers

The final velocity of the truck is 6.33 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: the total momentum of the truck-car system must be conserved before and after the collision (if there are no external forces), so we can write

[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]

where:

[tex]m_1 = 3000 kg[/tex] is the mass of the truck

[tex]u_1 = 8 m/s[/tex] is the initial velocity of the truck

[tex]v_1[/tex] is the final velocity of the truck

[tex]m_2 = 500 kg[/tex] is the mass of the car

[tex]u_2 = 0[/tex] is the initial velocity of the car

[tex]v_2 = 10 m/s[/tex] is the final velocity of the car

And by solving the equation for [tex]v_1[/tex], we find the velocity of the truck after the collision:

[tex]v_1 = \frac{m_1 u_1-m_2 v_2}{m_1}=\frac{(3000)(8)-(500)(10)}{3000}=6.33 m/s[/tex]

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A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 3.0 N/cm. The block becomes attached to the spring and compresses the spring 11 cm before momentarily stopping.

(a) While the spring is being compressed, what work is done on the block by the gravitational force on it?
(b) What work is done on the block by the spring force while the spring is being compressed?
(c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)
(d) If the speed at impact is doubled, what is the maximum compression of the spring?

Answers

Final answer:

The work done by gravity on a block dropped onto a spring is 0.22638 J, the spring does 1.815 J of work compressing, and if the speed at impact is doubled, the maximum spring compression becomes 22 cm. The speed of the block before impact cannot be determined without the drop height.

Explanation:

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 3.0 N/cm. The block becomes attached to the spring and compresses it 11 cm before momentarily stopping.

(a) To calculate work done by gravity, use Work = mgh, where m is the mass of the block, g is the acceleration due to gravity (9.8 m/s²), and h is the height (11 cm = 0.11 m, since we need consistent units). The work done by gravity is (0.21 kg)(9.8 m/s²)(0.11 m) = 0.22638 J (since potential energy lost by the block is equal to the work done by gravity).

(b) The work done by the spring is Work = 1/2 kx², converting k to N/m gives us 300 N/m. The compression x is 0.11 m, so the work done is (1/2)(300 N/m)(0.11 m)² = 1.815 J.

(c) Since we're assuming friction is negligible and using energy conservation, the potential energy (mgh) at the beginning will be equal to the kinetic energy (1/2 mv²) just before impact. Solving for v gives v = √(2gh), where h is the drop height. However, without the drop height, we cannot calculate the exact velocity.

(d) Doubling the speed will increase the kinetic energy by a factor of four (since KE = 1/2 mv²). To find the new compression distance, we set the new kinetic energy equal to the spring potential energy (1/2 kx²) and solve for x. The maximum compression x will be twice the original compression, or 22 cm.

How bad is the heavy​ traffic? You can walkwalk 1212 miles in the same time that it takes to travel 3232 miles by car. If the​ car's rate is 55 milesmiles per hour faster than your walkingwalking ​rate, find the average rate of each.

Answers

Answer:

Speed by walking is 33 miles per hour

And speed of by car is 88 miles per hour

Explanation:

We have given that it takes same time to walk 1212 miles as 3232 miles by car

Now let the speed by walk is x

As speed by car is 55 miles per hour faster than by walk = x+55

As time is same and we know that time is given as [tex]time=\frac{distance}{speed}[/tex]

So [tex]\frac{1212}{x}=\frac{3232}{x+55}[/tex]

[tex]1212(x+55)=3232x[/tex]

[tex]1212x+66660=3232x[/tex]

x = 33 miles per hour

So speed by walking is 33 miles per hour

And so speed of car = 33+55 =88 miles per hour

Before railroad were invented, goods often traveled along canals, with mules pulling barges from the bank. If a mule is exerting a 12,000N force for 10km, and the rope connecting the mule to the barge is at a 20 degree angle from the direction of travel, how much work did the mole do on the barge?
A. 12MJ
B. 11MJ
C. 4.1MJ
D. 6MJ

Answers

Answer:

W = 112.76MJ

Explanation:

the work is:

[tex]W = F_xD[/tex]

where [tex]F_x[/tex] is the force executed in the direction of the displacement and the d the displacement.

so:

W = 12000Ncos(20)(10000)

we use the cos of the angule because it give us the proyection in the axis x of the force, that means the force in the direction of the displacement.

W = 112.76MJ

A man cleaning his apartment pushes a vacuum cleaner with a force of magnitude 84.5 N. The force makes an angle of 33.9 ◦ with the horizontal floor. The vacuum cleaner is pushed 2.62 m to the right along the floor. Calculate the work done by the 84.5 N force.

Answers

Answer:

183.75641 Joules

Explanation:

F = Force of the vacuum cleaner = 84.5 N

s = Displacement of the vacuum cleaner = 2.62 m

[tex]\theta[/tex] = Angle the force makes with the horizontal = 33.9°

Work done is given by

[tex]W=F\times scos\theta\\\Rightarrow W=84.5\times 2.62\times cos33.9\\\Rightarrow W=183.75641\ J[/tex]

The work done by the force of the vacuum cleaner is 183.75641 Joules

Final answer:

The work done by the man pushing the vacuum cleaner with a force of 84.5 N at an angle of 33.9° over a distance of 2.62 m is approximately 184.8 joules.

Explanation:

To calculate the work done by a force, you can use the formula W = F × d × cos(θ), where W is the work done, F is the magnitude of the force, d is the distance the object moves, and θ is the angle the force makes with the horizontal direction of movement. In this case, the man pushes a vacuum cleaner with a force of 84.5 N at an angle of 33.9° over a distance of 2.62 m. We multiply the force by the distance and the cosine of the angle to find:

W = 84.5 N × 2.62 m × cos(33.9°)



Calculating cosine of 33.9 degrees and multiplying with the force and distance, we get:

W = 84.5 × 2.62 × 0.8326 ≈(joules)

The man does approximately 184.8 joules of work pushing the vacuum cleaner.

A box has a weight of 150 N and is being pulled across a horizontal floor by a force that has a magnitude of 110 N. The pulling force can point horizontally, or it can point above the horizontal at an angle θ. When the pulling force points horizontally, the kinetic frictional force acting on the box is twice as large as when the pulling force points at the angle θ. Find θ.

Answers

Final answer:

When the pulling force points at an angle θ above the horizontal, the frictional force acting on the box is -20N. By using the equation for frictional force and the weight of the box, we can determine that the coefficient of friction is 0.133. To find the angle θ, we use trigonometric ratios and find that it is 150°.

Explanation:

Given that the weight of the box is 150N and the pulling force has a magnitude of 110N, we can determine the angle θ at which the pulling force is directed. Let's assume the angle θ is above the horizontal. The weight of the box, 150N, is equal to the normal force acting on the box. The frictional force between the box and the floor can be calculated as the difference between the force of the pulling and the weight of the box, which is 110N - 150N = -40N. Since the kinetic frictional force acting on the box is twice as large when the pulling force points horizontally, the frictional force when the pulling force points at an angle θ is -20N.

We can use the equation for frictional force, which is F_friction = μN, where F_friction is the frictional force, μ is the coefficient of friction, and N is the normal force. As the frictional force is -20N, we can substitute this value into the equation and solve for the coefficient of friction. Therefore, -20N = μ(150N), which gives us μ = -20N/150N = -0.133. Since the coefficient of friction is always positive, the actual value of μ is 0.133.

Now, let's use trigonometric ratios to find the angle θ. Since the weight of the box acts vertically downward and the pulling force has a horizontal component of 110N and a vertical component of -150N × sin(θ), the vertical components of the weight and the pulling force must cancel each other. Therefore, -150N × sin(θ) = 150N, which simplifies to sin(θ) = -1/2. Taking the inverse sine of -1/2, we get θ = -30° or 150°. However, since the pulling force is directed above the horizontal, the angle must be 150°.

A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800.0 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating backward at 1.20m/s2.

a. What is the force of friction between the losing player’s feet and the grass?
b. What force does the winning player exert on the ground to move forward if his mass plus equipment is 110.0 kg?

Answers

Answer:

Part a)

[tex]F_f = 692 N[/tex]

Part b)

[tex]F_f = 932 N[/tex]

Explanation:

Part A)

As we know by Force equation on the losing player

[tex]F - F_f = ma[/tex]

so we will have

[tex]800 - F_f = 90\times 1.20[/tex]

[tex]800 - F_f = 108[/tex]

[tex]F_f = 692 N[/tex]

Part b)

As we know that the winning player is also moving with same acceleration

so we will have

[tex]F_f - F = ma[/tex]

[tex]F_f - 800 = 110\times 1.20[/tex]

[tex]F_f = 932 N[/tex]

In Fig. 10-37, two particles, each with mass m 0.85 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d ! 5.6 cm and mass M ! 1.2 kg. The combination rotates around the rotation axis with the angular speed v ! 0.30 rad/s. Measured about O, what are the combination’s (a) rotational inertia and (b) kinetic energy

Answers

Answer:

(a). The total moment of inertia of the system is 0.0233 kg-m².

(b). The kinetic energy is 0.0011 J.

Explanation:

Given that,

Mass of each particle m= 0.85 kg

Length = 5.6 cm

Mass of each rod M= 1.2 kg

Angular speed = 0.30 rad/s

The moment of inertia of the rod  between  axis of rotation and mass  is

[tex]I_{1}=\dfrac{Md^2}{3}[/tex]

The moment of inertia of the rod  between masses about center of mass is

[tex]I_{cm}=\dfrac{Md^2}{12}[/tex]

Moment of inertial of the rod between masses about point O is

[tex]I_{2}=M(d+\dfrac{d}{2})^2+\dfrac{Md^2}{12}[/tex]

Moment of inertia of two masses is

[tex]I_{m}=md^2+m(2d)^2[/tex]

(a). We need to calculate the total moment of inertia of the system

Using formula of moment of inertia

[tex]I_{t}=I_{1}+I_{2}+I_{m}[/tex]

Put the value into the formula

[tex]I_{t}=\dfrac{Md^2}{3}+M(d+\dfrac{d}{2})^2+\dfrac{Md^2}{12}+md^2+m(2d)^2[/tex]

[tex]I_{t}=\dfrac{Md^2}{3}+\dfrac{9Md^2}{4}+\dfrac{Md^2}{12}+md^2+4md^2[/tex]

[tex]I_{t}=\dfrac{32Md^2}{12}+5md^2[/tex]

[tex]I_{t}=2.67Md^2+md^2[/tex]

Put the value into the formula

[tex]I_{t}=2.67\times1.2\times(5.6\times10^{-2})^2+5\times0.85\times(5.6\times10^{-2})^2[/tex]

[tex]I_{t}=0.0233\ kg-m^{2}[/tex]

(b). We need to calculate the kinetic energy

Using formula of kinetic energy

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

Put the value into the formula

[tex]K.E=\dfrac{1}{2}\times0.0233\times(0.30)^2[/tex]

[tex]K.E=0.0011\ J[/tex]

Hence, (a). The total moment of inertia of the system is 0.0233 kg-m².

(b). The kinetic energy is 0.0011 J.

Final answer:

The rotational inertia of the combination is 0.041 kg·m² and the kinetic energy is 80.93 J.

Explanation:

The rotational inertia (moment of inertia) of the combination can be calculated using the formula I = 2mL², where m is the mass and L is the length of the rod. Substituting the given values, the rotational inertia is 0.041 kg·m².

The kinetic energy of rotation can be calculated using the formula K = ½Iω², where I is the moment of inertia and ω is the angular velocity. Substituting the given values, the kinetic energy is 80.93 J.

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A 5.90-g lead bullet traveling at 410 m/s is stopped by a large tree. If half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree, what is the increase in temperature of the bullet?

Answers

Answer:

rise in temperature = 333.53°C

Explanation:

mass of lead bullet, m = 5.9 g

initial velocity, u = 410 m/s

specific heat of lead = 0.126 J/gm - K = 126 J/kg K

Initial kinetic energy of the bullet

K = 1/2 mu^2 = 0.5 x 5.9 x 10^-3 x 410 x 410 = 495.895 J

Half of the kinetic energy is used to raise the temperature of lead bullet

K / 2 = mass of bullet x specific heat of lead x rise in temperature

247.95 = 5.9 x 10^-3 x 126 x rise in temperature

rise in temperature = 333.53°C

Thus, the rise in temperature of lead bullet is 333.53°C.

Which vector below goes from (0,0) to (-2,3)?

A. d
B. c
C. b
D. a

Answers

Answer:

The vector a, the answer is the point D

Explanation:

We need to make the make the subtraction from the x,y points. It will be the points of the head minus the points of the tail.

x=(-2 - 0) and y = (3-0) vector = -2 + 3

The Answer is—————————D. a

A motor does 30 kJ of work and gains 4 kJ as heatfrom the surroundings. What is the change in the internal energy of the motor?

Answers

Answer:

ΔU= *-26 KJ

Explanation:

Given that

Work done by motor W= 30 KJ

Heat gains by motor Q= 4 KJ

Sign convention:

 If heat is added to the system then it is taken as positive and if heat is rejected from the system then it is taken as negative.

If work done by the system then it is taken as positive and if work is done on the system then it is taken as negative.

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in internal energy

Q=Heat transfer

W=Work

Now by putting the values

4 = 30 + ΔU

ΔU= -26 KJ

Answer:

Internal energy ∆U=-26KJ

Explanation:

Given that:

Work done by the motor=+30KJ

Heat gained by the motor=+4KJ

In solving thermodynamical questions it is reasonable to use the sign convention this

Heat is positive if it is added to a system,but becomes negative if the system rejects heats.

Work is positive if the system does work,but becomes negative if work is done on the system.

Using the thermodynamics first law

∆U=Q-W

∆U= 4-30=-26KJ

A shallow groove on the surface of the cortex is called a ________.

Answers

Answer:

sulcus

Explanation:

A sulcus is an indentation or depression in the brain that causes it to look like it  ridges or folds

Cerebral sulci and fissures are grooves between the adjacent gyri on the surface of the cerebral hemispheres.

Sulci can be basically can be divided into three basic function

limiting sulcus: This happens to  develop between areas differing in structure and function, for example central sulcus

axial sulcus: This develops along the axis of a rapidly growing/developing area (e.g. calcarine sulcus)

operculated sulcus: a sulcus may be between two structurally-different areas and a third sulcus may lie in its wall and does not appear on the surface (e.g. lunate sulcus)

An iceberg (density=917 kg/m^3) is floating in seawater (density=1025 kg/m^3). The iceberg has a volume of 8000 m^3. What volume of the iceberg is below the water line? (Unit=m^3)

Answers

Answer:

7,157

Explanation:

Hi!

Use this equation to solve this problem:

Fb=ρfl*g*Vfl

where

Fb=buoyant force

ρfl=density of the fluidg=gravity (9.80)Vfl =volume of the iceberg (here)

What do we know?

The known values are highlighted in.

this now solve

An iceberg (density=917 kg/[tex]m^3[/tex]) is floating in seawater, the volume of the iceberg below the waterline is approximately 7168 [tex]m^3[/tex].

To calculate the volume of the iceberg below the waterline, we must apply the concept of buoyancy. The buoyant force acting on an item that floats in a fluid is equal to the weight of the fluid displaced by the object.

The buoyant force (F_b) may be computed using the formula:

F_b = ρ_fluid * V_submerged * g

W_iceberg = ρ_iceberg * V_iceberg * g

F_b = W_iceberg

ρ_fluid * V_submerged * g = ρ_iceberg * V_iceberg * g

Now,

V_submerged = (917 / 1025 ) * 8000

V_submerged = 0.896 * 8000 [tex]m^3[/tex]

V_submerged ≈ 7168 [tex]m^3[/tex]

Therefore, the volume of the iceberg below the waterline is approximately 7168  [tex]m^3[/tex].

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A uniform ladder of length L and mass m leans against a frictionless vertical wall, making an angle of 54° with the horizontal. The coefficient of static friction between the ladder and the ground is 0.32. If your mass is four times that of the ladder, what percentage of the way up the ladder can you climb before the ladder begins to slip?

Answers

Answer:

h=0.425 L

Explanation:

Given that

θ =  54°

Coefficient of friction μ = 0.32

Mass of rod = m

Lets take mass of man = M  = 4 m

C is the center of mass of the rod.

By balancing force in y and x direction

R= Fr

R = Fr=  μ N

N = mg + Mg = mg + 4 m g                     ( M =4m)

N = 5 m g

Lets take distance cover by man is h along rod before sliding

Now taking moment about the lower end

M g  h cosθ + m g cosθ L/2  =  R L sinθ

2 M g  h cosθ + m g cosθ L  = 2 R L sinθ

Now by putting the value of R  and M

8 m g  h cosθ + m g cosθ L  = 2  μ N L sinθ

8 m g  h cosθ + m g cosθ L  = 10 m g μ  L sinθ

8   h cosθ +  cosθ L  = 10  μ  L sinθ

8 h + L = 10 μ L tanθ

Now putting the value of θ  and μ

8 h + L = 10 x 0.32 x tan54° x L

8 h + L = 4.4 L

8 h = 3.4 L

h=0.425 L

The amount of light entering the eye is regulated by the

Answers

Answer:

Iris

Explanation:

The pupil is where the light enters the eye, however, the iris is pupil tissue and is the one who regulates the amount of light that it lets through.

The iris opens or closes to allow a greater or lesser flow of light through the pupil.

In summary, the iris is responsible for regulating the amount of light that enters the eye.

Final answer:

The amount of light entering the eye is mainlly regulated by the pupil, which adjusts its size based on the surrounding light levels. This is achieved by the action of muscles connected to the iris. Furthermore, the light that enters the eye is focused on the retina and processed to the brain via the optic nerve.

Explanation:

The amount of light entering the eye is majorly managed by the pupil. The pupil is the small opening in the center of the eye that adjusts its size based on the light levels in the environment. This adjustment is achieved by the contraction and relaxation of muscles connected to the iris, which is the colored part of the eye.

When the light levels are high, the pupil contracts (or constricts) to limit the amount of light that enters the eye. On the other hand, when the light levels are low, the pupil expands (or dilates) to allow more light to enter the eye. This process is regulated by specific nerves and structures, specifically the optic nerve and the oculomotor nerve in a process called the Pupillary Light Response.

Furthermore, light waves cross the cornea and focus on the retina, a light-sensitive layer lining the back of the eye. The image processing is then carried forward to the brain through the optic nerve.

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A 100 W electric heater (1 W = 1 J/s) operates for 11 min to heat the gas in a cylinder. At the same time, the gas expands from 1 L to 6 L against a constant external pressure of 3.527 atm. What is the change in internal energy of the gas? (1 L·atm = 0.1013 kJ)

Answers

Answer:

[tex]\Delta U = 64218.9 J[/tex]

Explanation:

As we know that power of the heater is given as

P = 100 W

now the energy given by the heater for 11 min of time

[tex]E = P \times t[/tex]

[tex]E = 100 \times 11 \times 60[/tex]

[tex]E = 100\times 11 \times 60[/tex]

[tex]E = 66000 J[/tex]

now from 1st law of thermodynamics we know that

[tex]E = \Delta U + W[/tex]

work done under constant pressure condition we have

[tex]W = P \Delta V[/tex]

[tex]W = (3.527 \times 1.01 \times 10^5)(6 - 1) \times 10^{-3}[/tex]

[tex]W = 1781.13 J[/tex]

now from first equation we have

[tex]66000 = 1781.13 + \Delta U[/tex]

[tex]\Delta U = 64218.9 J[/tex]

The change in the internal energy of the system is 67.787 kJ.

The given parameters;

power of the electric heater; P = 100 Wtime of operation, t = 11 min = 660 sinitial volume of the gas, = 1 Lfinal volume of the gas, = 6 Lpressure of the gas, P = 3.527 atm

The heat added to the system by the heater;

Q = Pt = 100 x 660 = 66,000 J = 66 kJ

The work done on the system is calculated as follows;

W = PΔV

W = 3.527(6 - 1)

W = 17.64 L.atm

1 Latm = 0.1013 kJ

17.64 Latm = 1.787 kJ

The change in the internal energy of the system is calculated by applying the first law of thermodynamics as follows;

ΔU = Q + W

ΔU = 66 kJ  +  1.787 kJ

ΔU = 67.787 kJ.

Thus, the change in the internal energy of the system is 67.787 kJ.

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The Explorer VIII satellite, placed into orbit November 3, 1960, to investigate the ionosphere, had the following orbit parameters: perigee, 459 km; apogee, 2 289 km (both distances above the Earth's surface); period, 112.7 min. Find the ratio vp/va of the speed at perigee to that at apogee

Answers

Answer:

[tex]\frac{v_p}{v_a}=1.268[/tex]

Explanation:

1) Basic concepts

Apogee: is the maximum distance for an object orbiting the Earth. For this case the distance for the apogee would be the sum of radius for the earth and 2289 km.

Perigee: Minimum distance of an object orbiting the Earth. For this case the distance for the perigee would be the sum of radius for the earth and 459 km.

A good approximation for these terms are related to the figure atached.

Isolated system: That happens when the system is not subdued to an external torque, on this case the change of angular momentum would be 0.

2) Notation and data

[tex]r_{earth}[/tex] radius for the Earth, looking for this value on a book we got [tex]r_{earth}=6.37x10^{3}km[/tex]

[tex]r_p =459km +r_{earth}=459km +6.37x10^3 km=6.829x10^3 km[/tex], represent the radius for perigee

[tex]r_a =2289km +r_{earth}=2289km +6.37x10^3 km=8.659x10^3 km[/tex], represent the radius for perigee

[tex]T=112.7min[/tex], period

[tex]v_p[/tex] velocity of the perigee

[tex]v_p[/tex] velocity of the apogee

[tex]r=\frac{v_p}{v_a}[/tex] is the variable of interest represented the ratio for the speed of the perigee to the speed of the apogee

3) Formulas to use

We don't have torque from the gravitational force since is a centered force. So the change of momentum is 0

[tex]\Delta L=0[/tex]   (1)

[tex]L_a=L_p[/tex]   (2)

From the definition of angular momentum we have [tex]L_i=mv_i r_i[/tex], replacing this into equation (2) we got:

[tex]mv_a r_a =mv_p r_p[/tex]   (3)

We can cancel the mass on both sides

[tex]v_a r_a =v_p r_p[/tex]   (4)

And solving [tex]\frac{v_p}{v_a}[/tex] we got:

[tex]\frac{v_p}{v_a}=\frac{r_a}{r_p}[/tex]   (5)

And replacing the values obtained into equation (5) we got:

[tex]\frac{v_p}{v_a}=\frac{8.659x10^3 km}{6.829x10^3 km}=1.268[/tex]  

And this would be the final answer [tex]\frac{v_p}{v_a}=1.268[/tex]

A disk-shaped space station 175 m in diameter spins uniformly about an axis perpendicular to the plane of the disk through its center. How many rpm (rev/min) must this disk make so that the acceleration of all points on its rim is g/2?

Answers

The angular velocity of the disk must be 2.25 rpm

Explanation:

The centripetal acceleration of an object in circular motion is given by

[tex]a=\omega^2 r[/tex]

where

[tex]\omega[/tex] is the angular velocity

r is the distance of the object from the axis of rotation

For the space station in this problem, we have

[tex]a=\frac{g}{2}=\frac{9.8}{2}=4.9 m/s^2[/tex] is the centripetal acceleration

The diameter of the disk is

d = 175 m

So the radius is

[tex]r=\frac{175}{2}=87.5 m[/tex]

So, a point on the rim has a distance of 87.5 m from the axis of rotation. Therefore, we can re-arrange the previous equation to find the angular velocity:

[tex]\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{4.9}{87.5}}=0.237 rad/s[/tex]

And this is the angular velocity of any point along the disk. Converting into rpm,

[tex]\omega=0.236 \frac{rad}{s}\cdot \frac{60 s/min}{2\pi rad/rev}=2.25 rpm[/tex]

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A box slides down a 31° ramp with an acceleration of 0.99 m/s2. Determine the coefficient of kinetic friction between the box and the ramp.μk=______.

Answers

Answer:[tex]\mu [/tex]=0.48

Explanation:

Given

inclination [tex]\theta =31^{\circ}[/tex]

Acceleration of object[tex]=0.99 m/s^2[/tex]

Now using FBD

[tex]mg\sin \theta -f_r=ma[/tex]

[tex]mg\sin \theta -\mu mg\cos \theta =ma[/tex]

[tex]a=g\sin \theta -\mu g\cos \theta [/tex]

[tex]0.99=5.04-\mu 8.4[/tex]

[tex]\mu 8.4=4.057[/tex]

[tex]\mu =0.48[/tex]

A rectangular wire loop is pulled out of a region of uniform magnetic field B at a constant speed v. What is true about the induced emf in the loop while the loop is pulled out of the region of uniform magnetic field

Answers

Answer:

There is a constant emf induced in the loop.

Explanation:

In the uniform magnetic field suppose the rectangular wire loop of length L and width b is moved out with a uniform velocity v. suppose any instance x length of the loop is out of the magnetic field and L-x length is inside the loop.

Area of loop outside the field = b(L-x)

we know that flux φ= BA

B= magnitude of magnetic field , A=  area

and emf [tex]\epsilon= \frac{d\phi}{dt}[/tex]

[tex]\epsilon=B\frac{dA}{dt}[/tex]

[tex]\epsilon=B\frac{db(L-x)}{dt}[/tex]

[tex]\epsilon=Bb\frac{d(L-x)}{dt}[/tex]

B,b and L are constant and dx/dt = v

⇒ε = -Bbv

which is a constant hence There is a constant emf induced in the loop.

Amber asked her roommate to turn down the radio because she was trying to study. Her roommate had increased the volume from a volume level of 14 to 15. This was just enough for Amber to detect the increase and subsequent decrease. Amber's detection of the increase and decrease of volume is an example of ________

Answers

Answer:

Difference threshold or also Just Noticeable Difference

Explanation:

The above mentioned case between room mates, where one room mate was able to detect a minute change in volume shows an instance of the difference threshold.

Difference threshold can be defined as stimulation at its minimum level that can be detected by an individual almost 50 % of the times.

It is the lowest possible level of sound that is detectable by a person.

That is what happened in the mentioned case that when the volume was increased from 14 to 15, Amber was able to detect it.

It is thought that bonding of adhesives occurs at the molecular level. What is the technical name of the force that holds glue to its bonding materials?

Answers

Answer:

Van der waals forces.

Explanation:

When we spread glue to stick any two substances as A and B with adhesives C. then there are adhesive force between substance A and C and adhesive force between substance B and C and cohesive force between C itself will act. In all adhesive and cohesive forces van der waals forces will apply at molecular level because there is no chemical bonding between adhesive and surface but lots of small attractive forces.

Answer:

Van der Waals force

Explanation:

The technical name given to the force that holds glue to its bonding materials is called Van der Waals force.

The forces of Van der Waals is defined by  attraction and repulsion between atoms, molecules, and surfaces and other intermolecular forces. They differ from covalent and ionic bond in that they are caused by correlations in the varying polarizations of the nearby particles (as a result of quantum dynamics).

Approximately what core temperature is required before hydrogen fusion can begin in a star?

Answers

Answer:

The temperature required is near about 3 million kelvin

Explanation:

The brilliance of the star results from the nuclear reaction that take place in the core of the star and radiate a huge amount of thermal energy resulting from the fusion of hydrogen into helium.

For this reaction to take place, the temperature of the star's core must be near about 3 million kelvin.

The hydrogen atoms collide and starts and the energy from the collision results in the heating of the gas cloud. As the temperature comes to near about [tex]1.5\times 10^{7 {\circ}C[/tex], the nuclear fusion reaction takes place in the core of the gas cloud.

The huge amount of thermal energy from the nuclear reaction gives the gas cloud a brilliance resulting in a protostar.

Final answer:

The core temperature necessary for hydrogen fusion to begin in a star is approximately 15,000,000 Kelvin, marking its entry into the main sequence. With hydrogen's exhaustion, fusion of helium and other more complex elements happen, each needing significantly higher temperatures.

Explanation:

Hydrogen fusion, also known as thermonuclear fusion, initiates within a star when its core temperature reaches approximately 15,000,000 Kelvin (K). At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. This process of hydrogen fusion occurring in the core of a star is critical for a star's energy balance and longevity.

A star reaches the main sequence when its core temperature is high enough (about 12 million K) to fuse hydrogen into helium. However, once the hydrogen fuel is exhausted in a star's core, fusion of helium, and later, other more complex elements can occur, with each requiring significantly higher temperatures than previous fusion reactions.

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Why is Saturn almost as big as Jupiter, despite its smaller mass?
a. Saturn's rings make the planet look bigger. Jupiter's strong magnetic field constrains its size.
b. Saturn has a larger proportion of hydrogen and helium than Jupiter, and is therefore less dense.
c. Jupiter's greater mass compresses it more, thus increasing its density.
d. Saturn is further from the Sun, thus cooler, and therefore less compact.

Answers

Answer:

c. Jupiter's greater mass compresses it more, thus increasing its density.

Explanation:

The mass of Jupiter is greater in its interior, this mass compresses Jupiter to some extent. Thus, its density is increased. Now, more mass is compressed in the lesser volume. Hence, its size does not increase enormously. On the other hand the mass of Saturn is lesser  and also density lower. this gives Saturn a reasonably higher volume.

Hence, option C is correct.

Saturn appears almost as large as Jupiter due to its lower mass leading to less gravitational compression of its hydrogen and helium composition, resulting in a larger volume for its mass. The correct answer from the provided choices is c. Jupiter's greater mass compresses it more, thus increasing its density.

The reason why Saturn appears almost as large as Jupiter despite having a smaller mass is largely due to its composition and structure. We see that Jupiter, which is 318 times more massive than Earth, has a significantly higher density because its greater mass compresses the planets' internal hydrogen and helium more than Saturn, which is about 25% as massive as Jupiter. In comparison, Saturn's lower mass results in less compression and hence a larger volume for its mass. So, the correct answer is c. Jupiter's greater mass compresses it more, thus increasing its density.

Jupiter and Saturn are both composed mainly of hydrogen and helium, which become liquid at great depths due to their large size and the resulting high pressure. However, Saturn, with its lower density, is the least dense planet in the solar system, even less dense than water. Contrary to the choices provided, Saturn's rings do not play a significant role in making the planet appear larger in terms of its physical size, and the effect of distance from the Sun on the compactness of the planets is negligible compared to the impact of their massive gravitational compression.

A 0.140-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. The glider has a head-on collision with a 0.299-kg glider that is moving to the left with a speed of 2.28 m/s. Find the final velocity (magnitude and direction) of each glider if the collision is elastic.

Answers

Answer:

v1 = 2.76 m/s and v2 = - 0.32 m/s

Explanation:

m1 = 0.140 kg

m2 = 0.299 kg

u1 = 0.80 m/s

u2 = - 2.28 m/s

Let the speed after collision is v1 and v2.

Use conservation of momentum

m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

0.140 x 0.80 - 0.299 x 2.28 = 0.140 x v1 + 0.299 x v2

0.112 - 0.68 = 0.14 v1 + 0.299 v2

0.14 v1 + 0.299 v2 = - 0.568 ..... (1)

By the use of coefficient of restitution, the value of e = 1 for elastic collision

[tex]e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}[/tex]

u2 - u1 = v1 - v2

- 2.28 - 0.8 = v1 - v2

v1 - v2 = 3.08

v1 = 3.08 + v2

Put in equation (1)

0.14 (3.08 + v2) + 0.299 v2 = - 0.568

0.43 + 0.44 v2 = - 0.568

v2 = - 0.32 m/s

and

v1 = 3.08 - 0.32 = 2.76 m/s

Thus, v1 = 2.76 m/s and v2 = - 0.32 m/s

A 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 Nm. What is the angel (assumed to be less than 90 degrees) between the wrench handle and the direction the force is applied?

Answers

Answer:

9.1°

Explanation:

Torque = distance from pivot * perpendicular force

15 Nm = 0.5 m * perpendicular force

perpendicular force  = 30 N

So the bertical component of the applied force that caused a turning  effect (torque) was 30 N. Now we use this information to find the angle that would produce a vertical component of  30N from an applied force of 95 N.

Fy = FSinθ

15 = 95 Sinθ

θ = 9.1°

person throws a ball horizontally from the top of a building that is 24.0 m above the ground level. The ball lands 100 m down range from the base of the building. What was the initial velocity of the ball? Neglect air resistance and use g = 9.81 m/s2.

Answers

Answer:45.24 m/s

Explanation:

Given

Height of Building h=24 m

Range of ball R=100 m

Considering Vertical motion of ball

using [tex]y=u_yt+\frac{a_yt^2}{2} [/tex]

initial vertical velocity is zero therefore [tex]u_y=0[/tex]

[tex]24=0+\frac{9.8\times t^2}{2}[/tex]

[tex]t=\sqrt{\frac{48}{9.8}}[/tex]

[tex]t=2.21 s[/tex]

Now considering Horizontal Motion

[tex]R=u_xt+\frac{a_xt^2}{2}[/tex]

[tex]100=u_x\times 2.21+0[/tex]  , as there is no horizontal acceleration

[tex]u_x=45.24 m/s[/tex]

                       

Final answer:

The physics problem can be addressed by using the principles of projectile motion. The time of flight determined by the vertical motion is used to calculate the horizontal initial velocity. The initial velocity of the ball is approximately 45.24m/s.

Explanation:

This is a problem in Physics based on the principles of Projectile Motion. We need to determine the initial velocity of the ball. The key point in this problem is that the horizontal motion of the projectile (in this case, the ball) is determined purely by the initial horizontal velocity, and is unaffected by the vertical motion. This is called the independence of the horizontal and vertical motions.

The time the ball is in the air is governed entirely by its vertical motion. Thus, we can find the time of flight by using the equation for vertical motion: y = 1/2gt², where y is the vertical displacement (24m in this case), g is the acceleration due to gravity (9.81 m/s²), and t is the time. So, t = sqrt(2y/g) = sqrt(2*24/9.81) = 2.21s.

Using this time, we can find the initial horizontal velocity using the equation for horizontal motion: x = vxt where x is the horizontal displacement (100m in this case), vx is the horizontal velocity, and t is the time. Rearranging the equation we get: vx = x/t which is approximately 45.24m/s . So, the initial velocity of the ball is around 45.24m/s

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