**Answer:**

w=2160kJ

**Explanation:**

work=power × Time

w=pt

p=600W

t=60min×60sec=3600

w=600 × 3600

w=2160000joules(J)

w=2160kJ

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How much work does a 12-V battery do in pushing 2 mC of charge through a circuit containing onelight bulb?

Given data:

[tex]\begin{gathered} V=12\text{ V} \\ Q=2\text{ mC} \end{gathered}[/tex]The work done is given as,

[tex]W=QV[/tex]Putting all known values,

[tex]\begin{gathered} W=2\times10^{-3}\text{ C}\times12\text{ V} \\ =24\text{ mJ} \end{gathered}[/tex]Therefore, **24 mJ of the work is done.**

A. Thermal to Chemical to Electrical to MechanicalB. Mechanical to Chemical to Electrical to Thermal C. Electrical to Chemical to Mechanical to ThermalD. Chemical to Thermal to Mechanical to Electrical

**Given:**

Coal-fired power plant

**Required:**

Energy transformation in coal-fired power plant.

**Explanation:**

Coal has chemical energy.

When the coal is heated, the chemical energy transforms into thermal energy.

Thermal energy is used to rotate the blade of the turbine, so thermal energy transforms into mechanical energy.

The rotating blades produce electricity, so mechanical energy transforms into electrical energy.

**Final Answer: **The energy transformation of a coal-fired power plant is chemical to thermal to mechanical to electrical.

Pls answer. Will rate 5 start and give a Thanks or Brainliest.

What are the characteristics of the Sun? (Select all that apply.)

Group of answer choices

The Sun is the center of our solar system and radiates energy.

The Sun is a medium-sized star compared to other stars in the universe.

The Sun has planets and dwarf planets that orbit it.

The Sun is composed of helium and hydrogen.

The Sun is near the end of its life.

The** Sun** is composed of** helium** and hydrogen. Option C.

The **photosphere** chromosphere and **corona** are all part of the Sun's atmosphere. Although the corona is sometimes called the solar atmosphere it is actually the sun's upper atmosphere. The sun's atmosphere includes features such as sunspots, coronal holes, and solar flares the sun is a star.

It's a huge spinning ball of hot gas. The **sun **is like a star in the night sky. Because we are so close it appears much larger and brighter than other stars. The sun is the center of the solar system and accounts for most of the mass of the **solar system**. Compared to the billions of other stars in the universe, the sun stands out. But for the Earth and the other planets orbiting around it, the Sun is a powerful focal point.

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what rays gives more heat from the sun?

**ANSWER**

**Vertical rays**

**EXPLANATION**

We have two types of rays from the sun: vertical rays (direct sunlight), which are those rays that hit the surface of Earth vertically, and slanted rays (indirect sunlight), which are those that hit the surface with an angle less than 90° with the surface. We can appreciate both in the following diagram,

As is shown in the diagram, direct sunlight hits the surface in a smaller area than indirect sunlight. This way, the vertical rays are more concentrated than slanted rays and, therefore, give more heat - note that the energy provided by those rays will be concentrated in a smaller area.

Hence, **vertical rays**** give more heat from the sun.**

**Answer:**

slanted rays that is correct

Can an object have a net negative charge of 2.00 x 10^-19

We will have that **as long as the object is a conductor it will be able to have a net negative charge of 2.00*10^-9 C**.

A box weighing 51.1 N is pulled horizontally until it slides uniformly over a level floor. If the applied force is 6.30 N, what is the coefficient of friction between the box and the floor?

**Given data**

*A box weighs is **N** = 51.1 N

*The applied force is **F** = 6.30 N

The formula for the coefficient of friction between the box and the floor is given as

[tex]\mu=\frac{F}{N}[/tex]Substitute the known values in the above expression as

[tex]\begin{gathered} \mu=\frac{6.30}{51.1} \\ =0.123 \end{gathered}[/tex]Hence, the coefficient of friction between the box and the floor is **0.123**

8. Climbing a mountain, your water bottle falls and hits the ground. If it is moving at 30m/s upon impact. how far did it fall?

**Answer:**

45.9 m

**Explanation:**

FIRST:

vf = a t

30 = 9.81 (t) shows t = 3.06 s

THEN:

df = 1/2 a t^2

= 1/2 ( 9.81)(3.06)^2 = 45.9 m

I am not sure if this is true or false. Please help

We have the formula of Newton's second law

[tex]F=ma[/tex]Where F is the force, m is the mass and a is the acceleration.

If we isolate the acceleration

[tex]a=\frac{F}{m}[/tex]As we can see if the mass increases the acceleration will decrease

Therefore the answer is **FALSE**

Solve for the impulse imparted on the golf ball.

t = 0.001 s

F = 2000 N

**Answer:**

2 N s

**Explanation:**

Impulse = force * time

= 2000 N * .001 s = 2 N s

A hockey stick 1.1 m long completes a swing in 0.15 seconds through a range of 85°. Assume a uniform angular velocity. What is the linear distance moved by the end of the stick? What is the tangential velocity of the end of the stick?

The** linear distance **and the** tangential velocity **of the end of the stick are 1.63 m and 10.9 m/s respectively.

**Rotational motion **is the motion of a body around a circular path, in a fixed orbit. The **dynamics** for rotational motion are entirely analogous to linear or translational dynamics. The equations for rotating objects are similar to the motion equations for **linear motion**.

Given, the **length** of the hockey stick, r = 1.1 m

The hockey stick completes one swing in time, t = 0.15 sec

The **angle** θ = 85° = 1.483 rad

The **average angular velocity** of the hockey stick,

ω = θ/t = 1.483rad/0.15s = 9.9 rad/s

The** linear distance **moved by the end of the hockey stick:

L = rθ = 1.1× 1.483 = 1.63 m

The **tangential velocity **of the end of the hockey stick:

v = rω = 1.1 × 9.9 = 10.9 m/s

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Find the equation of a line passing through (8, 43.6) and (4.5, 26.3)

5x + y = 3.6 **equation** of a line going between (8, 43.6) and (4.5, 26.3).

Having independent properties for both size and direction, a quantity or phenomenon is referred to as a** vector**. The representation of a quantity in mathematics or geometry is another use of the term. In nature**, vectors** include things like force, momentum, weight, electromagnetic fields, and velocity.

**Unit vectors** are those that have a magnitude that is exactly one unit. They are quite useful for a variety of reasons. The** unit vectors** [0,1] and [1,0] in particular can be combined to create any other **vector**.

The slope of the line joining points A(8,43.6) and B(4.5,26.3) is given by the equation m= y2-y1/x2-x1=26.3-43.6/4.5-8 = 5.

Consequently, the equation for the line that is required is given by y-y1 = m. (x-x1)

y- 43.6 = 5(x-8) =5x + y = 3.6.

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Consider a house with a floor space of 200 m2and anaverage height of 3 m at sea level, where thestandard atmosphericpressure is 101.3 kPa.Initially the house is at a uniformtemperature of 10°C.Now the electric heater is turned on, andthe heater runs until the air temperature in the house risesto anaverage value of 22°C. Determine how much heat is absorbedby the air assuming some airescapes through the cracks asthe heated air in the house expands at constant pressure. Also,determine the cost of this heat if the unit cost of electricity inthat area is $0.075/kWh.

9038 kJ of **heat **is absorbed by the air assuming some air escapes through the cracks as the heated air in the house expands at constant **pressure**, and cost of this **heat **if the unit cost of **electricity **in that area is $0.075/kWh is $0.19.

In the physical sciences, **pressure **is defined as the perpendicular force per unit area or the stress at a point within a confined fluid.

A 42-pound box with a bottom area of 84 square inches will press down on a surface with a **pressure **equal to the force divided by the area over which it is applied, or half a pound per square inch.

The specific **heat **of air at room temperature is

[tex]C_p[/tex] = 1.007 kJ/kg⋅°C

The volume and mass of the air in the house are

V = Floor space × height

= (200 m² )(3m)

= 600 m ³

Mass = PV/RT

= [tex]\frac{(1013 kPa)(600 m^3 )}{(0287 kPa.m/kg.K)(10+273.15 K)}[/tex]

= 7479 kg

The amount of **heat **that must be transferred to the air in the house as it is heated from 10 to 22°C is calculated, taking into account that the **pressure **in the house remains constant during heating, as follows:

Q = [tex]mC_p[/tex] [tex](T_2 - T_1)[/tex]

= (747.9 kg)(1.007 kJ/kg⋅ °C)(22 – 10) °C

= 9038 kJ

Noting that 1 kWh = 3600 kJ, the cost of this electrical energy at a unit cost of $0.075/kWh is

Enegy Cost = (Energy used)(Unit cost of energy)

= (9038/3600 kWh)($0.075/kWh)

= $0.19

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This defense mechanism offers self-justifying explanations in place of the real, more threatening, unconscious reasons for one's actions. This is known as:

This **defense** **mechanism** offers self-justifying explanations in place of the real, more threatening, unconscious reasons for one's actions. This is known as** ****Rationalization.**

According to the question, rationalization is a defense mechanism, and the phrase refers to the act of justifying through the creation of numerous sorts of **justifications**, **explanations**, and reasons for any behavior.

A coherent, logically ordered system that also serves to rationalize an individual's unconscious **motivation** for acting is referred to as a rationalization.

One method of reorganizing a business to improve operational effectiveness and alter **organizational policy** is rationalization.

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IfWhile shopping in England, Sally notices the cost of bananas is £2.10/kg. Sally isn't familiar with Great Britain Pounds (GBP) or kilograms, and needs to convert the unit price into dollars per pound. You may assume the conversion rate from dollars to GBP is $1 to £0.77, and there are 2.2lbs to 1kg. Round your answer to the nearest cent.

Each **pound** costs $1.2 in** unit pricing**.

Sally notes that the **price **of bananas in England is £2.10/kg.

Applying the **exchange rate** from **dollars **to** British pounds** (GBP), if $1 = £0.77, then £2.10 would be calculated as 2.10/0.77, or 2.72.

Using the pounds to kilograms **conversion rate**, 1 kilogram equals 2.2 pounds.

Consequently, the price **per unit** in dollars would be 2.72/2.2 = $1.236.

If we round to the **nearest cent**, the price per pound is $1.2.

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Electronegativity can best be defined as a) relative measure showing how many bonds an atom will formb) a relative measure showing the degree to which electrons can be removed from an atom c) the degree to which an electron will leave an atom d) the degree to which an atom wants to gain more electrons

The electronegativity of an element is the tendency of an atom to attract electrons to itself in a chemical compound.

The electronegativity decreases down the group and increases from left to right across the period.

Therefore,** electronegativity can best be defined as the degree to which an atom wants to gain more electrons. Hence, option (d) is the correct choice.**

Which of the following scenarios where a ball is in motion describes balanced forces? (choose only one answer choice) Select one or more: a. 2, 4, and 5 b. 1, 3, and 4 c. 2 and 4 d. 1, 2, and 5

The scenario where a ball in **motion** describes balanced **forces** is 1, 3, and 4. That is **option B.**

**Motion** is defined as the movement of an object from one position to another with respect to time.

There are different type of **motion** which include the following:

When an object is in **motion**, the forces acting on them are said to be balanced when they are equal in magnitude and in opposite directions.

1N + 3N = 4N

This shows that the **forces** 1N + 3N which is equal to 4N are equal and opposite.

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A tetherball leans against the smooth, frictionless

post to which it is attached (Figure 1). The string is

attached to the surface of the ball such that a line

along the string passes through the center of the

ball. The string is 1.40 m long, and the ball has a

radius of 0.110 m with mass 0.280 kg.

After thoroughly calculating we have come to find that, the **tension **in the **string **is 2.752 N

In physics, **tension **is defined as the pulling force transmitted axially by a **string**, cable, chain, or other similar object, or by each end of a rod, truss member, or other similar three-dimensional object.

**Tension **can also be defined as the action-reaction pair of forces acting at each end of the aforementioned elements. Compression's opposite, **tension**, is possible.

We know that

**Tension **= (Mass × gravity) / cos∅

Here given that

Mass = 0.280kg

Gravity = 9.8 m/s²

∅ = 4.2°

Lets substitute the values and we get

**Tension **= (280g × 9.8 m/s²) / cos4.2°

**Tension **= 2.752 N

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Full question

A tetherball leans against the smooth, frictionless

post to which it is attached (Figure 1). The string is

attached to the surface of the ball such that a line

along the string passes through the center of the

ball. The string is 1.40 m long, and the ball has a

radius of 0.110 m with mass 0.280 kg. What is the

tension in the string?

Julie flew 450 miles to Connecticut to visit her sister and then returned three days later. At the end of the trip, what was her displacement and distance traveled?

a. 450 miles; 0 miles

b. 900 miles; 0 miles

c. 0 miles; 900 miles

d. 450 miles, 450 miles

Submission

VIEW GRADE DETAILS

Julie travelled **displacement **of 0 miles **distance **of 900 miles. So the correct answer is option (C) 0 miles; 900 miles.

According to question, Julie travelled from Connecticut to visit her sister is 450 miles and return back to Connecticut .

Here, total journey = **distance **travelled by Julie = 450 + 450 = 900 miles.

**Displacement **is shortest distance from point one to other, but here Julie return back to her initial point. So, the displacement is zero.

**Distance **is the total journey travelled between two points from one point to other. Its **magnitude **cannot be zero. Its S.I. unit is **meter**.

**Displacement **is the shortest distance between two points. Its **magnitude **may be zero. Its S.I. unit is **meter**.

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What is the mass of a 37.0 Nweight?m =?] kgEnter

**Answer:**

**m = 3.8 kg**

**Explanation:**

The weight can be calculated as follows:

[tex]W=m\cdot g[/tex]Where m is the mass and g is the gravity. So, replacing the weight by 37 N and the gravity by 9.8 m/s², we get:

[tex]37\text{ N = m}\cdot9.8m/s^2[/tex]So, solving for m, we get:

[tex]\begin{gathered} \frac{37}{9.8}=\frac{m\cdot9.8}{9.8} \\ 3.8\operatorname{kg}\text{ = m} \end{gathered}[/tex]Therefore, the answer is m = 3.8 kg

A stone is thrown upword from the top of a 59.4m high cliff with an upword velocity component of 19.6m/s how long is stone in the air?

The** time taken** up to which the stone will be in the air is ** 4.01 seconds**.

Here we are dealing with **free falling** which is referred to as A free falling object is an object that's falling beneath the sole impact of **gravity**. Any object that's being acted upon as it were by the force of gravity is said to be in a state of **free fall**. As we are given the** velocity ** which is 19.6 m/s and the height which is 59.4 m.

So the in the first case when the stone travels for a certain **distance **until its velocity becomes zero.

The formula we are referring to for calculating the maximum height which is** v² = u² - 2gh , **where v is the final velocity , u is the intial velocity ,g is the acceleration due to gravity and h is the height.

So at v=0

=> h= u²/2g

=> h = (19.6)2/( 2*9.8)= 19.6 m

Whereas when the stone starts to freely fall, so

**h = 1/2*gt², **here the time taken will be, **t = √(2h/g)**

where h = sum of the height of the building and the height traveled after the throw

which is 19.6 m+59.4 =79

So , t = √(2h/g)

=>t = √2*(79) /9.8 = 4.01 second

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Which of the Figures correctly represents the force diagram to solve the problem?

**ANSWER**

Figure 1

**EXPLANATION**

We want to identify the correct force diagram acting on the object on the inclined plane.

When an object is moving on an inclined plane, there are four forces acting on it:

1. Friction force, Fr

2. Normal force, FN

3. Component of weight acting parallel to the inclined plane = Wsinθ

4. Component of weight acting perpendicular to the inclined plane = Wcosθ

Therefore, the correct option is **Figure 1**.

2. A saucepan containing 2 kg of water is heated. 400 kJ of energy is transferred. Calculate the temperature

increase of the water.

The specific heat capacity of water is 4200 J/kg °C. Give your answer to four significant figures.

The **temperature increase** of the 2 kg water if 400 kJ of energy is transferred is 47.62 °C

**q = m c ΔT**

q = Heat

m = Mass

c = Specific heat

ΔT = Change in temperature

q = 400 KJ

m = 2 kg

c = 4200 J/kg °C

ΔT = q / m c

ΔT = 400 * 10³ / 2 * 4200

ΔT = 47.62 °C

The quantity of heat energy required to change the temperature of particle by 1 °C is known as **heat capacity**. Specific heat capacity is the heat capacity is calculated for 1 gram of matter.

Therefore, the temperature increase of water is **47.62 °C**

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The oscilloscope can be used to measure AC frequency by counting the number of _____________ per unit time.A. PeaksB. Complete wavesC. ValleysD. Partial wavesJust please give me the answer, no explanation.

The **oscilloscope **can be used to measure AC frequency by counting the number of **complete waves** (Option B).

An **oscilloscope **is a tool used to measure voltage by observing how the electrical signals from different types of waves may change over a given unit or period of time, which is useful to monitor a client's heartbeat and brain waves.

Therefore, with this data, we can see that an oscilloscope is a device used to measure waves (in this case **brain waves**) in order to indicate a given outcome associated with electrical signals.

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A student made the table shown to list some contact and non-contact forces .

Examples of forces

Contact force - the upward force applied by a coach on a student sitting on it .

The force applied by a student in a box to move it .

Non- contact forces

The force applied by an electromagnetic on a wire

The force which opposes of an object on a ruff surface .

Which statement best explains why the table is incorrect?

A- Normal force is a force acting at a distance.

B- The student applies a non- contact force on the box .

C- The electromagnetic and wire have to be in contact to apply the force .

D- Frictional force acts between the object and the rough surface when they are In contact.

The statement that best explains why the table is incorrect is that **frictional force** acts between the object and the rough surface when they are In contact ( D ).

The statements in the table that describes contact forces are correct and one of the statements that describes a non-contact force is not correct. A contact force is a force applied by one mass on another due to **being in contact **with each other. A non contact force is the opposite of a contact force.

The force which **opposes the motion** of an object on a rough surface is frictional force. Frictional force is opposing between two surfaces in contact with each other. Since the two surfaces have to be in contact with each other it falls under the description of contact force.

Therefore, the statement that best explains why the table is incorrect is that frictional force acts between the object and the rough surface when they are **in contact **( D ).

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for equation solve for x given 0

**x=2.61 radians**

**Explanation**

[tex]2\cos (x)+\sqrt[]{3}=0[/tex]

**Step 1**

Let's isolate x

a)

[tex]\begin{gathered} 2\cos (x)+\sqrt[]{3}=0 \\ \text{subtract }\sqrt[]{3}\text{ in both sides} \\ 2\cos (x)+\sqrt[]{3}-\sqrt[]{3}=0-\sqrt[]{3} \\ 2\cos (x)=-\sqrt[]{3} \end{gathered}[/tex]

**Step 2**

b) now, divide both sides by 2

[tex]\begin{gathered} 2\cos (x)=-\sqrt[]{3} \\ \text{divide both sides by 2} \\ \frac{2\cos(x)}{2}=\frac{-\sqrt[]{3}}{2} \\ \cos (x)=\frac{-\sqrt[]{3}}{2} \\ \end{gathered}[/tex]**Step 3**

c) finally,Inverse cosine in both sides ( remember we are looking for an angle)

[tex]\begin{gathered} \cos (x)=\frac{-\sqrt[]{3}}{2} \\ \text{ Inverse cosine in both sides} \\ \cos ^{-1}(\cos (x))=\cos ^{-1}(\frac{-\sqrt[]{3}}{2}) \\ x=\cos ^{-1}(\frac{-\sqrt[]{3}}{2}) \\ x=\cos ^{-1}(-0.86) \\ x=2.61\text{ radians} \end{gathered}[/tex]therefore, the answer is

**x=2.61 radians**

I hope this helps you

. In one day, the temperatures of a desert range from 110°F in the afternoon to 55°F at night. What

is this temperature range on the Kelvin scale?

**Answer:**

30.5555°K

Can be rounded to 30.56°K

**Explanation:**

The relationship between °C and °K is

°K = °C + 273.15

To convert °F to °K, first convert °F to °C and then add 273.15

°C = (°F - 32) x 5/9

So combining the two conversions we get

°K = (°F - 32) x 5/9 + 273.15

55°F = (55-32) x 5/9 + 273.15 = 285.9278°K

110°F = (110-32) x 5/9 + 273.15 = 316.4833°K

Temperature range in Kelvin scale

= 316.4833°K - 285.9278°K = 30.5555°K

Given 2 concentric wheels; what weight will an applied force FA of 40.0N down and tangent to the 18.0 cm radius wheel will balance an unknown weight tangent and down on the 10.0 cm radius wheel? Draw the situation

The given problem can be exemplified using the following diagram:

We will determine the magnitude of the force "F" that will balance the concentric wheels.

To do that we will add the torques produced by the forces. If the torque is counterclockwise we will set it a positive and if it is clockwise it will be negative.

[tex]T_{40}-T_F=0[/tex]Where:

[tex]\begin{gathered} T_{40}=\text{ torque produced by the 40 N force} \\ T_F=\text{ torque produced by the unknown force. } \end{gathered}[/tex]The sum of toques adds up to zero because we want to determine the force "F" when the system is in equilibrium.

Now, we substitute the formula for the torques:

[tex]F_{40}r_{18}-Fr_{10}=0[/tex]Where:

[tex]\begin{gathered} r_{18}=\text{ 18 cm radius} \\ r_{10}=\text{ 10 cm radius} \end{gathered}[/tex]Now, we solve for the force "F":

First, we will add "Fr10" to both sides:

[tex]F_{40}r_{18}=Fr_{10}[/tex]Now, we divide both sides by r10:

[tex]\frac{F_{40}r_{18}}{r_{10}}=F[/tex]Now, we plug in the values:

[tex]\frac{(40N)(18\operatorname{cm})}{10\operatorname{cm}}=F[/tex]Solving the operations:

[tex]72N=F[/tex]Therefore, the required force is 72N.

Oil having a density of 924 kg/m? floats onwater. A rectangular block of wood 3.59 cmhigh and with a density of 975 kg/m° floatspartly in the oil and partly in the water. Theoil completely covers the block.How far below the interface between thetwo liquids is the bottom of the block?Answer in units of m.

A diagram of the given problem is the following:

To determine the distance "x" from the interface between the water and the oil of the block we need to add the forces that are acting on the system. We have the following:

Where:

[tex]\begin{gathered} F_b=\text{ force of the block} \\ F_w=\text{ force of the water} \\ F_{o\text{ }}=\text{ force of the oil} \end{gathered}[/tex]Since the system is in equilibrium this means that the total sum of forces adds up to zero:

[tex]F_b-F_w-F_0=0[/tex]Now, the force is the product of the pressure by the area, therefore, we have:

[tex]P_bA-P_wA-P_0A=0[/tex]We can cancel out the area:

[tex]P_b-P_w-P_0=0[/tex]The pressure is the hydrostatic pressure and is given by:

[tex]P=\rho gh[/tex]Where:

[tex]\begin{gathered} \rho=\text{ density} \\ g=\text{ acceleration of gravity} \\ h=\text{ height} \end{gathered}[/tex]Substituting in the equation we get:

[tex]\rho_bgh_b-\rho_wgh_w-\rho_ogh_0=0_{}[/tex]Now, we substitute the values of the heights according to the first diagram:

[tex]\rho_bgh_b-\rho_wgx-\rho_og(h_b-x)=0_{}[/tex]Now, we solve for "x". To do that we will apply the distributive property on the parenthesis:

[tex]\rho_bgh_b-\rho_wgx-\rho_ogh_b+\rho_ogx=0_{}[/tex]Now we associate terms according to the value of the height:

[tex]\rho_bgh_b-\rho_ogh_b-\rho_wgx+\rho_ogx=0_{}[/tex]Now, we take common factors:

[tex](\rho_b-\rho_o)gh_b+(-\rho_w+\rho_o)gx=0_{}[/tex]We can cancel out the gravity:

[tex](\rho_b-\rho_o)h_b+(-\rho_w+\rho_o)x=0_{}[/tex]Now, we bring the terms with the height of the block to the right side:

[tex](-\rho_w+\rho_o)x=-(\rho_b-\rho_o)h_b[/tex]Now, we divide both sides by the factor of "x":

[tex]x=-\frac{(\rho_b-\rho_o)}{(-\rho_w+\rho_o)}h_b[/tex]Now, we plug in the values:

[tex]x=-\frac{(975\frac{kg}{m^3}-924\frac{kg}{m^3})}{(-1000\frac{kg}{m^3}+924\frac{kg}{m^3})}(0.0359m^)[/tex]We converted the height of the block using the following conversion factor:

[tex]100cm=1m[/tex]Now, we solve the operations, we get:

[tex]x=0.024m[/tex]Therefore, the distance between the interface and the bottom of the block is 0.024 meters.

If the lines in problem below, have a resistance of 100-Ω, calculate the change of voltage along each line. We are asked to determine the current given the power and the voltage. To do that we will use the following formula:P=IVWhere:P= powerI= currentV= voltageNow, we divide both sides by "V":PV=INow, we convert the power from "kW" to "W" using the following conversion factor:1kW=1000WMultiplying by the conversion factor we get:200kW×1000W1kW=200000WNow, we plug in the values in the formula:200000W48000V=ISolving the operations:4.2A=ITherefore, the current is 4.2 Amp.

Resistance 100

Current 4.2A

Based on the information of the last point, we know the current and now we have the resistance. With these two numbers we can calculate the drop of voltage generated by the resistance.

Is important to highlight that we asume the same current for each line because the current is constant in all the parallel lines.

[tex]\begin{gathered} V=IR \\ V=100\Omega\cdot4.2A \\ V=420V \end{gathered}[/tex]How many hours would it take a plane to fly 1,500 miles if it was moving at a constant speed of 300 miles/hour? Include units!

Best answer gets brainliest!

Answer:

5 hours

Explanation:

distance/speed= time

1500/300= 5

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