Warm air is less dense than cold air. The same is true of water. You decide to take a bath and turn on both the hot and cold faucets. When the tub is one-third full, you dip your hand in and discover the water is too cold. You turn the cold water faucet off and keep the hot water turned on. You leave the hot water running for ten minutes. When you return and dip your hand in, you discover that the water in the tub is different temperatures in different places. It is too hot close to the faucet and in a top layer across the entire tub. When you sit down, however, the water in the bottom of the tub is cold. Explain how this happens and why: • how the hot water ended up in a layer on the top and near the faucet. • why this is or is not a transfer of heat. • how the temperature of the water in the tub will change over time. • why you will eventually want to get out of the tub, considering the air and your porcelain enameled steel tub.

Answers

Answer 1

Answer:

Given that the warm water is less dense than the cold water, we have;

The cold water which is more dense than the warm water will remain settled in the lower part of the tub while particles of the less dense warm water will move above the cold water not being able to penetrate lower due of difference in their gravitational potential energiesThis is a case of (lack of) heat transfer due to inability of heat to penetrate into the lower layers and transfer the heat by conduction, from particle to particle, and therefore, one layer, the top remains warm while the other layer, the lower layer, remains coldWith time (over an hour) more and more warm water units will be able to penetrate deeper and make the lower layer warmer while the warm water at the top loses heat by evaporation to the atmosphere, until the entire water reaches equilibriumHeat will continue to be lost by evaporation to the air and by conduction through the porcelain to the steel tub until the water in the tub becomes cold

Explanation:


Related Questions

A car start moving from the rest.If the acceleration of the car is 2m/2 for 10 seconds what will be it final velocity

Answers

Answer:

20 m/s

Explanation:

Applying,

a = (v-u)/t.................... Equation 1

Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

make v the subject of the equation

v = u+at.............. Equation 2

From the question,

Given: u = 0 m/s(start from rest), a = 2 m/s², t = 10 seconds

Substitute these values into equation 2

v = 0+(2×10)

v = 20 m/s

A 6.93*10-4 C charge has a
potential energy U = -3.09 J at a
point in space. What is the electric
potential V at that point?
Include the sign, + or -
(Unit = V)

Answers

Answer:

P = V * Q      potential energy = potential * charge

V = =3.09 J / 6.93 * 10E-4 C = 4460 Joules / Coulomb

The electric potential, V at the point given the data from the question is 4458.87 V

What is electric potential?

The electric potential or electromotive force (EMF) is defined as the energy supplied by a battery per unit charge. Mathematically, it can be expressed as:

Electromotive force (EMF) = Work (W) / charge (Q)

V = EMF = W / Q

How to determine the Electric potentialwork (W) = –3.09 JCharge on electron = 6.93×10⁻⁴ CElectric potential (V) =?

V = W / Q

V = –3.09 / 6.93×10⁻⁴

V = 4458.87 V

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The series circuit depicts three resistors connected to a voltage
source. The voltage source (AVtot) is a 110-V source and the resistor
values are 7.2 (R1), 6.2 A2 (R2) and 8.6 22 (R3).
b. Determine the current in the circuit.
A
c. Determine the voltage drops across each individual resistor.

Answers

Answer:

B. Current in the circuit is 5.

Ci. Voltage across 7.2 Ω (R₁) is 36 V

Cii. Voltage across 6.2 Ω (R₂) is 31 V

Ciii. Voltage across 8.6 Ω (R₃) is 43 V

Explanation:

We'll begin by calculating the number equivalent resistance in the circuit. This can be obtained as follow:

Resistor 1 (R₁) = 7.2 Ω

Resistor 2 (R₂) = 6.2 Ω

Resistor 3 (R₃) = 8.6 Ω

Equivalent Resistance (R) =?

Since the resistors are in series connection, the equivalent resistance can be obtained as follow:

R = R₁ + R₂ + R₃

R = 7.2 + 6.2 + 8.6

R = 22 Ω

B. Determination of the current.

Voltage (V) = 110 V

Resistance (R) = 22 Ω

Current (I) =?

V = IR

110 = I × 22

Divide both side by 22

I = 110 / 22

I = 5 A

Therefore, the current in the circuit is 5.

Ci. Determination of the voltage across 7.2 Ω (R₁)

Resistor 1 (R₁) = 7.2 Ω

Current (I) = 5 A

Voltage 1 (V₁) =?

V₁ = IR₁

V₁ = 5 × 7.2

V₁ = 36 V

Therefore, the voltage across 7.2 Ω (R₁) is 36 V

Bii. Determination of the voltage across 6.2 Ω (R₂)

Resistor 2 (R₂) = 6.2 Ω

Current (I) = 5 A

Voltage 2 (V₂) =?

V₂ = IR₂

V₂ = 5 × 6.2

V₂ = 31 V

Therefore, the voltage across 6.2 Ω (R₂) is 31 V

Ciii. Determination of the voltage across 8.6 Ω (R₃)

Resistor 3 (R₃) = 8.6 Ω

Current (I) = 5 A

Voltage 3 (V₃) =?

V₃ = IR₃

V₃ = 5 × 8.6

V₃ = 31 V

Therefore, the voltage across 8.6 Ω (R₃) is 43 V


A circuit has a current of 3 amps and is using a 9 volt battery. The circuit has a resistance of ____
ohms.

Answers

Answer:

so 9/3=3 current is 3 amperes

Explanation:

The fomula to calculate resistance is:

voltage/cutrent

9 V/3 A= 3 ohms

Two charges, each q, are separated by a distance r, and exert mutual attractive forces of F on each other. If both charges become 2q and the distance becomes 3r, what are the new mutual forces

Answers

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

          F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

       

we substitute

          F = k 4q² / 9 r²

          F = k q² r² 4/9

          F = ⅔ F₀

The equation provided (from the textbook) first defines the elastic potential energy of a spring as ΔUsp = −(WB + WW), where WB is work the spring does on an attached block and WW is work the spring does on the wall to which it is attached. But WW is ignored in the next step. Why?

Answers

Answer:

The given potential energy of the spring is expressed as follows;

ΔUsp = -(WB + WW)

Where;

WB = Th work done by the spring on the block to which it is attached

WW = The work done by the spring on the wall

We recall that work done, W = Force applied × Distance moved in the direction of the force

The work done by the spring on the block, WB = The spring force × The distance the block moves

The work done by the spring on the wall, WW = The spring force × The distance the wall moves

However, given that the wall does not move, we have;

The distance the wall moves = 0

∴ The work done by the spring on the wall, WW = The spring force × 0 = 0 J

Therefore, WW = 0 J, and the spring does not do work on the wall, and WW can be ignored in the next subsequent) steps

Explanation:

Numa máquina térmica uma parte da energia térmica fornecida ao sistema(Q1) é transformada em trabalho mecânico (τ) e o restante (Q2) é dissipado, perdido para o ambiente.



sendo:

τ: trabalho realizado (J) [Joule]
Q1: energia fornecida (J)
Q2: energia dissipada (J)


temos: τ = Q1 - Q2

O rendimento (η) é a razão do trabalho realizado pela energia fornecida:

η= τ/Q1

Exercícior resolvido:
Uma máquina térmica cíclica recebe 5000 J de calor de uma fonte quente e realiza trabalho de 3500 J. Calcule o rendimento dessa máquina térmica.

solução:

τ=3500 J
Q1=5000J

η= τ/Q1
η= 3500/5000
η= 0,7 ou seja 70%

Energia dissipada será:



τ = Q1 - Q2
Q2 = Q1- τ

Q2=5000-3500
Q2= 1500 J

Exercicio: Qual seria o rendimento se a máquina do exercicio anterior realizasse 4000J de trabalho com a mesma quantidade de calor fornecida ? Quanta energia seria dissipada agora?



obs: Entregar foto da resolução ou o cálculo passo a passo na mensagem

Answers

Numa máquina térmica uma parte da energia térmica fornecida ao sistema(Q1) é transformada em trabalho mecânico (τ) e o restante (Q2) é dissipado, perdido para o ambiente.



sendo:

τ: trabalho realizado (J) [Joule]
Q1: energia fornecida (J)
Q2: energia dissipada (J)


temos: τ = Q1 - Q2

O rendimento (η) é a razão do trabalho realizado pela energia fornecida:

η= τ/Q1

Exercícior resolvido:
Uma máquina térmica cíclica recebe 5000 J de calor de uma fonte quente e realiza trabalho de 3500 J. Calcule o rendimento dessa máquina térmica.

solução:

τ=3500 J
Q1=5000J

η= τ/Q1
η= 3500/5000
η= 0,7 ou seja 70%

Energia dissipada será:



τ = Q1 - Q2
Q2 = Q1- τ

Q2=5000-3500
Q2= 1500 J

Exercicio: Qual seria o rendimento se a máquina do exercicio anterior realizasse 4000J de trabalho com a mesma quantidade de calor fornecida ? Quanta energia seria dissipada agora?



obs: Entregar foto da resolução ou o cálculo passo a passo na mensagem

A voltage source provides ____ required for electric current. A. the electrons B. the potential difference C. the resistance D. the pathway

Answers

The answer is c bro I’m guessing

Define emf of a battery? ​

Answers

Answer:

Electromotive force or EMF is equal to the terminal potential difference when no current flows. EMF (ϵ) is the amount of energy (E) provided by the battery to each coulomb of charge (Q) passing through.

The two scientists who gave us a better understanding of the universe are?

Answers

Answer:

I'm pretty sure it is Edwin Powell Hubble and Albert Einstein

Explanation:

Transformar las siguientes unidades al Sistema Internacional: 30 km/h ; 37 Dm ; 750 g ; 4x10-6 km2 ; 7500 cm ; 600000 cm2 ; 520700000 mm3 ; 3,4 años.

Answers

Answer:

a)  3.0  10⁴ m / s, b) 3.7 10¹ m, c) 0.750 kg, d) 4 10¹² m²,  e)  75 m, f) 60 m²

g) 5.207 10³ m², e) 4.847 10⁷ s

Explanation:

The international system (SI) of measurements has as fundamental units the meter for length, the second for time and kilogram for mass.

Let's reduce the different magnitudes to the SI system

a) 30 km / h (1000m / 1 km) (1 h / 3600 s) = 3.0 10⁴ m / s

b) 37 Dm (10 m / 1 Dm) = 3.7 10¹ m

c) 750 g (1 kg / 10,000 g) = 0.750 kg

d) 4 10⁶ km² (1000 m / 1km) ² = 4 10¹² m²

e) 7500 cm (1 m / 100 cm) = 75 m

f) 600000 cm² (1m / 10² cm) ² = 60 m²

g) 520700000 mm³ (1 m / 10³ mm) ³ = 5.20700000 109/10 ^ 6

  = 5.207 10³ m²

e) 3.4 years (l65 days / 1 yr) (24 h / 1 day) (3600 s / 1h) = 4.847 10⁷ s

Lisa made the electromagnet shown. A nail with wire coiled around it has its head labeled S to the right and its point labeled N to the left. The end of the wire leading to the S is attached to the positive terminal of a battery. The end of the wire leading to the N is attached to the negative terminal of the battery. What can Lisa do to increase the strength of the electromagnet? She can use a nail with weaker magnetic properties. She can change the direction of the nail. She can increase the number of wire loops. She can reduce the current in the wire.

Answers

Answer:

C. She can increase the number of wire loops.

Explanation:

The more wire loops the more energy.

For a coil of wire, the magnetic field strength is increased by increasing the number of coils around the nail.

What is electromagnet?

An electromagnet is a soft metal core shaped into a magnet by the passing the electric current through a coil surrounding it.

The end of the wire leading to the S is attached to the positive terminal of a battery. The end of the wire leading to the N is attached to the negative terminal of the battery. The current begins to flow. Current cant be changed to increase magnetic field strength, but the no of coils will definitely increase it.

Thus, To increase the strength of the electromagnet, Lisa can increase the number of wire loops.

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1. If you use an applied force of 45N to slide a 12Kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?

Answers

Answer:

Coefficient of kinetic friction = 0.38 (Approx.)

Explanation:

Given:

Applied force = 45 N

Mass of wooden crate = 12 kg

Find:

Coefficient of kinetic friction

Computation:

Coefficient of kinetic friction = Applied force / (Mass)(Acceleration due to gravity)

Coefficient of kinetic friction = 45 / (12)(9.8)

Coefficient of kinetic friction = 45 / 117.6

Coefficient of kinetic friction = 0.3826

Coefficient of kinetic friction = 0.38 (Approx.)

A ball is dropped from rest out of a high window in a tall building for 5 seconds. Assuming the we ignore air resistance and assume upwards to be positive. A) What will be the final velocity of the ball B) What is the height of the building if it hits the ground after those 5 seconds. *​

Answers

Answer:

I am not sure if this is the answer

(B) what is the height of the building if it hits the ground after those 5 seconds.

Quanto tempo deve ficar ligado um ferro eletrico de 1000 w para que tenha o mesmo consumo de energia que um chuveiro de 4400 w que fica ligado 10 minutos

Answers

Answer:

Thus, the time for the first lamp is 44 minutes.

Explanation:

Power of first lamp, P' = 1000 W

Power of second lamp, P'' = 4400 W

time for second lamp, t'' = 10 minutes

Let the time for first lamp is t'.

As the energy is same, so,

P' x t' = P'' x t''

1000 x t' = 4400 x 10

t' = 44 minutes

Hello, can you help me? I have this question and I don't know how to answer it. Is a tire a conductor or an insulator? Thank you!

Answers

Answer: They're Conductors.

Explanation:

Suppose the height of object is +3cm and height of image is -12 cm. What is its magnification?

Answers

Answer:

magnification is 4

Explanation:

m= image height / object height

m= 12/3

m= 4

plsss plsss plsss helppppp ASAP thank u ❤️​

Answers

Answer:

7. (D) uniformly accelerated vertical motion

8. (A) zero

9. (A) zero

10. (C) parabolic

Answer:

7.Uniformly accelerated vertical motion

8.0m/s²

9.9.8m/s

10.parabolic

11.vertical component.

A solid cylinder has a mass of 5 kg and radius of 2 m and is fixed so that it is able to rotate freely around its center without friction. A 0.02 kg bullet is moving from right to left with an angular momentum of 9 kgm2s just before it strikes the cylinder near its bottom and gets stuck at the outer radius. What is the angular velocity (magnitude and direction) of the cylinder bullet system after the impact

Answers

Answer:

0.893 rad/s in the clockwise direction

Explanation:

From the law of conservation of angular momentum,

angular momentum before impact = angular momentum after impact

L₁ = L₂

L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)

L₂ = angular momentum of cylinder and angular momentum of bullet after collision.

L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision

So,

L₁ = L₂

L₁ = (I₁ + I₂)ω

ω = L₁/(I₁ + I₂)

ω = L₁/(1/2MR² + mR²)

ω = L₁/(1/2M + m)R²

substituting the values of the variables into the equation, we have

ω = L₁/(1/2M + m)R²

ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²

ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)

ω = + 9 kgm²/s/(2.52 kg)(4 m²)

ω = +9 kgm²/s/10.08 kgm²

ω = + 0.893 rad/s

The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.

A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, causing it to reverse its direction and giving it a velocity of +25 m/s the impulse the stick applies to the park is most nearly

Answers

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

[tex]I = m\cdot (\vec{v}_{2} - \vec{v_{1}})[/tex] (1)

Where:

[tex]I[/tex] - Impulse, in kilogram-meters per second.

[tex]m[/tex] - Mass, in kilograms.

[tex]\vec{v_{1}}[/tex] - Initial velocity of the hockey park, in meters per second.

[tex]\vec{v_{2}}[/tex] - Final velocity of the hockey park, in meters per second.

If we know that [tex]m = 0.2\,kg[/tex], [tex]\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right][/tex] and [tex]\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right][/tex], then the impulse applied by the stick to the park is approximately:

[tex]I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right][/tex]

[tex]I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right][/tex]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Which of the following represent units of capacitance? You may choose more than one correct answer:

Ampere
coulomb/volt
coulomb/second
Farad
volt/coulomb

Answers

Answer:

Capacitance is a derived physical quantity measured in farad

Answer:

"Farad" is another term for the coulomb/volt measurement of capacitance, so both of those options are the correct answer. Amperes and coulomb/second are measurements of electric flow, in other words, how strong a current is.

In a warehouse, the workers sometimes slide boxes along the floor to move them. Two boxes were sliding toward each other and crashed. The crash caused both boxes to change speed. Based on the information in the diagram, which statement is correct? In your answer, explain what the forces were like and why the boxes changed speed.
Box 1 has more mass than Box 2.
Box 1 and Box 2 are the same mass.
Box 1 has less mass than Box 2

Answers

Answer:

The second one.

Explanation:

It caused both to change speed because they have both the same mass.

La distancia por carretera de Chitré a Parita es de 12 km; exprese en pies ésta distanciaLa distancia por carretera de Chitré a Parita es de 12 km; exprese en pies ésta distancia

Answers

Answer:

La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.

Explanation:

La regla de tres es una forma de resolver problemas de proporcionalidad entre tres valores conocidos y un valor desconocido, estableciendo una relación de proporcionalidad entre todos ellos.

Si la relación entre las magnitudes es directa, es decir, cuando una magnitud aumenta, también lo hace la otra (o cuando una magnitud disminuye, también lo hace la otra), se debe aplicar la regla directa de tres. Para resolver una regla directa de tres, se debe seguir la siguiente fórmula, siendo a, b y c los valores conocidos y x el valor a determinar:

a ⇒ b

c ⇒ x

Entonces [tex]x=\frac{c*b}{a}[/tex]

La regla directa de tres es la regla que se aplica en este caso donde hay un cambio de unidades. Para realizar esta conversión de unidades, primero debes saber que 1 km = 3280,84 pies. Entonces, si 1 km son 3280,84 pies, ¿cuántos pies son 12 km?

1 km ⇒ 3280.84 pies

12 km ⇒ x

[tex]x=\frac{12 km*3280.84 pies}{1 km}[/tex]

x= 39370.08 pies

La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.

are Car Travels certain distance with a speed of 50 kilometre per hour and returns with a speed of 40 kilometre per hour what is the average speed for the whole journey
Best answer branlist and NO LINKS​

Answers

Answer:

The average speed for the whole journey is 44.[tex]\overline 4[/tex] kilometers per hour

Explanation:

The average speed with which the car travels the distance, v₁ = 50 kilometer per hour

The average speed with which the car returns, v₂ = 40 kilometer per hour

Average speed, v = (Total distance, d)/(Total time, t)

Let 'd' represent the distance travelled, we have;

The time it takes the car to travel to the distance = d/50 hours

The time it takes the car to return = d/40 hours

The total time = (d/50 hours + d/40 hours) = d·(40 + 50)/(40 × 50) hours= 9·d/200 hours

The total distance = d kilometers+ d kilometers = 2·d kilometers

The average speed for the whole journey, v = 2·d kilometers/(9·d/200 hours) = 400/9 kilometers per hours = 44.[tex]\overline 4[/tex] kilometer/hour.

a rocket with an initial velocity of 20 m/s fires another engine that gives it an acceleration of 4 m/s2 over 10 seconds. How far did the rocket travel during this time?

Answers

Answer: 400 m

Explanation:

Vf= 20 + (4*10)

Vf= 60 [m/s]

x= (60^2 - 20^2) / (2*4)

x= 400 m

Determine the applied force required to accelerate a 2.25 kg object rightward with a
constant acceleration of 1.50 m/s2 if the force of friction opposing the motion is 18.2 N.
(Neglect air resistance.)

Answers

Answer:

Explanation:

Im going to be using the rules for significant digits properly so I hope you're quite familiar with them. The equation we need for this is

F - f = ma where F is the applied force (our unknown), f is the frictional force, m is the mass, and a is the acceleration. Filling in:

F - 18.2 = 2.25(1.50) and

F = 2.25(1.50) + 18.2  Do the multiplication first and round to get

F = 3.38 + 18.2   The addition rules tell us that we will be rounding to the tenths place after we add to get

F = 21.6 N

3 - An object is being pushed with a net force of 15 N. If the net force is cut in third to 5 N, how will the acceleration be changed?​

Answers

Answer:

Explanation:

F = ma is a linear equation. This means that the Force change as the accleration changes. And vice versa. If the Force is cut in thirds, then the acceleration is also cut in thirds. Let's do some math on this just to prove it, shall we?

We know that at first, the F = 15. Let's give this object a mass of 5kg. That means that

15 = 5a so

a = 3

Then the F is cut into thirds, so

5 = 5a so

a = 1

The second acceleration is one-third of the first one, where the Force is 3 times greater.

A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height of 9m. Find the work required to empty the tank by pumping all of the water over the top of the tank. Use the fact that acceleration due to gravity is 9.8 m/sec2 and the density of water is 1000kg/m3. Round your answer to the nearest kilojoule.

Answers

Answer:

[tex]W=17085KJ[/tex]

Explanation:

From the question we are told that:

Height [tex]H=16m[/tex]

Radius [tex]R=3[/tex]

Height of water [tex]H_w=9m[/tex]

Gravity [tex]g=9.8m/s[/tex]

Density of water [tex]\rho=1000kg/m^3[/tex]

Generally the equation for Volume of water is mathematically given by

 [tex]dv=\pi*r^2dy[/tex]

 [tex]dv=\frac{\piR^2}{H^2}(H-y)^2dy[/tex]

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 [tex]dw=(pdv)g (H-y)[/tex]

Substituting dv

 [tex]dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)[/tex]

 [tex]dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy[/tex]

Therefore

 [tex]W=\int dw[/tex]

 [tex]W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy[/tex]

 [tex]W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)[/tex]

 [tex]W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0[/tex]

 [tex]W=3420.84*0.25[2401-65536][/tex]

 [tex]W=17084965.5J[/tex]

 [tex]W=17085KJ[/tex]

 

'

'

What is the net force acting on the airplane?
740 N right -->
700 N right -->
100 N left <--
760 N right -->

Answers

[tex]\huge{ \mathrm{  \underline{ Answer }\:  \:  ✓ }}[/tex]

Total force acting on right side = 800 N

Total force acting on left side :

60 N + 40 N

100 N

Now, equivalent force acting on the plane is :

greater force - minor force

800 N - 100 N 700 Newtons

And the direction of equivalent force will be the direction of greater force, that is right direction.

Hence, Correct option is :

700 N right -->

_____________________________

[tex]\mathrm{ \:TeeNForeveR\:}[/tex]

. A tennis ball rolls off the lab bench with an initial velocity of 3.0 m/s. The top of the lab bench is 1.5 m above the floor. How long will the tennis ball be in the air before it hits the ground

Answers

Answer:

[tex] { \huge{s}} = ut + \frac{1}{2} g {t}^{2} \\ 1.5 = 3t + \frac{1}{2} \times 10\times {t}^{2} \\ 1.5 = 3t + 5 {t}^{2} \\ 5 {t}^{2} + 3t - 1.5 = 0 \\ t = 0.3 \: seconds[/tex]

Other Questions
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