a metal with a specific heat of 0.780 requires 45 kj of heat to raise the temperature by 82 c what is the mass of the metal

Answers

Answer 1

To determine the mass of the metal, we can use the equation: mass = heat / (specific heat * temperature change).To find the mass of the metal, we can use the formula:

mass = heat / (specific heat * temperature change)

Given:

Specific heat (c) = 0.780 J/(g°C)

Heat (q) = 45 kJ = 45,000 J

Temperature change (ΔT) = 82°C

Using the formula, we can substitute the values:

mass = 45,000 J / (0.780 J/(g°C) * 82°C)

mass = 45,000 J / 63.96 J/g

mass ≈ 703.96 g or 0.704 kg

Therefore, the mass of the metal is approximately 703.96 grams or 0.704 kilograms.To find the mass of the metal, we can use the formula:

mass = heat / (specific heat * temperature change)

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Related Questions

.A water tank has dimensions
of 6m x 4m x 4m. Find the pressure exerted by
water at the base of dam when it is half
filled. ( Density of water - 1000 kg/ m2,
acceleration
due to to gravity is 9.8m/s2

Answers

Answer:

9800Nm⁻²

Explanation:

Given parameters:

Dimensions of the water tank = 6m x 4m x 4m

Density of water  = 1000kgm⁻³

Gravity  = 9.8m/s²  

Unknown:

Pressure exerted by the water at the base of the dam  = ?

Solution:

The pressure exerted by the water is given as:

  Pressure  = density of water x acceleration due to gravity x height of water

The height of the water here is half = 1m

   

Pressure  = 1000 x 9.8 x 1 = 9800Nm⁻²

If Earth orbited the Sun in 9 months instead of 12, then, compared with a sidereal day, a solar day would be

Answers

Answer:

longer

Explanation:

The relationship between the solar day and sidereal is as follows:

The length of the sidereal day is

= (Solar day) (solar year ÷ solar year +  1 day)

As it can be seen that the day i.e. solar its length would be greater as compared with the sidereal day.

In the case when the earn orbited the sun in 9 months rather than 12 so here the sidereal day would be the similar but the solar day would be longer as compared with the original one

For an Ultrasonic Vaporizer Us Duty Rate 4.2%, valued in USD$750, transported via Air and no valid CUSMA certificate available calculate: - Duty Charge - MPF - HMF

2. Same Ultrasonic Vaporizer Us Duty Rate 4.2%, valued in USD$750, transported via Ocean and no valid CUSMA certificate available calculate:

Duty Charge

MPF

HMF

Answers

1. The duty charge, MPF and HMF for the Ultrasonic Vaporizer transported via air are USD$31.50, USD$2.60 and USD$0.94, respectively.

2. The duty charge, MPF and HMF for the Ultrasonic Vaporizer transported via ocean are USD$31.50, USD$0.94 and USD$0.94, respectively.

How to calculate duty charge, MPF and HMF

For the Ultrasonic Vaporizer transported via air

Duty Charge:

Given information;

Duty rate = 4.2%

Value of goods = USD$750

Duty charge = Value of goods x Duty rate

Duty charge = USD$750 x 4.2%

Duty charge = USD$31.50

Merchandise Processing Fee (MPF)

MPF rate = 0.3464% (as of 2021)

Value of goods = USD$750

MPF = Value of goods x MPF rate

MPF = USD$750 x 0.3464%

MPF = USD$2.60

Harbor Maintenance Fee (HMF)

HMF rate = 0.125% (as of 2021)

Value of goods = USD$750

HMF = Value of goods x HMF rate

HMF = USD$750 x 0.125%

HMF = USD$0.94

Therefore, the total import fees for the Ultrasonic Vaporizer transported via air are:

Duty charge = USD$31.50

MPF = USD$2.60

HMF = USD$0.94

For the Ultrasonic Vaporizer transported via ocean

Duty Charge:

Given information:

Duty rate = 4.2%

Value of goods = USD$750

Duty charge = Value of goods x Duty rate

Duty charge = USD$750 x 4.2%

Duty charge = USD$31.50

Merchandise Processing Fee (MPF)

MPF rate = 0.125% (as of 2021)

Value of goods = USD$750

MPF = Value of goods x MPF rate

MPF = USD$750 x 0.125%

MPF = USD$0.94

Harbor Maintenance Fee (HMF):

HMF rate = 0.125% (as of 2021)

Value of goods = USD$750

HMF = Value of goods x HMF rate

HMF = USD$750 x 0.125%

HMF = USD$0.94

Therefore, the total import fees for the Ultrasonic Vaporizer transported via ocean are:

Duty charge = USD$31.50

MPF = USD$0.94

HMF = USD$0.94

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Duty Charge=$31.50

The MPF will be $27.23.

The HMF rate is $0.125 per metric ton.

Ultrasonic Vaporizer transported via Air:

Duty Charge: $31.50

MPF: $26.79

HMF: Not applicable

Ultrasonic Vaporizer transported via Ocean:

Duty Charge: $31.50

MPF: $27.23

HMF: HMF rate and weight of the vaporizer are needed to calculate the HMF.

Ultrasonic Vaporizer transported via Air:

Duty Charge: The duty charge is calculated by multiplying the value of the product ($750) by the duty rate (4.2%): Duty Charge = \(\$750 * 4.2\%\)

= \(\$31.50.\)

MPF: The MPF for air transport is based on the value of the merchandise. As the value of $750 is below the maximum, the MPF will be $26.79.

HMF: The HMF is not applicable for air transport.

Ultrasonic Vaporizer transported via Ocean:

Duty Charge: The duty charge is calculated in the same way as for air transport: Duty Charge = \(\$750 * 4.2\%\)

= \(\$31.50\).

MPF: The MPF for ocean transport is also based on the value of the merchandise. As the value of $750 is below the maximum, the MPF will be $27.23.

HMF: The HMF is applicable for ocean transport. The HMF rate is $0.125 per metric ton. The weight of the Ultrasonic Vaporizer is needed to calculate the HMF.

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A stationary charge Q is point charge. A particle of charge mores in the electric field created by Q from infinity to the distance R between them Part(a)_Findthepotentialatthe distance Rfrom the_chargeQ(ie_thepotential diffetence trarersedby_thetestcharge_ coming from infinity ) Part (6) If the charge q = 0.98 LC; distance R = 128. cm and Q = Ill nC; what is the Work (in J) done by the external force to move into the position R?'

Answers

Part (a):  The potential at a distance R from a point charge Q can be found using the equation:

V = kQ / R

Where k is the Coulomb constant (k = 9 x 10^9 Nm^2/C^2)

So in this case, the potential at a distance R from the point charge Q is:

V = (9 x 10^9 Nm^2/C^2) x (1.11 x 10^-7 C) / (0.128 m)

V = 7.23 x 10^5 V

Therefore, the potential difference traversed by the test charge coming from infinity is 7.23 x 10^5 V.

Part (b):
The work done by the external force to move the charge q to a distance of 128 cm from Q can be found using the equation:

W = q * V

Where q is the charge of the particle, and V is the potential difference (in volts) calculated in part (a).

So in this case, the work done is:

W = (0.98 x 10^-6 C) x (7.23 x 10^5 V)

W = 708 J

Therefore, the work done by the external force to move the charge q to a distance of 128 cm from Q is 708 J.


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A parallel-plate capacitor has square plates that are 7.40 cm on each side and 3.20 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 7.40 cm on a side and 1.60 mm thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is 84.0 V, find how much electrical energy (in nJ) can be stored in this capacitor.

Answers

Answer:

The energy that can be stored in the capacitor is 239 nJ

Explanation:

We first calculate the capacitance of each material. Let C₁ be the capacitance of pyrex glass and C₂ be the capacitance of polystyrene.

C₁ = κ₁ε₀A/d where κ₁ = dielectric constant of pyrex glass = 5, A = area of plates = L² where L = length of square plate = 7.40 cm = 7.40 × 10⁻² m and d = thickness of pyrex slab = 1.60 mm = 1.60 × 10⁻³ m and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

C₁ = κ₁ε₀A/d = κ₁ε₀L²/d = 5 × 8.854 × 10⁻¹² F/m × (7.40 × 10⁻² m)²/1.60 × 10⁻³ m = 2424.2252/1.60 × 10⁻¹¹ F = 1515.14 × 10⁻¹¹ F = 15.2 × 10⁻⁹ F = 15.2 nF

C₂ = κ₂ε₀A/d where κ₂ = dielectric constant of polystyrene = 3, A = area of plates = L² where L = length of square plate = 7.40 cm = 7.40 × 10⁻² m and d = thickness of polystyrene slab = 1.60 mm = 1.60 × 10⁻³ m and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

C₁ = κ₁ε₀A/d = κ₁ε₀L²/d = 3 × 8.854 × 10⁻¹² F/m × (7.40 × 10⁻² m)²/1.60 × 10⁻³ m = 1454.5351/1.60 × 10⁻¹¹ F = 909.08 × 10⁻¹¹ F = 9.09 × 10⁻⁹ F = 9.09 nF

Since the capacitors are in series, we find their effective capacitance C from

1/C = 1/C₁ + 1/C₂

C = C₁C₂/(C₁ + C₂)

= 15.2 × 10⁻⁹ F × 9.09 × 10⁻⁹ F/(15.2 × 10⁻⁹ F + 9.09 × 10⁻⁹ F)

= 138.168 × 10⁻¹⁸/24.29 × 10⁻⁹ F

= 5.69 × 10⁻⁹ F

The amount of electrical energy stored in a capacitor is given by W = 1/2CV² where C = capacitance and v = voltage applied. Now C = 5.69 × 10⁻⁹ F and V = 84.0 V for this capacitor

So W = 1/2 × 5.69 × 10⁻⁹ F × 84.0 V

= 238.98 × 10⁻⁹ J

≅ 239 × 10⁻⁹ J

= 239 nJ

So the energy that can be stored in the capacitor is 239 nJ

What's the difference between a tornado and a hurricane?

Answers

Answer:

hurricanes last longer that tornadoes

Apply the Pauli exclusion principle to determine the number of electrons that could occupy the quantum states described by the following.
(a)
n = 4,
ℓ = 1,
mℓ = −1
_____ electrons
(b)
n = 4,
ℓ = 3
____ electrons

Answers

The number of electrons that could occupy the quantum states described by n = 4, ℓ = 1, and mℓ = −1 is 2.

(a) The quantum numbers given represent the 4p orbital. According to the Pauli exclusion principle, each orbital can accommodate a maximum of two electrons with opposite spins.

The number of electrons that could occupy the quantum states described by n = 4 and ℓ = 3 is 14.

(b) The quantum numbers given represent the 4f subshell. The number of orbitals in the 4f subshell is 7, and each orbital can accommodate a maximum of 2 electrons with opposite spins.

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1) DeAnna, with a mass of 60.0 kg, climbs 3.5 m up a gymnasium rope. How much energy does
a system containing DeAnna and Earth gain from this climb?
2) An electric motor develops 65 kW of power as it lifts a loaded elevator 17.5 m in 35 s. How
much force does the motor exert?
3) A 25.0-kg shell is shot from a cannon at Earth’s surface. The reference level is Earth’s surface.
a. What is the shell-Earth system’s gravitational potential energy when the shell’s height is
425 m?
b. What is the change in the system's potential energy when the shell falls to a height of 225
m?
4) Shawn and his bike have a combined mass of 45.0 kg. Shawn rides his bike 1.80 km in 10.0
min at a constant velocity. What is the system's kinetic energy?
5) Mary weighs 505 N. She walks down a 5.50-m-high flight of stairs. What is the change in the
potential energy of the Mary-Earth system?
6) Katia and Angela each have a mass of 45 kg and are moving together with a speed of 10.0
m/s.
a. What is their combined kinetic energy.
b. What is the ratio of their combined mass to Katia’s mass?
c. What is the ratio of their combined kinetic energy to Katia’s kinetic energy? Explain how
this ratio relates to the ratio of their masses.
7) A 10.0-kg test rocket is fired vertically. When the engine stops firing, the rocket’s kinetic
energy is 1960 J. After the fuel is burned, to what additional height will the rocket rise?

Answers

1. As he climbs... He's gaining potential Energy.

PE=mgh = 60x9.8x3.5=2058J. This is the potential Energy gain.

2. Power = Workdone/Time.

65kW=65,000watts

65000=Workdone/35

Workdone=65,000 x 35=2,275,000J

Recall... Workdone=Force x Distance

given distance =17.5m

Force =Workdone/distance

=2,275,000/17.5

Force=130,000N or 130kN.

3.Potential Energy Increases or decreases as height increases or decreases respectively.

a) PE=mgh =25x9.8x425=104,125J.

b) change in PE=mg(h°-h) =25 x 9.8(225-425)

= -49000J. The Energy is negative because the system loses 49000J.

4. At Constant velocity; d=vt

where d=distance,v=velocity, t=time

v=d/t

1.8km=1800m

10min =10x60=600seconds

v=1800/600

v=3ms-¹

KE=1/2Mv²

=1/2 x 45 x 3² =202.5J.

5.Mary weighs 505N.

Weight=Mass x acceleration due to gravity

W=mg

505=m x 9.8

m=505/9.8 =51.53kg.

Since Mary is coming down... she's decreasing her potential energy cos Potential Energy Decreases as you decrease your height. So Her Initial height was 5.5m... she alighted to the ground which we'll say is the ground(h=0)

Change in PE=mg(h°-h)=51.53 x 9.8(0-5.5) = -2777.5J.

This energy is negative because she lost potential Energy.

6. Since each has a Mass of 45kg... and they're moving together.... Their combined Mass =45+45=90kg.

Velocity given as =10ms-¹

a)Combined KE=1/2MV² = 1/2 x 90 x 10² =4500J.

b)Ratio of combined Mass to Katias mass = CM/KM

Where CM=Combined Mass

KM=Katia's mass

=90/45 =2:1

c) Katia's kinetic E =1/2 x 45 x 10² =2250J.

Ratio of combined to Katia's = CK/KK

CK=Combiner Kinetic Energy

KK=Katia's kinetic energy

= 4500/2250 =2:1.

This answer should be the same cause from the Formula KE=1/2mv²

KInetic Energy varies Directly with Mass

so If the ratio of Masses is 2:1... That of KE should be the same because of their direct Relationship✌.

7.The height at which the Rocket Engine stops is not how far it went. Theres a Minute height that it'll still go before it starts going back down.

from the question... The energy when the engine stops is 1960J

PE=mgh

1960=10x9.8xh

h=1960/98 =20m.

The engine stops at this height

but The additional Minute height that was added before itll start falling cant be calculated from the insufficient info given in this question. If we had the Energy at the additional height-added... We can calculate that height and then subtract 20 from it. That will give the additional added height.

Thats pretty Much Everything✌

Spectro... Out

The energy in the system containing DeAnna and Earth gain from this climb will be 2058J.

How to calculate the energy?

From the information, DeAnna, with a mass of 60.0 kg, climbs 3.5 m up a gymnasium rope. The potential energy will be:

= mgh

= 60 × 9.8 × 3.5

= 2058J

When an electric motor develops 65 kW of power as it lifts a loaded elevator 17.5 m in 35s, the force that the motor exerts will be:

= Work done / Time

= 2275000/17.5

= 130000N

The shell-Earth system’s gravitational potential energy when the shell’s height is 425m will be:

= mgh

= 25 × 9.8 × 425

= 104125J

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a projectile with a mass of 20 grams is launched with a horizontal linear momentum of 15000 g cm/s; after it falls 250 cm vertically what would be its horizontal displacement? assume horizontal velocity remains constant after launch.

Answers

The horizontal displacement of the projectile is determined as 5.3 m.

What is the horizontal displacement of the projectile?

The horizontal displacement of the projectile is calculated using the following equation.

X = vt

where;'

v is the horizontal velocity if the projectilet is the time of motion

The time of motion of the projectile is calculated as follows;

t = √(2h/g)

where;

h is the height of fall of the projectile = 250 cm = 2.5 mg is acceleration due to gravity

t = √(2 x 2.5 / 9.8)

t = 0.71 s

The horizontal velocity of the projectile is calculated as follows;

P = mv

where;

P is momentum of the projectilem is massv is horizontal velocity

v = P/m

v = (15000 g cm/s) / (20 g)

v = 750 cm/s = 7.5 m/s

X = 7.5 x 0.71

X = 5.3 m

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what are the first three harmonics in a 2.45 m long pipe that is open at both ends? what are the first three harmonics of this pipe when one end of the pipe is closed? assume that the speed of sound in air is 345 m/s.

Answers

The frequencies for the first, second, and third harmonics would also be 70.41 Hz, 140.82 Hz, and 211.23 Hz, respectively, when one end of the pipe is closed.

In a 2.45 m long pipe that is open at both ends, the first three harmonics can be determined using the formula:

f = (n * v) / (2L),

where f represents the frequency, n is the harmonic number, v is the speed of sound in air (345 m/s), and L is the length of the pipe (2.45 m).

For the first harmonic (n = 1), the frequency is calculated as f = (1 * 345) / (2 * 2.45) = 70.41 Hz.

For the second harmonic (n = 2), the frequency is f = (2 * 345) / (2 * 2.45) = 140.82 Hz.

For the third harmonic (n = 3), the frequency is f = (3 * 345) / (2 * 2.45) = 211.23 Hz.

When one end of the pipe is closed, the length of the effective vibrating air column is halved.

Thus, the first three harmonics for this closed-end pipe can be calculated by substituting L = 2.45/2 = 1.225 m into the formula.

The frequencies for the closed-end pipe would be the same as the open-end pipe since the formula remains the same.

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if the earth stops rotating about its axis then what will be the change in the value of g at a place in the equator-plane? Assume that earth is a sphere with a radius of 6.4*10*6m.

Answers

Answer:

if the earth stops rotating the value of g increases

If u have a old dogs and it starts whining and it's legs lockup ,and wont let u pick it up. What do i do Im s scared?

Answers

dont panic maybe tell an adult do some research online ? suggest maybe going to the beg to be safe

The dog MUST be SEEN by a vet.

Not on the phone, not from your description. Not a web search, and not asking for advice online.

A qualified veterinarian has to have his hands on the animal, examine the dog, and make some suggestions. It might cost some money.

There's something missing here. A qualified real vet will never just say "I don't know, good bye!".

Which type of wave is most affected by gravity from the Sun and the moon?

Answers

Answer:

A tidal is a type of wave is most affected by gravity from the sun and the moon.

Explanation:

I hope it helps ❤❤

Tidal wave is the type of wave which is most affected by gravity from the

Sun and the moon

The sun and the moon's gravitational pull results to the tidal force being

generated. The tidal force from the tidal wave generated has some effects

on the earth.

It causes the  Earth and its water content to bulge out on two sides of the

earth. It bulges in the area with the highest proximity to the moon and areas

with the lowest proximity. This is responsible for the formation of high tides

in  water bodies.

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At which position, A or B, is the potential energy of the swing the greatest? As the swing moves from point B to point A, which form of energy is increasing, kinetic or potential?

Answers

Answer:

a)  Ua> Ub,  b) Em = K + U

Explanation:

A swing is an oscillatory system, which goes from the highest point of the trajectory to the lowest point and then continues to the highest point on the other side, with a curved trajectory.

Suppose point A is the highest point and point A the lowest point, as the potential energy is given by the formula

              U = m g h

if we define a reference system where the lowest point corresponds to y = or at this point the power energy is zero, therefore

              Ua> Ub

in this movement the mechanical or total energy of the system remains constant, therefore it can be written

            Em = K + U

therefore at the highest point the mechanical energy is only potential and at the lowest point it is only kinetic and at the intermediate points it has a part of each, but the value of the mechanical energy is constant and does not change.

Answer:

At which position, A or B, is the potential energy of the swing the greatest?

b

As the swing moves from point B to point A, which form of energy is increasing, kinetic or potential?

kinetic

Explanation:

Calculate the wavelength of a softball with a mass of 100 g traveling at a velocity of 35 m s^–1, assuming that it can be modeled as a single particle.

Answers

The wavelength of an object can be calculated using the de Broglie equation, which states that the wavelength is equal to Planck's constant divided by the momentum of the object. In this case, we can use the equation:

wavelength = h / p

where h is Planck's constant (6.626 x \(10^-34\) Js) and p is the momentum of the softball.

To calculate the momentum, we can use the equation:

momentum = mass x velocity

Given that the mass of the softball is 100 g (or 0.1 kg) and the velocity is 35 m/s, we can substitute these values into the equation to find the momentum:

momentum = 0.1 kg x 35 m/s
momentum = 3.5 kg m/s

Now we can substitute this momentum value into the wavelength equation:

wavelength = 6.626 x \(10^-34\)Js / 3.5 kg m/s

Calculating this, we find:

wavelength = 1.89 x \(10^-34\) m

The wavelength of the softball is approximately 1.89 x\(10^-34\) meters.

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Hallar la distancia que recorre una movil al cabo
de 8s si posee una velocidad de 5m/s y acelera a razón de 3 m/s

Answers

Answer:

Distancia, S = 136 metros

Explanation:

Dados los siguientes datos;

Aceleración, a = 3 m/s²

Velocidad inicial, u = 5 m/s

Tiempo, t = 8 segundos

Para encontrar la distancia recorrida, usaríamos la segunda ecuación de movimiento;

S = ut + ½at² Sustituyendo en la fórmula, tenemos;

S = 5 × 8 + ½ × 3 × 8²

S = 40 + 1,5 × 64

S = 40 + 96

Distancia, S = 136 metros

Write a paragraph comparing one bulb in series, two bulbs in series, and two bulbs in parallel.
What did you learn?

Answers

Answer:

Explanation:

The series circuit with two light bulbs will illuminate more than the series with one light bulb because two light bulbs will have more energy as compared to one light bulb.

The highest speed achieved by a standard non racing sports car is 3.60 x 10² km/h. Assuming that the car accelerates at a constant rate to reach this maximum speed for 25.0 s and covers a distance of 1.50 km while accelerating, what was the initial velocity of the car?

Answers

The car has a constant acceleration of 1.95m/s2 and reaches a top speed of 41.0m/s. The boat has a constant acceleration of 6.50m/s2 .

What is acceleration ?

When it comes to mechanics, acceleration refers to how quickly an object's velocity changes over time. Acceleration is a vector quantity (in that they have magnitude and direction). The direction of the net force acting on an item determines its acceleration.

According to Newton's Second Law, an object's acceleration is a sum of two factors:

The size of the net balance of all external forces acting on that thing is directly proportional to the net resultant force; the magnitude of the object's mass, depending on the materials from which it was created, is inversely related to the mass.

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Calculate a revised value of δhsoln for your unknown salt, using your average calorimeter constant for the styrofoam cups to account for qcalorimeter.

Answers

A revised value of δhsoln for unknown salt, using average calorimeter constant for the styrofoam cups to account for \(Q_{calorimeter\) is \(Q_{solution\) = δHsoln - \(Q_{calorimeter\)

To calculate the revised value of δHsoln for the unknown salt, we need to account for the heat absorbed or released by the calorimeter (\(Q_{calorimeter\)) using the average calorimeter constant for the styrofoam cups.

The general equation for the heat transfer in a solution is:

\(Q_{reaction\) = \(Q_{calorimeter\) + \(Q_{solution\)

where \(Q_{reaction\) is the heat absorbed or released by the reaction, \(Q_{calorimeter\) is the heat absorbed or released by the calorimeter, and \(Q_{solution\) is the heat absorbed or released by the solution (in this case, the dissolution of the salt).

We can rearrange the equation to solve for qsolution:

\(Q_{solution\) = \(Q_{reaction\) - \(Q_{calorimeter\)

Since we are interested in the enthalpy change (δHsoln) for the dissolution of the salt, we can substitute \(Q_{reaction\) with δHsoln:

\(Q_{solution\) = δHsoln - \(Q_{calorimeter\)

Now, we can calculate the revised value of δHsoln using the average calorimeter constant for the styrofoam cups.

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Write a paragraph comparing one bulb in series, two bulbs in series, and two bulbs in parallel. What did you learn?

Answers

the 2 lightbulbs in series on has one path so it produces  less light. The bulbs in parallel has two lightbulbs but two different paths so it produces more light. (i can make it longer if you want )

which best describes nuclear fission?​

which best describes nuclear fission?

Answers

Answer:

Third option: "A nucleus collides with a neutron and splits, releasing energy."

Explanation:

When nucleus that collides with a neutron splits (fission) releases energy in the process. This is a case of fission.

This is how a fission nuclear reactor works.

Answer:

It be C

Explanation:

Right on edg

An electromagnetic wave Group of answer choices a.Never moves b.Can travel through empty space or matter. c. Can travel only through empty space. d. Cannot travel through matter

Answers

Answer:

b

Explanation:

because electromagnetic waves can travel in vacuum as they don't require particles to transfer energy from one point to another. they can also travel through mediums such as the wall or air, if not how do radio waves transfer energy in this hyper advanced world? through the air

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When an object is moving towards us the waves compresses in a way that it has

Answers

Answer: The waves are compressed as the item moves closer to you, so their wavelength is shorter.

Explanation:

A satellite is orbiting Earth at a speed of approximately 4,000 m/s. Which of these distances is most likely the orbital radius of the satellite’s motion?

Answers

The radius of the satellite's motion is 2.49 × 10⁷ m

Since the centripetal force, F = mv²/r equals the gravitational force, F' = GMm/r² on the satellite,

F = F'

mv²/r = GMm/r²

v² = GM/r

Making r subject of the formula, we have

r = GM/v² where r = radius of satellite's motion, G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of earth = 5.972 × 10²⁴ kg and v = speed of satellite = 4,000 m/s.

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

r = GM/v²

r = 6.67 × 10⁻¹¹ Nm²/kg² × 5.972 × 10²⁴ kg/(4,000 m/s)²

r = 39.83324 × 10¹³ Nm²/kg ÷ 16 × 10⁶ m²/s²

r = 2.4895775 × 10⁷ m

r ≅ 2.49 × 10⁷ m

So, the radius of the satellite's motion is 2.49 × 10⁷ m.

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Answer:

25.0 million meters

Explanation:

force of earth is 6 into 10 to the power 24 Kg 1 kg-iii 6.4 find the force​

Answers

Answer:

Explanation:

Here is the answer---

Given Conditions ⇒

Mass of the Earth(m₁) = 6 × 10²⁴ kg.

Mass of the Moon(m₂) = 7.4 × 10²² kg.

Distance between the Earth and the Moon(d) = 3.84 × 10⁵ km.

= 3.84 × 10⁸ m.

Gravitational Constant(G) = 6.7 × 10⁻¹¹ Nm²/kg².

Using the Newton's law of Gravitation,

F = G × m₁×  m₂ × /d².

F is the Force of Gravitation between the Earth and the Moon.

Substituting the Given Values in the Formula,

∴ F = (6.7 × 10⁻¹¹ × 6 × 10²⁴ × 7.4 × 10²²) ÷ (3.84 × 10⁸)²

⇒ F = (6.7 × 6 × 7.4 × 10¹⁹) ÷ (14.7456)

⇒ F = 20.1741 × 10¹⁹ N.

⇒ F ≈ 20.2 × 10¹⁹ N.

Thus, the Gravitational Force of Attraction between the Earth and the Sun is 20.2 × 10¹⁹ N.

The small bunny exerts a force of 15 Newtons as it hops a distance of 3 meters to pass the large rabbit. How much work did the small bunny do?
45 Joules
0.2 Joules
18 Joules
5 Joules

Answers

Answer:

W = F × d

W = 15 × 3

W = 45 Joules

If I am in a Ferrari and I travelled for 3 hours at a speed of 270 km/h. How far did I travel?
Use the following formula:
distance velocity x time


a)810km

b)90km

c)810km/hr

Answers

it’s b sorry if it’s wrong

3. What current is drawn by a 90 W light bulb on a 110 V household circuit?
Please answer!!!

Answers

Answer:

So to find the current drawn, simply divide the power by the operating voltage. In the United States, most household appliances work at 110 Volts. So the calculation is divide the Wattage by 110.

Explanation:

i don't  know  if that correct because  i just copied and pasted  but it sounds right

An electric bulb is rated 220V and 100 watt. The power consumed by it when operated on
110 V is
i. 25 watt
ii. 50 watt
iii. 75 watt
iv. 90 wat

Answers

The power consumed by it when a bulb operated on 110 V is 50 watts. Thus, option B is correct.

Power is defined as the product of current and potential differences in a circuit. Current is defined as the flow of charges when the circuit is connected to the battery. The power in the circuit is Watt (W).

From the given,

potential difference(V) = 220V

Power (P) = 100 W

Current (I)=?

Power =? ( when the bulb has a voltage of 110V)

Power = Voltage×Current

100       = 220×I

100/220 = I

I = 0.454 A.

Thus, the current in the bulb is 0.454A.

Power = Voltage(V) × Current(I)

           = 110×0.454

           = 49.9 W

Thus, the power in the circuit is 50W.

Hence, the ideal solution is option B.

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times more energy than an earthquake with a magnituide of 2. Ori the Richter scale a magnitude 5 earthquake releases about
A. 30
B. 1000
C. 90
D. 27.000

What is the main caase of the spreading movement of the ocean crust
A. The convection current of the asthenosphere is drasging the ocean crust.
B. The riew magma forming at ocean ridges is pushing the ocean krure
C. There is a stearing motion on translorm plate boundaries
D. All of the above
consribute equally Respr Selestion
A. Increasing temperatures at the top of the asthenosphere
B. There is a greater water content in subducted oceanic crust
c. Decreasing pressure as the overlying crust thins
D. Higher viscosity as the silica content increases E. All of the above Reset Selection

Answers

The main cause of the spreading movement of the ocean crust is the new magma forming at ocean ridges, which pushes the oceanic plates apart. The other options listed do not contribute equally or directly to the spreading movement.

The spreading movement of the ocean crust is primarily driven by the formation of new magma at ocean ridges. As the mantle material beneath the Earth's surface rises and melts, it creates new magma. This molten material then pushes its way upward and fills the gap between the separating oceanic plates. As the new magma solidifies, it forms a new crust, expanding the ocean floor and causing the plates to move apart.

The other options listed do not directly contribute to the spreading movement of the ocean crust. Increasing temperatures at the top of the asthenosphere, the greater water content in subducted oceanic crust, decreasing pressure as the overlying crust thins, and higher viscosity as the silica content increases may have their own geological implications, but they are not the main causes of the spreading movement observed at ocean ridges.

Therefore, the correct answer is B. The new magma forming at ocean ridges is the main cause of the spreading movement of the ocean crust.

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The complete question is:

What is the main cause of the spreading movement of the ocean crust? A. As the ocean crust cools it becomes denser than the asthenosphere and the edge of the slab is pulled downward at subduction zones. B. The convection current of the asthenosphere is dragging the ocean crust C. The new magma forming at ocean ridges is pushing the ocean crust D. There is a shearing motion on transform plate boundaries E. All of the above contribute equally

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