How to calculate specific heat.

Answers

Answer 1
Need more information to answer this question

Related Questions

If the submarine intends to submerge to a depth of 150 m, what pressure must it be designed to withstand

Answers

Explanation:

Hydrostatic pressure, P = hdg where d is density if the liquid ( water)

P= 150m × 1000kgm-3 × 10ms-2

P= 1.5 ×10^6 Pa

What quantity is represented by the mass number minus the atomic number?.

Answers

Answer:

Mass = mass of nucleus

Atomic Number = number of protons (charge) of nucleus

Mass - Atomic Number = number of neutrons

State a hypothesis about your estimated daily water use.

Answers

Answer

Estimates vary, but, on average, each person uses about 80-100 gallons of water per day, for indoor home uses.

Explanation:

lonic Naming Practice
Write the name of the following chemical compound:
Na3Cl1

Answers

Answer:

sodium chloride. .....

Ionic naming practice: Na3Cl1= Sodium Chloride

An "energy bar" contains 26 ggof carbohydrates.How much energy is this in joules?

Answers

I’m it’s about 3.4878 energy in joules

Can u plz help me with my hw

Answers

Answer:

1) 4 x 25m = 100m

2) 0 because after 4 lengths, he's back at the starting block.

3) speed is distance over time so speed here is 100m/125s = 0.8m/s

4) ...

5) 100m / 1.25m/s = 80s

6.a) 100m / 0.5m/s = 200s

7.b) ... can't draw this now..

A block with a mass of 6.0 kg is
held in equilibrium on an incline
of angle θ = 30.0° by a horizontal
force, F, as shown in the figure.
Find the magnitudes of the
normal force on the block and of F. (Ignore friction.)

Answers

(a) The normal force on the block is 50.92 N.

(b) The horizontal force on block keeping it in equilibrium is 29.4 N.

The given parameters;

mass of the block, m = 6 kgangle of inclination, θ = 30.0°

The normal force on the block is calculated as follows;

[tex]F_n = W \times cos(\theta)[/tex]

where;

W is the weight of the block

[tex]F_n = mg \times cos(\theta)\\\\F_n = 6 \times 9.8 \times cos(30)\\\\F_n = 50 .92 \ N[/tex]

The horizontal force on block keeping it in equilibrium is calculated as follows;

[tex]F- F_x = 0\\\\F-mgsin\theta= 0\\\\F = mgsin\theta\\\\F = 6 \times 9.8 \times sin(30)\\\\F = 29.4 \ N[/tex]

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What symbols are these?

Answers

Answer:

the bottom one is wollsiegel

given. force of 88N and an acceleration of 4 m/s 2 what is the mass?

Answers

Mass, m=22 kg
It is given that Force, F = 88N
Acceleration, a=4 m/s 2

How does the radius of a string affect centripetal force.

Answers

Answer:

because a raduis is half of 25% of a cicrle.

Explanation:

The mercury in a barometer of a cross-sectional area 1 cm² stands at 75 cm, and the space above it is 9 cm in length. What
volume of air, measured at 3. The mercury in a barometer of a cross-sectional area 1 cm² stands at 75 cm, and the space above it is 9 cm in length. What
volume of air, measured at atmospheric pressure, would have to be admitted into the space to cause the column of the mercury
to drop to 59 cm?

Answers

The ideal gas equation and the pressure in barometer allows us to find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:

The  variation  of the volume is: ΔV = 7.67 cm³

Pressure is defined by the relationship between force and area.

       P = F / A

The ideal gas equation establishes a relationship between pressure, volume, and temperature of an ideal gas.

          PV = nR T

Where P is pressure, V is volume, and T is temperature.

Let's write this equation for two points assuming that the temperature has not changed.

          P₀ V₀ = P₁ V₁

          V₁ = [tex]\frac{P_o}{P_1} \ V_o[/tex]                 (1)

The subscript "o" is used for the start point and the subscript "1" for the end point.

The pressure in a barometer is:

         P = ρ g y

They indicate the initial height of the barometer y₀=75 cm, the distance from empty space y'₀ = 9 cm and the final height of the barometer y₁ = 59 cm.

 

The volume of the cylinder is

         V = π r² y

Let's calculate the initial volume.

         V₀ = π 1 9

         V₀ = 28.27 cm³

We substitute in equation 1.

         V₁ = [tex]\frac{\rho \ g \ y_o}{\rho \ g \ y_1} \ V_o[/tex]  

         V₁ = [tex]\frac{y_o}{y_1} \ V_o[/tex]  

Let's calculate.

        V₁ = [tex]\frac{75}{59} \ 27.27[/tex]  

        V₁ = 35.94 cm³

The volume to be incremented is

         ΔV = V₁ - V₀

         ΔV = 35.94 - 28.27

         ΔV = 7.67 cm³

Using the ideal gas equation and the pressure in barometer we can find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:

The change of the volume is: ΔV = 7.67 cm³

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Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, this gun accelerates a 5.6-kg projectile from rest to a speed of 3.5 × 103 m/s. The net force accelerating the projectile is 2.5 × 105 N. How much time is required for the projectile to come up to speed?

Answers

Answer:

7.84 s

Explanation:

Ok, one step at a time. We know the mass of the projectile, we know the force. Acceleration is easily obtained from Newton second law:

[tex]\vec F = m\vec a[/tex][tex]2.5\times 10^5 = 5.6a \rightarrow a = (2.5/5.6) \times 10^5 = 4.46 \times 10^4 ms^-^2[/tex].

At this point, we know the acceleration, we know initial and final velocity, we can time the ammount of time it took to get there.

[tex]v= v_0 +at \rightarrow 3.5\times10^3 = 0 + 4.46\times10^4 t\\t= \frac{3.5\times10^3}{4.46\times10^4} = 7.84 s[/tex]

. Quỹ đạo chuyển động của một vật là

Answers

Answer:

you are very kind Po lang

I need help in question 7, a and b.

Answers

Answer:

The graph for 7a is shown in the attachment. For question 7b she walks a distance of 16 meters. (m)

Explanation:

A 75.0 kg man pushes on a 500,000 kg wall for 250 s but it does not move.

Answers

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

The work done in the given case is 0

[tex] \large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}[/tex]

For calculating the work done by an object we apply the formula, that is ~

[tex]f \times s \times \cos( \theta) [/tex]

where f represents force applied, and s represents displacement.

but here, the displacement is zero since the wall didn't move at all, Therefore If we plug it's value as 0, we get

[tex]f = 0 \times s \times \cos( \theta) [/tex]

hence,

[tex]work \: done = 0[/tex]

Solution:Here, though the man is applying force on the wall, the work done is zero. This is because work is said to be done only when a force acting on a body produces motion in it in the direction of the force applied. Here, the man is trying to push the wall with a great force. He might fell exhausted after some time and say that he has done a lot of work. But in terms of physics, he has done no work because the force applied by the person has not produced any displacement. We know, work done = force × displacement. If displacement is zero, then work done is zero.

So therefore, we can conclude that the man hasn't done any work.

Hope it helps.

Do comment if you have any query.

which wave has a higher frequency and why?

Answers

Explanation:

the figure in the left side has higher frequency.

because it has more nos. of wave in 1sec.

how do i do this question in science?

Answers

simple subtract and ummmmmm.....

Without friction, what is the mass of an ball accelerating at 1.8 m/sec2 to which an
an unbalanced force of 42 Newtons has been applied?
A 75.60 kg
B 00.04 kg
C 23.33 kg
D 43.80 kg

Answers

Answer:

23.33 kg

Explanation:

The mass of the object can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force

a is the acceleration

From the question we have

[tex]m = \frac{42}{1.8} = 23.3333... \\ [/tex]

We have the final answer as

23.33 kg

Hope this helps you

Answer:

[tex]\boxed {\boxed {\sf C. \ 23.33 \ kg}}[/tex]

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

[tex]F=ma[/tex]

The mass of the ball is unknown. The ball is accelerating at 1.8 meters per second squared. An unbalanced force of 42 Newtons is applied to the ball.

Convert the units of force. 1 Newton is equal to 1 kilogram meter per second squared, so our answer of 42 Newtons is equal to 42 kg*m/s².

F= 42 kg*m/s² a= 1.8 m/s²

Substitute the values into the formula.

[tex]42 \ kg*m/s^1 = m * 1.8 \ m/s^2[/tex]

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 1.8 meters per second squared. The inverse operation of multiplication is division. Divide both sides by 1.8 m/s².

[tex]\frac {42 \ kg*m/s^2}{1.8 \ m/s^2} = \frac{a*1.8 \ m/s^2}{1.8 \ m/s^2}[/tex]

[tex]\frac {42 \ kg*m/s^2}{1.8 \ m/s^2} =m[/tex]

The units of meters per second squared cancel.

[tex]\frac {42 \ kg}{1.8 } =m[/tex]

[tex]23.3333333 \ kg=m[/tex]

Round to the hundredth place. The 3 in the thousandth place tells us to leave the 3 in the hundredth place.

[tex]23.33 \ kg \approx m[/tex]

The mass of the ball is approximately 23.33 kilograms.

what is Secular Music's instrument's?​

Answers

Answer:

Drums, harps, recorders, and bagpipes.

Explanation:

How fast is an object going if it travels from San Diego to Anaheim in 1.25 hours (hr)? The distance from San Diego to Anaheim is 93 miles (mi).

a) 74.4 mi/hr

b) 116.25 mi/hr

c) 0.013 mi/hr

d) 84.25 mi/hr

e) None of the answers

Answers

Answer:

74.4 mph

Explanation:

Since we have the distance in miles and the time it took in hours, we can divide the two to get miles per hour.

speed = mi/hr

speed = 93/1.25

speed = 74.4 mph

how have astronomers interpreted the unexpectedly fast rotation of galaxies

Answers

Answer:

There must be a lot of dark matter that can be felt but not seen

The astronomers interpreted the unexpectedly fast rotation of galaxies that there must be a large quantity of dark energy whose gravity is detectable yet invisible.

What is a galaxy?

Any system of stars plus interstellar material that makes up the cosmos is referred to as a galaxy. Such assemblages are common, and many of them are so massive that they hold tens of trillions of stars.

A vast variety of galaxies, from dim, hazy dwarf objects to spectacular spiral-shaped giants, have been created by nature. Almost all galaxies seem to have formed shortly after the universe started, and they are everywhere in space, even at the farthest limits of the universe that can be seen by the most advanced telescopes.

The majority of galaxies are found in clusters, many of which are further organized into clusters that span hundreds of billions of light-years.

Since there are almost empty spaces between these so-called super clusters, the universe's overall structure resembles a network of sheets or chains of galaxies.

To know more about Galaxy:

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Which of the following is NOT true of a hotspot?

Answers

Given what we know about hotspots and their characteristics, we can confirm that the option that is false regarding hotspots is "A. Hotspots are constantly moving forming chains of volcanoes".

In regards to the study of geology, we refer to hotspots as spots in an oceanic mantle where magma located below causes them to become intensely hot. This heat forces magma and therefore the mantle itself upwards as a tectonic plate moves across the spot, and therefore results in the formation of volcanic chains. Some very popular examples of volcanoes and volcanic islands formed by this method include:

Yellowstone National ParkHawaiiThe country of Iceland

Despite the other options being true, option A is false given that hotspots are stationary elements of the oceanic mantle. Although some scientists believe that there is evidence of the movement of hotspots, this movement is slow enough to not be taken into account.

This question was answered in regards to the complete question found online which states:

Which of the following is NOT TRUE about hotspots? a. Hotspots are constantly moving forming chains of volcanoes.

b. Hotspots are stationary.

c. Hotspots form volcanic island arcs

d. Hotspots lie along an oceanic plate

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4) An 873kg dragster, starting from rest, attains a speed of 26.3m/s in 0.590s.
A) What was the acceleration of the dragster?
B) What was the net force needed to accelerate the dragster?
C) Assume that the driver has a mass of 68.0kg. What horizontal force does the seat exert on
him?

Answers

Answer:

A.)a=44.57m/s

B.)f=38,909.61

C.)f=666.4

Explanation:

A.) A=v/t 26.3/0.590=44.57

B.) 873kg x 44.57= 38,909.61

C.) 68kg x 9.8=666.4

A transformer is being built to triple the voltage produced. Which data are collected to investigate whether the transformer can achieve the goal? (1 point)

A) the ratio of the number of turns in the primary and secondary coils

B) only the number of turns in the secondary coil and the magnetic field of the core

C) only the number of turns in the primary coil and the magnetic field of the core

D) the ratio of the number of turns in the primary and secondary coils and the magnetic field of the core

Answers

Answer:

the ratio of the number of turns in the primary and secondary coils.

Explanation:

hope i'm not too late!!

You throw a can of cranberry sauce straight up in the air with an initial velocity of 12 m/s. What is the maximum height the baseball reaches above your head.

Answers

Answer:

Δx=7.35m

Explanation:

This is a free fall problem (g= -9.8 m/s²).

Displacement: ? <--- what you're trying to find (max height)

Initial Velocity: 12 m/s

Final Velocity: 0 m/s <-- since the can will land back in your hands

Time: Missing <-- not needed to solve

Acceleration: -9.8 m/s²

The equation that you will use is vf²=vi²+2aΔx

Now plug the information into the equation!

(0m/s)²=(12m/s)²+2(-9.8m/s²)Δx

Multiply the 2 with the acceleration (-9.8m/s²)

(0m/s)²=(12m/s)²-19.6m/s²Δx

Now you need to square the final and initial velocity

0m/s=144m/s-19.6m/s²Δx

Move 144 to the other side of the equal sign

-144m/s=-19.6m/s²Δx

Divide both sides by -19.6m/s²

7.346m=Δx

You can round it to the tenth or hundreth place, up to you!

For this example, I'll round to the hundredth...

Δx=7.35m

I would double check this answer or review it yourself to see if it's correct.

Good luck with your studies!

when light enters water, it bends. what does the amount of bending depend on?

Answers

Answer:

the change of speed!

Explanation:

if light speeds up or slow down more it bend more

A particular roller coaster has a mass of 3500 kg, a height of 4.0, and a velocity of 12m/s. What is the potential energy? If needed, use g=10.m/s^2

Answers

[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]

Gravitational Potential Energy of an object is calculated by formula ~

[tex] \large\boxed{\sf P = mgh}[/tex]

where,

m = mass of the object = 3500 kg

g = Acceleration due to gravity = 12 m/s²

h = height attained by the object = 4 m

Now, let's calculate its potential energy ~

[tex]3500 \times 10 \times 4[/tex]

[tex]140000 \: \: joules[/tex]

[tex]140 \: \: kj[/tex]

Answer:

Potential Energy of an object is calculated by formula:

Potential Energy (P.E)=m×g×h

Where,

m=mass of bodyg=acceleration due to gravityh=height from the earth surface

Now, let's solve the question.

Given,

mass(m)=3500 kgheight (h)=4mvelocity (v)=10m/s²

Now,

We know that,

Potential Energy (P.E)=mgh

[tex] = 3500 \times 10 \times 4[/tex]

[tex] = 3500 0\times 4[/tex]

[tex] =140000 joules [/tex]

[tex]\mathfrak{\blue{DisneyPrincess29}}[/tex]

Suppose you were able to take measurements of current through the
resistor and voltage across the resistor. Your record of numerical values is written in a

tabular data below​

Answers

Answer:

  see attached

Explanation:

It will be easier to plot the data if you have grid lines to work with.

In the attached, we had to reverse the order of entries in the table, because the graphing program likes the independent variable listed first.

Calculate torque using angular momentum

Answers

Answer:

The equation net τ=ΔLΔt net τ = Δ L Δ t gives the relationship between torque and the angular momentum produced.

I'll give brainliest​

Answers

Good luck dude hahaha

Answer: figuratively speaking both the lines for the force Q and force P is equal.

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