An ice chest at a beach party contains 12 cans of soda at 3.78 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 6.48-kg watermelon at 29.4 °C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon in degrees Celsius.

Answers

Answer 1

Answer:

T = 13.25°C

Explanation:

From the law of conservation of energy:

Heat Lost by Watermelon = Heat Gained by Cans

[tex]m_wC_w\Delta T_w = m_cC_c\Delta T_c[/tex]

where,

[tex]m_w[/tex] = mass of watermelon = 6.48 kg

[tex]m_c[/tex] = mass of cans = (12)(0.35 kg) = 4.2 kg

[tex]C_w[/tex] = specific heat capacity of watermelon = 3800 J/kg.°C

[tex]C_c[/tex]  = specific heat capacity of cans = 4200 J/kg.°C

[tex]\Delta T_w[/tex] = Change in Temprature of watermelon = 29.4°C - T

[tex]\Delta T_c[/tex] = Change in Temperature of cans = T - 3.78°C

T = final temperature = ?

Therefore,

[tex](4.2\ kg)(3800\ J/kg.^oC)(29.4^oC-T)=(6.48\ kg)(4200\ J/kg^oC)(T-3.78^oC)\\469224\ J-(15960\ J/^oC)T = (27216\ J/^oC)T-102876.48\ J\\469224\ J + 102876.48\ J = (27216\ J/^oC)T+(15960\ J/^oC)T\\\\T = \frac{572100.48\ J}{43176\ J/^oC}[/tex]

T = 13.25°C


Related Questions

It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.
1. What is the passenger's apparent weight before the elevator starts moving?
2. What is the passenger's apparent weight whilethe elevator is speeding up?
3. What is the passenger's apparent weight afterthe elevator reaches its cruising speed?

Answers

Answer:

1. 588 N

2. 738 N

3. 588 N

Explanation:

time, t = 4 s

initial velocity, u = 0

final velocity, v = 10 m/s

mass, m= 60 kg

1.

Weight of passenger before starts

W =m g = 60 x 9.8 = 588 N

2.

When the elevator is speeding up

v = u + a t

10 = 0 + a x 4

a = 2.5 m/s2

Now the weight is

W' = m (a + g) = 60 (9.8 + 2.5) = 738 N

3.

When he reaches the cruising speed, the weight is

W = 588 N

When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What maximum wavelength of light could be used to ejectelectrons from aluminum

Answers

Answer:

The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm

Explanation:

Given;

wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m

maximum kinetic energy of the ejected electron, K.E = 0.92 eV

let the work function of the aluminum metal = Ф

Apply photoelectric equation:

E = K.E + Ф

Where;

Ф is the minimum energy needed to eject electron the aluminum metal

E is the energy of the incident light

The energy of the incident light is calculated as follows;

[tex]E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J[/tex]

The work function of the aluminum metal is calculated as;

Ф = E - K.E

Ф = 8.02 x 10⁻¹⁹  -  (0.92 x 1.602 x 10⁻¹⁹)

Ф =  8.02 x 10⁻¹⁹ J   -  1.474 x 10⁻¹⁹ J

Ф = 6.546 x 10⁻¹⁹ J

The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;

[tex]\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm[/tex]

In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-called cathode rays. One remarkable series of experiments with cathode rays, conducted by J. J. Thomson around 1897, led to the discovery of the electron.
With the idea that cathode rays were charged particles, Thomson used a cathode-ray tube to measure the ratio of charge to mass, q/m, of these particles, repeating the measurements with different cathode materials and different residual gases in the tube.
Part A
What is the most significant conclusion that Thomson was able to draw from his measurements?
He found a different value of q/m for different cathode materials.
He found the same value of q/m for different cathode materials.
From measurements of q/m he was able to calculate the charge of an electron.
From measurements of q/m he was able to calculate the mass of an electron.
Part B
What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.
Express your answer in terms of e, m, d, v0, L, and E0.
Part C
Now imagine that you place your entire apparatus inside a region of magnetic field of magnitude B0 (Figure 2) . The magnetic field is perpendicular to E⃗ 0 and directed straight into the plane of the figure. You adjust the value of B0 so that no deflection is observed on the screen.
What is the speed v0 of the electrons in this case?
Express your answer in terms of E0 and B0.

Answers

Answer:

a) He found the same value of q/m for different cathode materials.

b)      y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex] ,  c)  v = [tex]\frac{E_o}{B_o}[/tex]

Explanation:

In Thomson's experiments he was able to measure the deflection of the light beam under the effect of the magnetic field and with these results find the e / m relationship, which in all cases is the same, therefore the most important conclusion is that the value e E / m is constant for all materials.

b) In the part of the plates the electrons are accelerated by the electric field,

              F = ma

             - e E = m a

              a = - (e/m)  E₀

               

the distance traveled is          

X axis

          x = v₀ t

the separation of the plates is x = d

          t = vo / d

               

Y axis

          y = v_{oy} t + ½ to t²

          y = ½ a t²

          y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex]

           

c) In this case there is a magnetic field B₀ and the electrons have no deflection

         F = - e E + e v x B

       

if there is no deviation F = 0

         e E = e v B

         v = [tex]\frac{E_o}{B_o}[/tex]

A body of mass 2kg is released from from a point 100m above the ground level. calculate kinetic energy 80m from the point of released.​

Answers

Answer:

1568J

Explanation:

Since the problem states 80 m from the point of drop, the height relative to the ground will be 100-80=20m.

Use conservation of Energy

ΔUg+ΔKE=0

ΔUg= mgΔh=2*9.8*(20-100)=-1568J

ΔKE-1568J=0

ΔKE=1568J

since KEi= 0 since the object is at rest 100m up, the kinetic energy 20meters above the ground is 1568J

Two positive charges, 91 = 5 x 10-'[C] and q2 =1 x 10-9 [C], are
separated by a distance of d=0.05 m. At location 'P' between the
two charges, the net electric field is found to be zero.

b. [10 points] The distance between charge qı and location 'P' is
considered to be 'x'. Find the value of 'x' in [cm]

Answers

Answer:

wareffctgggyyggghhhh

Paauto A: Isulat sa papel ang alpabetong Ingles at bilang I hanggang 10 sa istilong
Roman ng pagleletra.​

Answers

Answer:

Explanation:

English alphabets numbered fro 1 to 26

and the numbers 1 to10 so they are written in roman numbers as

1 - I

2 - II

3 - III

4 - IV

5 -V

6 - VI

7 -VII

8 - VIII

9 - IX

10 -X

11 - XI

12 - XII

13 - XIII

14 - XIV

15 - XV

16 - XVI

17 - XVII

18 - XVIII

19 - XIX

20- XX

21 - XXI

22 - XXII

23 - XXIII

24 - XXIV

25 - XXV

26 - XXVI  

A uniform ladder of length 24 m and weight w is supported by horizontal floor at A and by a vertical wall at B. It makes an angle 45 degree with the horizontal. The coefficient of friction between ground and ladder is 1/2 and coefficient of friction between ladder and wall is 1/3. If a man whose weight is one-half than the ladder, ascends the ladder, how much length x of the ladder he shall climb before the ladder slips

Answers

Answer:

I could not find the answer or do it myself if I did find it I would defenetly share

If an object of a constant mass experiences a constant net force, it will have a constant what?

Answers

Explanation:

hope it helps !!!!!!!!!!!!!

If an object of a constant mass experiences a constant net force, it will have a constant acceleration.

What is force?

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.

According to Newton's second law of motion:

Applied force = mass × acceleration.

Hence, if an object of a constant mass experiences a constant net force, it will have a constant acceleration.

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Click Stop Using the slider set the following: coeff of restitution to 1.00 A velocity (m/s) to 6.0 A mass (kg) to 6.0 B velocity (m/s) to 0.0 Calculate what range can the mass of B be to cause mass A to bounce off after the collision. Calculate what range can the mass of B be to cause mass A to continue forward after the collision. Check your calculations with the simulation. What are the ranges of B mass (kg)

Answers

Answer:

[tex]M_b=6kg[/tex]

Explanation:

From the question we are told that:

Coefficient of restitution [tex]\mu=1.00[/tex]

Mass A [tex]M_a=6kg[/tex]

Initial Velocity of A [tex]U_a=6m/s[/tex]

Initial Velocity of B [tex]U_b=0m/s[/tex]

Generally the equation for Coefficient of restitution is mathematically given by

 [tex]\mu=\frac{V_b-V_a}{U_a-U_b}[/tex]

 [tex]1=\frac{v_B}{6}[/tex]

 [tex]V_b=6*1[/tex]

 [tex]V_b=6m/s[/tex]

Generally the equation for conservation of linear momentum  is mathematically given by

 [tex]M_aU_a+M_bU_b=M_aV_a+M_bV_b[/tex]

 [tex]6*6+=M_b*6[/tex]

 [tex]M_b=6kg[/tex]

What is true when an object floats in water? A. When an object floats, it exceeds the volume of water available. B. When an object floats, it displaces a volume of water equal to its own volume. C. When an object floats, it does not displace its entire volume.

Answers

Answer:

C. When an object floats, it does not displace its entire volume.

Explanation:

Buoyancy can be defined as an upward force which is created by the water displaced by an object.

According to Archimede's principle, it is directly proportional to the amount (weight) of water that is being displaced by an object.

Basically, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up. The buoyancy of an object is given by the formula;

[tex] Fb = pgV [/tex]

[tex] But, \; V = Ah [/tex]

[tex] Hence, \; Fb = pgAh [/tex]

Where;

Fb = buoyant force of a liquid acting on an object.

g = acceleration due to gravity.

p = density of the liquid.

v = volume of the liquid displaced.

h = height of liquid (water) displaced by an object.

A = surface area of the floating object.

The unit of measurement for buoyancy is Newton (N).

Additionally, the density of a fluid is directly proportional to the buoyant force acting on it i.e as the density of a liquid decreases, buoyancy decreases and vice-versa.

Furthermore, an object such as a boat, ship, ferry, canoe, etc, are able to float because the volume of water they displace weigh more than their own weight. Thus, if a boat or any physical object weighs more than the volume of water it displaces, it would sink; otherwise, it floats.

In conclusion, the true statement is that when an object floats, it does not displace its entire volume.

Which indicates the first law of thermodynamics ​

Answers

Answer:

(d)

Explanation:

because dU = Q -W so ,that the option d(D) is correct

Please help,it is urgent!)

Answers

Answer:

answer is 18.58 because

Answer:

My answer is d.25.1 because

Suppose the pucks start spinning after the collision, whereas they were not before. Will this affect your momentum conservation results

Answers

Answer:

No, it will not affect the results.

Explanation:

For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.

What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.

Which phase of matter makes up stars?
O liquid
O gas
O plasma

Answers

Answer:

The answer to this question is plasma

Answer:

Plasma

Explanation:

If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additional 4.0 cm

Answers

16 additional J of work

Astronauts in space move a toolbox from its initial position ????????→=<15,14,−8>m to its final position ????????→=<17,14,−1>m. The two astronauts each push on the box with a constant force. Astronaut 1 exerts a force ????1→=<18,7,−12>???? and astronaut 2 exerts a force ????2→=<16,−10,16>????.

Required:
What is the total work performed on the toolbox?

Answers

If both forces are measured in Newtons, then the net force is

F = (18, 7, -12) N + (16, -10, 16) N = (34, -3, 4) N

The toolbox undergoes a displacement (i.e. change in position) in the direction of the vector

d = (17, 14, -1) m - (15, 14, -8) m = (2, 0, -9) m

The total work done by the astronauts on the toolbox is then

F • d = (34, -3, 4) N • (2, 0, -9) m = (68 + 0 - 36) N•m = 32 J

The work done by the two astronauts is equal to 96 J.

What is work done?

work done?Work done is defined as the product of force applied and the distance moved by the force.

Work done = Force × Distance

The forces applied = 18+16 N, 7+ -10 N, and -12 + 16N

Forces = 34 N, -3 N, and 4N

Distances = (17 - 15, 14 - 14, -1 - - 8) m

Distances = 2, 0, 7

Work done = 34 × 2 + -3 × 0 + 4 × 7

Work done = 96 J

Therefore, the work done by the two astronauts is equal to 96 J.

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The primary coil in a transformer has 250 turns; the secondary coil has 500. Which is correct?

a. This is a step-down transformer.

b. The voltage in the secondary coil will be higher than in the primary.

c. The power in the secondary coil is greater.

d. The power in the primary coil is greater.

Answers

Explanation:

option b is the correct one

What is the total number of moles of products involved in the following reaction?
CaCO3 (s) + 2HCl (aq) - CaCl2 (aq) + CO2 (g) + H20 (g)
O 6
2.
3
5

Answers

Answer:

3

Explanation:

You must first make sure the equation is balanced. This one is. Then, you simply add up the coefficients of each compound on the products side of the equation. When the coefficient is not specified, you can assume it is 1 mole. So, in this equation, there is 1 mole of CaCl₂, 1 mole of CO₂, and 1 mole of H₂O = 3 moles.

The reactant side of the equation also has three moles:

1 mole of CaCO₃ and 2 moles of HCl.

. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the box hits the wall, stop the animation. What can you tell me about relative magnitudes of the frictional force and the applied force

Answers

Answer:

elative magnitude of the two forces is the same and they are applied in a constant direction.

Explanation:

Newton's second law states that the sum of the forces is equal to the mass times the acceleration  

              ∑ F = m a

in this case there are two forces on the x axis

             F_applied - fr = 0

since they indicate that the velocity is constant, consequently

             F_applied = fr

the relative magnitude of the two forces is the same and they are applied in a constant direction.

water contracts on freezing is it incorrect or conrrect

Answers

Answer:

hope it helps

much as you can

vector A has a magnitude of 8 unit make an angle of 45° with posetive x axis vector B also has the same magnitude of 8 unit along negative x axis find the magnitude of A+B?​

Answers

Answer:

45 × 8 units = A + B as formular

Choose the force diagram that best represents a ball thrown upward by Peter, at the
top of its path.

Diagram A
Diagram B
Diagram C
Diagram D

Answers

Answer:Diagram A

Explanation:

Since the air resistance is to be neglected, only the gravitational force acts on the ball ( and has acted all the way from the throw upward). Diagram A reflects this fact correctly indicating the gravity acting on the ball downward.

How do you find the product of gamma decay?

Answers

Answer:

The mass and atomic numbers don't change

Explanation:

An excited atom relaxes to the ground state emitting a photon...called a gamma ray.

The answer is that the mass and atomic numbers don't change.

In gamma decay, the product refers to the nucleus resulting from the emission of a gamma ray. Gamma decay occurs when an excited atomic nucleus releases excess energy in the form of a high-energy photon called a gamma ray.

To find the product of gamma decay, you need to identify the nucleus before and after the decay process. The product nucleus is determined by the parent nucleus that undergoes gamma decay.

During gamma decay, the number of protons and neutrons in the nucleus remains unchanged. Therefore, the identity of the element remains the same, but the energy state of the nucleus is altered.

The product nucleus is typically represented by the same chemical symbol as the parent nucleus, followed by a superscript indicating the mass number (total number of protons and neutrons) and a subscript indicating the atomic number (number of protons).

For example, if a parent nucleus with an atomic number of Z and a mass number of A undergoes gamma decay, the product nucleus will have the same atomic number Z and mass number A.

It's important to note that gamma decay does not involve the emission or absorption of any particles, only the release of electromagnetic radiation (gamma ray).

Thus, the product nucleus remains unchanged in terms of atomic number and mass number.

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Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car B is 100 from car A, car A begins to accelerate toward car B with a constant acceleration of 5 m/s/s. Let right be positive.
1) How much time elapses before the two cars meet? 2) How far does car A travel before the two cars meet? 3) What is the velocity of car B when the two cars meet?
4) What is the velocity of car A when the two cars meet?

Answers

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

[tex]A_a(t) = 5m/s^2[/tex]

To get the velocity, we integrate over time:

[tex]V_a(t) = (5m/s^2)*t + V_0[/tex]

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

[tex]V_a(t) = (5m/s^2)*t[/tex]

To get the position equation we integrate again over time:

[tex]P_a(t) = 0.5*(5m/s^2)*t^2 + P_0[/tex]

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

[tex]P_a(t) = 0.5*(5m/s^2)*t^2[/tex]

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

[tex]V_b(t) =20m/s[/tex]

To get the position equation, we can integrate:

[tex]P_b(t) = (20m/s)*t + P_0[/tex]

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

[tex]P_b(t) = (20m/s)*t + 100m[/tex]

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

[tex]P_a(t) = P_b(t)[/tex]

We can solve this for t.

[tex]0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0[/tex]

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

[tex]t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}[/tex]

We only care for the positive solution, which is:

[tex]t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s[/tex]

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

[tex]P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m[/tex]

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

[tex]V_a(t) = (5m/s^2)*11.48s = 57.4 m/s[/tex]

Two projectiles A and B are fired simultaneously from a level, horizontal surface. The projectiles are initially 62.2 m apart. Projectile A is
fired with a speed of 19.5 m/s at a launch angle 30° of while projectile B is fired with a speed of 19.5 m/s at a launch angle of 60°. How long
it takes one projectile to be directly above the other?​

Answers

Let the point where A is launched act as the origin, so that the horizontal positions at time t of the respective projectiles are

• A : x = (19.5 m/s) cos(30°) t

• B : x = 62.2 m + (19.5 m/s) cos(60°) t

These positions are the same at the moment one projectile is directly above the other, which happens for time t such that

(19.5 m/s) cos(30°) t = 62.2 m + (19.5 m/s) cos(60°) t

Solve for t :

(19.5 m/s) (cos(30°) - cos(60°)) t = 62.2 m

t = (62.2 m) / ((19.5 m/s) (cos(30°) - cos(60°))

t8.71 s

A boy throws a ball upward with a velocity of 4.50 m/s at 60.0o. What is the maximum height reached by the ball?

Answers

Answer:

3.1m

Explanation:

Since we only care about the y direction we only need to find vy. Once u draw your vector you will realize that vy= 4.5sin60=3.897m/s.

use vf²=v²+2a(y)

At the maximum height the velocity is 0 and since the object is in freefall, a=-g

Plug in all values

0=15.1875-2*9.8(y)

solve for y

-15.1875*2/-9.8=y

y=3.1m

Answer:

0.774m

Explanation:

The formula for maximum height is given by:

hmax = ∨₀² ₓ Sin (α)² / 2 × g

where;

∨₀ = initial velocity

Sin (α) = angle of launch

g = gravitational acceleration which is equal to 9.8m/s²

Plugging in our values, we will have:

hmax = (4.50m/s)² × (Sin 60.0)² / 2 × 9.8m/s²

hmax= 20.25m/s × 0.75 / 19.8m/s²

hmax = 15.1875 / 19.8

hmax = 0.774m

Which level of government relies the most on income tax?
OA.
federal
state
OC.
local

Answers

Answer:

Its the Federal government

Federal government tax

if 145kl of energy is added to water, what mass of water can be heated from 35C to 100C then vaporized at 100C

Answers

Answer:

m = 0.057 kg = 57 g

Explanation:

Energy Added to Water = Heat added to raise the temperature of water + Heat used to vaporize water

[tex]E = mC\Delta T + mH\\E = m(C\Delta T + H)[/tex]

where,

E = Energy added to water = 145 KJ

m = mass of water = ?

C = specific heat capacity of water = 4.2 KJ/kg.°C

ΔT = change in temperature = 100°C - 35°C = 65°C

H = Latent heat of vaporization of water = 2260 KJ/kg

Therefore,

[tex]145\ KJ = m[(4.2\ KJ/kg.^oC)(65^oC)+2260\ KJ/kg]\\\\145\ KJ = m(2533\ KJ/kg)\\\\m = \frac{145\ KJ}{2533\ KJ/kg}[/tex]

m = 0.057 kg = 57 g

The mass of water that can be heated is equal to 0.527 kilograms.

Given the following data:

Quantity of energy = 145 kJ = 145,000 Joules.Initial temperature = 35.0°CFinal temperature = 100.0°C

Scientific data:

Specific heat capacity of water = 4200 J/kg°CLatent heat of vaporization of water = 2260 KJ/kg

To calculate the mass of water that can be heated:

The quantity of energy and heat.

Note: The quantity of energy added to water is equal to the quantity of heat used to vaporize water and the quantity of heat that is added to raise the temperature of water.

Mathematically, this is given by this expression:

[tex]E=mc\theta + mH\\\\E= m(c\theta + H)[/tex]

Making m the subject of formula, we have:

[tex]m=\frac{E}{c\theta + H}[/tex]

Substituting the parameters into the formula, we have;

[tex]m=\frac{145000}{[42000\times (100-35)] + 2260}\\\\m=\frac{145000}{(4200\times 65) + 2260}\\\\m=\frac{145000}{273000 + 2260}\\\\m=\frac{145000}{275260}[/tex]

Mass, m = 0.527 kilograms.

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A stopped organ pipe of length L has a fundamental frequency of 220 Hz. For which of the following organ pipes will there be a resonance if a tuning fork of frequency 660 Hz is sounded next to the pipe?

a. a stopped organ pipe of length L
b. a stopped organ pipe of length 2L
c. an open organ pipe of length L;
d. an open organ pipe of length 2L.

Answers

Answer:

Option (a), (d) are correct.

Explanation:

Frequency, f = 220 Hz

Resonant frequency = 660 Hz

The next frequency of stopped organ pipe is

2f, 3 f, 4 f ....

= 2 x 220 , 3 x 220 , 4 x 220 ....

= 440 Hz, 660 Hz, 880 Hz

So, the option (a) is correct.

The next resonant frequency of open organ pipe is

3 f, 5 f,...

= 3 x 220, 5 x 220 , ..

= 660 Hz, 1100 Hz,...

So, option (d) is correct.

A scientist who studies fossils of ancient life forms .O ornithologist O Paleontologist O Ichthyologist O Marine Biologist .

Hurry !! First one to answer gets points !

Answers

Answer:

paleontologist

Explanation:

Paleontologists are scientists that investigate the fossils of extinct life forms. Thus, the correct option is B.

What is Fossil?

A fossil is defined as the preserved trace, imprint, or proof of a once-living entity from a past geological era. Exoskeletons, bones, shells, impressions of animals or microbes in stone, objects preserved in amber, hair, petrified wood, and genetic traces are a few examples. The collection of all the fossils is called as the fossil record.

An organism from a past geologic era that has been preserved in the Earth's crust is referred to as a fossil. Paleontologists are scientists that investigate the fossils of extinct life forms. The intricate system of fossil records is the main source of information about the evolution of life on Earth.

Therefore, the correct option is B.

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