How is gravity involved while swinging?
​

Answer:

wavelength in the water = 1.311  m

Explanation:

givendata

wavelength of the sound in air =  5.67 m

temperature = 20°C

solution

as we know velocity c in air = 343.3 m/s and velocityˇc in water= 1484m/s

and we take wavelength = w

and

wavelength=velocity ÷ frequency     .................1a

as we know here that In air and in water if the frequency not change then there wavelength depend on speed of sound in air and water

so

wavelength of sound in water = w in air × c in water ÷ c in air

wavelength in the water in water = 5.67 ×  343.2 ÷ 1484

wavelength in the water = 1.311  m

The Hubble sphere forms a sort of edge to our observable universe. This isn't a real edge to the universe because of its expanding nature. For this reason also,  this edge doesn't violate the cosmological principle or the Copernican principle.

In cosmology, the Hubble sphere is a spherical region of the observable universe that surrounds an observer and beyond which, as the universe expands, objects move away from that observer at a faster pace than the speed of light. Approximately 10^31 cubic light years make up the Hubble volume (or about 10^79 cubic meters).

Since the Hubble parameter varies across cosmological models, the Hubble limit typically does not coincide with a cosmological event horizon. For instance, in a decelerating Friedmann universe, the Hubble sphere expands with time and its border passes over light from more distant galaxies, allowing light that was released earlier by objects outside the Hubble volume to finally reach the sphere and be detected by observers.

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Answer:

Explanation: The moon of the Earth is much lighter in mass than the planet itself. In addition to being smaller than Earth, the Moon is also only approximately 60% as dense. A human weighs less on the Moon because there is less gravitational attraction there than there is on Earth.

Moon has lesser mass as conpared to earth, therefore gravitational force exerted by moon on any object is lesser than that of gravitational force exerted by earth on the same object, hence we can say that astronauts weight less on moon, i.e approximately 1/6 th of their weight on earth.

The acceleration, in m/s² , that riders experience is 50.3 m/s²

Since the riders move in a circle, they undergo circular motion and will have a centripetal acceleration.

So, the centripetal acceleration, is given by a = v²/r where

Given that a rider in one swing is moving at 34 m/s with respect to the ground in a 46- m -diameter circle. We have that

So, substituting the values of the variables into the equation for the acceleration, we have

a = v²/r

a = (34 m/s)²/23 m

a = 1156 m²/s²/23 m

a = 50.26 m/s²

a ≅ 50.3 m/s²

So, the acceleration, in m/s² , that riders experience is 50.3 m/s²

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Wind is the most frequent cause waves. The friction between the wind and the surface of the water produces wind-driven waves, also known as surface waves. A wave crest is produced when wind continuously disturbs the water's surface in an ocean or lake.

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Answer:

The predominantly single-celled organisms of the domains Bacteria and Archaea are classified as prokaryotes (pro– = before; –karyon– = nucleus). Animal cells, ...

Answer:

0.3 newtons

Explanation:

If the coefficient of kinetic friction between a crate and the floor is 0.20, then the fore needed to slide a 92 kg crate uniformly across the floor would be 180.504 N

Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.

The friction force prevents any two surfaces of objects from easily sliding over each other or slipping across one another. It depends upon the force applied to the object.

As given in the problem If the coefficient of kinetic friction between a crate and the floor is 0.20

The force needed = μN

where μ is the coefficient of the kinetic friction

μ =0.2

N is the normal force

N = mg

N = 92×9.81 N

Force needed to slide = 0.20× 92×9.81

                                     =180.504 N

Thus, The force needed to slide a 92 kg crate uniformly across the floor would be 180.504 N

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a) To determine if the car will hit the object before coming to a stop, we need to calculate the distance required to stop the car, assuming maximum braking deceleration. We can use the following formula:

d = (v^2) / (2a)

where:

d = distance required to stop

v = initial velocity

a = acceleration/deceleration

In this case, v = 100 km/h = 27.78 m/s (converted from km/h to m/s)

a = -7 m/s^2 (negative sign indicates deceleration)

We know that the car's headlights extend out to a range of 30 m, so if the distance required to stop the car is greater than 30 m, the car will hit the object before coming to a stop.

Plugging in the values to the formula, we get:

d = (27.78^2) / (2 x -7) = 108.61 m

Since 108.61 m is greater than 30 m, the car will hit the object before coming to a stop.

b) To calculate the time required to stop, we can use the following formula:

t = v / a

where:

t = time required to stop

v = initial velocity

a = acceleration/deceleration

Plugging in the values, we get:

t = 27.78 / 7 = 3.97 s

Therefore, it will take 3.97 seconds to stop the car.

Answer:

A. 0.199 J

B. 0.0663 C

C = 0.0221 F

D. 12.68 ohms

Explanation:

From the question:

time duration, t = 0.28 seconds

Average power, P = 0.71 W

Average voltage, V = 3 V

A) Energy is given as:

E = P * t

=> E = 0.71 * 0.28 = 0.199 J

B) Electrical energy is also given as:

E = qV

where q = charge

=> q = E / V

∴ q = 0.199 / 3 = 0.0663 C

C) Capacitance is given as charge over voltage:

C = q / V

=> C = 0.0663 / 3 = 0.0221 F

D) Electrical power, P, can also be given as:

P = \(V^2 / R\)

where R = resistance

=> R = \(V^2 / P\)

R = \(3^2 / 0.71 = 9 / 0.71 = 12.68 ohms\)

It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.

2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.

3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.

4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.

5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.

6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".

7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).

8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).

9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.

10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.

11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).

12. Evaluating the expression, we find: t ≈ 546.

13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

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Answer:

the correct answer is up and down

Answer:

Energy was lost as frictional energy. Hence when trying to move up to the top for the second time, the skateboarder was unable to do so.

The others provided additional energy to make up for the lost energy by dending their knees

Explanation:

According to law of conservation of energy, energy can niether be created nor destroyed but can change its form.

Here kinetic energy of skate boarder was converted into frictional energy and lost in the surroundings(energy was not destroyed but simply changed its form to energy of particles in the air). She was therefore unable to reach to the top the second time.

Other skateboarders faced the same, but the added energy to their movement by converting energy stored in their bodies to their kinetic energies. They were therefore abole to reach to the top.

Answer:

E

Explanation:

It's velocity and wavelength change, but its frequency does not

Answer:

4.246 light years

Explanation:

4.246 light years

The closest star, Proxima Centauri, is 4.24 light-years away. A light-year is 9.44 trillion km, or 5.88 trillion miles. That is an incredibly large distance. Walking to Proxima Centauri would take 950 million years.

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Answer:

4.25 light years

Explanation:

a light year is the distance light travels in one year it is equal to 9.461 x 1012 km. alpha centauri A & B are roughly 4.35 light years away from us. proxima centauri is slightly closer at 4.25 light years.

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Un proyectil se define como cualquier objeto que traza una trayectoria parabólica. Los parámetros importantes en el movimiento de proyectiles son; Tiempo de vuelo, rango y altura máxima.

a) 1 segundo

b) 1,25 m

c) 10 m

Deje que el tiempo de vuelo sea T

T = 2usinθ / g

u = velocidad inicial

g = aceleración debida a la gravedad

θ = ángulo

T = 2 * 10 m / s * sin (30) / 10

T = 1 segundo

Sea la altura máxima H

H = u ^ 2 sin ^ 2 θ / 2g

H = (10) ^ 2 (sin30) ^ 2/2 * 10

Alto = 1,25 m

Sea el rango R

R = u ^ 2sin 2θ / g

R = (10) ^ 2 sin 2 (30) / 10

R = 8.66 m

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a) El tiempo que tarda el proyectil en llegar al piso es:

\(t_{vuelo}=1\: s\)  

b) La máxima altura alcanzada por el proyectil es:

\(y_{max}=1.25\: m\)  

c) La máxima distancia alcanzada por el proyectil es:

\(x_{max}=8.66\: m\)

a)

Para calcular el tiempo que le toma al proyectil llegar al piso, se puede usar la siguiente ecuación de tiro parabólico.

\(y=y_{i}+v_{iy}t-0.5gt^{2}\) (1)

Donde:

Recordemos que la componente de la velocidad incial (v(i)=10 m/s) en el eje y esta dada por:

\(v_{iy}=v_{i}sin(30)\)

\(v_{iy}=10sin(30)\)

Sabemos que y(i) = 0 y ademas haciendo que y = 0, sabremos el tiempo total de vuelo. De la ecuación (1):

\(0=0+10sin(30)t-0.5(10)t^{2}\)

Despejando t.

\(10sin(30)=0.5(10)t\)

\(t=\frac{10sin(30)}{5}\)  

El tiempo de vuelo sera:

\(t_{vuelo}=1\: s\)  

b)

Para calcular la máxima altura, usamos la siguiente ecuación.

\(v_{fy}^{2}=v_{iy}^{2}-2gy\)

Si hacemos que veocidad final en y v(fy) sea 0, encontraremos la máxima altura.

\(0=v_{iy}^{2}-2gy_{max}\)

\(0=(v_{i}sin(30))^{2}-2gy_{max}\)

Despejamos y(max):

\(y_{max}=\frac{(v_{i}sin(30))^{2}}{2g}\)

\(y_{max}=\frac{(10sin(30))^{2}}{2(10)}\)  

La máxima altura alcanzada por el proyectil será:

\(y_{max}=1.25\: m\)  

c)

Sabemo que la velocidad del proyectil en la dirección x es constante, por lo tanto, la ecuacion cinemática será:

\(x=vt\)

La máxima distancia se determina con el tiempo de vuelo:

\(x_{max}=v_{ix}t_{vuelo}\)

La componente de la velocidad en la direccion x es:

\(v_{ix}=v_{i}cos(30)\)

Por lo tanto, el máximo desplazamiento será:

\(x_{max}=10cos(30)*1\)

\(x_{max}=8.66\: m\)

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The relative permittivity of the insulator is εr = 1 / 2 = 0.5.

The relative permittivity of the insulator is calculated by applying the equation for the Coulomb force.

F = (kq²) / r²

where;

The initial force equation becomes;

3 = (kq²) / r²

Let's denote the relative permittivity as εr.

The new force equation becomes;

1.5 = (kq² x εr) / r²

Divide the two force equations;

(3 / 1.5) = (kq²) / (kq² x εr)

2 = 1 / εr

εr = 1/2 = 0.5

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Newton's laws of motion describe the behaviour of objects when they are at rest or in motion.

The 5 lbs bag of sugar is at rest in this situation, and you are exerting a 5 lbs upward force on it. The sugar bag will continue to be at rest unless an outside force acts upon it, in accordance with the first law of motion. It is being subjected to an external force that you are applying.

According to the second law of motion, theforce exerted on the bag of sugar will have a direct proportional effect on how fast it accelerates. The mass of the bag of sugar is 5 pounds, and the force you are exerting on it is 5 lbs. Because\(\textsf{$\frac{\textsf{5 lbs (force)}}{\textsf{ 5 lbs (mass)}}$ = 1 (unit of accleration)}\), the bag of sugar will accelerate by 1 (unit of acceleration).

The third law of motion states that there is an equal and opposite reaction to every action. In this case, the sugar bag pushes down on your hand with a 5 lb force when you push up on it with a 1 lb force. The force you are providing to the bag of sugar will have an equal and opposite reaction here.

Therefore, this scenario illustrates all three of Newton's laws of motion

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Yes, this is an example of Newton's Third Law of Motion. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. In this example, when you push upwards on the 5 lb bag of sugar with a force of 5 lb, the bag of sugar also pushes downwards on you with a force of 5 lb.In other words, the action is you pushing upwards on the bag of sugar and the reaction is the bag of sugar pushing downwards on you with the same amount of force. So, this is an example of Newton's Third Law of Motion.

The average velocity of the car is 100 kilometers/hour south. This means that, on average, the car is traveling 100 kilometers per hour in the south direction relative to its starting point.

To determine the average velocity of the car, we need to calculate the displacement and divide it by the time taken. Velocity is defined as the rate of change of displacement with respect to time.

In this case, the car is traveling south, and its displacement is 200 kilometers from its starting point after 2 hours.

The average velocity is given by the formula:

Average velocity = Displacement / Time

The displacement is 200 kilometers south, and the time is 2 hours. Therefore, we have:

Average velocity = 200 kilometers south / 2 hours

Simplifying the calculation:

Average velocity = 100 kilometers/hour south

Hence, the correct answer is B. 100 kilometers/hour south. This indicates that the car's average velocity is 100 kilometers per hour towards the south direction.

It's important to note that velocity is a vector quantity and includes both magnitude (speed) and direction. In this case, the direction is specified as south, which indicates that the car is moving towards the south relative to its starting point.

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Newton's Laws of motion are: an object at rest remains at rest, and an object in motion remains in motion at constant speed and in a straight line. The acceleration of an object is determined by its mass and the amount of force applied. When one object applies a force to another, the second object applies an equal and opposite force to the first.

Sir Isaac Newton contributed to physics and mathematics in a variety of ways. At the age of just 23, he created the theories of gravitation in 1666. In the "Principia Mathematica Philosophiae Naturalis," published in 1686, he presented his three laws of motion.

Newton revolutionised science by formulating his three laws of motion. Planets move in elliptical orbits as opposed to circles because of Newton's laws and Kepler's laws.

Newton's Laws of motion are

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Answer:

17 km

Explanation:

Given values:

As the unit of time for speed and time is different, convert the time into hours by dividing the minutes by 60:

\(\implies \textsf{Time }=\dfrac{12}{60}\; \sf =0.2 \; h\)

Therefore:

\(\boxed{\sf Speed=\dfrac{Distance}{Time}}\)

Substitute the speed and time into the formula and solve for distance:

\(\implies \sf 85=\dfrac{\sf Distance}{0.2}\)

\(\implies \sf Distance=85\times0.2\)

\(\implies \sf Distance=17\;km\)

Therefore, the distance recorded in 12 minutes is 17 km.

Answer:

The distance recorded by the motorcyclist in 12 minutes is 17 km.

Explanation:

We can use the formula:

where:

We know that 12 minutes is equal to 12/60 = 0.2 hours.

Substituting the values, we get:

\(\therefore\) The distance recorded by the motorcyclist in 12 minutes is 17 km.

\(\blue{\overline{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}\)

Answer:

paying too much on the black market instead of getting a prescription

Explanation:

i just took the quiz

Answer:

Paying too much on the black market instead of getting a prescription

Explanation:

The rest of the options are risks associated with using legal drugs without medical supervision.

Given that hotter blackbodies produce more energy than cooler blackbodies,  cooler red giants have much higher luminosities than much hotter white dwarfs.

The hotter blackbodies emits more light as compared to the cooler blackbodies over all the wavelength. A blackbody emits light over all wavelength. the red giant is a star , hydrogen is fused to helium core to stat the fusion reactions. this hydrogen fusion caused the star to expand. the sun becomes red giant it will swell up to size of earth orbit.  This gives a red color and make red giant much hotter. while the white dwarf comes after the red giant . white dwarf is also very hot but it is not hot enough to star fusion reaction in carbon and oxygen.

Thus, Given that hotter blackbodies produce more energy than cooler blackbodies,  cooler red giants have much higher luminosities than much hotter white dwarfs.

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The specific heat capacity of silver is {c} = 0.24 J/(kg °C).

The specific heat capacity is defined as the quantity of heat (J) absorbed per unit mass (kg) of the material when its temperature increases by 1 °C. Its units are J/(kg °C).

Given is that 55g of silver absorbs 197.9 joules of heat and the temperature rises by 15 °C.

Mathematically, we can write the specific heat capacity as -

Q = mcΔT

where -

{Q} → heat energy

{m} → mass

{c} → specific heat capacity

{ΔT} → change in temperature

We can write -

Q = mcΔT

197.9 = 55 x (15 - 0) x c

197.9 = 55 x 15 x c

c = 197.9/825

{c} = 0.24 J/(kg °C)

Therefore, the specific heat capacity of silver is {c} = 0.24 J/(kg °C).

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Answer:

1) Adenine (A)   2. Uracil (U)   4) Cytosine (C)

Explanation:

"RNA consists of four nitrogenous bases: adenine, cytosine, uracil, and guanine."

Answer:

200 meters

Explanation:

20 x 10 = 200