Which type of indirect evidence can substitute for direct measurements of past climate change?

Waves can transfer energy and momentum from one place to another without a permanent shift in the mass. They propagate via different oscillation modes and can impact the propagation direction when they encounter an interface between two media.

The ability to move or change an object or what a wave carries, essentially describes the energy transfer capability of waves. According to Big Idea 6, waves can transfer energy and momentum from one place to another without permanently relocating the mass. This aspect characterizes waves and differentiates them from other phenomena.

Waves can propagate through various oscillation modes, such as transverse and longitudinal. For instance, when you shake one end of a rope, the perturbation moves in the form of a wave up to the other end of the rope, thus, facilitating energy transfer without a physical shift of the rope's parts.

Another important aspect to remember about waves is their ability to affect the path of propagation when they encounter an interface between two different media, as articulated by Enduring Understanding 6.E. This property explains various natural phenomena, such as the bending of light when it moves from air to water.

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Answer:

[tex]X_2=25.27m[/tex]

Explanation:

Here we will call:

1. [tex]E_1[/tex]: The energy when the first spring is compress

2. [tex]E_2[/tex]: The energy after the mass is liberated by the spring

3. [tex]E_3[/tex]: The energy before the second string catch the mass

4. [tex]E_4[/tex]: The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. [tex]E_1 = E_2[/tex]

2. [tex]E_2 -E_3= W_f[/tex]

3.[tex]E_3 = E_4[/tex]

where [tex]W_f[/tex] is the work of the friction.

1. equation 1 is equal to:

[tex]\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2[/tex]

where K is the constant of the spring, x is the distance compressed, M is the mass and [tex]V_2[/tex] the velocity, so:

[tex]\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2[/tex]

Solving for velocity, we get:

[tex]V_2[/tex] = 65.319 m/s

2. Now, equation 2 is equal to:

[tex]\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd [/tex]

where M is the mass, [tex]V_2[/tex] the velocity in the situation 2, [tex]V_3[/tex] is the velocity in the situation 3, [tex]U_k[/tex] is the coefficient of the friction, N the normal force and d the distance, so:

[tex]\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2) [/tex]

Volving for [tex]V_3[/tex], we get:

[tex]V_3 = 65.27 m/s[/tex]

3. Finally, equation 3 is equal to:

[tex]\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2[/tex]

where [tex]K_2[/tex] is the constant of the second spring and [tex]X_2[/tex] is the compress of the second spring, so:

[tex]\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2[/tex]

solving for [tex]X_2[/tex], we get:

[tex]X_2=25.27m[/tex]

The second spring will compress by 25.27 meters to bring the mass to a stop.

To solve this problem, we will go step-by-step considering energy conservation and the work done by friction.

1) Calculate the potential energy stored in the first spring:
The spring constant k is 20 N/m and the compression x is 8 m. The potential energy (PE) in the spring is calculated using the formula:

PE = (1/2) * k * x²

PE = (1/2) * 20 N/m * (8 m)²= 640 J

When the spring is released, this potential energy will convert to kinetic energy (KE) of the mass:

KE = 640 J

Calculate the work done by friction (Wf):
The frictional force (f) is calculated by:

f = μ * m * g

f = 0.16 * 0.3 kg * 9.8 m/s² = 0.4704 N

The work done by friction over a distance (d) of 2 m is:

Wf = f * d

Wf = 0.4704 N * 2 m = 0.9408 J

The kinetic energy after overcoming friction is:

Remaining KE = 640 J - 0.9408 J = 639.0592 J

Calculate the compression of the second spring (x₂):
The remaining kinetic energy will convert into potential energy in the second spring with a constant of 2 N/m.

PE = (1/2) * k₂ * x₂²

We set the remaining KE equal to this potential energy:

639.0592 J = (1/2) * 2 N/m * x₂²

639.0592 J = x₂²

x₂ = √639.0592 = 25.27 m

Therefore, the second spring will compress by 25.27 meters.

Answer:

[tex]l=0.068 m[/tex]

Explanation:

given,

inductance of solenoid = 2.07 μ H

winding of the wire = 1.19 m

Using formula of inductance

[tex]L = \dfrac{\mu_0N^2A}{l}[/tex]

L is the inductance

N is number of turns of the coil

μ₀ is permeability of free space

L is length of winding

N (2π r) = 1.19          

squaring both side

4π(N²(πr²))=1.19²    

N² A = 0.113            

now                              

[tex]2.07 \times 10^{-6}= \dfrac{4\pi\times 10^{-7}\times 0.113}{l}[/tex]

[tex]l= \dfrac{4\pi\times 10^{-7}\times 0.113}{2.07 \times 10^{-6}}[/tex]

[tex]l=0.068 m[/tex]

This question involves the concepts of the inductance of an inductor and the length of the winding.

The tube should be "6.84 cm" long.

The inductance of an inductor can be given by the following formula:

[tex]L=\frac{N^2\mu_oA}{l}[/tex]

where,

L = inductance = 2.07 μH = 2.07 x 10⁻⁶ H

N = No. of turns in coil = [tex]\frac{length\ of\ wire}{circumference\ of\ tube} = \frac{1..19\ m}{2\pi r}[/tex]

μ₀ = permeability of free space = 4π x 10⁻⁷ Wb/A.m

A = Area of tube = πr²

l = length of tube = ?

Therefore,

[tex]2.07\ x\ 10^{-6}\ H = \frac{(\frac{1.19\ m}{2\pi r})^2(\pi r^2)(4\pi\ x\ 10^{-7}\ Wb/A.m)}{l}\\\\l = \frac{1.4161\ x\ 10^{-7}}{2.07\ x\ 10^{-6}}\\\\[/tex]

l = 0.0684 m = 6.84 cm

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The attached picture shows the inductance formula.

Answer:

m = 4.65 kg

Explanation:

As we know that the mass of the water that evaporated out is given as

[tex]m = 0.0250 kg[/tex]

so the energy released in form of vapor is given as

[tex]Q = mL[/tex]

[tex]Q = (0.0250)(2.25 \times 10^6)[/tex]

[tex]Q = 56511 J[/tex]

now the heat required by remaining water to bring it from 15 degree to 100 degree

[tex]Q_2 = ms\Delta T[/tex]

[tex]Q_2 = (4 - 0.025)(4186)(100 - 15)[/tex]

[tex]Q_2 = 1.41\times 10^6J[/tex]

total heat required for above conversion

[tex]Q = 56511 + 1.41 \times 10^6 = 1.47 \times 10^6 J[/tex]

now by heat energy balance

heat given by granite = heat absorbed by water

[tex]m(790)(500 - 100) = 1.47 \times 10^6[/tex]

[tex]m = 4.65 kg[/tex]

Answer:

[tex]d=627.9\ m[/tex]  is the distance from the obstacle of reflection.

wavelength [tex]\lamb=0.5279\ m [/tex]

Explanation:

Given that:

We know,

[tex]\rm distance = speed \times time[/tex]

During an echo the sound travels the same distance back and forth.

[tex]2d=v.t[/tex]

[tex]2d=322\times 3.9[/tex]

[tex]d=627.9\ m[/tex]  is the distance from the obstacle of reflection.

Now the wavelength of sound waves:

[tex]\lambda=\frac{v}{f}[/tex]

[tex]\lambda=\frac{322}{610}[/tex]

[tex]\lamb=0.5279\ m [/tex]

Answer:

Bulb will remain at constant brightness for 6 hour

Explanation:

We have given resistance of the bulb R = 4 ohm

Potential difference of the battery V = 2 volt

So current [tex]i=\frac{V}{R}=\frac{2}{4}=0.5A[/tex]

Capacity of the battery is given as 3 amp hour

We have to find the time by which bulb will remain at constant brightness

So time will be [tex]=\frac{3}{0.5}=6hour[/tex]

So bulb will remain at constant brightness for 6 hour

The new tangential speed of the ball is four times the original speed.

When the length of the string is decreased to R/4, the radius of the circular path becomes R/4. The tension force exerted on the ball remains constant because it only depends on the mass of the ball and the gravitational force acting on it. The tangential speed is given by the formula v = ΩR, where Ω is the angular velocity. Since the radius is now 1/4 of the original radius, the new tangential speed is 4 times the original speed (option a).

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Answer:before throwing and after catching the ball

Explanation:

When basketball is in the hand of player net force on it zero as holding force is canceled by gravity Force. During its entire motion gravitational force is acting on the ball which is acting downward. Even at highest point gravity is constantly acting downwards.

After catching the ball net force on it zero as holding force is canceled by gravity force and ball is continue to be in stationary motion.          

The point in time where the zero net force acted on the ball is before throwing and after catching the ball

At the time of basketball is in the hand of the player, the net force should be zero since holding force is canceled via gravity Force. At the time of overall motion,  gravitational force should be acted on the ball which is acting downward. After catching the ball net force on it zero since holding force should be canceled via gravity force and ball should be continue in stationary motion.    

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Answer:

The original frequency of the guitar string is 437.0 Hz

Explanation:

It is an interesting phenomenon that when two waves of slightly different frequencies (f1 and f2) interfere they produce beats, and the frequency (fb) of those beats is:

[tex] f_{b}=\mid f_{1}-f_{2}\mid [/tex] (1)

Note that because the equation takes the absolute value of the difference between f1 and f2 we have two possible results of [tex] f_{1}-f_{2} [/tex] that satisfy the equation, those are:

[tex] 440.0-f_{2}=3\Longrightarrow f_{2}=437.0\,Hz [/tex]

[tex] -(440.0-f_{2})=3\Longrightarrow f_{2}=443.0\,Hz [/tex]

Because the frequency of the beat is heard to decrease when the frequency of the guitar string is increased, that means the guitar string was at first under the pitch, so the original frequency of the guitar string is 437.0 Hz

Final answer:

The original frequency of the guitar string was 437 Hz, since the string was out of tune and produced a beat frequency of 3 Hz with a 440 Hz tuning fork, and the beat frequency decreased when the string was tightened.

Explanation:

The original frequency of the guitar string can be determined using the concept of beat frequency. The beat frequency is the absolution difference between the frequencies of two sounds. In this case, the beat frequency was 3 Hz when the guitar string was sounded with the 440.0 Hz tuning fork.

Since the beat frequency decreased when the string tension was increased, the original frequency of the string must have been lower than the tuning fork. Therefore, the original frequency of the guitar string could be either 437 Hz or 443 Hz. However, since tightening the string increases its frequency and resulted in a decrease in the beat frequency, this indicates that the original frequency was 437 Hz (440 Hz - 3 Hz).

Answer:

0 N

Explanation:

The elevator is under free fall. So, when a body is under free fall, the acceleration is only due to gravity. So, the acceleration of the elevator or the woman inside it, is acceleration due to gravity in the downward direction.

The spring scale gives the value of the normal force acting on the woman and doesn't give the exact weight of the woman. Under normal conditions, when the spring scale is at rest, then the upward normal force equals the weight and hence weight of a body is equal to the normal force acting on the body.

But, here, the body is not at rest. Weight\tex](mg)[/tex] acts in the downward direction and normal force[tex](N)[/tex] acts in the upward direction. The woman is moving down with acceleration equal to acceleration due to gravity[tex](g)[/tex]

So, we apply Newton's second law on the woman.

[tex]\textrm{Net force} = \textrm{mass}\times \textrm{acceleration}\\F_{net}=ma[/tex]

Net force is equal to the difference of the downward force and upward force.

[tex]F_{net}=mg-N[/tex]

Now, [tex]F_{net}=ma[/tex]

[tex]mg-N=mg\\N=mg-mg=0[/tex]

Therefore, the reading on the spring scale is 0 N.

Answer:

 v = 13.75 m/s

Explanation:

mass of rod (m) = 0.147 kg

spring constant (k) = 32800 N/m

initial compression (Y1) = 4.44 cm = 0.0444 m

final compression (Y2) = 1.53 cm = 0.0153 m

find the speed at the instant of contact

kinetic energy of a spring = elastic potential energy

0.5m[tex]v^{2}[/tex] = 0.5k[tex]x^{2}[/tex]

[tex]v^{2}[/tex] = [tex]\frac{0.5k[tex]x^{2}[/tex]}{0.5m} \[/tex]

where x = Y1 - Y2

x = 0.0444 - 0.0153 = 0.0291 m

[tex]v^{2}[/tex] = [tex]\frac{0.5 x 32800 x [tex]0.0291^{2}[/tex]}{0.5 x 0.147} \[/tex]

        [tex]v^{2}[/tex] = 188.95

       v = 13.75 m/s

Answer:

it will need 0.06 hectare of land

So option (c) is correct

Explanation:

We have given that 12000000 kilo calories of chicken can be produced per hectare

It is given that someone with 2000 kilo calories / day for a year

Total number of days in a year = 365

So total kilo calories of chicken = 365×2000 = 730000 kilo calories

As 12000000 kilo calories of chicken can be produced per hectare

So 730000 kilo calories need a land of [tex]=\frac{730000}{12000000}=0.06hectare[/tex]

So it will need 0.06 hectare of land

Final answer:

To provide someone with 2,000 kilocalories/day for a year, 0.06 hectares of land used for chicken production is needed, as the closest correct answer.

Explanation:

The question asks how much land is needed to provide someone with 2,000 kilocalories/day for a year if 12,000,000 kilocalories of chicken can be produced per hectare. First, calculate the total kilocalories needed for a year by multiplying 2,000 kcal/day by the number of days in a year (365), which equals 730,000 kcal/year. Then, find how much land is needed by dividing the total annual kilocalorie requirement (730,000 kcal) by the amount of kilocalories produced per hectare (12,000,000 kcal).

This calculation gives us:

730,000 kcal / 12,000,000 kcal/ha = 0.0608333 ha

Therefore, the closest answer is c. 0.06 ha.

Answer:

A) Bohr’s work with atomic spectra led him to say that the electrons were limited to existing in certain energy levels, like standing on the rungs of a ladder.

Explanation:

The change to the atomic model that helped solve the problem seen in Rutherford's model was the discovery of the strong nuclear force.

Rutherford's model required that the electrons be in motion. Positive and negative charges attract each other, so stationary electrons would fall into the positive nucleus. However, Rutherford's model failed to explain why electrons were not pulled into the atomic nucleus by this attraction. The change to the atomic model that helped solve this problem was the discovery of the strong nuclear force, which is much stronger than electrostatic interactions and holds the protons and neutrons together in the nucleus.

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Answer:

353 s

Explanation:

Energy               = energy needed to raise the temperature of pot to 100°C +

required              energy needed to raise the temperature of water to 100°C

                    = MCΔT + mcΔT                      

                     = 0.65 kg * 4186 J/kg-°C * 87 °C + 0.36 kg * 900 J/kg-°C * 87 °C

                     = 264 906.3 J

Energy required = energy delivered * time

264 906.3 J = 750 J/s * time

time = 353 s

The study of chemicals and bonds is called chemistry. There are two types of elements these rare metals and nonmetals.

The correct answer is [tex]5787*10^{-6}[/tex]

The balanced reaction is:-

[tex]2H_2O_2(aq) ----> 2H_2O(l) + O_2(g).[/tex]

The rate law of decomposition:-

[tex]rate =-\frac{1}{2}\frac{d[H_2O_2]}{dt}[/tex]

Where d[H₂O₂] is the change in concentration of H₂O₂ (between 0s and 2,16x10⁴s) is (1,000M-0,500M) and dt is (0s-2,16x10^4s). Replacing:

[tex]rate= -\frac{1}{2} \frac{0.5}{2.16*10^4}[/tex]

rate =[tex]1157*10^{-5}[/tex]

As this rate is = d[O₂]/dt(Rate of production of O₂), the rate of production of O₂(g) is 1,157x10⁻⁵M/s

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Answer:

A

Explanation:

v = change of X / change of T

v = 200/19.3

Answer:

A. 10.36 m/s

Explanation:

The Olympic record for running the 200 m dash is 19.3 seconds. What is the average speed for this record?

A. 10.36 m/s

B. 3960 m/s

C. 219.3 m

D. 0.0965 m/s

speed is the change in distance per time

speed is scalar quantity and hence as no direction but only magnitude. time  and distance are also  scalar quantity

200/19.3

speed=10.36m/s

Answer:

4.20m/s^2

Explanation:

The mass of 210g stretch the spring by 26.9cm, 9 oscillation took 14.3 s.

Period (T) = time of oscillation / number of oscillation = 14.3/ 9 = 1.59s

Using the formula for period of a spring;

T = 2π√(m/k)

Divide both side by 2π

T/2π = √(m/k)

Square both side

T^2/ (2π)^2 = m/k

Make k subject of the formula

K = 4π^2 * m/ T^2

Also using Hooke's law

Since the spring was suspended from the ceiling,

F (mg) = k * DL ( where g is the new gravity in m/s^2, k is the force constant of the spring in N/m and DL is the change in length in meter

Make k subject of the formula

Mg/DL = k

Since both equation equal to K then

4π^2 * m/ T^2 = mg/DL

Cancel m on both side

4π^2 * DL / T^2 = g

DL = 26.9cm = 26.9/100 m = 0.269 m

T = 1.59s and π = 3.142

Substitute these into the equation

g = ( 4 * 3.142 * 3.142* 0.269) / (1.59^2) = 4.20 m/s^2

Final answer:

To determine the acceleration due to gravity on Planet X, the period T of the spring-mass system's oscillations is used along with the mass m and the spring constant k, which is calculated using the initial stretch of the spring.

Explanation:

To find the acceleration due to gravity (g) on Planet X, we can use the properties of simple harmonic motion for a spring-mass system. The formula for the period T of a mass m attached to a spring with spring constant k is T = 2π√(m/k). Since the spring stretches 26.9 cm when the 210 g mass is attached, the force due to gravity must be equal to the force the spring exerts at that stretch, which is F = kx, where x is the stretch distance. Thus, mg = kx, and we can solve for k using k = mg/x (with g as the gravity on Earth, 9.81 m/s²). Then we can calculate the period using the measured time of oscillations. From the period, we can solve for the gravity on Planet X.

Considering 9 oscillations took 14.3 seconds, the period T for one oscillation is 14.3 s / 9. With T and m known, one can solve for k and then use that to find g. Since the initial stretch was caused by the force of gravity, the spring constant k can be found using Earth's gravity, and subsequently used in the equation T = 2π√(m/g) rearranged to solve for g on Planet X.

Answer:

(a). The normal force exerted by the ground on the front wheel is [tex]4.67\times10^{5}\ N[/tex].

(b). The normal force exerted by the ground on each of the two rear wheels is [tex]7.02\times10^{5}\ N[/tex]

Explanation:

Given that,

Weight of jet [tex]W=1.87\times10^{6}\ N[/tex]

Distance = 16 m

Second distance = 12.0 m

We need to calculate the normal force exerted by the ground on the front wheel

Using formula of torque

[tex]\sum\tau=-Wl_{W}+F_{f}l_{f}=0[/tex]

Where, W = weight of jet

[tex]l_{f}[/tex]=lever arms for the forces [tex]F_{f}[/tex]

[tex]l_{w}[/tex]=lever arms for the forces W

Put the value into the formula

[tex]-(1.87\times10^{6})\times(16.0-12.0)+F_{f}\times16.0=0[/tex]

[tex]F_{f}=\dfrac{(1.87\times10^{6})\times(16.0-12.0)}{16.0}[/tex]

[tex]F_{f}=4.67\times10^{5}\ N[/tex]

(b). We need to calculate the normal force exerted by the ground on each of the two rear wheels

The sum of vertical forces equal to zero.

[tex]\sum F_{y}=F_{f}+2F_{r}-W=0[/tex]

We using 2 for two rear wheels

[tex]\sum F_{y}=0[/tex]

[tex]F_{f}+2F_{r}-W=0[/tex]

[tex]F_{r}=\dfrac{F_{f}-W}{2}[/tex]

Put the value into the formula

[tex]F_{r}=\dfrac{-4.67\times10^{5}+1.87\times10^{6}}{2}[/tex]

[tex]F_{r}=7.02\times10^{5}\ N[/tex]

Hence, (a). The normal force exerted by the ground on the front wheel is [tex]4.67\times10^{5}\ N[/tex].

(b). The normal force exerted by the ground on each of the two rear wheels is [tex]7.02\times10^{5}\ N[/tex]

Answer:

Explanation:

by Doppler effect

F apparent = F real x (Vair +- Vobserver) / (Vair +- Vsource)

case1 :if observer is approaching a source then put a + in numerator and - in denominator

case 2 : if observer is going away from source then do opposite.

it is case1 cos observer is moving toward wall . actually wall is acting as a source for observer

so

F apparent = 240 x (344 + vp ) / (344 - vp )

now F apparent - Freal = beat frequency of 6.00 Hz

so {240 x (344 + vp ) / (344 - vp )} - 230 = 6

240 x (344 + vp -344 + vp) / (344 - vp ) = 6

240 x 2 x vp = 6 x (344 - vp )

480 vp = 2064 - 6 x vp

486 x vp = 2064

vp = 4.24 m/s

Answer: (2) Use the Momentum Principle.

Explanation:

In fact, it is called the Conservation of linear momentum principle, which establishes the initial momentum [tex]p_{i}[/tex] of the asteroids before the collision must be equal to the final momentum [tex]p_{f}[/tex] after the collision, no matter if the collision was elastic or inelastic (in which the kinetic energy is not conserved).

In this sense, the linear momentum [tex]p[/tex] of a body is defined as:

[tex]p=mV[/tex]

Where [tex]m[/tex] is the mass and [tex]V[/tex] the velocity.

Therefore, the useful approach in this situation is option (2).

Answer: M = 118.9g

Explanation:

Using the principle of moment to solve the equilibrium question,

Sum of clockwise moments is equal to the sum of anticlockwise moments

*Check attachment for the diagram

Note that the mass of the body is always located at the center i.e 50cm from the end.

To get M, taking moment about the fulcrum,

Since moment is force × perpendicular distance and according to the above principle, we have;

49×26.2= M×10.8

M= 49×26.2/10.8

M = 118.9g

Final answer:

To find the final temperature of the system, we can use the principle of conservation of energy. The heat lost by the hot metal is equal to the heat gained by the water. Using the given equation, we can solve for the specific heat capacity of the metal, which is found to be 3.75 J/g°C.

Explanation:

To find the final temperature of the system, we can use the principle of conservation of energy. The heat lost by the hot metal is equal to the heat gained by the water. We can use the equation:m1c1(T1 - T) = m2c2(T - T2)

where:

m1 is the mass of the metal (34.5 g)

c1 is the specific heat capacity of the metal

T1 is the initial temperature of the metal (75°C)

T is the final temperature of the system (39°C)

m2 is the mass of the water (64.0 g)

c2 is the specific heat capacity of water (4.18 J/g°C)

T2 is the initial temperature of the water (25°C)

Plugging in the given values, we can solve for c1: 34.5g × c1 × (75°C - 39°C) = 64.0g × 4.18 J/g°C × (39°C - 25°C)

Simplifying the equation gives:34.5g × c1 × 36°C = 64.0g × 4.18 J/g°C × 14°C

Dividing both sides by 34.5g × 36°C gives:c1 = (64.0g × 4.18 J/g°C × 14°C) / (34.5g × 36°C)

Calculating the value gives:c1 = 3.75 J/g°C

The final temperature for both the aluminum and water at thermal equilibrium is 17.3 °C. This is calculated using the principle of conservation of energy. The heat lost by aluminum equals the heat gained by water.

To find the final temperature of a 32.5 g cube of aluminum and 105.3 g of water at thermal equilibrium, we use the principle of conservation of energy. The heat lost by the aluminum will be equal to the heat gained by the water.

Let us simplify: (32.5 g) x (0.900 J/g°C) x (T_final - 45.8 °C) = - (105.3 g) x (4.186 J/g°C) x (T_final - 15.4 °C)

Therefore, the final temperature of both substances at thermal equilibrium is **17.3 °C**.

Answer:

8684.2 kg/m³

Explanation:

Tension in the rope as a result of the weight = 8.86 N

Tension in the rope when submerge in water = 7.84

upthrust = 8.86 - 7.84 =1.02 N = mass of water displaced × acceleration due to gravity

Mass of water displaced = 1.02 / 9.81 = 0.104 kg

density of water = mass of water / volume of water

make volume subject of the formula

volume of water displaced = mass / density ( 1000) = 0.104 / 1000 = 0.000104 m³

volume of the object = volume of water displaced

density of the object = mass of the object / volume of the object = (8.86 / 9.81) / 0.000104 = 0.9032 / 0.000104 = 8684.2 kg/m³

Answer: The answer is 225 joules

Final Answer:

The mass of the water is approximately 48.4 grams.

Explanation:

1. To solve this problem, we can use the principle of conservation of energy, specifically the principle of heat exchange.

2. We can calculate the heat gained by the iron rod and the heat lost by the water, and equate them since there is no heat exchange with the surroundings.

3. The formula for heat exchange is given by: [tex]\[ Q = mc\Delta T \][/tex]

  where Q is the heat exchanged, m is the mass, c is the specific heat capacity, and [tex]\( \Delta T \)[/tex] is the change in temperature.

4. We can set up the equation for heat exchange for the iron rod and the water:

  [tex]\[ mc_{\text{iron}}\Delta T_{\text{iron}} = mc_{\text{water}}\Delta T_{\text{water}} \][/tex]

5. Rearranging the equation and solving for the mass of water:

  [tex]\[ m_{\text{water}} = \frac{m_{\text{iron}}c_{\text{iron}}\Delta T_{\text{iron}}}{c_{\text{water}}\Delta T_{\text{water}}} \][/tex]

6. Substituting the given values, we can calculate the mass of water to be approximately 48.4 grams.

Therefore, the mass of the water is approximately 48.4 grams.

The mass of the water is approximately 199.8 grams.

Here's how to find the mass of the water:

1. Define the variables:

m_iron = mass of iron rod = 32.5 g

c_iron = specific heat capacity of iron = 0.49 J/g°C

T_iron_initial = initial temperature of iron rod = 21.8°C

T_water_initial = initial temperature of water = 63.3°C

T_final = final temperature of the mixture = 58.9°C

m_water = mass of water (unknown)

c_water = specific heat capacity of water = 4.186 J/g°C

2. Apply the principle of heat transfer:

Heat lost by iron = Heat gained by water

3. Set up the equation:

m_iron * c_iron * (T_iron_initial - T_final) = m_water * c_water * (T_final - T_water_initial)

4. Substitute the known values and solve for m_water:

32.5 g * 0.49 J/g°C * (21.8°C - 58.9°C) = m_water * 4.186 J/g°C * (58.9°C - 63.3°C)

5. Solve for m_water:

m_water ≈ 199.8 g

Therefore, the mass of the water is approximately 199.8 grams.

Answer:

Length, l = 0.866 meters

Explanation:

Given that,

Frequency of sound, f = 99 Hz

Speed of sound in air, v = 343 m/s

To find,

The length of a wind instrument.

Solution,

The standing wave will gets formed in wind instrument. For the closed tube, the closed tube the frequency is given by :

[tex]f=\dfrac{v}{4l}[/tex]

Where

l is the length of the instrument

[tex]l=\dfrac{v}{4f}[/tex]

[tex]l=\dfrac{343\ m/s}{4\times 99\ Hz}[/tex]

l = 0.866 meters

So, the length of a wind instrument is 0.866 meters. Hence, this is the required solution.

Answer:

Explanation:

Decrease in kinetic energy of the ball

= 1/2 m ( v² - u² )

= 1/2 x .6 ( 8.3² - 7.1² )

= 5.544 J

This energy will be converted into potential energy

mgH = 5.544 ( H is maximum height attained )

H = 5.544 /( .6 x 9.8 )

= .942 m

= 94.2 cm

b ) In a ) we used the relation

mgH = 1/2 m ( v² - u² )

H =  ( 1/2g ) x ( v² - u² )

Here we see H or height attained does not depend upon mass as m cancels out on both sides.

Answer:

The magnitude of the net force is 2.48294 N and the angle is 25.02° with respect to the vertical

Explanation:

Consider the gravitational force as [tex]\vec{F_g}=2.25\ N[/tex]

Consider the wind force as [tex]\vec{F_w}=1.05\ N[/tex]

The angle between the gravitational force and wind force is 90°

From the parallelogram law we have

[tex]F=\sqrt{F_g^2+F_w^2+2F_gF_wcos\theta}\\\Rightarrow F=\sqrt{2.25^2+1.05^2+2\times 2.25\times 1.05\times cos90}\\\Rightarrow F=2.48294\ N[/tex]

The angle between the resultant and the gravitational force would be

[tex]tan^{-1}\frac{1.05}{2.25}=25.02^{\circ}[/tex]

The magnitude of the net force is 2.48294 N and the angle is 25.02° with respect to the vertical

Answer:

All these is caused by the repulsion force.

Explanation:

The electroscope produces a series of electric charges that produce a repulsion force when is putted in contact with a electric charged object.

As the physics law mentions, two different forces are repealed, the electrocospe is charged negatively and the object positively, causing a repulsion force that avoids that both objects touch the other.

Answer:D

Explanation:

Given

mass of A is twice the mass of B half the velocity of B

Suppose [tex]F_a[/tex] and [tex]F_b[/tex] be the average force exerted on A and B respectively

and According to Newton third law of motion Force on the body A is equal to Force on body B but opposite in direction as they are action and reaction force.

Thus [tex]F_a=-F_b[/tex]  and option d is correct

According to Newton's third law of motion, the forces exerted by each object on the other are the same.

The correct answer is (C) The forces exerted by each object on the other are the same because inter objects cannot exert forces of different magnitudes on each other.

According to Newton's third law of motion, whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force it exerts. This means that the forces exerted by object A on object B and by object B on object A are equal in magnitude.

Here, object A has twice the mass of object B, but because both objects have the same speed before the collision, the forces exerted on each other will be the same.

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Answer:

The terminal speed of this object is 12.6 m/s

Explanation:

It is given that,

Mass of the object, m = 80 kg

The magnitude of drag force is,

[tex]F_{drag}=12v+4v^2[/tex]

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

[tex]F_{drag}=mg[/tex]

[tex]12v+4v^2=80\times 9.8[/tex]

[tex]4v^2+12v=784[/tex]

On solving the above quadratic equation, we get two values of v as :

v = 12.58 m/s

v = -15.58 m/s (not possible)

So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.