All categories

3 years ago

Sound level B in decibels is defined as

Physics

1 answer:

Lelechka [254]3 years ago

3 0

Answer:

The approximate combined sound intensity is $I_{T}=1.1\times10^{-4}W/m^{2}$

Explanation:

The decibel scale intensity for busy traffic is 80 dB. so intensity will be

$10log(\frac{I_{1}}{I_{0}} )=80$ , therefore $I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}$

In the same way for the loud conversation having a decibel intensity of 70 dB.

$10log(\frac{I_{2}}{I_{0}} )=70$ , therefore $I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}$

Finally we add both of them $I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}$ , is the approximate combined sound intensity.

You might be interested in

Kinetic Energy HELP! Brainly included

Talja [164]

Answer:

Explanation:

When you dropping the swing, it increases the kinetic energy. In the middle, that is when it is at it's fullest potential. As it makes its way up, it pushes again the gravitational force, and soon, the swing will stop at the top for a VERY SHORT AND SMALL period of time. That is when it is at it's very weakest point, making the answer, indeed, A.

6 0

3 years ago

1. A friend measures the length of the school

shusha [124]

Answer:

option A

Explanation:

A soccer field is about 100 meters of length, with the other instruments is possible to make the measure, but it will be very complex and it will take time. Therefore with the 50-m tape, the student will need only two measures.

6 0

2 years ago

A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperat

Nostrana [21]

Answer:

Approximately $13\; \rm g$ of steam at $100\; \rm ^\circ C$ (assuming that the boiling point of water in this experiment is $100\; \rm ^\circ C\!$ .)

Explanation:

Latent heat of condensation/evaporation of water: $2260\; \rm J \cdot g^{-1}$ .

Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to $\rm J \cdot g^{-1}$ .

Specific heat of water: $4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}$ .

Specific heat of copper: $0.39\; \rm J \cdot g^{-1}\cdot K^{-1}$ .

The temperature of this calorimeter and the $250\; \rm g$ of water that it initially contains increased from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ . Calculate the amount of energy that would be absorbed:

$\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J \end{aligned}$ .

$\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J \end{aligned}$ .

Hence, it would take an extra $585\; \rm J + 31500\; \rm J = 32085\; \rm J$ of energy to increase the temperature of the calorimeter and the $250\; \rm g$ of water that it initially contains from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ .

Assume that it would take $x$ grams of steam at $100\; \rm ^\circ C$ ensure that the equilibrium temperature of the system is $50\; \rm ^\circ C$ .

In other words, $x\; \rm g$ of steam at $100\; \rm ^\circ C$ would need to release $32085\; \rm J$ as it condenses (releases latent heat) and cools down to $50\; \rm ^\circ C$ .

Latent heat of condensation from $x\; \rm g$ of steam: $2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J$ .

Energy released when that $x\; {\rm g}$ of water from the steam cools down from $100\; \rm ^\circ C$ to $50\; \rm ^\circ C$ :

$\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J \end{aligned}$ .

These two parts of energy should add up to $32085\; \rm J$ . That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from $20\; \rm ^\circ C$ to $50\; \rm ^\circ C$ .

$(2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J$ .

Solve for $x$ :

$x \approx 13$ .

Hence, it would take approximately $13\; \rm g$ of steam at $100\; \rm ^\circ C$ for the equilibrium temperature of the system to be $50\; \rm ^\circ C$ .

4 0

2 years ago

Define Newton’s three laws of motion and how they apply to everyday situations.

son4ous [18]

1. By Ilkka Cheema2. Newton’s 1st Law  The first law of motion sates that an object will not change its speed or direction unless an unbalanced force (a force which is distant from the reference point) affects it. Another name for the first law of motion is the law of inertia. If balanced forces act on an object it doesn’t accelerate or change direction. This means it doesn’t change its velocity and it doesn’t have momentum.3. Examples of Newton’s 1st Law  If you slide a hockey puck on ice, eventually it will stop, because of friction on the ice. It will also stop if it hits something, like a player’s stick or a goalpost.  If you kicked a ball in space, it would keep going forever, because there is no gravity, friction or air resistance going against it. It will only stop going in one direction if it hits something like a meteorite or reaches the gravity field of another planet.  If you are driving in your car at a very high speed and hit something, like a brick wall or a tree, the car will come to an instant stop, but you will keep moving forward. This is why cars have airbags, to protect you from smashing into the windscreen.4. Newton’s 2nd Law  The second law of motion states that acceleration is produced when an unbalanced force acts on an object (mass). The more mass the object has the more net force has to be used to move it.5. Examples of Newton’s 2nd Law  If you use the same force to push a truck and push a car, the car will have more acceleration than the truck, because the car has less mass.  It is easier to push an empty shopping cart than a full one, because the full shopping cart has more mass than the empty one. This means that more force is required to push the full shopping cart.6. Newton’s 3rd Law The third law of motion sates that for every action there is a an equal and opposite reaction that acts with the same momentum and the opposite velocity.7. Examples of Newton’s 3rd Law  When you jump off a small rowing boat into water, you will push yourself forward towards the water. The same force you used to push forward will make the boat move backwards.  When air rushes out of a balloon, the opposite reaction is that the balloon flies up.  When you dive off of a diving board, you push down on the springboard. The board springs back and forces you into the air.

3 0

2 years ago

A 6 V battery is connected to a 24 ohm resistor to create a circuit. The 6 V battery is then replaced with a 12 V battery. How d

11Alexandr11 [23.1K]

Answer:

1.) The current is doubled

2.) moving a magnet up and down near the wire

3.) An electric current in the wire produces a magnetic field.

4.) Distance between Particles (m)

Explanation:

1.) When 6 V battery is connected to a 24 ohm resistor to create a circuit, using ohms law, V = IR

Current I = 6/24 = 0.25 A

When the The 6 V battery is replaced with a 12 V battery,

Current I = 12/24 = 0.5 A

Therefore, The current is doubled

2.) Electric current will be induced when moving a magnet up and down near the wire

3.) An electric current in the wire produces a magnetic field.

4.) Distance between Particles (m)

The force of attraction between two different masses is inversely proportional to the square of the distance between them.

6 0

3 years ago