Use your understanding of heat loss to ESTIMATE the cost of the lost energy through one standard window during an average summer day in Maryland. Use $0.17 per kWh as your energy cost.

Answer:

required distance is 233.35 m

Explanation:

Given the data in the question;

Sound intensity [tex]I[/tex] = 1.62 × 10⁻⁶ W/m²

distance r = 165 m

at what distance from the explosion is the sound intensity half this value?

we know that;

Sound intensity [tex]I[/tex] is proportional to 1/(distance)²

i.e

[tex]I[/tex] ∝ 1/r²

Now, let r² be the distance where sound intensity is half, i.e [tex]I[/tex]₂ = [tex]I[/tex]₁/2

Hence,

[tex]I[/tex]₂/[tex]I[/tex]₁ = r₁²/r₂²

1/2 = (165)²/ r₂²

r₂² = 2 × (165)²

r₂² = 2 × 27225

r₂² = 54450

r₂ = √54450

r₂ = 233.35 m

Therefore, required distance is 233.35 m

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s [tex]w_o[/tex] = 12 rad/s

Angular Velocity at time 0.15s [tex]w_f[/tex] = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( [tex]w_f[/tex] - [tex]w_o[/tex] ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Answer:

[tex](b)\ t_1 - t_0[/tex]

[tex](d)\ t_2 - t_1[/tex]

[tex](e)\ \frac{t_2 - t_0}{2}[/tex]

Explanation:

Given

See attachment for complete question

Required

How long to reach the ground from the maximum height

First, calculate the time of flight (T)

[tex]T =t_2 - t_0[/tex]

The time taken (t) from maximum height to the ground is:

[tex]t = \frac{1}{2}T[/tex]

So, we have:

[tex]t = \frac{t_2 - t_0}{2}[/tex]

Another representation is:

At ymax, the time is: t1

On the ground, the time is t2

The difference between these times is the time taken.

So;

[tex]t = t_2 - t_1[/tex]

Since air resistance is to be ignored, then

[tex]t_2 - t_1 = t_1 - t_0[/tex] --- i.e. time to reach the maximum height from the ground equals time to reach the ground from the maximum height

Answer:

b. is the correct answer ....

Answer:

the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB

Explanation:

Given the data in the question;

sound intensity reduced by the factor, m = 305

the sound intensity level experienced by the crew members wearing protective earplugs, L = 79 dB

Now, using the expression of sound intensity level;

L = 10log( [tex]I_0[/tex] )

where  [tex]I_0[/tex] is the intensity at L level

so we substitute

79 = 10log( [tex]I_0[/tex] )

[tex]I_0[/tex] = [tex]10^{7.9[/tex]

Now, expression for actual intensity;

[tex]I[/tex] = m[tex]I_0[/tex]

where [tex]I[/tex] is the actual intensity

so we substitute

[tex]I[/tex] = 305 × [tex]10^{7.9[/tex]

Next, we write the expression of sound intensity level for reduced intensity;

L' = 10log( [tex]I[/tex] )

So we substitute

L' = 10log( 305 × [tex]10^{7.9[/tex] )

L' = 10log( 24227011159.09058 )

L' = 103.8 dB

Therefore, the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB

Answer:

C

Explanation:

The answer is C because if you look at the 1 hour mark it shows 10km

Answer:It will be 10km/hour

Explanation:

Answer:

The centripetal acceleration is 6.95 m/s²

Explanation:

Given;

angular displacement of the blade, θ = 90.08⁰

duration of motion of the blade, t = 0.4 s

radius of the circle moved by the blade, r = 0.45 m

The angular speed of the blade in radian is calculated as;

[tex]\omega = \frac{\theta}{t} \times \frac{\pi \ radian}{180^0} \\\\\omega = \frac{90.08 ^0}{0.4 \ s} \times \frac{\pi \ radian}{180^0} \\\\\omega = 3.93 \ rad/s[/tex]

The centripetal acceleration is calculated as;

a = ω²r

a = (3.93)² x 0.45

a = 6.95 m/s²

Answer:

I don't know hhaha ammmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm I wish I could help

Answer:

The distance traveled in 1 year is: [tex]3.143*10^{16}ft[/tex]

Explanation:

Given

[tex]s = 982,080,000 ft/s[/tex] --- speed

[tex]t = 32,000,000 s[/tex] --- time

Required

The distance traveled

This is calculated as:

[tex]Speed = \frac{Distance}{Time}[/tex]

So, we have:

[tex]Distance = Speed * Time[/tex]

This gives:

[tex]Distance = 982,080,000 ft/s * 32,000,000 s[/tex]

[tex]Distance = 982,080,000 * 32,000,000ft[/tex]

[tex]Distance = 3.143*10^{16}ft[/tex] -- approximated

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

Given the following data;

Radius, r = 2.6 km

Time = 360 seconds

Conversion:

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

[tex] Circular \; speed (V) = \frac {2 \pi r}{t}[/tex]

Where;

Substituting into the formula, we have;

[tex] Circular \; speed (V) = \frac {2*3.142*2600}{360} [/tex]

[tex] Circular \; speed (V) = \frac {16338.4}{360} [/tex]

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

[tex] Centripetal \; acceleration = \frac {V^{2}}{r}[/tex]

Where;

Substituting into the formula, we have;

[tex] Centripetal \; acceleration = \frac {45.38^{2}}{2.6}[/tex]

[tex] Centripetal \; acceleration = \frac {2059.34}{2600}[/tex]

Centripetal acceleration = 0.79 m/s²

Answer: [tex]3590\ kJ[/tex]

Explanation:

Given

Paddle wheel work is [tex]W=-5090\ kJ\quad \text{work is done on the system}[/tex]

Heat transfer from the tank is [tex]Q=-1500\ kJ\quad \text{heat taken from the system}[/tex]

From the first law of thermodynamics

Change in the internal energy of the system is equal to the difference of heat and work .

[tex]\Rightarrow \Delta U=Q-W\\\Rightarrow \Delta U=-1500-(-5090)\\\Rightarrow \Delta U=3590\ kJ[/tex]

Therefore, the change in internal energy is [tex]3590\ kJ[/tex]

Answer:

Explanation:

From the missing image attached below, it is obvious that there no external force. This implies that they cannot change their position by merely just pulling the ropes. As a result, there will be no movement and no net force will exist.

So, if there is no external force;

The center of mass of the two people is:

[tex]X_{cm}= \dfrac{m_1x_1+m_2x_2}{m_1+m_2} \\ \\ X_{cm}= \dfrac{m(-3m)+m(+3m)} {m+m}\\ \\ X_{cm}= \dfrac{0}{2m} \\ \\ X_{cm} =0[/tex]

Thus, In the system, no movement occurs and all forces remain the same.

Answer: 55 kmph

Explanation: Divide 110 by 2

Answer:  I think that its b, they would have difficulty sprinting in a race

Explanation:

Answer:

Explanation:

Answer:

B is excellent answer..............

The model of the brain that is shown here is the experimental model that is present in Option B, as it is used to study the brain's parts and its function, which is helpful for a better understanding of the brain.

There are various experimental models of the brain that have been developed to better understand its functions and mechanisms, such as Animal models, such as mice, rats, and primates, have been widely used to study the brain due to their similarity to the human brain in terms of structure and function. Computer models can simulate brain function and behavior at various levels of detail, from individual neurons to large-scale brain networks. These models are useful for testing hypotheses and predicting outcomes, as well as for designing new experiments.

Hence, the model of the brain that is shown here is the experimental model that is present in Option B.

Learn more about the experimental model of the brain here.

https://brainly.com/question/23802617

#SPJ7

Answer:

a circular path

Explanation:

In a magnetism field if a charged particle having a charge of magnitude '' enters  such that its velocity vector V is 90° to the direction of the magnetic field "B'', then it will experience a force, called Lorentz force F

[tex]F = V\times B[/tex]

According to the property of cross-product, the Lorentz force (F) acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path.

Impact printers involve mechanical components for conducting printing. It is a type of printer that works by direct contact of an ink ribbon with paper.

In Non-Impact printers, no mechanical moving component is used.

mark me brainliesttt :)))

Answer:

8760 N

Explanation:

think this is the right answer :)

Answer:

in this case around the loop the field points downwards on the outside and upwards on the inside

Explanation:

To find the direction of the magnetic field in a wire you must use the right hand rule.

The thumb points in the direction of the current flow and the other created fingers point in the direction of the magnetic field.

Therefore in this case around the loop the field points downwards on the outside and upwards on the inside

Answer:

50 watts

Explanation:

Applying,

Power (P) = Workdone (W)/Time(t)

But,

Work done (W) = Force (F)×distance(d)

Therefore,

P = Fd/t..................... Equation 1

Where P =  power of the weightlifter, F = Force applied, d = distance, t = time.

From the question,

Given: F = 200 N, d = 0.5 m, t = 2 s

Substitute these values into equation 1

P = (200×0.5)/2

P = 100/2

P = 50 watts

Answer:

F = 10.78 N

Hence q₂ will move away from the charge q₁ towards right side.

Explanation:

The force between two charged particles can be found by using Colomb's Law:

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

where,

F = Force = ?

k = Colomb Constant = 8.99 x 10⁹ N.m²/C²

q₁ = charge on first particle = 6 μC = 6 x 10⁻⁶ C

q₂ = charge on second particle = 2 μC = 2 x 10⁻⁶ C

r = distance between particles = 0.1 m

Therefore,

[tex]F = \frac{(8.99\ x\ 10^9\ N.m^2/C^2)(6\ x\ 10^{-6}\ C)(2\ x\ 10^{-6}\ C)}{(0.1\ m)^2}[/tex]

F = 10.78 N

Since both particles have a positive charge. Therefore this force will be the force of repulsion.

Hence q₂ will move away from the charge q₁ towards right side.

Answer:

Explanation:

E2020

Answer:

Answer is 183.6 J

Explanation:

Using the Physics reference sheet the formula for Potential energy is

(mass) x (gravity) x (height)

Mass= 1.2

Gravity I used is 9.81 (use 10 to get the answer most schools use)

Height= 15.6

Answer:

272° C

Explanation:

Given :

Volume of the balloon, V = 500 [tex]m^3[/tex]

The temperature of the surrounding air, [tex]T_{air} = 15^\circ C[/tex]

Total load, [tex]m_{T}[/tex] = 290 kg

Density of the air, [tex]$\rho_{air} = 1.23 \ kg/m^3$[/tex]

We known buoyant force,

[tex]$F_B = \rho_{air} V$[/tex]

For a 290 kg lift,  [tex]$m_{hot} = \frac{F_B}{g} = 290 \ kg$[/tex]

[tex]$m=\rho V$[/tex]

∴ [tex]$m_{hot}=\rho_{hot} V ; \ \ \ \ \ \frac{F_B}{g}-m_{hot} = 290 \ kg$[/tex]

  [tex]$(\rho_{air} - \rho_{hot}) V= 290 \ kg$[/tex]

  [tex]$\rho_{hot} = \rho_{air}- \frac{290}{V} \ kg = 1.23 \ kg/m^3 - \frac{290 \ kg}{500 \cm^3}$[/tex]

  [tex]$\rho_{hot}= 0.65 \ kg/m^3 =\frac{\rho M}{R T_{hot}}$[/tex]

 ∴ [tex]$\rho_{hot} T_{hot}= \rho_{air} T_{air}$[/tex]

   [tex]$T_{hot}= T_{air}\left[\frac{\rho_{air}}{\rho_{hot}}\right]$[/tex]

          [tex]$=288 \ K \times \frac{1.23 \ kg/m^3}{0.65 \ kg/m^3}$[/tex]

         = 545 K

          [tex]$=272^\circ C$[/tex]

Therefore, temperature of the air in the balloon is 272 degree Celsius.

To lift a load more than the weight of the balloon, the temperature of the air in the balloon has to be higher than the air in the surrounding.

Reasons:

Given information are;

Volume of the balloon = 500.0 m³

Temperature of the surrounding air = 15.0°C

Density of air at 15.0°C = 1.23 kg/m³

Required:

The temperature required to lift 290kg.

Solution:

Let, [tex]\rho _{air , b}[/tex], represent the density of the air in the balloon, we have;

[tex]\rho _{air , b}[/tex] × 500.0 + 290 = 1.23 × 500

Therefore;

[tex]\displaystyle \rho _{air , b} = \frac{1.23 \times 500- 290}{500} = 0.65[/tex]

According to the Ideal Gas Law, we have;

ρ₁ × R × T₁ = ρ₂ × R × T₂

Therefore;

[tex]\displaystyle T_2 = \mathbf{\frac{\rho_1 \times T_1}{\rho_2}}[/tex]

Therefore;

[tex]\displaystyle T_2 = \frac{1.23\times288.15}{0.65} \approx 545.27[/tex]

The temperature of the balloon, T₂ ≈ 545.27 -  273.15 = 272.12

The temperature of the air in the balloon, T₂ ≈ 272.12 °C

Learn more hear:

https://brainly.com/question/11236279

There's no multiple answers that you added if that's what you meant but it possibly could be Magma or radioactive decay of particles from the earths core if those two are any of the options