Answer:
The cost of energy is $ 0.34.
Explanation:
The energy is the capacity to do work.
The energy is a scalar quantity and its SI unit is Joule.
The commercial unit of energy is kWh.
Cost of 1 kWh energy = $ 0.17
energy loss by standard window is 2 kWh .
So, the cost of lost of energy is
Cost = $ 0.17 x 2 = $ 0.34
A 1.2 kg basketball is thrown upwards. What is the potential energy of the basketball at the top of its path if it reaches a height of 15.6 m?
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
Jet aircraft maintenance crews are required to wear protective earplugs. Members of a particular crew wear earplugs that reduce the sound intensity by a factor of 305. If, when the jet is fired up, the sound intensity level experienced by the crew members wearing protective earplugs is 79 dB, determine the sound intensity level they would experience without the earplugs.
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
Two people that have identical weight are holding onto a massless pole while standing on horizontal frictionless ice. 1)If the guy on the left starts to pull on the pole, where do they meet
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.
If a second ball were dropped from rest from height ymax, how long would it take to reach the ground
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
Which of the following would likely happen if a person’s lactic acid system had difficulty breaking down glycogen in the muscles?
The person would have difficulty swimming across a lake.
The person would have difficulty sprinting in a race.
The person would have difficulty cycling down a hill.
The person would have difficulty running a marathon.
Answer: I think that its b, they would have difficulty sprinting in a race
Explanation:
What kind of model is shown below?
O A. A mathematical model
B. An experimental model
O C. A computer model
D. A physical Model
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.
What is an experimental model 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
A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vector. Ignoring the particle's weight, what type of path will the particle follow
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.
please help! will mark brainliest
Answer:
27.D.28.AExplanation:
THE ANSWERME MY ANSWERWAH IS ME AND MY ANSWER WAHHHWALA NA SIYA WALA NA SIYA MASAYA MASAYA MASAYA WAHHtinapon na siya tipon na siya wahhsa basurahan wahh wahh ...............A tank containing a fluid is stirred by a paddle wheel. The work input to the paddle wheel is 5090 kJ. The heat transfer from the tank is 1500 kJ. Consider the tank and the fluid inside a control surface and determine the change in internal energy, in kJ, of this control mass.
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]
ytyfgdvvdgjuvsthvhfcj
A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is the magnitude
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;
r represents the radius and t is the time.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;
V is the circular speed (velocity) of an object.r is the radius of circular path.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
An illustration with two positive spheres 0.1m apart. The one on the left is labeled q Subscript 1 baseline = 6 microcoulombs and the sphere on the right is labeled q Subscript 2 baseline = 2 microcoulombs.
Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.
Recall that k = 8.99 × 109 N•meters squared per Coulomb squared.
What is the force applied between q1 and q2?
In which direction does particle q2 want to go?
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
A weightlifter presses a 200 N weight 0.5 m over his head in 2 s. What is the power of the weightlifter
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
The maximum amount of pulling force a truck can apply when driving on
concrete is 8760 N. If the coefficient of static friction between a trailer and
concrete is 0.8, what is the heaviest that the trailer can be and still be pulled
by the truck?
O A. 8760 N
O B. 12,680 N
O C. 10,950 N
O D. 7240 N
Answer:
8760 N
Explanation:
think this is the right answer :)
consider a circular loop of wire carrying a counterclockwise current as shown. Indicate the direction of the magnetic field at points both inside and outside of the loop.
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
At what speed was object A moving ?
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:
5N
5 N
19 N
19 N
Pls help look at the pic
Answer:
b. is the correct answer ....
What are the relationships between the temperature scales of Fahrenheit, Kelvin, Celsius, and Rankine
If the ball is 0.60 mm 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
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
The blade of a windshield wiper moves through an angle of 90.08 in 0.40 s. The tip of the blade moves on the arc of a circle that has a radius of 0.45 m. What is the magnitude of the centripetal acceleration of the tip of the blade
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
Activity 1
The equation for the combustion of butane gas is given below.
1.1
1.2
AH < 0
butane(g) + 1302(g) → 8CO2(g) + 10H2O(g)
Define the term activation energy.
Is the combustion reaction of butane exothermic or endothermic? Give
reason for the answer.
Draw a sketch graph of potential energy versus course of reaction for
reaction above.
3
Clearly indicate the following on the graph:
o
Activation energy
Heat of reaction (AH)
Reactants and products
Determine the empirical formula of butane gas if it consists of 82,76%
and 17,24% hydrogen.
Answer:
I don't know hhaha ammmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm I wish I could help
If you move 10 times farther away from a source of light, then how will the
apparent brightness of that source change?
it will become 10 times less bright
it will become 2 times less bright
its brightness will not change
O it will become 100 times less bright
Please help! ❤️
I’ll make you the Brainlyest, I can’t get this one wrong.
Convert 387.1 K to °C
What are impact and non-impact printers?
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 :)))
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.62 10-6 W/m2 at a distance of 165 m from the explosion, at what distance from the explosion is the sound intensity half this value
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
A car drives 110 km in 2 hours. Calculate the speed of the car
Answer: 55 kmph
Explanation: Divide 110 by 2
Which of these is a source of thermal energy inside earth
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
light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance light travels in one year.
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
A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. If the volume of the balloon is 500.0 m^3 and the surrounding air is at 15.0°C. What must the temperature of the air in the balloon be for it to lift a total load of 290 kg (in addition to the mass of the hot air)? The density of air at 15.0°C and atmospheric pressure is 1.23kg/m^3.
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.
The temperature of the air in the balloon to lift a total load of 290 kg is approximately 272.12°C.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