Answer: [tex]85.46\ kJ[/tex]
Explanation:
Given
Volume of air [tex]V=105\ m^3[/tex]
Temperature of air [tex]T=305\ K[/tex]
Increase in temperature [tex]\Delta T=0.7^{\circ}C[/tex]
Specific heat for diatomic gas is [tex]C_p=\dfrac{7R}{2}[/tex]
Energy required to increase the temperature is
[tex]\Rightarrow Q=nC_pdT\\\\\Rightarrow Q=n\times \dfrac{7R}{2}\times \Delta T\\\\\Rightarrow Q=\dfrac{7}{2}nR\Delta T\\\\\Rightarrow Q=\dfrac{7}{2}\times \dfrac{PV}{T}\times \Delta T\quad [\text{using PV=nRT}][/tex]
Insert the values
[tex]\Rightarrow Q=\dfrac{7}{2}\times \dfrac{1.01325\times 10^5\times 105}{305}\times 0.7\\ \text{Assuming air pressure to be atmospheric P=}1.01325\times 10^5\ N/m^2\\\\\Rightarrow Q=0.8546\times 10^5\\\Rightarrow Q=85.46\ kJ[/tex]
if a 230V television takes a current of 3A. calculate it's power
Answer:
Formula
P = I x V
Information
V = 230
A = 3
P = x
Calculation
x = 230 x 3
x = 690 W
Explanation:
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PLEASE HELP ME WITH THIS ONE QUESTION
The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x 10^-19, 1 eV = 1.6 x 10^-19 J)
A) 2.81 eV
B) 3.89 eV
C) 2.10 eV
D) 2.78 eV
The color orange has a wavelength of 590 nm. The energy of an orange photon is approximately 0.337 eV.
The correct answer is option E.
To calculate the energy of a photon, we can use the equation:
E = (hc) / λ
where E is the energy of the photon, h is the Planck's constant (6.626 x [tex]10^-^3^4[/tex]J·s or 6.626 x[tex]10^-^1^9^[/tex] eV·s), c is the speed of light (3.00 x [tex]10^8[/tex] m/s), and λ is the wavelength of the light.
Given that the wavelength of orange light is 590 nm (or 590 x [tex]10^-^9[/tex]m), we can substitute the values into the equation:
E = [(6.626 x[tex]10^-^1^9^[/tex] eV·s) x (3.00 x [tex]10^8[/tex] m/s)] / (590 x[tex]10^-^9[/tex]m)
E = (1.9878 x [tex]10^-^1^0[/tex]eV·m) / (590 x [tex]10^-^9[/tex] m)
E = 3.3695 x [tex]10^-^1[/tex] eV
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The question probable may be:
The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x [tex]10^-^1^9^[/tex], 1 eV = 1.6 x[tex]10^-^1^9^[/tex]J)
A) 2.81 eV
B) 3.89 eV
C) 2.10 eV
D) 2.78 eV
E) 0.337 eV
Which type of energy is stored in a battery?
A. Nuclear energy
B. Electromagnetic energy
C. Chemical energy
D. Electrical energy
SUBMI
Answer:
c
Explanation:
in food and batteries chemical energy is stored :) hope this helped
If a person pulls back a rubber band on a slingshot without letting to go of it, what type of energy will the rubber band have? A. Potential energy B. Rotational energy C. Kinetic energy D. Translational energy
Answer:
Potential energy. Releasing it, the potential energy would convert into motion, kinetic energy.
Potential energy is when an object has some sort of potential eg. for motion such as in this example.
Answer:
A. Potential energy
Explanation:
Potential energy can be thought of as stored energy, which has the potential of becoming kinetic energy once it has been released.
A motor lifts a weight of 5N up to the height of 2m in 4 seconds. What will be the power of motor?
2.5 watt
Explanation:
we know
power = work done /time
work done gainst gravity is = mgh
as mg = weight = force = 5 N
and height = 2 m
so work done = mgh = 5*2 = 10N
therefore ,
Power = 10/4 = 2.5 watt
PLEASE HELP!! URGENTT!
Answer:
120 Newton
Explanation:
Given the following data;
Mass = 12 kg
Angle = 4°
We know that acceleration due to gravity is equal to 10 m/s
To find the minimum force to stop the block from sliding;
Force = mgCos(d)
Where;
m is the mass of an object.
g is the acceleration due to gravity.
d is the angle of inclination (theta).
Substituting into the formula we have;
F = 12*10*Cos(4°)
F = 120 * 0.9976
F = 119.71 ≈ 120 Newton
g is incident on 3 successive sheets of polarizing material. The transmission axis of the first sheet is vertical. The transmission axis of the second sheet is at 30 degrees from vertical. The transmission axis of the third is horizontal. What is the intensity of the light emerging from the third sheet
Answer:
The intensity of light passing from the third polarizer is 3Io/16.
Explanation:
The law of Malus is given by
[tex]I=I_o cos^2\theta[/tex]
Let the incident intensity of light is Io.
The intensity of light passing from the first polarizer is
[tex]I' = \frac{I_o}{2}[/tex]
The intensity of light passing from the second polarizer is
[tex]I''=\frac{I_o}{2}\times cos^230 =\frac{3I_o}{8}[/tex]
The intensity of light passing from the third polarizer is
[tex]I''' = \frac{3I_o}{8}\times cos^2 60\\\\\\I''' = \frac{3I_o}{16}[/tex]
An object is in free fall Group of answer choices Anytime it has an acceleration equal to 9.8 m/s/s Anytime it feels the force of gravity Only if the weight is the only force acting on the object Only when it's moving vertically down
Answer:
Anytime it feels the force of gravity.
Explanation:
Gravity is considered to be a universal force of attraction which acts between all objects that has both mass, energy and can occupy space. Therefore, it acts in such a way as to bring objects together i.e causing the objects to fall down towards the Earth.
This ultimately implies that, an object is in free fall anytime it feels the force of gravity i.e regardless of how fast the object moves or the direction it moves, the only force acting on the object is the force of gravity (g).
For example, when you throw any object up, it will naturally fall down due to the gravitational force between the Earth and the object.
Additionally, the gravity of earth makes it possible for all physical objects to possess weight.
On planet Earth, the acceleration due to gravity that all physical objects experience is 9.8 meters per seconds square.
A slingshot obeys Hooke's law and is used to launch pebbles vertically into the air. You observe that if you pull a pebble back 20 cm against the elastic
band, the pebble goes 6 m high. Assuming that air drag is negligible, if you pull a pebble that is twice as heavy back 20 cm, how high will it go?
4 m
5 m
3 m
6 m
Answer:
24 m
Explanation:
By applying the law of conservation of energy.
K.E = P.E
[tex]\dfrac{1}{2}kx^2 = mgh[/tex]
By rearrangement;
[tex]\dfrac{k}{m} = \dfrac{2*g*h}{x^2}[/tex]
where;
k = spring constant
m = mass of pebble
h = height reached
x = stretched length
[tex]\dfrac{k}{m} = \dfrac{2*9.8*6}{(20 \times 10^{-2})^2}[/tex]
[tex]\dfrac{k}{m} =0.2940 \times 10^4[/tex]
Since the drag force is negligible;
[tex]\dfrac{1}{2}kx^2 = mgh[/tex]
[tex]h = \dfrac{k}{m}\dfrac{x^2}{g}[/tex]
[tex]h = \dfrac{1}{2}\times 0.2940 \times 10^4 \times \dfrac{(40\times 10^{-2})^2}{9.8}[/tex]
h = 24 m
Erica (37 kg ) and Danny (45 kg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is moving upward past her at 4.7 m/s . At that instant he grabs hold of her. What is their speed just after he grabs her?
Answer:
V = 2.58 m/s
Explanation:
Below is the calculation:
Given the weight of Erica = 37 kg
The weight of Danny = 45 kg
Danny's speed to move upward = 4.7 m/s
Use below formula to find the answer.
m1 * u1 = (m1+m2) * V
V = m1*u1 / (m1+m2)
Here, m1 = 45
u1 = 4.7
m1 = 45
m2 = 37
Now plug the values in formula:
V = m1*u1 / (m1+m2)
V = (45*4.7)/(45+37)
V = 2.58 m/s
En 2.0 s, una particula con aceleración constante a lo largo del eje x se mueve desde x =10 m
hasta x =50 m. La rapidez al final del recorrido es de 10 m/s. ¿Cuál es la aceleración de la partícula?
Consider a sample containing 1.70 mol of an ideal diatomic gas.
(a) Assuming the molecules rotate but do not vibrate, find the total heat capacity of the sample at constant volume. nCv = J/K
(b) Assuming the molecules rotate but do not vibrate, find the total heat capacity of the sample at constant pressure. nCp = J/K
(c) Assuming the molecules both rotate and vibrate, find the total heat capacity of the sample at constant volume. nCv = J/K
(d) Assuming the molecules both rotate and vibrate, find the total heat capacity of the sample at constant pressure. nCp = J/K
I don't know
because I don't know
A rock is thrown from the top of a building 146 m high, with a speed of 14 m/s at an angle 43 degrees above the horizontal. When it hits the ground, what is the magnitude of its velocity (i.e. its speed).
Answer:
time is 32 s and speed is 304.3 m/s
Explanation:
Height, h = 146 m
speed, u = 14 m/s
Angle, A = 43 degree
Let it hits the ground after time t.
Use second equation of motion
[tex]h = u t +0.5 at^2\\\\- 146 =14 sin 43 t - 4.9 t^2\\\\4.9 t^2 - 9.5 t - 146 =0 \\\\t =\frac{9.5\pm\sqrt {90.25 + 2861.6}}{9.8}\\\\t=\frac{9.5\pm 54.3}{9.8}\\\\t = 32.05 s, - 22.4 s[/tex]
Time cannot be negative so the time is t = 32 s .
The vertical velocity at the time of strike is
v' = u sin A - g t
v' = 14 sin 43 - 9.8 x 32 = 9.5 - 313.6 = - 304.1 m/s
horizontal velocity
v'' = 14 cos 43 =10.3 m/s
The resultant velocity at the time of strike is
[tex]v=\sqrt{v'^2 + v''^2}\\\\v = \sqrt{304.1^2 +10.3^2 }\\\\v = 304.3 m/s[/tex]
A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the truck is 12.7 m/s^2. The mass of the truck is 2490 kg and the mass of the car is 890 kg. Assume that the only horizontal forces on the vehicles during the collision are the forces they exert on one another.
Required:
What is the acceleration of the car at the same moment?
Answer:
The magnitude of the acceleration of the car is 35.53 m/s²
Explanation:
Given;
acceleration of the truck, [tex]a_t[/tex] = 12.7 m/s²
mass of the truck, [tex]m_t[/tex] = 2490 kg
mass of the car, [tex]m_c[/tex] = 890 kg
let the acceleration of the car at the moment they collided = [tex]a_c[/tex]
Apply Newton's third law of motion;
Magnitude of force exerted by the truck = Magnitude of force exerted by the car.
The force exerted by the car occurs in the opposite direction.
[tex]F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2[/tex]
Therefore, the magnitude of the acceleration of the car is 35.53 m/s²
Answer:
The magnitude of the acceleration of the car is 35.53 m/s²
Explanation:
Given;
acceleration of the truck, = 12.7 m/s²
mass of the truck, = 2490 kg
mass of the car, = 890 kg
let the acceleration of the car at the moment they collided =
Apply Newton's third law of motion;
Magnitude of force exerted by the truck = Magnitude of force exerted by the car.
The force exerted by the car occurs in the opposite direction.
Therefore, the magnitude of the acceleration of the car is 35.53 m/s²
Two objects are separated by a distance D. Which force, in theory, will act on the objects for any value of D?
tension
gravity
normal
friction
Answer:
here is ur answer...
Explanation:
gravity .......is ur answer
and the value of D is DE³S NU + s
An attractive force that exist between all object with mass attract another objects the magnitude of the force is directly proportional..
Hope this help!
a 1600 kg car rounds a curve of radius 71 m banked at an angle of 15, What is the magnitude of the friction force required for the car to travel at 86 km/h
Answer:
The frictional force required for the car to travel is 8,365.01 N
Explanation:
Given;
mass of the car, m = 1600 kg
radius of the curved road, r = 71 m
banking angle, θ = 15⁰
velocity of the car, v = 86 km/h = 86/3.6 = 23.89 m/s
The two forces acting on the are:
1. the parallel force to the banked plane
2. the centripetal force pushing the car up the banked plane
To keep the car traveling at 86 km/h;
frictional force + parallel force to the plane = centripetal force pushing the car up the banked plane
The parallel force to the banked plane:
F = mgsinθ
F = 1600 x 9.8 x sin(15⁰)
F = 4,057.98 N
The centripetal force pushing the car up the banked plane:
[tex]F_c= (\frac{mv^2}{r} )cos(\theta)\\\\F_c = (\frac{1600 \times 23.89^2}{71} )cos(15^0)\\\\F_c = 12,422.99 \ N[/tex]
The frictional force required for the car to travel:
[tex]F_k = F_c - F\\\\F_k = 12,422.99 \ N - 4,057.98 \ N\\\\F_k = 8,365.01 \ N[/tex]
Therefore, the frictional force required for the car to travel is 8,365.01 N
Komal found that her vial of isopropyl alcohol showed a much better surface
tension bubble shape (a higher bubble) than her vial of water. Her peer group
suggested some experimental errors that may have caused this to happen.
Which three experimental errors are most likely to have occurred?
Answer:
A B C
Explanation:just took the test
Answer:
Komal could have mixed up the labels on the vials.
Someone might have jiggled the table and made the water surface tension bubble spill over.
The water could actually have been saltwater instead of pure water.
Explanation:
...........................
A toy car of mass 600g moves through 6m in 2 seconds. The average kinetic energy of the toy car is
Answer:
12
Explanation:
I'm a beginner so am not sureeeeee
planet a takes one year to go around the sun at a distance of one au. .planet b is three a u. from the sun. how many years does planet b take to orbit?
Answer:
3 years
Explanation:
Find the circumference of each orbit in AU.
2xπx1=6.283185307
2xπx3=18.84955592
Divide them.
18.84955592/6.283185307=3
3 years
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
a) 8.03 x 10^16 nuclei
b) 4.01 x 10^16 nuclei
c) 2.02 x 10^16 nuclei
d) 1.61 x 10^17 nuclei
Answer:
c) 2.02 x 10^16 nuclei
Explanation:
The isotope decay of an atom follows the equation:
ln[A] = -kt + ln[A]₀
Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope
[A] = Our incognite
k is constant decay:
k = ln 2 / Half-life
k = ln 2 / 4.96 x 10^3 s
k = 1.40x10⁻⁴s⁻¹
t is time = 1.98 x 10^4 s
[A]₀ = 3.21 x 10^17 nuclei
ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]
ln[A] = 37.538
[A] = 2.01x10¹⁶ nuclei remain ≈
c) 2.02 x 10^16 nucleiIs the following chemical reaction balanced?
2H202-H2O + O2
yes
no
Which statement describes an action-reaction pair?
O A. You push on a car, and the car pushes back on you.
B. A book pushes down on a table, and the table pushes down on the
Earth.
C. The Moon pulls on Earth, and Earth pulls on the Sun.
D. You push down on your shoe, and Earth's gravity pulls down on the
shoe.
Answer:
A
Explanation:
a pex
A car is moving at 10 m/s on a horizontal road with friction on a dry day. The car can travel around a traffic circle with a minimum radius of 4.8 meters. It rains and the car around a traffic circle with a minimum radius of 11.8 meters. What is the percentage of the coefficient of static friction on the rainy day compared to the dry day
Answer:
[tex]\mu_w=86\%[/tex]
Explanation:
From the question we are told that:
Velocity on Dry road [tex]V_d=10m/s[/tex]
Radius Dry [tex]r_d=4.8[/tex]
Radius wet [tex]r_w=11.8[/tex]
Generally the equation for coefficient of static friction on the dry day is mathematically given by
[tex]\mu mg=\frac{mv^2}{r_d}[/tex]
[tex]\mu g=\frac{v^2}{r_d}[/tex]
[tex]\mu_d 9.8=\frac{10^2}{4.8}[/tex]
[tex]\mu_d=2.125[/tex]
Generally the equation for the relationship between Radius & coefficient of static friction is mathematically given by
[tex]\frac{\mu_d}{\mu_w}=\frac{r_d}{r_w}[/tex]
[tex]\frac{\mu_w}{2.125}=\frac{4.8}{11.8}[/tex]
[tex]\mu_w=0.86[/tex]
Therefore
[tex]\mu_w=86\%[/tex]
how to calculate sound of an echo
by an echo meter
please flw me and thank my answers
#Genius kudi
You want to calculate how long it takes a ball to fall to the ground from a
height of 20 m. Which equation can you use to calculate the time? (Assume
no air resistance.)
O A. vz? = v? +2aAd
B. a =
V₂-vi
At
O c. At=V1
4
a
O D. At=
2Ad
a
If a person wants to calculate the length of time it takes for a ball to fall from a height of 20m, the correct equation that they should use is:
D. Δt= √2Δd/a
What is the equation for finding the length of time for a free fall?The free fall formula should be used to obtain the length of time that it takes for a ball to fall from a given height. This formula also factors the height or distance from which the fall occurred and this is denoted by the letter d. The small letter 'a' is denotative of acceleration due to gravity and this is a constant pegged at -9.98 m/s².
So, the change in height is obtained and multiplied by two. This is further divided by the acceleration and the square root of the derived answer translates to the time taken for the ball to fall from the height of 20m. Of all the options listed, option D represents the correct equation.
Learn more about free fall here:
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Please need in hurry
Explanation:
i) center of gravity (or mass)
ii) m = W/g = (160 N)/(9.8 m/s^2)
= 16.3 kg
can anyone explain soil erosion
Answer:
Soil erosion is the act of removing the top layer of the soil either by rain wind or human activities
Answer:
l erosion is a gradual process that occurs when the impact of water or wind detaches and removes soil particles, causing the soil to deteriorate. ... The impact of soil erosion on water quality becomes significant, particularly as soil surface runoff. Sediment production and soil erosion are closely related.
Explanation:
need in hurry please
Using pascals principle, F1/A1=F2/A2, solve this like a proportion.
A force of 50 N is applied to an area of 200 sq feet, how much force will be applied of the area to be covered is 50 sq feet?
Answer:
NO CLUE
Explanation:
GOOD LUCK THOUGH
violations in a responsible manner in a democratic society?
Activity 5:
Ending
From your findings, what conclusions and recommendations can you make on the
issue of human rights violations to:
5.1
Government
Answer:
Kindly check explanation
Explanation:
The trampling and violation of human rights individuals, groups and corporate organizations is really alarming ad as such, the government who are charged to protect the right and interest of its citizen. In other to curtail the trending issues of human right violation, it is imperative if sensitization programmes could be organiz d in other to keep people informed of the various ways in which people's right may be trampled upon. With these education, the ignorance can be expunged leaving only those who genuinely decides to relate and threaten the right of his fellow country person.
The laws on human right violation should be reviewed and capital punishment metted on violators in other to send a strong warning to those who still nurture the intention.