Answer:
Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature); dT/dt = -K(T - Tₐ) where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute),
Explanation:
Give the number of protons and the number of neutrons in the nucleus of each of the following isotopes Aluminum 25 :13 protons and 12 neutrons
Answer:
No of proton is 13 and nucleus is 13
Calculate the heat energy conducted per hour through the side walls of a cylindrical steel
boiler of 1.00 m diameter and 3.0 m long if the internal and external temperatures of the
walls are 140 °C and 40 °C respectively and the thickness of the walls is 6.0 mm. (Thermal
conductivity of steel, k = 42 Wm-4°C-4)
Explanation:
heat caoacity and heat is difference
The heat energy conducted per hour through the side walls of the cylindrical steel boiler is 27708847 kJ.
What is thermal conductivity?The rate at which heat is transported by conduction through a material's unit cross-section area when a temperature gradient exits perpendicular to the area is known as thermal conductivity.
In the International System of Units (SI), thermal conductivity is measured by Wm⁻¹K⁻¹.
Diameter of the cylindrical steel boiler: d = 1.00m.
Length of the cylindrical steel boiler: l = 3.00m.
thickness of the walls is = 6.0 mm = 0.006 m
Temperature gradient is = (140-40) °C/0.006 m = 1666.67 °C/m
Thermal conductivity of steel, = 42 W/m-°C.
Hence, the heat energy conducted per hour through the side walls of the cylindrical steel boiler = 42×3600×1666.67 ×2π×0.5(0.5+3.0) Joule
= 27708847 kJ
Learn more about thermal conductivity here:
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what are 2 ways that we can express to show our connection to our culture
Answer:
Food, clothes, language, and belief
The food calorie, equal to 4186 J, is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 160 food calories per bar.
Part A
If a 67.0 kg hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy?
Express your answer in meters.
Part B
If, as is typical, only 20.0 % of the food calories go into mechanical energy, what would be the answer to Part A? (Note: In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)
Express your answer in meters.
Answer: 1 cal is 4.186 J, 1 kcal = 4186 J A : 1014 m , B 200 m
Explanation: A) Work done by climber is change in potential energy.
W = ΔEp = mgh = 67.0 kg· 9.81 m/s²· h = 160 kcal · 4186 J / kcal.
Solve h = 160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 1014 m
B Energy is only 20 % : Then h = 0.20 ·160 kcal · 4186 J / kcal /67.0 kg· 9.81 m/s² = 200 m.
Actually, muscles also produce heat from most of the energy provided by food.
A hockey puck is sliding across the ice with an initial velocity of 25 m/s. If the coefficient of friction between the hockey puck and the ice is 0.08, how much time (in seconds) will it take before the hockey puck slides to a stop
Answer: 31.89seconds
Explanation:
Based on the information given, we are meant to calculate deceleration which will be:
t = V/a
where, a = mg
Therefore, t = V/mg
t = 25/0.08 × 9.8
t = 25/0.784
t = 31.89seconds
Therefore, the time that it will take before the hockey puck slides to a stop is 31.89seconds.
Find the weight of a man whose mass is 40 kg on earth.
(also
write complete data plus proper formula).
I am sure it help you with that much ☺️
Explanation:
pleasae give me some thanks please good morning sister
20 pts.
A man forgets that he set his coffee cup on top of his car. He starts to drive and the coffee CUP rolls off the car onto the road. How does this scenario demonstrate the first law of motion? Be specific and use the words from the law in your answer.
Answer:
The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.
Explanation:
Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."
Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.
The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at the moments of time: 0, T, 2T, 3T and 4T.
Explanation:
The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum
How can i prove the conservation of mechanical energy?
Answer:
We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved
Explanation:
A charge Q is transferred from an initially uncharged plastic ball to an identical ball 24 cm away.The force of attraction is then 17 mN. How many electrons were transferred from one ball to the other?
Answer:
The number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
Explanation:
Given;
magnitude of the attractive force, F = 17 mN = 0.017 N
distance between the two objects, r = 24 cm = 0.24 m
The attractive force is given by Coulomb's law;
[tex]F = \frac{1}{4\pi \epsilon _0} \times \frac{Q^2}{r^2} = \frac{kQ^2}{r^2} \\\\Q^2 = \frac{Fr^2}{k} \\\\Q = \sqrt{ \frac{Fr^2}{k}} \\\\Q = \sqrt{ \frac{(0.017)(0.24)^2}{9\times 10^9}} \\\\Q = 3.298 \times 10^{-7} \ C[/tex]
The charge of 1 electron = 1.602 x 10⁻¹⁹ C
n(1.602 x 10⁻¹⁹ C) = 3.298 x 10⁻⁷
[tex]n = \frac{3.298 \times 10^{-7}}{1.602 \times 10^{-19}} = 2.06 \times 10^{12} \ electrons[/tex]
Therefore, the number of electrons transferred from one ball to the other is 2.06 x 10¹² electrons
List the 5 theoretical perspectives that underlie much of the research on human development. Also, name an individual strongly associated with each perspective.
Answer:
25
Explanation:
5 theroical name indvivdual perspective asssssoitive each persp
In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer spce. From stationary positions, they push against each other. Bonzo flies off with a velocity of 1.1 m/s, while Ender recoils with a velocity of -4.3 m/s. Determine the ratio Bonzo/mEnder of the masses of these two enemies.
Answer:
the ratio Bonzo/mEnder of the masses of these two enemies is 3.91
Explanation:
Given the data in the question;
Velocity of Bonzo [tex]V_{Bonzo[/tex] = 1.1 m/s
Velocity of Ender [tex]V_{Ender[/tex] = -4.3 m/s
the ratio Bonzo/mEnder of the masses of these two enemies = ?
Now, using the law of conservation of momentum.
momentum of both Bonzo and Ender are conserved
so
Initial momentum = final momentum
we have
0 = [tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] + [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex]
[tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] = -[ [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex] ]
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = -[ [tex]V_{Ender[/tex] / [tex]V_{Bonzo[/tex] ]
we substitute
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = -[ -4.3 m/s / 1.1 m/s ]
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = -[ -3.9090 ]
[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex] = 3.91
Therefore, the ratio Bonzo/mEnder of the masses of these two enemies is 3.91
Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes are designed to have a large drag coefficient. One model expands to a square 1.8 mm on a side, with a drag coefficient of 1.4. A runner completes a 240 mm run at 6.0 m/s with this chute trailing behind.
Required:
How much thermal energy is added to the air by the drag force?
Answer:
by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.
Explanation:
Given the data in the question;
drag coefficient of Cd = 1.4
speed v = 6.0 m/s
One model expands to a square 1.8 mm on a side
Area A = 1.8 × 1.8 = 3.24 mm² = 3.24 × 10⁻⁶ m²
distance travelled s = 240 mm = 0.24 m
we know that; density of air e = 1.225 kg/m³
Now,
Dragging force F[tex]_D[/tex] = ( Cd × e × v² × A ) / 2
thermal energy = F[tex]_D[/tex] × s
so
thermal energy = ( 1.4 × 1.225 × (6)² × (3.24 × 10⁻⁶) × 0.24 ) / 2
thermal energy = ( 4.8009024 × 10⁻⁵ ) / 2
thermal energy = 2.4004512 × 10⁻⁵ J
Therefore, by the drag force, 2.4004512 × 10⁻⁵ J is added to the air.
A rectangular field is of length 42 cm and breadth 25 m. Find the area of the field in SI unit. EXPLAIN STEP BY STEP
Answer:
the area of the rectangular field is 10.5 m²
Explanation:
Given;
length of the rectangular field, L = 42 cm = 0.42 m
breadth of the rectangular field, b = 25 m
The area of the rectangular field is calculated as follows;
Area = Length x breadth
Area = 0.42 m x 25 m
Area = 10.5 m²
Therefore, the area of the rectangular field is 10.5 m²
A motorist travels due North at 90 km/h for 2 hours. She changes direction and travels West at 60 km/for 1 hour.
a) Calculate the average speed of the motorist [4]
b) Calculate the average velocity of the motorist.
Answer:
a. 50km/hr.
b. 10km/hr
Explanation:
Average speed, which is calculated by dividing the total distance travelled by the time interval as follows:
Average speed = total distance travelled ÷ time
Average velocity is calculated by dividing the total displacement by the time interval as follows:
Average velocity = change in displacement (∆x) ÷ time (t)
According to this question, a motorist travels due North at 90 km/h for 2 hours. She then changes direction and travels West at 60 km/for 1 hour.
Total distance of this journey is 90 + 60 = 150
Total time taken = 1 + 2 = 3hours
Average speed = 150/3
= 50km/hr.
b.) Average velocity = x2 - x1/t
Average velocity = 90 - 60/3
= 30/3
= 10km/hr
A light-emitting diode (LED) connected to a 3.0 V power supply emits 440 nm blue light. The current in the LED is 11 mA , and the LED is 51 % efficient at converting electric power input into light power output. How many photons per second does the LED emit?
Answer:
3.73 * 10^16 photons/sec
Explanation:
power supply = 3.0 V
Emits 440 nm blue light
current in LED = 11 mA
efficiency of LED = 51%
Calculate the number of photons per second the LED will emit
first step : calculate the energy of the Photon
E = hc / λ
=( 6.62 * 10^-34 * 3 * 10^8 ) / 440 * 10^-9
= 0.0451 * 10^-17 J
Next :
Number of Photon =( power supply * efficiency * current ) / energy of photon
= ( 3 * 0.51 * 11 * 10^-3 ) / 0.0451 * 10^-17
= 3.73 * 10^16 photons/sec
1. Lifting an elevator 18m takes 100kJ. If doing so takes 20s, what is the average power of the elevator during the process?
2. How much work can a 0.4 hp electric mixer do in 15 s?
Answer:
1. Power = 5000 Watts
2. Workdone = 11185.5 Joules
Explanation:
Given the following data;
1. Distance = 18 m
Energy = 100 KJ = 100,000 Joules
Time = 20 seconds
To find the average power of the elevator;
Power = energy/time
Power = 100000/20
Power = 5000 Watts
2. Power = 0.4 HP
Time = 15 seconds
Conversion:
1 horsepower = 745.7 Watts
0.4 horsepower = 0.4 * 745.7 = 298.28 Watts
To find the amount of work done by the electric mixer;
Work done = power * time
Workdone = 745.7 * 15
Workdone = 11185.5 Joules
semiconductor have negative temperature coefficient of resistance why
Answer:
As the number of free electrons increases, the resistance of this type of non-metallic material decreases with increasing temperature.
Explanation:
Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0 = 1.68×10-21 J and R0= 3.82×10 the frequency of small oscillations of one Ar atom about its equilibrium position.
Answer:
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Explanation:
From the given information:
The elastic potential energy can be calculated by using the formula:
[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]
Making K the subject;
[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]
[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]
k = 2.3 × 10⁻² N/m
Now; the frequency of the small oscillation can be determined by using the formula:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
where;
m = mass of each atom = 1.66 × 10⁻²⁶ kg
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
45. Pressure in air undergoes a decrease when the air
a) rises to higher altitudes.
b) accelerates to higher speed.
c) fills a greater space.
d) All of these.
What happens if you move a magnet near a coil of wire?
A) current is induced
B)power is consumed
C)the coil becomes magnetized
D) the magnets field is reduced
I’ve been stuck please help !!
Answer:
The slope of the position time graph gives the velocity.
Explanation:
The slope of the position time graph gives the value of velocity.
In first graph,
The slope is constant in both the parts but positive . So the velocity is also constant and positive for both the parts. and more than the second part, so the initial velocity is more than the final velocity.
In the second graph,
The slope is constant in both the parts but negative. So, the velocity is constant but negative for both the parts. Initial velocity is more negative than the final velocity.
A circus performer stretches a tightrope between two towers. He strikes one end of the rope and sends a wave along it toward the other tower. He notes that it takes 0.9 s for the wave to travel the 26 m to the opposite tower. If one meter of the rope has a mass of 0.28 kg, find the tension in the tightrope.
Answer:
the tension in the tightrope is 233.68 N
Explanation:
Given the data in the question;
Time taken to reach the opposite tower t = 0.9 s
Distance between the two towers S = 26 m
mass per one meter length = 0.28 kg
First we calculate the velocity;
Velocity V = Distance / time
we substitute
Velocity V = 26 m / 0.9 s
Velocity V = 28.889 m/s
We know that Velocity V can also be expressed as;
V = √( T / m )
we make T the subject of formula
V² = T / m
T = mV²
we substitute
T = 0.28 × ( 28.889 )²
T = 233.68 N
Therefore, the tension in the tightrope is 233.68 N
A professional quarterback throws a 0.40 kg football. what is the force of weight?
Answer:
3.92N
Explanation:
Force= mass×accelerarion due gravity
But mass= 0.40kg
acceleration due to gravity = 9.8 m/s^2
Force = 0.40×9.8
Force=3.92N
From 2 King 6:1-6, one of the disciples of Elisha was cutting a tree and the ax head fell into the water. While we do not know how high the ax head was when it fell into the water, we will work through a physics example of the ax head's vertical motion as if it were dropped into the water. ( Due date 09/07)
Write your name and date. The due date of this assignment is the height the ax head falls from in meters into the water. For example, if the due date is July 15, then the ax head fell 15 meters to the water.
Write Newton’s 2nd Law in Equation Form.
Write the quantity and units of average gravitational acceleration on the surface of Earth.
Given the ax head mentioned in the opening portion with the height being equal in numerical value of the due day of this assignment. How long does it take for the ax to fall to the river surface?
Compute the final speed of the ax when it hits the water.
Answer:
time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.
Explanation:
Height, h = 15 m
Newton's second law
Force = mass x acceleration
The unit of gravitational force is Newton and the value is m x g.
where, m is the mas and g is the acceleration due to gravity.
Let the time of fall is t.
Use second equation of motion
[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]
Let the final speed is v.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]
At the start of a basketball game, a referee tosses a basketball straight into the air by giving it some initial speed. After being given that speed, the ball reaches a maximum height of 4.35 m above where it started. Using conservation of energy, find the height of the ball when it has a speed of 2.5 m/s.
Answer:
0.32 m.
Explanation:
To solve this problem, we must recognise that:
1. At the maximum height, the velocity of the ball is zero.
2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.
With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:
Mass (m) = constant
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) = 2.5 m/s
Height (h) =?
PE = KE
Recall:
PE = mgh
KE = ½mv²
Thus,
PE = KE
mgh = ½mv²
Cancel m from both side
gh = ½v²
9.8 × h = ½ × 2.5²
9.8 × h = ½ × 6.25
9.8 × h = 3.125
Divide both side by 9.8
h = 3.125 / 9.8
h = 0.32 m
Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.
A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that
the acceleration due to gravity is 9.8 m/s2, what is the crate's potential energy
at this point?
Answer:
[tex]\boxed {\boxed {\sf 29,400 \ Joules}}[/tex]
Explanation:
Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.
[tex]E_P= m \times g \times h[/tex]
The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.
m= 150 kg g= 9.8 m/s²h= 20 mSubstitute the values into the formula.
[tex]E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m[/tex]
Multiply the three numbers and their units together.
[tex]E_p=1470 \ kg*m/s^2 \times 20 m[/tex]
[tex]E_p=29400 \ kg*m^2/s^2[/tex]
Convert the units.
1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.
[tex]E_p= 29,400 \ J[/tex]
The crate has 29,400 Joules of potential energy.
Answer:
29,400 J
Explanation:
did the quiz <3
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 550 lines/mm , and the light is observed on a screen 1.7 m behind the grating.
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.
Answer:
Δd = 7.22 10⁻² m
Explanation:
For this exercise we must use the dispersion relationship of a diffraction grating
d sin θ = m λ
let's use trigonometry
tan θ = y / L
how the angles are small
tant θ = sinθ /cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
let's use direct ruler rule to find the distance between two slits
If there are 500 lines in 1 me, what distance is there between two lines
d = 2/500
d = 0.004 me = 4 10⁻⁶ m
diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1
let's calculate for each wavelength
λ = 656 nm = 656 10⁻⁹ m
d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶
d₁ = 2.788 10⁻¹ m
λ = 486 nm = 486 10⁻⁹ m
d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶
d₂ = 2.066 10⁻¹ m
the distance between the two lines is
Δd = d1 -d2
Δd = (2,788 - 2,066) 10⁻¹
Δd = 7.22 10⁻² m
An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is 8.2 times the magnitude of the tangential acceleration. What is the angle?
Answer:
The angle is 4.1 rad.
Explanation:
The centripetal acceleration (α) is given by:
[tex] \alpha = \omega^{2} r [/tex] (1)
Where:
ω: is the angular velocity
r: is the radius
And the tangential acceleration (a) is:
[tex] a = \alpha r [/tex] (2)
Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:
[tex] \omega^{2} r = 8.2\alpha r [/tex]
[tex] \omega^{2} = 8.2\alpha [/tex] (3)
Now, we can find the angle with the following equation:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex]
Where:
[tex] \omega_{f}[/tex]: is the final angular velocity [tex] \omega_{0}[/tex]: is the initial angular velocity = 0 (it starts from rest)
[tex]\Delta \theta[/tex]: is the angle
[tex] \omega^{2} = 2\alpha \Delta \theta [/tex] (4)
By entering equation (3) into (4) we can calculate the angle:
[tex] 8.2\alpha = 2\alpha \Delta \theta [/tex]
[tex] \Delta \theta = 4.1 rad [/tex]
Therefore, the angle is 4.1 rad.
I hope it helps you!
What are 3 artificial and 2 natural sources of electromagnetic radiation?
Answer: its b bro
Explanation:
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