Answer:
The answer is "[tex]55.05 \ \frac{J}{K}[/tex]"
Explanation:
Given value:
[tex]Q_m=4.184 \times 10^3 \ J\\T_0=100^{\circ}\\T_1=24^{\circ}[/tex]
Calculating the heat capacity of this chunk of metal:
Using formula:
[tex]C_m=\frac{Q_m}{T_0-T_1}\\[/tex]
[tex]=\frac{4.184 \times 10^3}{100 -24}\\\\=\frac{4.184 \times 10^3}{76}\\\\=55.05 \ \frac{J}{K}[/tex]
The wings of the blue-throated hummingbird, which inhabits Mexico and the southwestern United States, beat at a rate of up to 900 times per minute.
Required:
a. Calculate the period of vibration of the bird’s wings.
b. Calculate the frequency of the wings’ vibratio
c. Calculate the angular frequency of the bird’s wingbeats.
Answer:
a) the period of vibration of the bird’s wings is 0.0667s
b) the frequency of the wings’ vibration is 15 Hz
c) the angular frequency of the bird’s wingbeats is 94.2 rad/s
Explanation:
Given the data in the question;
beat at a rate of up to 900 times per minute.
The number of vibrations per unit time can be be determined as follows;
frequency f = number of vibration / time ---------- let this be equation 1
Also Time period T is;
T = 1/f { time period and frequency are reciprocal to each other }
T = 1/f ---------- Let this be equation 2
Then, the angular frequency ω can be determined using the formular;
ω = 2π × f -------------- Let this be equation 3.
Now, given that; The wings of the blue-throated hummingbird beat at a rate of up to 900 times per minute; Hence,
from equation 1, frequency f = number of vibration / time
f = 900 / 1 min
f = 900 / 60 sec
f = 15 Hz
Therefore, the frequency of the wings’ vibration is 15 Hz
a) the period of vibration of the bird’s wings
from equation 2, T = 1/f ,
we substitute in the value of f
T = 1 / 15 Hz
T = 0.0667s
Therefore, the period of vibration of the bird’s wings is 0.0667s
c) the angular frequency of the bird’s wingbeats;
from equation 3, ω = 2π × f
we substitute
ω = 2π × 15 Hz
ω = 94.2 rad/s
Therefore, the angular frequency of the bird’s wingbeats is 94.2 rad/s
Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is sitting three meters away from the fulcrum at the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. The mass of the board is evenly distributed so that its center of mass is over the fulcrum. What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground?
a. 4m
b. 5m
c. 2m
d. 3m
NO LINKS.
Answer:
x = 4 m
Explanation:
For this exercise we must use the rotational equilibrium relationship, where we place zero at the turning point and counterclockwise rotations we will consider positive
as it indicates that the bar is in equilibrium, its center of mass coincides with the turning point, so the distance is zero and does not create torque on the system
∑τ = 0
W 3 - w x = 0
x = 3W / w
x = 3 Mg / mg
x = 3 M / m
let's calculate
x = 3 60/45
x = 4 m
135g of an unknown substance gains 9133 J of heat as it is heated from 25⁰C to 100⁰C. Using the chart below, determine the identity of the unknown substance.
Answer:
The unknown substance is Aluminum.
Explanation:
We'll begin by calculating the change in the temperature of substance. This can be obtained as follow:
Initial temperature (T₁) = 25 ⁰C
Final temperature (T₂) = 100 ⁰C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 100 – 25
ΔT = 75 ⁰C
Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:
Change in temperature (ΔT) = 75 ⁰C
Mass of the substance (M) = 135 g
Heat (Q) gained = 9133 J
Specific heat capacity (C) of substance =?
Q = MCΔT
9133 = 135 × C × 75
9133 = 10125 × C
Divide both side by 10125
C = 9133 / 10125
C = 0.902 J/gºC
Thus, the specific heat capacity of substance is 0.902 J/gºC
Comparing the specific heat capacity (i.e 0.902 J/gºC) of substance to those given in the table above, we can see clearly that the unknown substance is aluminum.
A truck was carrying a substance in a tank. The molecules of that substance were moving away from each other. The truck parked overnight in a place where energy transferred out of the substance. In the morning, the substance was a gas. How were the molecules moving in the morning? Explain why the molecules were moving that way after energy was transferred out of them.
(Make it ask long as you need)
Or (short)
Answer:
In the morning the molecules were moving away from each other with a smaller speed than when the truck was carrying the substance.
Explanation:
A loudspeaker, mounted on a tall pole, is engineered to emit 75% of its sound energy into the forward hemisphere, 25% toward the back. You measure an 85 dB sound intensity level when standing 3.5 m in front of and 2.5 m below the speaker. What is the speaker’s power output?
Answer:
"0.049 W" is the correct answer.
Explanation:
According to the given question,
[tex]r = \sqrt{(3.5)^2+(2.5)^2}[/tex]
[tex]=\sqrt{8.5}[/tex]
[tex]SL=85[/tex]
As we know,
⇒ [tex]SL=10 \ log(\frac{I}{I_o} )[/tex]
[tex]85=10 \ log(\frac{I}{10^{-12}} )[/tex]
[tex]I=3.162\times 1^{-4} \ W/m^2[/tex]
Now,
⇒ [tex]P_{front} = I(2\pi r^2)[/tex]
[tex]=(3.162\times 10^{-4})(2\pi\times 18.5)[/tex]
[tex]=0.0368 \ W[/tex]
[tex]=0.75 \ P[/tex]
or,
[tex]=0.049 \ W[/tex]
74Be decays with a half-life of about 53 d. It is produced in the upper atmosphere, and filters down onto the Earth's surface.
If a plant leaf is detected to have 350 decays/s of 74Be, how long do we have to wait for the decay rate to drop to 25 per second?
Answer:
201.8 days
Explanation:
The activity of a radioactive sample as a function of times is :
[tex]$R=R_0e^{\frac{0.693t}{T_{1/2}}}$[/tex]
Here, [tex]$R_0$[/tex] = the initial activity
[tex]$T_{1/2}$[/tex] = half life
t = elapsed time
Now rearranging the equation for time, t, we get:
[tex]$\frac{R}{R_0}=e^{-\frac{0.693t}{T_{1/2}}}$[/tex]
[tex]$\ln\left(\frac{R}{R_0}\right)=-\frac{0.693t}{T_{1/2}}$[/tex]
[tex]$t=\frac{-\ln\left(\frac{R}{R_0}\right)T_{1/2}}{0.693}$[/tex]
[tex]$t=\frac{-\ln\left(\frac{25}{350}\right)\times 53}{0.693}$[/tex]
= 201.8 days
Therefore, the required time is 201.8 days
A book is sitting on a table. The book is
applying a 65 N down on the table.
How much force is the table exerting on the
book?
Answer:
65 Newtons
Explanation:
A book is kept on the table. The book is applying 65 N of force down on the table. So the table will also apply 65 N force to the book.
What is Force?According to physics, a force is an effect that has the power to change an object's motion. A force can cause a massed item to accelerate or modify its velocity. A push or a pull is a straightforward method to explain force. Considering that a force has both a magnitude and a direction, it is a vector quantity.
The pace at which an object's direction changes when it is traveling is referred to as its velocity, and is measured by a particular unit of time and from a particular point of view.
When the book is applying force on the table that is mg, and it is 65 N the table will also generate a normal reaction force which will be equal to the force applied by the book which is 65 N.
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How does a generator use the effect you noticed in the pickup coil to generate electrical energy? What energy transformations are taking place? Why does a generator make alternating current? What change would have to be made to make direct current?
Answer:
Movement of Electrons in opposite direction of the rotation of the pickup coil leads to the direction of Electrical energy ( Alternating current ).
Explanation:
For a Generator to generate electricity, Electrons that is found in the rotating coil of the Generator will experience some sort of force that makes them to start moving in a direction that is perpendicular to the direction of the rotating/pickup coil found in the Generator.
The conversion of mechanical energy (Rotation of the pickup coil ) to electrical energy takes place in the Generator
The generator makes an alternating current because electrons move in opposite direction of the rotating coil
In other to generate a Direct current using a generator we have to replace the slip rings with commutator.
So this helicopter pilot dropped me in the middle of an absolutely smooth frictionless
lake. He thought it was really funny. The only thing the pilot let me take is a bowling
ball. I tried to claw my way to the shore but was unable to get any horizontal force on
the super slippery zero friction ice. Explain how I can get to shore and why it will
work before I freeze to death. This test is about momentum. I wonder if my solution
has anything to do with that ...
The FitnessGram™ Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. [beep] A single lap should be completed each time you hear this sound. [ding] Remember to run in a straight line, and run as long as possible. The second time you fail to complete a lap before the sound, your test is over. The test will begin on the word start. On your mark, get ready, start.
A tension of 35.0N is applied downward on a 2.50kg mass. What is the acceleration of the mass?
A. -59.5 m/s2
B. -14.0 m/s2
C. -4.20 m/s2
D. -23.8 m/s2
Answer:
D. -23.8 m/s2
Explanation:
Answer:
[tex]\text{B. }-14.0\:\mathrm{m/s^2}[/tex]
Explanation:
From Newton's 2nd Law, we have [tex]\Sigma F=ma[/tex].
Substituting given values, we get:
[tex]-35.0=2.50a,\\a=\frac{-35.0}{2.50},\\a=\boxed{-14.0\:\mathrm{m/s^2}}[/tex]
*Note the negative sign simply implies that the mass is travelling downwards (direction).
Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father exerts a force on the merry-go-round perpendicular to its radius to achieve an angular acceleration of 4.44 rad/s^2.
Required:
a. How long (in s) does it take the father to give the merry-go-round an angular velocity of 1.53 rad/s? (Assume the merry-go-round is initially at rest.)
b. How many revolutions must he go through to generate this velocity?
c. If he exerts a slowing force of 270 N at a radius of 1.20 m, how long (in s) would it take him to stop them?
Answer:
Explanation:
Given that:
the initial angular velocity [tex]\omega_o = 0[/tex]
angular acceleration [tex]\alpha[/tex] = 4.44 rad/s²
Using the formula:
[tex]\omega = \omega_o+ \alpha t[/tex]
Making t the subject of the formula:
[tex]t= \dfrac{\omega- \omega_o}{ \alpha }[/tex]
where;
[tex]\omega = 1.53 \ rad/s^2[/tex]
∴
[tex]t= \dfrac{1.53-0}{4.44 }[/tex]
t = 0.345 s
b)
Using the formula:
[tex]\omega ^2 = \omega _o^2 + 2 \alpha \theta[/tex]
here;
[tex]\theta[/tex] = angular displacement
∴
[tex]\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }[/tex]
[tex]\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }[/tex]
[tex]\theta =0.264 \ rad[/tex]
Recall that:
2π rad = 1 revolution
Then;
0.264 rad = (x) revolution
[tex]x = \dfrac{0.264 \times 1}{2 \pi}[/tex]
x = 0.042 revolutions
c)
Here; force = 270 N
radius = 1.20 m
The torque = F * r
[tex]\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm[/tex]
However;
From the moment of inertia;
[tex]Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}[/tex]
given that;
I = 84.4 kg.m²
[tex]\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2[/tex]
For re-tardation; [tex]\alpha=-3.84 \ rad/s^2[/tex]
Using the equation
[tex]t= \dfrac{\omega- \omega_o}{ \alpha }[/tex]
[tex]t= \dfrac{0-1.53}{ -3.84 }[/tex]
[tex]t= \dfrac{1.53}{ 3.84 }[/tex]
t = 0.398s
The required time it takes= 0.398s
What is the definition of Heat Transfer
Answer:
Heat transfer is an engineering discipline that concerns the generation, use, conversion, and exchange of heat (thermal energy) between physical systems.
Explanation:
Answer:
Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy between physical systems. Heat transfer is classified into various mechanisms, such as thermal conduction, thermal convection, thermal radiation, and transfer of energy by phase changes.
Explanation:
hope that helps
The Nardo ring is a circular test track for cars. It has a circumference of 12.5 km. Cars travel around the track at a constant speed of 100 km/h. A car starts at the easternmost point of the ring and drives for 30 minutes at this speed.
Required:
a. What distance, in km, does the car travel?
b. What is the magnitude of the car's displacement, in km, from its initial position?
c. What is the speed of the car in m/s?
Answer: 50 km, 0, 27.78 m/s
Explanation:
Given
Circumference of the track is [tex]12.5\ km[/tex]
Speed of car is [tex]100\ km/h[/tex]
Car drives for [tex]30\ \text{minute}\ or\ 0.5\ hr[/tex]
(a)Distance traveled is
[tex]\Rightarrow D=100\times 0.5\\\Rightarrow D=50\ km[/tex]
(b)displacement of the car
It can be observed that 12.5 is a multiple of 50, that is, 50 km can be interpreted as 4 complete rounds of the track.
Therefore, the displacement of the car is zero.
(c)To convert kmph to m/s, multiply the entity by [tex]\frac{5}{18}[/tex]
[tex]\Rightarrow 100\times \dfrac{5}{18}\\\\\Rightarrow 27.78\ m/s[/tex]
If a car has a force of 900 N and has a mass of 650 kg. What would be the acceleration of the
car as it moves down the street?
The total resistance of a series circuit is 15.0 ohms what is the second resistance of the first resistance is 10.0 ohms?
A. less than 5.0 ohms
B. 5.0 ohms
C. 15 ohms
D. 25 ohms
A scientist wants to create a new material using nanotechnology. In which size range will she be working? 1 nm to 100 nm 1 to 10,000 nm 100 to 1000 nm 100 to 10,000 nm
Answer:
I believe it’s A
Explanation:
Answer:
A is correct (1 nm to 100 nm)
Explanation:
I just took the exam
What is the net force on a skydiver falling with a constant velocity of 0 m/s downward?
please select all that would be categorized as vectors
Answer:
1. 10 ft/s² straight up
2. 4 Km north-west
3. 20 m/s South
Explanation:
To successfully categorize the quantities given above as vectors, we must understand what vector quantity is.
A vector quantity is a quantity that has both magnitude and direction.
Considering the question given above, the vector quantities are:
1. 10 ft/s² straight up
2. 4 Km north-west
3. 20 m/s South
The remaining quantities in the question are not considered as vectors because they only have magnitude but no direction. They are therefore referred to as scalar quantities.
Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with the radius of RE). Use the gravitational acceleration on the surface of the Earth is 9.8m/s2.
Answer:
The magnitude of the free-fall acceleration at the orbit of the Moon is [tex]2.728\times 10^{-3}\,\frac{m}{s^{2}}[/tex] ([tex]\frac{2.784}{10000}\cdot g[/tex], where [tex]g = 9.8\,\frac{m}{s^{2}}[/tex]).
Explanation:
According to the Newton's Law of Gravitation, free fall acceleration ([tex]g[/tex]), in meters per square second, is directly proportional to the mass of the Earth ([tex]M[/tex]), in kilograms, and inversely proportional to the distance from the center of the Earth ([tex]r[/tex]), in meters:
[tex]g = \frac{G\cdot M}{r^{2}}[/tex] (1)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of the Earth, in kilograms.
[tex]r[/tex] - Distance from the center of the Earth, in meters.
If we know that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex] and [tex]r = 382.26\times 10^{6}\,m[/tex], then the free-fall acceleration at the orbit of the Moon is:
[tex]g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(382.26\times 10^{6}\,m)^{2}}[/tex]
[tex]g = 2.728\times 10^{-3}\,\frac{m}{s^{2}}[/tex]
The yearly whole-body dose caused by radioactive 40K absorbed in our tissues is 17 mrem. Note: 40K also emits gamma rays, many of which leave the body before being absorbed. Because fatty tissue has low potassium concentration and muscle has a higher concentration, gamma ray emissions indicate indirectly a person's fat content. Assuming that 40K undergoes beta decay with an RBE of 1.4 , determine the absorbed dose in rads. How much beta ray energy does an 75-kg person absorb in one year? Use that 1rad=10^−2J/kg .
Answer:
The beta ray Energy (E) absorbed = 0.009105 Joules
Explanation:
The absorbed dose rate in the tissue can be determined by using the known formula:
Dose equivalent = absorbed dose × relative biological effective (RBE)
[tex]absorbed \ dose = \dfrac{Dose \ equivalent}{RBE}[/tex]
[tex]absorbed \ dose = \dfrac{17 \times 10^{-3} \ rem}{1.4}[/tex]
absorbed dose = 12.14 × 10⁻³ rad
absorbed dose = 12.14 m rad
However, since the absorbed dose is determined, the energy absorbed in one year is can be expressed as:
Energy (E) = mass × absorbed rate
Energy (E) = 75 kg × 12.14 × 10⁻³ rad
Energy (E) = 75 kg × 12.14 × 10⁻³× 10⁻² J/kg
Energy (E) = 75 × 12.14 × 10⁻³× 10⁻² Joules
Energy (E) = 0.009105 Joules
When two objects are at the same temperature we say they are at
Answer:
equal temperatures
Explanation:
As a galaxy evolves and becomes more massive, what is most likely to happen?
Answer:
Explanation:
The most likely thing that would happen is that the galaxies continue becoming more and more massive, eventually becoming part of a galaxy cluster. These are massive but cluttered parts of the universe that hold many galaxies extremely close to one another. This also leads to galaxies colliding with one another, although when this happens they usually seem to pass right through each other as if they were ghosts. This is simply due to their sheer size and distance between their bodies of mass.
Answer:
It will merge with other galaxies
Explanation:
Pearson Connexus 2023
Does a ball rolling on an inclined plane have the same acceleration on the way up as it does on the way down?
Answer:
No, it does not.
Explanation:
According to the Law of Gravitation, something going down has more kinetic energy than something going up because it attracts pressure from around it when going down. When it goes up, it has less gravitational force and inertia also stops the ball from rolling upward. Therefore, without the amount of kinetic energy, it will not have the same amount of acceleration.
Magnets on the tracks
How can levitating trains support the claim that magnetic fields exist between objects exerting magnetic force on each other even when there is no physical contact?
Answer:
Explanation: They levitate sort of because when to opposite forces of magnets cant go together they repelin a simple machine the input work is 187.5 N and output work is 125N. If load distance is 25 cm and effort distance is 75 cm calculate value of load and effort
Answer:
1. Laod = 500 N
2. Effort = 250 N
Explanation:
From the question given above, the following data were obtained:
Input work = 187.5 J
Output work = 125 J
Load distance = 25 cm
Effort distance = 75 cm
1. Determination of the load.
Output work = 125 J
Load distance = 25 cm = 25/100 = 0.25 m
Load =?
Output work = Load × Load distance
125 = Load × 0.25
Divide both side by 0.25
Load = 125 / 0.25
Laod = 500 N
2. Determination of the effort.
Input work = 187.5 J
Effort distance = 75 cm = 75/100 = 0.75 m
Effort =?
Input work = Effort × Effort distance
187.5 = Effort × 0.75
Divide both side by 0.75
Effort = 187.5 / 0.75
Effort = 250 N
Which is one way to determine whether a reaction was a chemical reaction or a nuclear reaction?
Answer:
Check how much energy was released during the reaction or check for a change in total mass
which one hurry i need help pls dont get me wrong ..
Answer:
OPEN
Explanation:
An intentionally open circuit would be the circuit to the lights in the room that are turned off. There is no closed path available for the electricity to flow to the lights because the switch is in the "off" position which "opens" the path the electricity would normally flow through.
At what rate is thermal energy being generated in the 2R-resistor when ε = 12 V and R = 3.0 Ω?
Answer:
6 W
Explanation:
From the given information:
The resistance in Parallel for 2R is:
[tex]R_p = \dfrac{2R\times 2R}{2R+2R} \\ \\ R_p= R[/tex]
The equivalent resistance:
[tex]R_{eq} = R_p + R = 2R[/tex]
[tex]R_{eq} = 2(3)[/tex]
[tex]R_{eq} = 6 \ \ ohms[/tex]
The current through the circuit in R is:
[tex]= \dfrac{12}{R+R} \\ \\ = \dfrac{12}{2\times 3} \\ \\ = 2 A[/tex]
The current through the circuit in 2R is:
[tex]I_2R = (2A) \times \dfrac{2R}{2R+2R}[/tex]
[tex]I_2R = 2A \times \dfrac{1}{2} \\ \\ I_2R = 1A[/tex]
Finally, the thermal energy:
[tex]P_{2R} = (1)^2 (2R)[/tex]
[tex]P_{2R} = (1)^2 (2\times 3)[/tex]
[tex]P_{2R} = 6W[/tex]
You are designing an airplane propeller that is to turn at 2400 rpm (Fig. 9.13a). The forward airspeed of the plane is to be 75.0 m/s 75.0 m/s, and the speed of the propeller tips through the air must not exceed 270 m/s 270 m/s. (This is about 80% of the speed of sound in air. If the propeller tips moved faster, they would produce a lot of noise.) What is the maximum possible propeller radius?
Answer:
r = 1.07 m
Explanation:
The maximum radius of the propeller that can be allowed is given by the following formula:
[tex]v = r\omega\\\\r = \frac{v}{\omega}[/tex]
where,
r = maximum possible radius of the propeller = ?
v = maximum possible linear speed of the propeller = 270 m/s
ω = angular speed of the propeller = (2400 rpm)(2π rad/1 rev)(1 min/60 s)
ω = 251.33 rad/s
Therefore,
[tex]r = \frac{270\ m/s}{251.33\ rad/s}[/tex]
r = 1.07 m
Which of the following does not represent a significant negative environmental impact of using nuclear fuel? a. air pollution b. mining c. radiation d. waste production Please select the best answer from the choices provided A B C D
A significant negative environmental impact of using nuclear fuel is mining. The correct option is b.
What is mining?Mining is the process or industry of obtaining coal or other minerals from a mine.
Using nuclear fuels for mining will have no negative impact on the environment. All other terms represent negative impact of nuclear fuel Thus, The correct option is b.
Learn more about mining.
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