Answer:
Bottom line: When charged particles from the sun strike atoms in Earth's atmosphere, they cause electrons in the atoms to move to a higher-energy state. When the electrons drop back to a lower energy state, they release a photon: light. This process creates the beautiful aurora, or northern lights.Explanation:
^-^I hope it's help u100 POINTS AND BRAINLIEST!!! How does the egg sucked into a glass bottle trick work?
Answer: The egg will get sucked into the bottle. To get the egg out of the bottle, turn the bottle upside down and blow into it, so that the egg acts as a one-way valve. The increased air pressure in the bottle will cause the egg to pop back out.
Explanation:
Quickly place the egg over the mouth of the bottle. The egg will get sucked into the bottle. To get the egg out of the bottle, turn the bottle upside down and blow into it, so that the egg acts as a one-way valve. The increased air pressure in the bottle will cause the egg to pop back out.
If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force between them?
Answer:
5.619×10⁶ N
Explanation:
Applying,
F = kqq'/r²................... Equation 1
Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges
From the questiion,
Given: q = 2.5 C, q' = 2.5 C, r = 100 m
Constant: 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.5×2.5×8.99×10⁹)/100²
F = 56.19×10⁵
F = 5.619×10⁶ N
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 electric field has a positive test charge of 5.00 C placed in it. The force on the test charge is
6.000 N. The magnitude of the electric field at the location of the test charge is
o 30.0 NVC
0 1.20 N/C
0 120, NVC
O 3.00 N/C
01.02 N/C
Answer:
yes
Explanation:
this means the answer is yes
different between pressure and force
Force is mass into acceleration
and pressure is force applied per unit area.
Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s. Calculate his:
a. acceleration in his direction of motion
b. acceleration opposite to his direction of motion.
Answer: [tex]56.4\ m/s^2, 201.42\ m/s^2[/tex]
Explanation:
Given
Rocket attain a velocity of [tex]v=282\ m/s[/tex] in a time period of [tex]t=5\ s[/tex]
It was brought jarringly back to rest in only [tex]t'=1.4\ s[/tex]
Acceleration is the change in velocity of the object over a period of time
(a) Acceleration in his direction of motion
[tex]\Rightarrow a=\dfrac{v-0}{t}\\\\\Rightarrow a=\dfrac{282}{5}\\\\\Rightarrow a=56.4\ m/s^2[/tex]
(b) acceleration opposite to his direction of motion i.e. deceleration is
[tex]\Rightarrow a_d=\dfrac{0-v}{t'}\\\\\Rightarrow a_d=\dfrac{-282}{1.4}\\\\\Rightarrow a_d=-201.42 \ m/s^2\\\Rightarrow a_d=201.42\ \text{decelration}[/tex]
One of the earliest vertebrate animal groups that evolved in the early Paleozoic Era
are
Given this relationship, if you and your twin sibling (assuming you have the same mass) were to be separated by three times your original distance, what is the new gravitational force between you?
Answer:
The new force becomes (1/9)th of the original force.
Explanation:
The gravitational force between two masses is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
Where
r is the distance between masses,
If the new distance is, r' = 3r
The new force is given by :
[tex]F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(3r)^2}\\\\F'=\dfrac{1}{9}\times G\dfrac{m_1m_2}{r^2}\\\\F'=\dfrac{F}{9}[/tex]
So, the new force becomes (1/9)th of the original force.
You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/sm/s. Part A What is the speed of the putty just before it strikes the ceiling
Answer: [tex]4.65\ m/s[/tex]
Explanation:
Given
Distance putty has to travel is 3.5 m
The initial speed of putty is 9.50 m/s
Using equation of motion to determine the velocity of putty just before it hits ceiling
[tex]v^2-u^2=2as[/tex]
[tex]\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s[/tex]
So, the velocity of putty just before hitting is [tex]4.65\ m/s[/tex]
Suppose an astronomer observes a binary star system where the stars are separated by 2.0 AU , and they have an orbital period of 7.0 years . Using Newton's version of Kepler's Third Law, find the combined mass of the stars.
Answer:
4.408 [tex]\mathsf{M_{sun}}[/tex]
Explanation:
According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:
[tex](m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}[/tex]
combined mass (m₁+m₂) = 4.408 [tex]\mathsf{M_{sun}}[/tex]
A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the battery supplies energy at a rate of W, how large is the resistance
Answer:
Resistance is as large as 2.8 ohm
Explanation:
Complete question
A 3.0 A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the battery supplies energy at a rate of 25 W, how large is the resistance?
Solution -
The relation between Power and current is as follows
P = I^2*R
R = P/I^2
Were P = Power
R = resistance and
I = current
Given-
P = 25 W
I = 3.0 A
Substituting the given values, in above equation, we get -
R = 25/3.0^2
R = 2.8 ohm
How much work can a motor with a power output of 0.70 hp do in 2 s?
Answer:
the work done by the motor is 1,044 J.
Explanation:
Given;
the output power of the motor, P = 0.7 hp
duration of the work, t = 2 s
The relationship between horse-power and watt is given as;
1 hp = 745.7 W
0.7 hp = ?
0.7 hp = 522 W = 522 J/s
The work done by the motor is calculated as;
W = Power x time
W = 522 J/s x 2 s
W = 1,044 J
Therefore, the work done by the motor is 1,044 J.
In an experiment, a student brings up the rotational speed of a piece of laboratory apparatus to 24 rpm. She then allows the apparatus to slow down uniformly on its own, and counts 236 revolutions before the apparatus comes to a stop. The moment of inertia of the apparatus is known to be 0.076 kg m2. What is the magnitude of the torque on the apparatus
Answer:
T = 6.43 x 10⁻⁵ N.m
Explanation:
First, we will calculate the deceleration of the apparatus by using the third equation of motion:
[tex]2\alpha \theta = \omega_f^2-\omega_i^2[/tex]
where,
α = angular decelration = ?
θ = angular displacement = (236 rev)(2π rad/rev) = 1482.83 rad
ωi = initial angular speed = (24 rpm)(2π rad/1 rev)(1 min/ 60 s) = 2.51 rad/s
ωf = final angular speed = 0 rad/s
Therefore,
[tex]2\alpha(1482.83\ rad) = (0\ rad/s)^2-(2.51\ rad/s)^2\\\\\alpha = -\frac{(2.51\ rad/s)^2}{2965.66\ rad} \\\\\alpha = - 8.46\ x\ 10^{-4}\ rad/s^2[/tex]
negative sign shows deceleration
Now, for torque:
T = Iα
where,
T = Torque = ?
I = moment of inertia = 0.076 kg.m²
Therefore,
T = (0.076 kg.m²)(8.46 x 10⁻⁴ N.m)
T = 6.43 x 10⁻⁵ N.m
True or false, wrrect the false
statement:
• The magnetic field created by a flat coil is
uniform.
• Inside a solenoid, the lines of field are
oriented from the north face to the south
face.
• The magnetic field outside Helmholtz
coils is uniform.
• Le champ B à l'intérieur d'un solénoïde
est uniforme.
• The magnitude of B, created by a flat coil
of radius R, at any point in its plane is
B= 2m x 10-NI
R
• The designation of the faces of a wil
depend the sense of the current
traversing it.
Answer:
false
Explanation:
How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W
Complete question:
How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W. Assume the average energy per photon in 78.0 keV.
Answer:
The number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec
Explanation:
Given;
power of the flux produced, P = 1 W = 1 J/s
energy per photon, E = 78 keV
Convert the energy per photon to J
E = 78 x 10³ x 1.6 x 10⁻¹⁹ = 1.248 x 10⁻¹⁴ J / photon
let the number of photons = n
n(1.248 x 10⁻¹⁴ J / photon) = 1 J/s
[tex]n = \frac{1 \ J/s}{1.248 \times 10^{-14}\ J/photon } = 8.01 \times 10^{13} \ photons/s[/tex]
Therefore, the number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec
A train with mass 3.3 x 107 kg starts from rest and accelerates to a speed of 42
m/s. What is the initial kinetic energy of the train?
Answer:
kinetic energy of the train = 2,910.6 x 10⁷ joule
Explanation:
Given:
Mass of train = 3.3 x 10⁷ kg
Speed of train = 42 m/s
Find:
kinetic energy of the train
Computation:
kinetic energy = (1/2)(m)(v²)
kinetic energy of the train = (1/2)(3.3 x 10⁷)(42²)
kinetic energy of the train = (1/2)(3.3 x 10⁷)(1,764)
kinetic energy of the train = (3.3 x 10⁷)(882)
kinetic energy of the train = 2,910.6 x 10⁷ joule
Answer: The initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].
Explanation:
Given: Mass = [tex]3.3 \times 10^{7} kg[/tex]
Speed = 42 m/s
Kinetic energy is the energy acquired by an object due to its motion.
Formula to calculate kinetic energy is as follows.
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
where,
m = mass of object
v = speed of object
Substitute the values into above formula as follows.
[tex]K.E = \frac{1}{2}mv^{2}\\= \frac{1}{2} \times 3.3 \times 10^{7} kg \times (42 m/s)^{2}\\= 2910.6 \times 10^{7} kg m^{2}/s^{2} (1 J = 1 kg m^{2}/s^{2})\\= 2910.6 \times 10^{7} J[/tex]
Thus, we can conclude that the initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].
Walking at a brisk pace, you cover 10 m in 5.0 s . How many seconds will you need to cover 50 m ?
A cart weighing 40 pounds is placed on a ramp incline 15 degrees to the horizon. The cart is held in place by a rope inclined 60 degrees to the horizontal. find the force that the rope must exert on the cart to keep it from rolling down the ramp.
Answer: [tex]14.64\ N[/tex]
Explanation:
Given
Inclination of ramp is [tex]\theta=15^{\circ}[/tex]
Rope is inclined [tex]\phi=60^{\circ}[/tex] to the horizontal
Weight of cart [tex]W=40\ lb[/tex]
from the diagram, rope is at angle of [tex]45^{\circ}[/tex] w.r.t ramp
Sine component of weight pulls down the cart Cosine component of force applied through rope held it at the position
[tex]\Rightarrow 40\sin 15^{\circ}=F\cos 45^{\circ}\\\\\Rightarrow F=40\cdot \dfrac{\sin 15^{\circ}}{\cos 45^{\circ}}\\\\\Rightarrow F=40\times 0.366\\\Rightarrow F=14.64\ N[/tex]
A deer with a mass of 156 kg is running head on toward you with a speed of 10 m/s. Find the momentum of the deer
Hi there!
[tex]\large\boxed{1560 kgm/s}[/tex]
Recall that:
P = m · v, where:
P = momentum
m = mass (kg)
v = velocity (m/s)
Thus:
P = 156 · 10
P = 1560 kgm/s
Find the value of T1 if 1 = 30°, 2 = 60°, and the weight of the object is 139.3 newtons.
A.
69.58 newtons
B.
45.05 newtons
C.
25 newtons
D.
98.26 newtons
Answer:
Option A (69.56 newtons) is the appropriate solution.
Explanation:
According to the question,
On the X-axis,
⇒ [tex]T_1Cos30^{\circ}-T_2Cos60^{\circ}=0[/tex]
or,
[tex]T_1Cos 30^{\circ}=T_2Cos60^{\circ}[/tex]
On substituting the values, we get
[tex]T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}[/tex]
[tex]T_1\times \sqrt{3} =T_2[/tex]....(equation 1)
On the Y-axis,
⇒ [tex]T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N[/tex]
[tex]\frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N[/tex]
[tex]T_1+\sqrt{3}T_2=139.2\times 2[/tex]
From equation 1, we get
[tex]T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N[/tex]
[tex]T_1+3T_1=278.4 \ N[/tex]
[tex]4T_1=278.4 \ N[/tex]
[tex]T_1=\frac{278.4}{4}[/tex]
[tex]=69.6 \ N[/tex]
Answer:
69.58
Explanation:
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:
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
Q5: An ice skater moving at 12 m/s coasts
to a halt in 95m on an ice surface. What is the coefficient
of (kinetic) friction between ice and skates?
u = 0.077
Explanation:
Work done by friction is
Wf = ∆KE + ∆PE
-umgx = ∆KE,. ∆PE =0 (level ice surface)
-umgx = KEf - KEi = -(1/2)mv^2
Solving for u,
u = v^2/2gx
= (12 m/s)^2/2(9.8 m/s^2)(95 m)
= 0.077
Kinetic friction is the ratio of the friction force to the normal force experienced by a body in moving state.The coefficient of kinetic friction between the ice and skates is 0.077.
Given-
velocity of the ice skater is 12 m/ sec.
Work done by the friction is the sum of the change of the kinetic energy and the change in potential energy.
[tex]W_{f}=\bigtriangleup KE +\bigtriangleup PE[/tex]
The value for the potential energy will be equal to Zero in this case. Therefore the work done by the friction is,
[tex]W_{f}=\bigtriangleup KE +0[/tex]
Kinetic energy is directly proportional to the mass of the object and to the square of its velocity and work done can be given as,
[tex]W_{f} =u_{f} mgx[/tex]
Here, [tex]u_{f}[/tex] is friction force, [tex]m[/tex] is mass, [tex]g[/tex] is gravity and x is the distance .
Equate the value of kinetic energy and work done of friction for further result, we get,
[tex]u_{f} mgx=\dfrac{1}{2} \times mv^2[/tex]
[tex]u_{f} =\dfrac{1}{2gx} \times v^2[/tex]
[tex]u_{f} =\dfrac{1}{9.8\times 95} \times 12^2[/tex]
[tex]u_{f} =0.077[/tex]
Hence, the coefficient of kinetic friction between the ice and skates is 0.077.
For more about the friction, follow the link below-
https://brainly.com/question/13357196
A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A small tack with charge Q is lowered on a silk thread through the hole into the interior of the shell.
Required:
a. What is the charge on the inner surface of the shell?
b. What is the charge on the outer surface of the shell?
Answer:
(a) Negative Q
(b) Positive Q
Explanation:
Charge is the inherent property of matter due to the transference of electrons.
There are three methods of charging a body.
(i) Charging by friction: When two uncharged bodies rubbed together, then one body gets positive charged and the other is negatively charges it is due to the transference of electrons form one body to another.
(ii) Conduction: when a charged body comes in contact with the another uncharged body, the uncharged body gets the same charge and the charge is distributed equally.
(iii) Induction: When a uncharged body keep near the charged body, the uncharged body gets the same amount of charge but opposite in sign.
(a) When a small tack of charge Q is lowered into the hole, then due to the process of induction, the charge on the inner surface of the shell is - Q.
(b) Due to the process of conduction, the charge on the outer surface of the shell is Q.
The charge on the inner surface of the shell is negative whereas the charge on the outer surface of the shell is positive.
Reasons for change of charge on a body
Due to the process of induction the inner surface of the shell creates negative charge because when a uncharged body bring near to the charged body, the uncharged body gets the same amount of charge but opposite in sign.
While on the other hand, there is no charge interaction with the outer surface so it remains positively charge so we can conclude that the charge on the inner surface of the shell is negative whereas the charge on the outer surface of the shell is positive.
Learn more about charge here: https://brainly.com/question/18102056
which statement regarding the idealized model of motion called free fall is true?
a. the effect of air resistance is factored in the equation of motion in the idealized model called free fall.
b. free fall only models motion for objects that do not have an initial velocity in the upward direction.
c. the idealized model of the motion called free fall applies in cases where distance of the fall is large compared with the radius of the astronomical body on which the fall occurs.
d. a freely falling object has a constant acceleration due to gravity.
What is the speed acquired by a freely falling object 4 seconds after being dropped from a rest position? Use units of meter per second (m/s) and assume acceleration from gravity is 10 m/s2.
speed = 40 m/s
Explanation:
Since the object is dropped, V0y = 0.
Vy = V0y - gt
= -(10 m/s^2)(4 s)
= -40 m/s
This means that its velocity is 40 m/s downwards. Its speed is simply 40 m/s.
The speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
What are the three equations of motion?There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
Keep in mind that these calculations only apply to uniform acceleration.
As given in the problem, we have to find the speed acquired by a freely falling object 4 seconds after being dropped from a rest position,
By using the first equation of motion,
v = u + at
initial velocity(u) = 0 m/s
acceleration(a) = 10 m/s²
v = 0 + 10×4
v = 40 meters/seconds
Thus, the speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
Learn more about equations of motion from here,
brainly.com/question/5955789
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A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 6600 m. You can ignore any effects of air resistance.
Required:
a. What was the rocket's acceleration during the first 16s?
b. What is the rocket's speed as it passes through acloud 5100 m above the ground?
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f = [tex]\frac{x-x_1}{t}[/tex]
we substitute the values
v_f = [tex]\frac{ 6600 -x_1}{4}[/tex]
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f = [tex]\frac{6600 - x_1}{4}[/tex]
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a = [tex]\frac{6600 -128 a}{4}[/tex]
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s
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
HELP ME PLEASE !!!!!!!!!!!!
Answer:
D
Explanation:
Because the y axis is meter. If it is straight line at time and meter graph then it velocity and speed is 0
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.
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An athlete training for an event does 6.53 104 J of work during a workout and gives off 5.97 105 J of heat. Consider the athlete to be like a heat engine. (a) Determine the magnitude of the change in internal energy of the athlete. J (b) What is the efficiency of the athlete
Answer:
(a) The magnitude of the change in internal energy is 6.623 x 10⁵ J
(b) the efficiency of the athlete is 10.94 %
Explanation:
Given;
work done by the athlete (system), W = 6.53 x 10⁴ J
the heat given off by the athlete (system), Q = 5.97 x 10⁵ J
The simple diagram below will be used to illustrate the direction of the energy flow assuming a heat engine.
Q← ⊕ →W
The work, W, points away from the system since the system does the work
The heat, Q, points away from the system since heat is given off
Apply first law of thermodynamic;
ΔU = Q + W
where;
q is the heat flowing into or out of the system
(+q if the heat is flowing into the system
(-q if the heat is leaving the system
w is the work done by or on the system
(+w if the work is done on the system by the surrounding
(-w if the work is done by the system to the surrounding
Thus, from the above explanation, the change in internal energy of the system is calculated as;
ΔU = -Q - W
ΔU = - 5.97 x 10⁵ J - 6.53 x 10⁴ J
ΔU = -6.623 x 10⁵ J
The magnitude of the change in internal energy = 6.623 x 10⁵ J
(b) the efficiency of the athlete;
[tex]Efficiency = \frac{W}{Q} \times 100\%\\\\Efficiency = \frac{6.53 \times 10^4}{5.97 \times 10^5} \times 100\%\\\\Efficiency = 10.94 \ \%[/tex]