Answer:
stopping potential is the negative potential applied to the circuit to stop the moving electrons so as to stop the flow of current
for high current high negative potential is applied
A model airplane with mass 1.0 kg is held by a wire so that it flies in a horizontal circle with radius 20.0 m. The airplane engine provides a net thrust of 1.0 N perpendicular to the wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in this horizontal flight.
Answer:
330
Explanation:
(a) The torque the net thrust produces about the center of the circle is of 20 N-m.
(b) The angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².
Given data:
The mass of model airplane is, m = 1.0 kg.
The radius of horizontal circle is, r = 20.0 m.
The magnitude of net thrust by engine is, F = 1.0 N.
(a)
The effort made to turn any object is known as the torque. The mathematical expression for the torque is given as,
T = F × r
Solving as,
T = 1.0 × 20.0
T = 20 N-m
Thus, we can conclude that the torque the net thrust produces about the center of the circle is of 20 N-m.
(b)
The expression for the angular acceleration of airplane during the horizontal flight is given as,
[tex]T = I \times \alpha[/tex]
Here, I is the moment of inertia of airplane and its value is,
[tex]I = \dfrac{1}{2}mr^{2}\\\\\\I = \dfrac{1}{2} \times 1.0 \times 20^{2}\\\\\\I =200 \;\rm kg.m^{2}[/tex]
So, the angular acceleration is,
20 = 200 × α
α = 20/200
α = 0.1 rad/s²
Thus, we can conclude that the angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².
Learn more about the torque here:
https://brainly.com/question/19247046
Can someone please help?
Answer:
Gold = 19.3g/cm³
Explanation:
Given the following data;
Mass = 38.6g
Volume = 2cm³
To find the density;
Density can be defined as mass all over the volume of an object.
Simply stated, density is mass per unit volume of an object.
Mathematically, density is given by the equation;
[tex]Density = \frac{mass}{volume}[/tex]
Substituting into the equation, we have;
[tex]Density = \frac{38.6}{2}[/tex]
Density = 19.3g/cm³
Therefore, from the table the material is Gold.
Gold is a chemical element and it is the element 79 on the periodic table. Thus, it has an atomic number of 79. The chemical symbol for Gold is Au, it is chemically classified as a transition metal and as a solid at room temperature.
Generally, the chemical element Gold is known to have the following physical properties;
I. Malleable.
II. Ductile.
III. A good conductor of electricity and heat.
Also, it is a non-toxic chemical element with a beautiful lustrous sheen.
Helping me please What can smart modern biosensors be used in heath applications?
Answer: Biosensors are modern electronic devices which are made up of biological recognition system and a transducer, for processing of signals and to quantify a particular analyte.
Explanation:
In modern medicine, a Biosensors are used to replace some clinical laboratory investigations of biological fluids with the advantages of being easy to use, rapid and robust as well as offering multianalyte testing capability. They are classified based on their biological recognition elements which includes:
--> enzymatic biosensors
--> immuno biosensors
--> DNA and whole cell biosensors.
They can also be classified according to their signal transduction methods. These includes:
--> electrochemical Biosensors,
--> optical biosensors,
--> thermal and mass-based biosensors;
Some of the HEALTH APPLICATIONS of a modern biosensor includes:
--> Monitoring of glucose levels in diabetic patients which makes use of enzyme based sensors to monitor concentration of glucose in patients.
--> Detection of general metabolic status of bacteria, fungi, yeast, animal or plant cells.
--> Monitoring of cholesterol levels to prevent cardiovascular diseases. Cardiac troponin and. C - Reactive Proteins are biomarkers that are easily detected using biosensors.
--> They are used for Cancer clinical testings
calculate the aceleration of a vehicle wich start with a zero meter per second, and acelerates to 34 m/s in 21 s.
HELP PLSSS
Answer:
For the aceleration we have:
Vf = Vo + a * t
Clearing "a":
a = (Vf - Vo) / t
Replacing and resolving:
a = (34 m/s - 0 m/s) / 21 s
a = 34 m/s / 21 s
a = 1,61 m/s^2
The aceleration of the vehicle is 1,61 meters per second squared
Which water would you use to make salt dissolve the slowest?
Boiling water
Cold water
Hot water
Room temperature water
Answer:
boiling water because salt dissolves quicker in hot Water and the hottest is boiling
A 0.0600-kilogram ball traveling at 60.0 meters per second hits a concrete wall. What speed must a 0.0100-kilogram bullet have in order to hit the wall with the same magnitude of momentum as the ball?
Answer:
the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s
Explanation:
The computation of the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is as follows:
The ball momentum is
[tex]\Delta Q = mv\\\\\Delta Q = 6 \times 1^-2 \times 60\\\\\Delta Q = 3.6 kg \times m/s[/tex]
As there is a similar momentum
So,
[tex]\Delta Q = MV\\\\3.6 = 10^-2V\\\\V = 360 m/s[/tex]
Hence, the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s
Find the resultant of the following displacement:
A = 20 Km 30° south of east;
B = 50 Km due west;
C = 40 Km north east;
D = 30 Km 60° south of west.
Answer:
Explanation:
Basically you just have to find the left vectors. To do so divide A, C and D into horizontal and vertical vector. A: 10km to south and 10root3 to east. Just sine and cosine of 30 at 20km. D: 15 km to west and 15root3 to south. Again sine and cosine of 60 at 30 km. C: 45 degrees so 20root2 to north and east. Add all these up with B. Then you have 7.696 km due south and 19.395 km due west. Resultant displacement magnitude = root(7.696^2+19.395^2)=20.866 south of west with angle=arctan(7.696/19.395)=21.644 degrees
An ambulance is traveling east at 62.4 m/s. Behind it a car travels along the same direction at 34.5 m/s. The ambulance driver hears his siren with a wavelength of 0.47 m. What wavelength would a stationary observer behind the ambulance measure for the sound? The velocity of sound in air is 343 m/s.
Answer:
The answer is "0.5555 m"
Explanation:
Where the reference leaves the list and the viewer is at rest:
[tex]\lambda'=\frac{v-v_s}{v} \times \lambda\\\\[/tex]
[tex]=\frac{343 \frac{m}{s} - (-62.4 \frac{m}{s})}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{343 \frac{m}{s} + 62.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m[/tex]
[tex]=0.5555 \ m[/tex]
During the pitching motion, a baseball pitcher exerted an average horizontal force of 90 N against the 0.1 kg baseball while moving it through a horizontal displacement of 2.0 m before release. (1) what was the amount of work performed by the pitcher on the baseball (2) If the velocity at the start of the pitching motion was zero, at release the ball was traveling horizontally at which velocity
Answer:
(1) 180 J
(2) 60 m/s
Explanation:
(1) From the question,
Amount of work performed by the pitcher on the baseball = Force × distance.
W = F×d............... Equation 1
Given: F = 90 N, and d = 2.0 m.
Substitute into equation 1
W = 90×2
W = 180 Joules.
(2)
F = ma........... Equation 2
Where a = acceleration of the ball.
a = F/m
Given: F = 90 N, m = 0.1 kg.
therefore,
a = 90/0.1
a = 900 m/s².
Using,
v² = u² + 2as ............ Equation 3
Where u = 0 m/s, a = 900 m/s², s = 2 m.
substitute into equation 3
v² = 0² + 2×900×2
v² = 3600
v = √3600
v = 60 m/s
A red laser with a wavelength of 670 nm and a blue laser with a wavelength of 450 nm emit laser beams with the same light power. How do their rates of photon emission compare
E=hf C=wavelength*F
E=hC/wavelength
E=(6.626*10^-34)*(3.00*10^8)/670*10^-9
E=(6.626*10^-34)*(3.00*10^8)/450*10^-9
A brick is moving at a speed of 3 m/s and a pebble is moving at a speed of 5 m/s. If both objects have the same kinetic energy, what is the ratio of the brick's mass to the rock's mass
Answer:
let M be the mass of brick
let m be the mass of the pebble
.5M(3)^2 =KE
.5m(5)^2= KE
.5M9 = .5m25
9M = 25m
(9/25)M = m
Answer:
A
Explanation:
e=1/2M[tex]V^{2}[/tex]
1/2[tex]M_{1}[/tex]9=1/2[tex]M_{2}[/tex]25
[tex]\frac{M_{1} }{M_{2} }[/tex] =[tex]\frac{25}{9}[/tex]
Find the quantity of heat needed
to melt 100g of ice at -10 °C
into water at 10 °C. (39900 J)
(Note: Specific heat of ice is
2100 Jkg 'K', specific heat
of water is 4200 Jkg K',
Latent heat of fusion of ice is
336000 Jkg ').
Answer:
Approximately [tex]3.99\times 10^{4}\; \rm J[/tex] (assuming that the melting point of ice is [tex]0\; \rm ^\circ C[/tex].)
Explanation:
Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.
[tex]\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}[/tex]
The energy required comes in three parts:
Energy required to raise the temperature of that [tex]0.100\; \rm kg[/tex] of ice from [tex](-10\; \rm ^\circ C)[/tex] to [tex]0\; \rm ^\circ C[/tex] (the melting point of ice.)Energy required to turn [tex]0.100\; \rm kg[/tex] of ice into water while temperature stayed constant.Energy required to raise the temperature of that newly-formed [tex]0.100\; \rm kg[/tex] of water from [tex]0\; \rm ^\circ C[/tex] to [tex]10\;\ rm ^\circ C[/tex].The following equation gives the amount of energy [tex]Q[/tex] required to raise the temperature of a sample of mass [tex]m[/tex] and specific heat capacity [tex]c[/tex] by [tex]\Delta T[/tex]:
[tex]Q = c \cdot m \cdot \Delta T[/tex],
where
[tex]c[/tex] is the specific heat capacity of the material,[tex]m[/tex] is the mass of the sample, and[tex]\Delta T[/tex] is the change in the temperature of this sample.For the first part of energy input, [tex]c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:
[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].
Calculate the energy required to achieve that temperature change:
[tex]\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}[/tex].
Similarly, for the third part of energy input, [tex]c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:
[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].
Calculate the energy required to achieve that temperature change:
[tex]\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}[/tex].
The second part of energy input requires a different equation. The energy [tex]Q[/tex] required to melt a sample of mass [tex]m[/tex] and latent heat of fusion [tex]L_\text{f}[/tex] is:
[tex]Q = m \cdot L_\text{f}[/tex].
Apply this equation to find the size of the second part of energy input:
[tex]\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}[/tex].
Find the sum of these three parts of energy:
[tex]\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}[/tex].
Which is not true about waves?
O Compression and rarefaction are found in longitudinal waves.
O Sound waves are transverse waves.
O Amplitude, wavelength, and troughs all describe parts of a wave.
O Electromagnetic waves are transverse waves.
Answer:
sound waves are transverse waves
A typical donation of whole blood is equal to a pint (473 ml). If the average density of blood rhoBLOOD = 1060 kg/m^3, what is the mass of donated blood?
a. 5.01 kg
b. 0.501 lb
c. 0.501 kg
d. 0.501 g
e. 5.01 g
Answer:
c. 0.501 kg
Explanation:
Given;
volume of the denoted blood, V = 473 ml = 0.000473 m³
density of the denoted blood, ρ = 1060 kg/m³
The mass of donated blood is given as;
mass of the donated blood = density of the denoted blood x volume of the denoted blood
m = 1060 x 0.000473
m = 0.501 kg
Therefore, the mass of donated blood is 0.501 kg
how to calculate the speed using time and distance
Answer:
speed = distance/time
Explanation:
distance -> s
speed -> v
time -> t
The calculation of speed using time and distance shows that speed is equal to distance/time.
What is speed?Speed refers to the rate of movement of a vehicle or person.
Thus, the speed can be computed by dividing the total distance by the time consumed.
Learn more about speed, distance, and time at https://brainly.com/question/4931057
The arrow strikes a deer in the woods with the speed of 55 m/sec at an angle of 315 degrees. Calculate the Horizontal and vertical components of the arrow’s velocity.
Answer:
100 m
Explanation:
A block has mass of 20 gm and volume of 4 m3 , calculate the density *
Answer:
Density = 5 g/m³
Explanation:
Given:
Mass of block = 20 gm
Volume of block = 4 m³
Find:
Density
Computation:
Density = Mass / Volume
Density = 20 / 4
Density = 5 g/m³
5 kg/m3
Explanation:
The formula to find density is (d) = mass (m) ÷ volume (v)
a girl weighing 350N runs up hill upto 5m in 20sec. calculate her work done and power
Work done against gravity becomes potential energy of the object-Earth system. Its given by:
W = Ep
W = mgh but weigh(w) = mg
W = wh
W = 350 N × 5 m
W = 1750 JPower is the amount of energy transferred or converted per unit time. This is:
P = W/t
P = 1750 J / 20 s
P = 87.5 WDetermine the weight of a 5.1 kg scooter that moves with a constant acceleration of 3.0 ms2. (Make sure you use the weight equation: W or Fg = mg)
Answer:
15.3 sorry if i got it wrong
Explanation:
what is the acceleration of a softball if it has a mass of 0.50 kg and hits the catcher’s glove with a force of 25 newtons
Answer:
a = 50 [m/s²]
Explanation:
This type of problem can be solved using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = force =25 [N] (units of Newtons)
m = mass = 0.5 [kg]
a = acceleration [m/s²]
[tex]a=F/m\\a=25/0.5\\a= 50 [m/s^{2}][/tex]
A
B
с
4 teaspoons sugar
2 cups water
6 teaspoons sugar
2 cups water
4 teaspoons sugar
3 cups water
Which of these lemonade glasses is the most concentrated for sugar?
Ос
Oь
Oa
If your parasympathetic nervous system was activated what might you be doing?
A. Relaxing under a tree
B. Running a marathon
C. Skiing
D. Driving home in traffic
Which of the following types of energy are present at some point in the energy transfer process in a nuclear power plan? Select all that apply.
Heat energy
Geothermal energy
Tidal energy
Hydroelectric energy
Nuclear energy
Electrical energy
Solar Energy
Solar Energy is the answer to the question tell me if i`m wrong
A 0.7 kg block attached to a spring with force constant 160 Nm is free to move on a frictionless, horizontal surface. The block is released from rest when the
spring is stretched + 0.15 m as shown in figure below. At the instant the block is released the force vector on the block is.
(Consider Right direction is positive direction and Left direction is negative direction)
- 34.29 N
+ 34.29 N
+ 24 N
Оe
h Oa-241
Answer:
+ 24 N
Explanation:
the computation is shown below:
Given that
Mass of the block = m = 0.7 kg
Sprint constant = k = 160 N / m
x = 0.15m
Now the force on the block is
F = kx
= (160) (0.15)
= 24 N
As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction
Therefore the third option is correct
6. a. find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm. B. the electron makes a transition from the n
Complete Question
(A) Find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm .
(B) The electron makes a transition from the n=1 to n= 4 level by absorbing a photon. Calculate the wavelength of this photon.
Answer:
A
[tex]\Delta E = 337 \ eV[/tex]
B
[tex]\lambda = 3.439 *10^{-9} \ m[/tex]
Explanation:
Considering question a
From the question we are told that
The width of the box is [tex]w = 0.125 \ nm = 0.125 *10^{-9} \ m[/tex]
Generally the energy level of a particle confined to a box is mathematically represented as
[tex]E_n = \frac{n^2 h^2}{8 m L^2 }[/tex]
Generally the excitation energy is mathematically represented as
[tex]\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ][/tex]
From the question [tex]n_2 = 3\ (Third \ excited \ level ) \ \ and \ \ n_1 = 1[/tex]
Here h is the Planck's constant with a value of [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
m is the mass of electron with value [tex]m = 9.11 *10^{-31} \ kg[/tex]
So
[tex]\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [3^2 - 1^2 ][/tex]
=> [tex]\Delta E = 539 *10^{-19} \ J[/tex]
=> [tex]\Delta E = \frac{539 *10^{-19}}{1.60 *10^{-19}} \ J[/tex]
=> [tex]\Delta E = 337 \ eV[/tex]
Considering question b
Generally the energy level of a particle confined to a box is mathematically represented as
[tex]E_n = \frac{n^2 h^2}{8 m L^2 }[/tex]
Generally the excitation energy is mathematically represented as
[tex]\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ][/tex]
From the question [tex]n_2 = 4 \ \ and \ \ n_1 = 1[/tex]
Here h is the Planck's constant with a value of [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
m is the mass of electron with value [tex]m = 9.11 *10^{-31} \ kg[/tex]
So
[tex]\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [4^2 - 1^2 ][/tex]
=> [tex]\Delta E = 578 *10^{-19} \ J[/tex]
=> [tex]\Delta E = \frac{ 578 *10^{-19}}{1.60 *10^{-19 }}[/tex]
=> [tex]\Delta E = 361.45 \ eV[/tex]
Gnerally the wavelength is mathematically represented as
[tex]\lambda = \frac{hc}{\Delta E }[/tex]
=> [tex]\lambda = \frac{ 6.626 *10^{-34} * (3.0 *10^{8})}{578 *10^{-19} }[/tex]
=> [tex]\lambda = 3.439 *10^{-9} \ m[/tex]
The total yearly world consumption of energy is approximately 4.0 × 1020 J. How much mass would have to be completely converted into energy to provide this amount of energy?
Answer:
m = 4.4 × 10³ kg
Explanation:
Given that:
The total yearly energy is 4.0 × 10²⁰ J
The amount of mass that provides this energy can be determined by using the formula:
E = mc²
where;
c = speed of light in free space = (3 × 10⁸)
4.0 × 10²⁰ = m × (3 × 10⁸)²
[tex]m = \dfrac{4.0 \times 10^{20} }{(3\times 10^8)^2}[/tex]
m = 4.4 × 10³ kg
from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa at point B. Friction between the water and the pipe walls is negligible. What is the rate of discharge at point B
Answer:
1.8 m/s
Explanation:
Here is the complete question
The diameters of a water pipe gradually changes from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa and point B. Friction between the water and the pipe walls is negligible.
What is the rate of discharge at point B?
Solution
Using Bernoulli's equation,
P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂² where P₁ = pressure at point A = 700 kPa, h₁ = height at point A, v₁ = speed at point A and P₂ = pressure at point B = 664 kPa, h₂ = height at point B, v₂ = speed at point B, ρ = density of water = 1000 kg/m³
P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²
P₁ - P₂ - ρg(h₂ - h₁) = 1/2ρ(v₂² - v₁²)
(h₂ - h₁) = 5 m
Substituting the values of the variables, we have
700000 Pa - 664000 Pa - (1000 kg/m³ × 9.8 m/s² × 5 m) = 1/2 × 1000 kg/m³(v₂² - v₁²)
36000 Pa - 49000 Pa = 500 kg/m³(v₂² - v₁²)
- 13000 Pa = 500 kg/m³(v₂² - v₁²)
(v₂² - v₁²) = - 13000 Pa/500 kg/m³
(v₂² - v₁²) = -26 m²/s²
By mass flow rate, A₁v₁ = A₂v₂ where A₁ = cross-sectional area at point A and A₂ = cross-sectional area at point B
πd₁²v₁/4 = d₂²v₂/4 where d₁ = diameter at point A = 5 cm = 0.05 m and d₂ = diameter at point B = 15 cm = 0.15 m
d₁²v₁ = d₂²v₂
v₁ = v₂(d₂/d₁)²
= v₂(0.15/0.05)²
= v₂(3)²
= 9v₂
So, substituting v₁ = 9v₂ into the above equation, we have
(v₂² - v₁²) = -26 m²/s²
v₂² - 9v₂² = -26 m²/s²
- 8v₂² = -26 m²/s²
v₂² = -26 m²/s² ÷ (-8)
v₂² = 3.25 m²/s²
taking square root of both sides,
v₂ = √(3.25 m²/s²)
= 1.8 m/s
So, the rate of discharge at point B is 1.8 m/s
A boat is moving due east in a region where the earth's magnetic field is 5.0×10⁻⁵ NA⁻¹ m⁻¹ due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms⁻¹ , the magnitude of the induced emf in the wire of aerial is:
Answer: Induced emf is given by:
ε= Bvl
On putting the values we get
=5×10
−5
×1.50×2
=0.15mV
Explanation:
Hope these helped have a wonderful Christmas break ❄️
a car accelerates from rest at 2 m/s. what is the speed after 8 sec?
Answer:
16m/s
Explanation:
[tex]v_{f}=v_{i}+at[/tex]
[tex]v_{f}=0+2\cdot8[/tex]
[tex]v_{f}=16\ \frac{m}{s}[/tex]
Therefore, the speed after 8 seconds is 16m/s
If you are holding a stack of books at your waist and then you walk across the room your hands did?
Answer: 3) work
Explanation:
Work is doing something that requires you to expend energy. In walking across the room while your hands were carrying the stack of books, you expended energy through them which means that they did work.
We can neither tell the efficiency nor the power taken to do the work above from the given information. And by moving and holding books, there was definitely work done which means that all options apart from 3 are false.