Answer:
The object is not still as it observe from the outer space.
Explanation:
Our earth is not stationary.
It rotates about its own axis with the period of 24 hours and it revolves around the sun with the period of 1 year.
An object is said to be at rest if the position of the object with respect to time changes.
An object is said to be in motion when the position of the object changes with respect to time.
Both the terms rest and motion are relative to each other.
If an observer observe an object on the earth from the outer space, he observe that the object s not still it is also moving as the earth is moving.
So, the object is not still as it observe from the outer space.
A 2800-lbm car climbs a 1160 ft. long uphill road which is inclined at 15o (to the horizontal) and covers the distance in 12 sec. Determine the power required if the car covers the distance (a) at constant velocity, (b) from an initial velocity, at the bottom of the hill, of 10 mph to a final velocity of 50 mph at the top of the hill and (c) from an initial velocity of 45 mph to a final velocity of 15 mph. Neglect the effects of friction and wind resistance.
Answer:
a) P = 70054.3 W, b) P = 18820 W, c) P = 14116.7 W
Explanation:
Power is defined as work per unit of time
P = W / t = F x / t
P = F v
a) in this case the velocity is constant, let's use the equilibrium relation to find the force.
Let's set a reference system with the x axis parallel to the plane
F - Wₓ = 0
F = Wₓ
with trigonometry let's decompose the weight
sin θ = Wₓ / W
Wₓ = W sin θ
F = W sin 15
F = 2800 sin 15
F = 724.7 lb
we look for the speed, as it rises with constant speed we can use the relations of uniform motion
v = x / t
v = 1160/12
v = 96.67 ft / s
we calculate the power
P = 724.7 96.67
P = 70054.3 W
b) In this case, the speed of the vehicle changes during the ascent, so we use the relationship between work and the change in kinetic energy
W = ΔK
W = ½ m v_f² - ½ m v₀²
let's reduce to the SI system
v₀ = 10 mph (5280 ft / 1 mile) (1h / 3600 s = 14.67 ft / s
v_f = 50 mph (5280 ft / 1 mile) (1 h / 3600s) = 73.33 ft.s
mass : m = w / g
W = ½ 2800/32 (73.33² - 14.67²)
W = 225841 J
we calculate the average power
P = W / t
P = 225841/12
P = 18820 W
c) we repeat the previous procedure
v₀ = 45 mph = 66 ft / s
v_f = 15 mph = 22 ft / s
W = ½ 2800/32 (22² - 66²)
W = -169400 J
P = W / t
P = 169400/12
P = 14116.7 W
A 5.0 A electric current passes through an aluminum wire of 4.0~\times~10^{-6}~m^2 cross-sectional area. Aluminum has one free electron per atom. The density of aluminum is 2.7~grams/cm^3, and the aluminum molar mass is 27 g. What is the electron number density (the number of electrons per unit volume) in the wire
Answer: The electron number density (the number of electrons per unit volume) in the wire is [tex]6.0 \times 10^{28} m^{-3}[/tex].
Explanation:
Given: Current = 5.0 A
Area = [tex]4.0 \times 10^{-6} m^{2}[/tex]
Density = 2.7 [tex]g/cm^{3}[/tex], Molar mass = 27 g
The electron density is calculated as follows.
[tex]n = \frac{density}{mass per atom}\\= \frac{\rho}{\frac{M}{N_{A}}}\\[/tex]
where,
[tex]\rho[/tex] = density
M = molar mass
[tex]N_{A}[/tex] = Avogadro's number
Substitute the values into above formula as follows.
[tex]n = \frac{\rho \times N_{A}}{M}\\= \frac{2.7 g/cm^{3} \times 6.02 \times 10^{23}/mol}{27 g/mol}\\= \frac{16.254 \times 10^{23}}{27} cm^{3}\\= 0.602 \times 10^{23} \times \frac{10^{6} cm^{3}}{1 m^{3}}\\= 6.0 \times 10^{28} m^{-3}[/tex]
Thus, we can conclude that the electron number density (the number of electrons per unit volume) in the wire is [tex]6.0 \times 10^{28} m^{-3}[/tex].
In which direction of the wave motion do longitude waves transfer energy ?
Answer:
Hello There!!
Explanation:
The answer is O parallel.
hope this helps,have a great day!!
~Pinky~
40ml of Liquid A are poured into a beaker, and 40.0ml of Liquid B are poured into an identical beaker. Stirrers in each beaker are connected to motors, and the forces FA and FB needed to stir each liquid at a constant rate are measured.
a. FA will be greater than F B
b. FA will be less than FB
c. FA will be equal to FB
d. It's impossible to predict whether FA or FB will be greater without more information.
Answer:
d
Explanation:
there is not enough information about the liquid to know the force required for each. ie. stirring a cup of water is different than stirring a cup of pudding.
PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
Answer:
Energy = 1.5032*10^(-10) Joules
Explanation:
By Einstein's relativistic energy equation, we know that the energy of a given particle is given by:
Energy = rest energy + kinetic energy.
= m*c^2 + (γ - 1)*mc^2
Where γ depends on the velocity of the particle.
But if the proton is at rest, then the kinetic energy is zero, and γ = 1
Then the energy is just given by:
Energy = m*c^2
Where we know that:
mass of a proton = 1.67*10^(-27) kg
speed of light = c = 2.9979*10^(8) m/s
Replacing these in the energy equation, we get:
Energy = ( 1.6726*10^(-27) kg)*( 2.9979*10^(8) m/s)^2
Energy = 1.5032*10^(-10) kg*m^2/s^2
Energy = 1.5032*10^(-10) J
A point source of light illuminates an aperture 1.70 m away. A 11.0 cm -wide bright patch of light appears on a screen 0.800 m behind the aperture. Part A How wide is the aperture
Answer:
7.48 cm
Explanation:
The diagrammatic representation of information is shown in the image below:
From the image, Δ X₁X₂ and Δ Y₁Y₂ are equivalent triangles.
where;
X₁X₂ = width of the aperture
Using equivalent triangles, It implies that:
[tex]\implies \dfrac{X_1X_2}{Y_1Y_2} = \dfrac{1.70}{1.70+0.800} \\ \\ \implies X_1X_2=\dfrac{1.70}{1.70+0.800}\times 11.0[/tex]
[tex]X_1X_2= \dfrac{1.70}{2.50}\times 11.0 cm[/tex]
X₁X₂ = 7.48 cm
What is the magnitude of the current in the R= 6 Ω resistor?
kirchhoff
Answer:
Here's an explanation but not the answer
Explanation:
When a resistor is traversed in the same direction as the current, the ... Traversing the internal resistance r1 from c to d gives −I2r1. ... I1 = I2 + I3 = (6−2I1) + (22.5− 3I1) = 28.5 − 5I1.
motto that will help in reaching a goal in life that related the projectile motion
Explanation:
hmm! very good questions I hope u understand
How does the thickness of a planets atmosphere affect a planets average temperature
Answer:
The atmosphere of planets will block or enable heat coming from the sun
John and Caroline go out for a walk one day. This graph represents the distance they traveled over time.
During what part of thelr walk are they not moving?
A) From 0 to 0.75 hours
B)From 1.25 to 2.0 hours
From 0.75 to 1.25 hours
C) from 1.25 to 2.0 hours
D) they are moving the entire time
Answer:
From 0.75 to 1.25 hours
Explanation:
Given
See attachment for graph
Required
Point where they didn't move
This means that we identify the point where the distance didn't change.
Given that the distance is plotted on the y-axis, we simply check for the end points of any horizontal line on the graph
The horizontal line on the graph represents 30km and the time interval is: 0.75 to 1.25 hours.
Hence, (c) is correct
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each a distance 0.0429 m from the origin. Where should a third particle with charge 9.03 µC be placed so that the magnitude of the electric field at the origin is zero?
Answer:
The third particle should be at 0.0743 m from the origin on the negative x-axis.
Explanation:
Let's assume that the third charge is on the negative x-axis. So we have:
[tex]E_{1}+E_{3}-E_{2}=0[/tex]
We know that the electric field is:
[tex]E=k\frac{q}{r^{2}}[/tex]
Where:
k is the Coulomb constant q is the charger is the distance from the charge to the pointSo, we have:
[tex]k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0[/tex]
Let's solve it for r(3).
[tex]\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0[/tex]
[tex]r_{3}=0.0743\: [/tex]
Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.
I hope it helps you!
A 4.38 kg sphere makes a perfectly inelastic collision with a second sphere that is initially at rest. The composite system moves with a speed equal to one third the original speed of the 4.38 kg sphere. What is the mass of the second sphere?
Answer: The mass of the second sphere is 8.76 kg
Explanation:
The equation for a perfectly inelastic collision follows:
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
where,
[tex]m_1\text{ and }u_1[/tex] are the mass and initial velocity of first sphere
[tex]m_2\text{ and }u_2[/tex] are the mass and initial velocity of second sphere
v = final velocity of the system
We are given:
[tex]m_1=4.38kg\\u_2=0m/s\\v=\frac{u_1}{3}[/tex]
Rearranging the above equation, we get:
[tex]m_1u_1-m_1v=m_2v\\\\m_2=m_1\frac{u_1-v}{v}\\\\m_2=m_1(\frac{u_1}{v}-1)[/tex]
Plugging values in the above equation, we get:
[tex]m_2=4.38(\frac{3u_1}{u_1}-1)\\\\m_2=(4.38\times 2)=8.76kg[/tex]
Hence, the mass of the second sphere is 8.76 kg
How much heat is absorbed by 60g of copper when it is heated from 20°C to 80°C
Answer:
I HOPE THIS IS CORRECT
Explanation:
It is heated from 20°C to 80°C. We need to find the heat absorbed. It can be given by the formula as follows : So, 1386 J of heat is absorbed.
A 50 Ω resistor in a circuit has a current flowing through it of 2.0 A. What is the power dissipated by the resistor?
A. 50 W
B. 200 W
C. 100 W
D. 25 W
Explanation:
A nichrome (ρ = 100x10-8 Ωm) wire has a diameter of 0.40 mm. ... A power line with a resistance of 2 ohms has a current of 80 A in it
1/8 Watt 50 kΩ resistor has 70 V applied.
A 50 Ω resistor in a circuit has a current flowing through it of 2.0 A then the power dissipated by the resistor is 200 Watt. Therefore the correct option is B.
What is power?The rate of doing work is known as power. The Si unit of power is the watt.
Power =work/time
The mathematical expression for the power is as follows
P = VI
where P is the power
V is the voltage
I is current passing through the circuit
The expression of power can be represented in many different forms by using the ohms law
As we know from Ohm's law that V= IR
Power = VI
by substituting V=IR in place of the voltage in power formula
P = I²R
As given in the problem a 50 Ω resistor in a circuit has a current flowing through it of 2.0 A so by substituting the values of the current and the resistance in the formula of power
P = 2²×50
= 200 Watt
Therefore the correct answer is B.
Learn more about power from here
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Two kg of water is contained in a piston-cylinder assembly, initially at 10 bar and 200°C. The water is slowly heated at constant pressure to a final state. If the heat transfer for the process is 1740 kJ, determine the temperature at the final state, in °C, and the work, in kJ. Kinetic and potential energy effects are negligible.
Answer:
Explanation:
Given that:
mass of water = 2kg
Pressure P₁ = 10 bar
temperature T₁ = 200⁰ C
Obtaining the following value from superheated tables at P₁ = 10 bar and T₁ = 200⁰ C
v₁ = 0.2060 m³/kg
h₁ = 2827.9 kJ/kg
The heat transfer Q₁₋₂ = 1740 kJ
Using the energy balance equation:
Q₁₋₂ = m(h₂ - h₁)
1740 = 2(h₂ - 2827.9)
1740 = 2h₂ - 5655.8
1740 + 5655.8 = 2h₂
7395.8 = 2h₂
h₂ = 7395.8/2
h₂ = 3697.9 kJ/kg
Since P₁ = P₂ = 10 bar
Using the superheated table at P₂ = 10 bar and h₂ = 3697.9 kJ/kg;
v₂ = 0.4011 m³/kg
The temperature at final state = 600 °C
The work done [tex]W_{1-2} = \int P dv[/tex]
[tex]W_{1-2} = P\times m (V_2-V_1)[/tex]
= (10×10² kPa)×2kg (0.4011 - 0.2060) m³/kg
= (10×10² kPa)×2kg × 0.1951 m³/kg
= 390.2 kJ
A car moves round a circular track of radius 0.3m of two revolution per/sec find its angular velocity.
Answer:
the angular velocity of the car is 12.568 rad/s.
Explanation:
Given;
radius of the circular track, r = 0.3 m
number of revolutions per second made by the car, ω = 2 rev/s
The angular velocity of the car in radian per second is calculated as;
From the given data, we convert the angular velocity in revolution per second to radian per second.
[tex]\omega = 2 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 4\pi \ rad/s = 12.568 \ rad/s[/tex]
Therefore, the angular velocity of the car is 12.568 rad/s.
Appliances A, B, and C consume 250, 480 and 1450 watts of power, respectively. The system voltage is 120V, and the circuit breaker is rated at 15 amps. Which combinations of the three appliances can be on at the same time, and which combinations will trip the circuit breaker
Answer:
Appliance A and B can work together without tripping
Explanation:
We will calculate the amount of current consumed by each appliances.
Appliance A
P = VI
I = P/V
I = 250/120 = 2.08 A
Appliance B
I = 480 /120 = 4 A
Appliance C
I = 1450/120
I = 12.08 A
Hence, appliance C will trip the circuit as it consumes a lot of electricity.
Question 24 of 33 Which of the following is an example of uniform circular motion? A. A car speeding up as it goes around a curve O B. A car slowing down as it goes around a curve 2 C. A car maintaining constant speed as it goes around a curve D. A car traveling along a straight road
Answer:
Uniform Circular Motion is the Movement or Rotation of an Object along a circular Path at constant speed.
OPTION C IS YOUR ANSWER!.
WILL GIVE 50 POINTS TO CORRECT ANSWER‼️
A capacitor has 4.33 x 10-7 C of
charge on it when 3.45 V is applied.
How much energy is stored in the
capacitor?
Answer:
7.47×10¯⁷ J
Explanation:
From the question given above, the following data were obtained:
Charge (Q) = 4.33×10¯⁷ C
Potential difference (V) = 3.45 V
Energy (E) =?
The energy stored in a capacitor is given by the following equation:
E = ½QV
Where
E => is the energy.
Q => is the charge.
V => is the potential difference.
With the above formula, we can obtain the energy stored in the capacitor as follow:
Charge (Q) = 4.33×10¯⁷ C
Potential difference (V) = 3.45 V
Energy (E) =?
E = ½QV
E = ½ × 4.33×10¯⁷ × 3.45
E = 7.47×10¯⁷ J
Thud, the energy stored in the capacitor is 7.47×10¯⁷ J
Answer:
The answer would be 7.47x10^-7J
Explanation:
Put in 7.47 then -7
A basketball player shoots toward a basket 4.9 m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor at an angle of 60 o above the horizontal, what must the initial speed be if it were to go through the basket
Answer:
v₀ = 6.64 m / s
Explanation:
This is a projectile throwing exercise
x = v₀ₓ t
y = y₀ + v_{oy} t - ½ g t²
In this case they indicate that y₀ = 1.8 m and the point of the basket is x=4.9m y = 3.0 m
the time to reach the basket is
t = x / v₀ₓ
we substitute
y- y₀ = [tex]\frac{ v_o \ x \ sin \theta }{ v_o \ cos \theta} - \frac{1}{2} g \ \frac{x^2 }{v_o^2 \ cos^2 \theta }[/tex]
y - y₀ = x tan θ - [tex]\frac{ g \ x^2 }{ 2 \ cos^2 \theta } \ \frac{1}{v_o^2 }[/tex]
we substitute the values
3 -1.8 = 3.0 tan 60 - [tex]\frac{ 9.8 \ 3^2 }{2 \ cos^2 60 } \ \frac{1}{v_o^2}[/tex]
1.2 = 5.196 - 176.4 1 / v₀²
176.4 1 / v₀² = 3.996
v₀ = [tex]\sqrt{ \frac{ 176.4}{3.996} }[/tex]
v₀ = 6.64 m / s
The area around a magnet containing all of magnetic Lines of force is called
Answer:
Magnetic Field
Explanation:
The area around a magnet containing all of magnetic Lines of force is called the magnetic field.
2. Why is it necessary for rotation to occur within a
generator?
Answer:
ON
Explanation:
Locate your computer's Power button.
Press and hold that button until your computer shuts down.
Wait until you hear the computer's fans shut off, and your screen goes completely black.
Wait a few seconds before pressing and holding the power button to initiate your computer's normal startup.
State the relation between acceleration and momentum
Answer:
Acceleration is the rate of change in velocity. Momentum is the mass times the velocity. So if you multiply the mass times the acceleration, you get the of change of momentum.
In the series circuit, if the potential difference across the battery is 20 V and the potential difference across R1 is 12 V, what is the potential difference across R2
Answer:
The correct answer is "8 V".
Explanation:
Given:
Potential difference across battery,
= 20 V
Potential difference across R1,
= 12 V
Now,
On applying the Kirchorff loop, we get
⇒ [tex]E-I_1R_1-I_2R_2=0[/tex]
⇒ [tex]E-V_1-V_2=0[/tex]
⇒ [tex]V_2=E-V_1[/tex]
On putting values, we get
⇒ [tex]=20-12[/tex]
⇒ [tex]= 8 \ V[/tex]
The potential difference across the resistance R2 will be "8 Volts.
What is Kirchoff;s law?According to the kirchoff's law in a loop of a circuit when there are number of the resistances so the sum of all the potential differences will be zero.
It is given that:
Potential difference across battery,= 20 V
Potential difference across R1,= 12 V
Now,
On applying the Kirchorff loop, we get
⇒ [tex]\rm E-I_1R_1-I_2R_2=0[/tex]
⇒ [tex]\rm E-V_1-V_2=0[/tex]
⇒ [tex]\rm V_2=E-V_1[/tex]
On putting values, we get
⇒ [tex]V_2=20-12=8\ volt[/tex]
Hence the potential difference across the resistance R2 will be "8 Volts.
To know more about Kirchoff's law follow
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It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft is dropping vertically at a speed of 293 m/s. The spacecraft is to make a soft landing -- that is, at the instant it reaches the surface of Mars, its velocity is zero. Assume the spacecraft undergoes constant acceleration from the elevation of 34.7 km until it reaches the surface of Mars. What is the magnitude of the acceleration
Answer: [tex]1.23\ m/s^2[/tex]
Explanation:
Given
At an elevation of [tex]y=34.7\ km[/tex], spacecraft is dropping vertically at a speed of [tex]u=293\ m/s[/tex]
Final velocity of the spacecraft is [tex]v=0[/tex]
using equation of motion i.e. [tex]v^2-u^2=2as[/tex]
Insert the values
[tex]\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2[/tex]
Therefore, magnitude of acceleration is [tex]1.23\ m/s^2[/tex].
A 1350 kg automobile has a kinetic energy 3.24 x 106 J. What is the velocity of the car?
Answer:
219 m/s
Explanation:
Kinetic energy is calculated using the formula:
KE = 1/2mv²Given:
m = 1350 kg KE = 3.24 · 10⁶ JWe can solve for the unknown variable v.
Substitute these known values into the formula.
3.24 · 10⁶ = 1/2(1350)v² 3.24 · 10⁶ = 675v² (3.24 · 10⁶) / 675 = v² v² = 48,000v = 219.09 m/sThe velocity of the car is around 219 m/s.
a stone is thrown vertically upward with an initial velocity of 14m/sc negleting in air resistance find (a)Maximum height reached and (b)Time taken before it reached the ground
Answer:
1/2 m v^2 = m g H conservation of energy
H = (v^2 / (2 * g))^1/2 = (14^2 / 19.6)^1/2 = 10 m
H = 1/2 g t^2 time to fall 10 m
t = (2 * H / g)^1/2 = (20 / 9.8)^1/2 = 1.43 sec
Total time in air = 2 * 1.43 = 2.86
You can also use 2 a S = V0^2 since final velocity = 1.43 * 9.8 m/s = 14 m / s
and 2 a S = 2 * 9.8 * 10 = 196 m^2 / s^2 which is V0 squared
if a seismic wave has a period of 0.0202s, find the frequency of the wave.
Answer:
49.5 Hz.
Explanation:
From the question given above, the following data were obtained:
Period (T) = 0.0202 s
Frequency (f) =?
The frequency and period of a wave are related according to the following equation:
Frequency (f) = 1 / period (T)
f = 1/T
With the above formula, we can obtain the frequency of the wave as follow:
Period (T) = 0.0202 s
Frequency (f) =?
f = 1/T
f = 1/0.0202
f = 49.5 Hz
Therefore the frequency of the wave is 49.5 Hz.
The work function of titanium metal is 6.93 x 10-19 J. Calculate the kinetic energy of the ejected electrons if light of frequency 2.50 X 1015 s-1 is used to irradiate the metal.
Answer:
g m t o k liye cbbhhhf to be be free and ear is not a short time to be a short of a week and I am a short kat key and ear to be free and ear is not a short time and ear buds and duster for a short of a week of the action is not by the action is not a short time to make sure to be free and duster and not by the way to
Explanation:
ahhhhhhh to be be free and ear is nothing to be be free and ear is not a short time to be be free and ear is not by the class is not a a short time to be be free and
A horizontal force of 40N is needed to pull a 60kg box across the horizontal floor at which coefficient of friction between floor and box? Determine it to three significant figures even through that's quite unrealistic. How much work is done in overcoming friction between the object and floor if the box slides 8m along horizontally on the floor?
Answer:
Coefficient of friction is [tex]0.068[/tex].
Work done is [tex]320~J[/tex].
Explanation:
Given:
Mass of the box ([tex]m[/tex]): [tex]60[/tex] kg
Force needed ([tex]F[/tex]): [tex]40[/tex] N
The formula to calculate the coefficient of friction between the floor and the box is given by
[tex]F=\mu mg...................(1)[/tex]
Here, [tex]\mu[/tex] is the coefficient of friction and [tex]g[/tex] is the acceleration due to gravity.
Substitute [tex]40[/tex] N for [tex]F[/tex], [tex]60[/tex] kg for [tex]m[/tex] and [tex]9.80[/tex] m/s² for [tex]g[/tex] into equation (1) and solve to calculate the value of the coefficient of friction.
[tex]40 N=\mu\times60 kg\times9.80 m/s^{2} \\~~~~~\mu=\frac{40 N}{60 kg\times9.80 m/s^{2}}\\~~~~~~~=0.068[/tex]
The formula to calculate the work done in overcoming the friction is given by
[tex]W=Fd..........................(2)[/tex]
Here, [tex]W[/tex] is the work done and [tex]d[/tex] is the distance travelled.
Substitute [tex]40[/tex] N for [tex]F[/tex] and [tex]8 m[/tex] for [tex]d[/tex] into equation (2) to calculate the work done.
[tex]W=40~N\times8~m\\~~~~= 320~J[/tex]