The work done by the gravitational field of the earth on the rock is 9.998 x 10⁸ J.
The given parameter include:
the mass of the object, m₁ = 16 kg
Note: the mass of the earth, m₂ = 5.972 x 10²⁴ kg
The work done by the gravitational field of the earth is given as;Work done = gravitational force (F) x radius of the earth (R)
[tex]Work \ done = \frac{Gm_1m_2}{R^2} \times R\\\\Work \ done = \frac{Gm_1m_2}{R} \\\\where;\\\\R \ is \ the \ radius \ of \ the \ earth = 6,378 \ km = 6,378,000 \ m\\\\G \ is \ the \ universal \ gravitation \ constant = 6.674 \times 10^{-11} Nm^2/kg^2\ \\\\Work \ done = \frac{(6.674 \times 10^{-11} ) \times (5.972\times 10^{24}) \times (16)}{6,378,000 } \\\\Work \ done = 9.998 \times 10^{8} \ J[/tex]
Therefore, the work done by the gravitational field of the earth on the rock is 9.998 x 10⁸ J.
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A vehicle is used to transport material down a straight aisle. The max acceleration of the vehicle is 1 m/s/s and the max speed of the vehicle is 5m/s. The vehicle starts at the beginning of the aisle. How long does it take to move down the aisle and come to a stop at the other end if:
a) the aisle is 100 meters long?
b) the aisle is 9 meters long?
Answer:
(a) 14.14 s
(b) 4.24 s
Explanation:
maximum acceleration, a = 1 m/s^2
maximum speed, v = 5 m/s
initial speed, u = 0 m/s
(a) distance, s = 100 m
Let the time is t.
Use second equation of motion
[tex]s = u t 0.5 at^2\\\\100 = 0 + 0.5 \times 1 \times t^2\\\\t = 14.14 s[/tex]
(b) distance, s = 9 m
Let the time is t'.
Use second equation of motion
[tex]s = u t + 0.5 at^2\\\\9= 0 + 0.5 \times 1 \times t'^2\\\\t' = 4.24 s[/tex]
your neighbour is throttling his recent bought motorbike to show off. the sound intensity measured at your window 16m away is 0.25W/m^2. what is the sound intensity level in dB at your friend's house, a distance of 28m away from the noisy bike?
Answer:
hi your pinterest I'd or Twitter I'd pls.
Two people, who have the same mass, throw two different objects at the same velocity. If the first object is heavier than the second, compare the velocities gained by the two people as a result of recoil.
a. The first person will gain more velocity as a result of recoll.
b. The second person will gain more velocity as a result of recoll.
c. Both people will gain the same velocity as a result of recoll.
d. The velocity of both people will be zero as a result of recoil
Answer:
The first person will gain more velocity as a result of recoil.
Explanation:
Let us recall that from Newton's third law of motion, action and reaction are equation and opposite. A consequence of this law is the proposition that ''momentum can neither be created nor destroyed.''
Hence, when two people who have the same mass, throw two different objects at the same velocity but the first object is heavier than the second, the first object possesses greater momentum than the second object hence the first person will gain more velocity as a result of recoil.
Explain why it takes much more effort to stop a freight train compared with a car?
Answer:
Train wheels and rails are both made of steel, and the steel-steel friction coefficient is around 0.25. As a result, the stopping time and distance will be three to four times that of a car.
In what kind of reaction is water (H20) broken down into hydrogen gas (H2) and oxygen gas (O2)?
A. Combination
B. Decomposition
C. Displacement
D. Combustion
Answer:
Answer is B (Decomposition)
Sorry I really see ur questions but I don't know the answer but next time I will try to answer sorry:(
A falcon is hovering above the ground, then suddenly pulls in its wings and begins to fall toward the ground. Air resistance is not negligible.
Identify the forces on the falcon.
a. Kinetic friction
b. Weight w
c. Static friction
d. Drag D
e. Normal force n
f. Thrust
g. Tension T
Answer:
Explanation:
When a falcon is hovering, the force of up thrust is balanced by the weight.
When it begins to fall towards the ground, the weight acts downwards, kinetic friction is upwards, drag is upwards, normal force is upwards, thrust is upwards.
3. Define 1 standard kilogram?
Answer:
standard kilogram is the SI unit of mass
Answer:
The total mass of platinum-irridum cylinder whose diameter is equal to its height and stored at 0°C in the bureau of weight and measure in France is called 1 standard kilogram
The best and most common way to measure the intensity of a cardiovascular exercise is to determine
O The person's heart rate
O The fatigue level of the person
O Amount of perspiration the person produces
The person's breathing rate
Answer:
the person's heart rate
question 1+1677-789909
Answer:
your answer is -788231
Explanation:
1+1677=1678
1678-789909=-788231
The upward normal force exerted by the floor is 710 N on an elevator passenger who weighs 720 N . You may want to review (Pages 107 - 110) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Weighing yourself in an elevator. Part A What is the reaction force to the upward normal force exerted by the floor
Answer:
If the person is to remain the floor the reaction force will be equal to the normal force exerted by the floor.
F(normal) - F(reaction) = 0
That means the person is not moving with respect to the elevator.
Expanding the applied forces we have:
Fw - Fn = 720 - 710 = 10 N where the positive direction is chosen as down
Fw is the weight of the person and Fn the force exerted on the person by the elevator,
The acceleration of the person the becomes F = m a = m * 10 N and will be downward agreeing with our choice of coordinate axes.
What is the maximum wavelength, in nm, of light that can eject an electron from a metal with Φ =4.50 x 10–19 J?
[tex]4.4×10^{-7}\:\text{m}[/tex]
Explanation:
The minimum energy needed to kick out an electron from a metal's surface is when the energy of the incident radiation is equal to the metal's work function [tex]\phi[/tex]:
[tex]E = h\nu - \phi = \dfrac{hc}{\lambda} - \phi = 0[/tex]
or
[tex]\dfrac{hc}{\lambda} = \phi[/tex]
Solving for the wavelength [tex]\lambda[/tex],
[tex]\lambda = \dfrac{hc}{\phi}[/tex]
[tex]\:\:\:\:\:=\dfrac{(6.62×10^{-34}\:\text{J-s})(3.0×10^8\:\text{m/s})}{4.5×10^{-19}\:\text{J}}[/tex]
[tex]\:\:\:\:\:= 4.4×10^{-7}\:\text{m}[/tex]
Note that as the radiation's wavelength increases, its energy decreases. So a radiation whose wavelength is longer than this maximum will lose its ability to kick out an electron from this metal.
The maximum wavelength, in nm, of light that can eject an electron from the metal, given the data is 441.73 nm.
To find the wavelength, the given values are,
Energy (E) = 4.50×10¯¹⁹ J
What is wavelength?The distance between two consecutive crests and troughs is called the wavelength of a wave.
Here, for the wavelength,
Energy (E) = 4.50×10¯¹⁹ J
Planck's constant (h) = 6.626×10¯³⁴ Js
Speed of light (v) = 3×10⁸ m/s
The wavelength of the light can be obtained as illustrated below:
E = hv / λ
Cross multiply λ,
E × λ = hv
Divide both sides by E,
λ = hv / E
Substituting all the values,
λ = (6.626×10¯³⁴ × 3×10⁸) / 4.50×10¯¹⁹
λ = 0.000000441733 m
λ = 441.73nm
λ - The maximum wavelength of light.
Thus, the wavelength of the light that can eject an electron from the metal is 441.73 nm
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(a) What is the efficiency of an out-of-condition professor who does 1.90 ✕ 105 J of useful work while metabolizing 500 kcal of food energy? % (b) How many food calories would a well-conditioned athlete metabolize in doing the same work with an efficiency of 25%? kcal
Answer:
a) The energy efficiency of the out-of-condition professor is 9.082 %.
b) The food calories needed by the well-conditioned athlete is 181.644 kilocalories.
Explanation:
a) The energy efficiency of the food metabolization ([tex]\eta[/tex]), no unit, is defined by following formula:
[tex]\eta = \frac{W}{E}\times 100\,\%[/tex] (1)
Where:
[tex]W[/tex] - Useful work, in joules.
[tex]E[/tex] - Food energy, in joules.
If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]E = 2.092\times 10^{6}\,J[/tex], the energy efficiency of the food metabolization is:
[tex]\eta = \frac{1.90\times 10^{5}\,J}{2.092\times 10^{6}\,J} \times 100\,\%[/tex]
[tex]\eta = 9.082\,\%[/tex]
The energy efficiency of the out-of-condition professor is 9.082 %.
b) If we know that [tex]W = 1.90\times 10^{5}\,J[/tex] and [tex]\eta = 25\,\%[/tex], then the quantity of food energy is:
[tex]E = \frac{W}{\eta}\times 100\,\%[/tex]
[tex]E = 1.90\times 10^{5}\,J\times \frac{100\,\%}{25\,\%}[/tex]
[tex]E = 7.60\times 10^{5}\,J[/tex]
[tex]E = 181.644\,kcal[/tex]
The food calories needed by the well-conditioned athlete is 181.644 kilocalories.
A CD is spinning on a CD player. In 12 radians, the cd has reached an angular speed of 17 r a d s by accelerating with a constant acceleration of 3 r a d s 2 . What was the initial angular speed of the CD
Answer:
The initial angular speed of the CD is equal to 14.73 rad/s.
Explanation:
Given that,
Angular displacement, [tex]\theta=12\ rad[/tex]
Final angular speed, [tex]\omega_f=17\ rad/s[/tex]
The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]
We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,
[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]
Put all the values,
[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]
So, the initial angular speed of the CD is equal to 14.73 rad/s.
A soap bubble was slowly enlarged from a radius of 4cm to 6cm. The amount of work necessary for enlargement was 1.5 x 10^-4 joules. Calculate the surface tension of the soap bubble.
Answer:
[tex]T=3*10^-3 N/m[/tex]
Explanation:
From the question we are told that:
Radius :
[tex]R_1=4=>0.04\\\\R_2=6=>0.06[/tex]
Work [tex]W=1.5 * 10^{-4}[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=TdA[/tex]
Where
[tex]dA=A_2-A_1\\\\dA=(2 \pi r_2^2)(2 \pi r_1^2)[/tex]
[tex]dA=8 \pi*(r_2^2-r_1^2)\\\\dA=8*3.142*(0.06^2-0.04^2)[/tex]
[tex]dA=0.050m^2[/tex]
Therefore
[tex]W=TdA[/tex]
[tex]T=\frac{1.5 * 10^{-4}}{0.05m^2}[/tex]
[tex]T=3*10^-3 N/m[/tex]
An inductive circuit contains resistance of 20 ohm and an inductance of 20 H. If an ac voltage of 120 V and frequency 60 Hz is applied to this circuit, the current would be
A 0.0159
A 0.017
A 0.02
A 0.16
Answer:
answer : option (b) 0.016 amp
explanation : resistance of resistor , R = 10 Ω
inductance of inductor , X_LX
L
= 20H
voltage of AC circuit , V = 120volts
frequency, ff =60Hz
so, angular frequency, \omega=2\pi fω=2πf = 2 × π × 60 = 120π rad/s
now, current , i=\frac{V}{\sqrt{R^2+\omega^2L^2}}i=
R
2
+ω
2
L
2
V
= 120/√{10² + (120π)² × 20²}
= 120/√{100 + 14400π² × 400}
after solving this we get, i = 0.016 amp
The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. A
Complete Question
The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20kg/m^2
Answer:
[tex]h=1614m[/tex]
Explanation:
From the question we are told that:
Initial Pressure [tex]P_1=980mbar=>98000Pa[/tex]
Final Pressure [tex]P_2=790mbar=>79000Pa[/tex]
Density [tex]\rho=1.20kg/m^2[/tex]
Generally the equation for Height climbed is mathematically given by
[tex]h=\frac{P_1-P_2}{\rho*g}[/tex]
[tex]h=\frac{P_1-P_2}{1.20*9.81}[/tex]
[tex]h=1614m[/tex]
How does an airpump work?
What process provides the sun with its energy
Answer:
nuclear fusion
The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom. During the fusion process, radiant energy is released.Answer:
nuclear fusion
Explanation:
The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom.
A supertrain with a proper length of 100 m travels at a speed of 0.950c as it passes through a tunnel having a proper length of 50.0 m. As seen by a trackside observer, is the train ever completely within the tunnel? If so, by how much do the train’s ends clear the ends of the tunnel?
Answer:
19m
Explanation:
we have proper length L = 100m
the speed of the train v = 0.95
the speed of light is given as = 3x10⁸
length of the tunnel is given as = 50 meters
we can solve for the lenght contraction as
LX√1-v²/c²
= 100 * √1-(0.95*3x10⁸)²/(3x10⁸ )
= 31.22 metres
the train would be well seen at
50 - 31.22
= 18.78
= this is approximately 19 metres
we conclude tht the trains ends clears the ends of the tunnel by 19 meters.
thank you!
A 77 turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.18 T field, starting with the normal of the plane of the coil perpendicular to the field. Assume that the positive max emf is reached first.
a. What is the peak emf?
b. At what time is the peak emf first reached?
c. At what time is the emf first at its most negative?
d. What is the period of the AC voltage output?
Answer:
a) fem = 5.709 V, b) t = 0.196 s, c) t = 0.589 s, d) T = 0.785 s
Explanation:
This is an exercise in Faraday's law
fem= - N [tex]\frac{d \Phi _B}{dt}[/tex]
fem = - N [tex]\frac{d \ (B A cos \theta)}{dt}[/tex]
The magnetic field and the area are constant
fem = - N B A [tex]\frac{d \ cos \ \theta}{dt}[/tex]
fem = - N B A (-sin θ) [tex]\frac{d \theta}{dt}[/tex]
fem = N B (π d² / 4) sin θ w
fem= [tex]\frac{\pi }{4}[/tex] N B d² w sin θ
with this expression we can correspond the questions
a) the peak of the electromotive force
this hen the sine of the angle is 1
sin θ = 1
fem = [tex]\frac{\pi }{4}[/tex] 77 1.18 0.10² 8.0
fem = 5.709 V
b) as the system has a constant angular velocity, we can use the angular kinematics relations
θ = w₀ t
t = θ/w₀
Recall that the angles are in radians, so the angle for the maximum of the sine is
θ= π/2
t = [tex]\frac{\pi }{2} \ \frac{1}{8}[/tex]
t = 0.196 s
c) for the electromotive force to be negative, the sine function of being
sin θ= -1
whereby
θ = 3π/ 2
t = [tex]\frac{3\pi }{2} \ \frac{1}{8}[/tex]
t = 0.589 s
d) This electromotive force has values that change sinusoidally with an angular velocity of
w = 8 rad / s
angular velocity and period are related
w = 2π / T
T = 2π / w
T = 2π / 8
T = 0.785 s
a fixed mass of gas occupies a volume of 1000 CM3 at 0 degree celsius if it is heated at constant pressure of 100 degree celsius calculate the new volume
Answer:
P V = N R T ideal gas equation
V1 = k * T1 if P is constant and also N and R will be constant
V2 = k * T2 where k is some constant
Or V2 = (T2 / T1) * V1 also known as "Charles Law" for expansion at
constant pressure
V2 = (373 / 273) * 1000 cm^3 = 1366 cm^3 where T is absolute temperature
What is the electric field strength between two parallel conducting plates separated by 10 cm and having a potential difference between them of 2000 V?
a.
2000 V/m
b.
200 V/m
c.
20 kV/m
d.
200000 V/m
Answer:
• Potential Difference (V) = 2000 V
• Distance b/w the two parallel plates (d) = 10 cm = 10/100 = 1/10 = 0.1 m
• Electric field (E) = ?
[tex]\implies V = E.d[/tex]
[tex]\implies E = \dfrac{V}{d} [/tex]
[tex]\implies E = \dfrac{2000}{0.1} [/tex]
[tex]\implies E = \dfrac{2000}{ {10}^{ - 1} } [/tex]
[tex]\implies E = 2000 \times {10}^{1} [/tex]
[tex]\implies\bf E = 20000 \:V/m[/tex]
[tex]\implies\bf E = 20\:kV/m[/tex]
Hence, option C) the correct answer.
what's impulse of
force
Answer:
The impulse experienced by the object equals the change in momentum of the object. In equation form, F.t = m. Δv. In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum.
Hannah wants to create a record keeping system to track the inventory needed to efficiently run her lawn and landscape business, such as spare parts, gas cans, string trimmers, etc. Her crew manager will also be using the system. Hannah is considering whether to use Excel or Access. Which one of the following is NOT a benefit of using Access?
a. More data storage
b. Multiuser capability
c. Easier setup
d. Additional reporting features
Answer:
c). Easier setup
Explanation:
As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.
Find the ratio of the Coulomb electric force Fe to the gravitational force Fo between two
electrons in vacuum.
Answer:
thus the coulomb force is F – 8.19x10-8N. this is also an attractive force, although it is traditionally shown as positive since gravitational force is always attractive. the ratio of the magnitude of the electrostatic force to gravitational force in this case is,thus,FFG – 2.27x1039 F F G – 2.27x 10 39.
gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be
Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:
[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]
where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,
[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]
d = 68.5 x 10⁻⁶ m = 68.5 μm
you decide to work part time at a local supermarket. The job pays eight dollars and 60 per hour and you work 20 hours per week. Your employer withhold 10% of your gross pay federal taxes, 7.65% for FICA taxes, and 5% for state taxes
I guess that we want to find how much money you get each week.
We know that the job pays $8.60 per hour.
We know that you work 20 hours per week.
Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:
20*$8.60 = $172.
Now we know that your employer witholds:
10% + 7.65% + 5% = 22.65%
Then your employer withholds 22.65% of your gross pay.
if the 100% of your gross pay is $172
Then the 22.65% will be:
(22.65%/100%)*$172 = 0.2265*$172 = $38.96
This means that your employer withholds $38.96 of your weekly gross pay.
Then each week you get:
$172 - $38.96 = $133.04
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The Sun is a type G2 star. Type G stars (from G0 to G9) have a range of temperatures from 5200 to 5900. What is the range of log(T) for G stars? Show your work
Answer:
log T = 3.72 to 3.77
Explanation:
Temperature range is
T = 5200 to 5900
Take the log
So,
log T = log 5200 to log 5900
log T = 3.72 to 3.77
A magnetic force acting on an electric charge in a uniform magnetic field what happend
Answer:
hgff
Explanation:
Answer:
The charge moves to equilibrium.
E.e = B.e.V
E is electric field force.
e is the charge.
B is magnetic field force.
V is acceleration voltage.
A frictionless spring with a 9-kg mass can be held stretched 1.8 meters beyond its natural length by a force of 80 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 1.5 m/sec, find the position of the mass after tt seconds. meters
Answer:
the required solution is; x(t) = 0.675sin( 2.222t )
Explanation:
Given the data in the question;
Using both Newton's and Hooke's law;
m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0, [tex]x[/tex](0) = 0, [tex]x^f[/tex](0) = 1.5
given that mass m = 9 kg
[tex]x[/tex] = 1.8 m
k is F / x
hence
k = F / x
given that, F = 80 N
we substitute
k = 80 / 1.8
k = 44.44
so
m[tex]x^{ff[/tex] + k[tex]x[/tex] = 0,
we input
9[tex]x^{ff[/tex] + 44.44[tex]x[/tex] = 0,
[tex]x^{ff[/tex] + 4.9377[tex]x[/tex] = 0
so auxiliary equation is,
r² + 4.9377 = 0
r² = -4.9377
r = √-4.9377
r = ±2.222i
hence, the solution will be;
x(t) = A×cos( 2.222t ) + B×sin( 2.222t )
⇒ [tex]x^t[/tex](t) = -2.222Asin( 2.222t ) + 2.222Bcos( 2.222t )
using initial conditions
x(0) = 0
⇒ 0 = A
[tex]x^t[/tex](t) = 1.5
1.5 = 2.222B
so
B = 1.5 / 2.222 = 0.675
Hence, the required solution is; x(t) = 0.675sin( 2.222t )
