Answer:
Normal force exerted on the rover would be greater at a point on the surface of the planet where the weight of the rover is experienced to be greater.
Explanation:
Since weight is a vector quantity, it can vary with position. Weight is the amount of force the planet exerts on the rover centered towards the planet.
Such a force is the result of gravitational pull and is quantified as:
[tex]F=G\times \frac{M.m}{R^2}[/tex]
and [tex]M=\rho\times \frac{4\pi.r^3}{3}[/tex]
where:
R = distance between the center of mass of the two bodies (here planet & rover)
G = universal gravitational constant
M = mass of the planet
m = mass of the rover
This gravitational pull varies from place to place on the planet because the planet may not be perfectly spherical so the distance R varies from place to place and also the density of the planet may not be uniform hence there is variation in weight.
Weight is basically a force that a mass on the surface of the planet experiences.
According to Newton's third law the there is an equal and opposite reaction force on the body (here rover) which is the normal force.
Notice that all the initial spring potential energy was transformed into gravitational potential energy. If you compressed the spring to a distance of 0.200 mm , how far up the slope will an identical ice cube travel before reversing directions
Answer:
The correct answer will bs "2.41 m".
Explanation:
According to the question,
M = 50 g
or,
= 0.050 kg
[tex]\Theta = 25^{\circ}[/tex]
k = 25.9 N/m
Δx = 0.200 m
Let the traveled distance be "x".
By using trigonometry, the height will be:
⇒ [tex]h = l Sin \Theta[/tex]
hence,
⇒ [tex]Potential \ energy \ at \ the \ top=Spring \ potential \ energy[/tex]
[tex]Mgh=\frac{1}{2} k(\Delta x)^2[/tex]
By putting the values, we get
[tex]0.050\times 9.8\times lSin 25^{\circ}=\frac{1}{2}\times 25.0\times (0.200)^2[/tex]
[tex]l=2.41 \ m[/tex]
Physics is killing me. Any help?
Answer:
The simplified expression is 3.833 x 10⁷ g
Explanation:
Given expression;
3.88 x 10⁷ g - 4.701 x 10⁵ g
The expression above is simplified as follows;
= (3.88 x 100 x 10⁵ )g - ( 4.701 x 10⁵) g
= (388 x 10⁵ )g - ( 4.701 x 10⁵) g
= (388 - 4.701 ) x (10⁵ )g
= 383.299 x 10⁵ g
In standard form, the simplified expression can be expressed as;
= (3.83299 x 100 x 10⁵) g
= 3.83299 x 10⁷ g
= 3.833 x 10⁷ g
Therefore, the simplified expression is 3.833 x 10⁷ g
Oxygen is obtained through various methods. Which of the following methods involves a chemical
change?
1. Electrolysis of water
2. Distillation of liquid air
3. Heating of KCIO,
02
1 and 2
1 and 3
Answer:
1
Explanation:
Electrolysis is the passing of an current through a conducting solution, when the occurs, a chemical reaction takes place.
Heating a chemical will always cause a chemical reaction, which is why 3 is also correct
Some information as to why 2 is NOT correct.
2 is NOT a chemical reaction, but rather a process of physical separation. It uses selective boiling and condensation, but is not considered a chemical reaction.
as with 3, heating is not considered a chemical reaction, but rather a physical temperature change. This is always what it is considered to be (e.g boiling water is a physical temperature change, not a chemical reaction)
Hope this helps.
Hope this helps.
What is the strength of the magnetic field a distance 4.4 mm above the center of a circular loop of radius 0.8 mm and current 474.1 A
Answer:
B = 0.118 T
Explanation:
From Biot-Savart Law:
[tex]B = \frac{\mu_o I}{2\pi r}[/tex]
where,
B = strength of magnetic field = ?
μ₀ = 4π x 10⁻⁷ Tm/A
I = current enclosed = 474.1 A
r = radius = 0.8 mm = 8 x 10⁻⁴ m
Therefore,
[tex]B = \frac{(4\pi\ x\ 10^{-7}\ Tm/A)(474.1\ A)}{2\pi(8\ x\ 10^{-4}\ m)}[/tex]
B = 0.118 T
When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What maximum wavelength of light could be used to ejectelectrons from aluminum
Answer:
The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm
Explanation:
Given;
wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m
maximum kinetic energy of the ejected electron, K.E = 0.92 eV
let the work function of the aluminum metal = Ф
Apply photoelectric equation:
E = K.E + Ф
Where;
Ф is the minimum energy needed to eject electron the aluminum metal
E is the energy of the incident light
The energy of the incident light is calculated as follows;
[tex]E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J[/tex]
The work function of the aluminum metal is calculated as;
Ф = E - K.E
Ф = 8.02 x 10⁻¹⁹ - (0.92 x 1.602 x 10⁻¹⁹)
Ф = 8.02 x 10⁻¹⁹ J - 1.474 x 10⁻¹⁹ J
Ф = 6.546 x 10⁻¹⁹ J
The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;
[tex]\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm[/tex]
If an object, initially at rest, accelerates at the rate of 25m/s2, what will the magnitude of the displacement be after 50s
Answer:
31250 meters
Explanation:
Given data
Intitially at rest, the velocity will be
u= 0m/s
acceleration a= 25m/s^2
Time= 50s
We know that the expression for the displacement is given as
S=U+ 1/2at^2
S= 0+ 1/2*25*50^2
S= 12.5*2500
S=31250 meters
Hence the displacement is 31250 meters
Two speakers in a stereo emit identical pure tones. As you move around in front of the speakers, you hear the sound alternating between loud and zero. This occurs because of
Answer:
Interference
Explanation:
When two traveling waves traveling waves along the same path are superimposed(combine). The superimposition of these two waves results in the production of a resultant wave which is defined by the net effect of the two waves. Wave interference occurs most types of waves including radio wave, light, acoustic waves and other wave types. Alternating sound between loud and Zero is heard as the two speakers emit identical pure tones because the resultant amplitude after the interference of the two sound waves is the vector sum of each of their amplitudes. A loud sound is heard, when the crest of both waves meets each other and a zero is heard if the crest of one meets the trough of the other as they cancel out.
When you stand on tiptoes on a bathroom scale, there is an increase in
A) weight reading.
B) pressure on the scale, not registered as weight.
C) both weight and pressure on the scale.
D) none of the above
Answer:
B) Pressure on the scale, not registered as weight.
Explanation:
This is because energy (derived from weight) becomes compiled on the tips of your toes, and therefore does not increase your weight, but simply the pressure at a smaller point
g A computer is reading data from a rotating CD-ROM. At a point that is 0.0189 m from the center of the disk, the centripetal acceleration is 241 m/s2. What is the centripetal acceleration at a point that is 0.0897 m from the center of the disc?
Answer:
the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²
Explanation:
Given the data in the question;
centripetal acceleration a[tex]_c[/tex]₁ = 241 m/s²
radius r₁ = 0.0189 m
radius r₂ = 0.0897 m
centripetal acceleration a[tex]_c[/tex]₂ = ? m/s²
since the rotational period will be the same for the two disk,
we use the centripetal acceleration formula a[tex]_c[/tex] = (4π²r/T²) to find the rotational period for the first disk.
a[tex]_c[/tex]₁ = (4π²r₁/T²)
make T² subject of formula
T² = 4π²r₁ / a[tex]_c[/tex]₁
we substitute
T² = ( 4 × π² × 0.0189 ) / 241
T² = 0.00309602528 s²
Now we use the same formula to find a[tex]_c[/tex]₂
a[tex]_c[/tex]₂ = ( 4π²r₂ / T² )
we substitute
a[tex]_c[/tex]₂ = ( 4 × π² × 0.0897 ) / 0.00309602528
a[tex]_c[/tex]₂ = 1143.8 m/s²
Therefore, the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²
When an automobile moves with constant velocity the power developed is used to overcome the frictional forces exerted by the air and the road. If the power developed in an engine is 50.0 hp, what total frictional force acts on the car at 55 mph (24.6 m/s)
P = F v
where P is power, F is the magnitude of force, and v is speed. So
50.0 hp = 37,280 W = F (24.6 m/s)
==> F = (37,280 W) / (24.6 m/s) ≈ 1520 N
Paauto A: Isulat sa papel ang alpabetong Ingles at bilang I hanggang 10 sa istilong
Roman ng pagleletra.
Answer:
Explanation:
English alphabets numbered fro 1 to 26
and the numbers 1 to10 so they are written in roman numbers as
1 - I
2 - II
3 - III
4 - IV
5 -V
6 - VI
7 -VII
8 - VIII
9 - IX
10 -X
11 - XI
12 - XII
13 - XIII
14 - XIV
15 - XV
16 - XVI
17 - XVII
18 - XVIII
19 - XIX
20- XX
21 - XXI
22 - XXII
23 - XXIII
24 - XXIV
25 - XXV
26 - XXVI
Choose the force diagram that best represents a ball thrown upward by Peter, at the
top of its path.
Diagram A
Diagram B
Diagram C
Diagram D
Answer:Diagram A
Explanation:
Since the air resistance is to be neglected, only the gravitational force acts on the ball ( and has acted all the way from the throw upward). Diagram A reflects this fact correctly indicating the gravity acting on the ball downward.
During a practice shot put throw, the 7.9-kg shot left world champion C. J. Hunter's hand at speed 16 m/s. While making the throw, his hand pushed the shot a distance of 1.4 m. Assume the acceleration was constant during the throw.
Required:
a. Determine the acceleration of the shot.
b. Determine the time it takes to accelerate the shot.
c, Determine the horizontal component of the force exerted on the shot by hand.
Answer:
a) a = 91.4 m / s², b) t = 0.175 s, c)
Explanation:
a) This is a kinematics exercise
v² = vox ² + 2a (x-xo)
a = v² - 0/2 (x-0)
let's calculate
a = 16² / 2 1.4
a = 91.4 m / s²
b) the shooting time
v = vox + a t
t = v-vox / a
t = 16 / 91.4
t = 0.175 s
c) let's use Newton's second law
F = ma
F = 7.9 91.4
F = 733 N
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz. Calculate the wavelength of the infrared radiation.
Answer:
[tex]\lambda=6.83\times 10^{-5}\ m[/tex]
Explanation:
Given that,
An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.
We know that,
1 THz = 10¹² Hz
So,
f = 4.39 × 10¹² Hz
We need to find the wavelength of the infrared radiation.
We know that,
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m[/tex]
So, the wavelength of the infrared radiation is [tex]6.83\times 10^{-5}\ m[/tex].
If a car generates 22 hp when traveling at a steady 100 km/h , what must be the average force exerted on the car due to friction and air resistance
Answer:
The average force exerted on the car is 590.12 N.
Explanation:
Given that,
The power generated, P = 22 hp = 16405.4 W
Speed of the car, v = 100 km/h = 27.8 m/s
We need to find the average force exerted on the car due to friction and air resistance.
We know that,
Power, P = F v
Where
F is force exerted on the car
[tex]F=\dfrac{P}{v}\\\\F=\dfrac{16405.4}{27.8}\\\\F=590.12\ N[/tex]
So, the average force exerted on the car is 590.12 N.
How many wavelengths of the radio waves are there between the transmitter and radio receiver if the woman is listening to an AM radio station broadcasting at 1180 kHz
Answer:
254 m
Explanation:
Applying,
v = λf............... Equation 1
Where v = velocity of radio wave, λ = wave length, f = frequency
make λ the subject of the equation
λ = v/f............ Equation 2
From the question,
Given: f = 1180 kHz = 1180000 Hz
Constant: v = 3×10⁸ m/s
Substitite into equation 2
λ = 3×10⁸/1180000
λ = 2.54×10²
λ = 254 m
A particle charge of 2.7 µC is at the center of a Gaussian cube 55 cm on edge. What is the net electric flux through the surface?
Answer:
3.05×10⁵ Nm²C⁻¹
Explanation:
According to Gauss' law,
∅' = q/e₀............... Equation 1
Where ∅' = net flux through the surface, q = net charge, e₀ = electric permittivity of the space
From the question,
Given: q = 2.7 μC = 2.7×10⁻⁶ C,
Constant: e₀ = 8.85×10⁻¹² C²/N.m²
Substituting these values into equation 1
∅' = (2.7×10⁻⁶)/(8.85×10⁻¹²)
∅' = 3.05×10⁵ Nm²C⁻¹
A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
Complete question:
A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
a) 60 ⁰
b) 90 ⁰
c) 120 ⁰
Answer:
(a) When the angle, θ = 60 ⁰, force = 4.07 N
(b) When the angle, θ = 90 ⁰, force = 4.7 N
(c) When the angle, θ = 120 ⁰, force = 4.07 N
Explanation:
Given;
length of the wire, L = 2.8 m
current carried by the wire, I = 5.6 A
magnitude of the magnetic force, F = 0.3 T
The magnitude of the magnetic force is calculated as follows;
[tex]F = BIl \ sin(\theta)[/tex]
(a) When the angle, θ = 60 ⁰
[tex]F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N[/tex]
(b) When the angle, θ = 90 ⁰
[tex]F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N[/tex]
(c) When the angle, θ = 120 ⁰
[tex]F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N[/tex]
NEED HELP ASAP- Please show work
The angular position of an object is given by θ = 4t3 +10t −40 , where θ is in radians and t is in seconds what is:
(a) (5 points) The angular velocity at t = 2 s?
(b) (5 points) The angular acceleration at t = 2 s?
Answer:
Look at work
Explanation:
Θ= 4t^3+10t-40
a) In order to find ω, we need to find displacement so plug in t=2 to find Θ.
Θ= 4*8+20-40=12
use ω=Θ/t
Plug in values
ω=6 rad/s
b) In order to find α we use ω/t.
Plug in values
α=6/2= 3 rad/s^2
A 10.0 L tank contains 0.329 kg of helium at 28.0 ∘C. The molar mass of helium is 4.00 g/mol . Part A How many moles of helium are in the tank? Express your answer in moles.
Answer:
82.25 moles of He
Explanation:
From the question given above, the following data were obtained:
Volume (V) = 10 L
Mass of He = 0.329 Kg
Temperature (T) = 28.0 °C
Molar mass of He = 4 g/mol
Mole of He =?
Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:
1 Kg = 1000 g
Therefore,
0.329 Kg = 0.329 Kg × 1000 g / 1 Kg
0.329 Kg = 329 g
Thus, 0.329 Kg is equivalent to 329 g.
Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:
Mass of He = 329 g
Molar mass of He = 4 g/mol
Mole of He =?
Mole = mass / molar mass
Mole of He = 329 / 4
Mole of He = 82.25 moles
Therefore, there are 82.25 moles of He in the tank.
ACCORDING TO NEWTON'S THIRD LAW EVERY ACTION HAS EQUAL AND OPPOSITE REACTION BUT THEN WHY DON'T WE FLY WHEN WE FART??
Answer:
Your fart only has so much force, not nearly enough to launch you into oblivion. Your fart and you still exert a force onto each other, so I guess, hypothetically, you could fly if you really, really try hard enough. Just make sure you don't try too hard and prolapse as a result :)
Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer section at a distance of 2.0 m from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 1.1 m from the center. The carousel takes 5.8 s to make each complete revolution.
Required:
a. What is Mary's angular speed %u03C9M and tangential speed vM?
b. What is Alex's angular speed %u03C9A and tangential speed vA?
Answer:
you can measure by scale beacause we dont no sorry i cant help u but u can ask me some other Q
Assuming the atmospheric pressure is 1 atm at sea level, determine the atmospheric pressure at Badwater (in Death Valley, California) where the elevation is 86.0 m below sea level.
Answer:
Atmospheric pressure at Badwater is 1.01022 atm
Explanation:
Data given:
1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa
Elevation (h) = 86m
gravity (g) = 9.8 m/s2
Density of air P = 1.225 kg/m3
Therefore pressure at bad water Pb = Pi + Pgh
Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)
Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa
hence:
Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm
A body of mass 450g changes it speed from 5ms¹ to 25ms¹. what is the work done by the body?
Answer:
135J
Explanation:
So we know ΔKinetic Energy= ΔWork
Kinetic energy=1/2mv²
So Kf-Ki=ΔK
ΔK=1/2*0.45(25²-5²)=135J
135J=ΔWork
Find a parametric representation for the surface. The plane through the origin that contains the vectors i - j and j - k
Answer:
parametric representation: x = u, y = v - u , z = - v
Explanation:
Given vectors :
i - j , j - k
represent the vector equation of the plane as:
r ( u, v ) = r₀ + ua + vb
where: r₀ = position vector
u and v = real numbers
a and b = nonparallel vectors
expressing the nonparallel vectors as :
a = i -j , b = j - k , r = ( x,y,z ) and r₀ = ( x₀, y₀, z₀ )
hence we can express vector equation of the plane as
r(u,v) = ( x₀ + u, y₀ - u + v, z₀ - v )
Finally the parametric representation of the surface through (0,0,0) i.e. origin = 0
( x, y , z ) = ( x₀ + u, y₀ - u + v, z₀ - v )
x = 0 + u ,
y = 0 - u + v
z = 0 - v
∴ parametric representation: x = u, y = v - u , z = - v
The slope of a d vs t graph represents velocity. Describe 3 ways you know this to be true.
Answer:
Look at explanation
Explanation:
I only know 1 way, there is another way you can rephrase this using derivatives but that's pretty much the same thing.
The slope is calculated by Δy/Δx so the slope of distance vs time graph is Δd/Δt which is the velocity
The image of an object placed 30cm from a diverging lens is formed 10cm in front of the lens.
Calculate the focal length of the lens.
Answer:
15cm
Explanation:
Since the lens is a diverging lens, the image distance is negative (virtual)
v = -30cm
u = 10cm
Required
focal length f
Using the lens formula;
1/u + 1/v = 1/u
1/10 - 1/30 = 1/f
(3-1)/30 = 1/f
2/30 = 1/f
f = 30/2
f = 15cm
Hence the focal length of the lens is 15cm
Explain what a circuit breaker is and how it helps protect your house?
Explanation:
A circuit breaker is an electrical switch designed to protect an electrical circuit from damage caused by overcurrent/overload or short circuit. Its basic function is to interrupt current flow after protective relays detect a fault.
Circuit breakers have been designed to detect when there is a fault in the electricity, so it will “trip” and shut down electrical flow. ... This detection is key to preventing surges of electricity that travel to appliances or other outlets, which can cause them to break down
Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resistance. But driving slowly in dry sand is another story. If a 1500 kg car is driven in sand at 4.9 m/s , the coefficient of rolling friction is 0.060. In this case, nearly all of the energy that the car uses to move goes to overcoming rolling friction, so you can ignore air drag in this problem.
Required:
a. What propulsion force is needed to keep the car moving forward at a constant speed?
b. What power is required for propulsion at 5.0 m/s?
c. If the car gets 15 mpg when driving on sand, what is the car's efficiency? One gasoline contains 1.4×10 ^8 J of chemical energy.
Answer:
a) [tex]F_p=882N[/tex]
b) [tex]P=4410W[/tex]
c) [tex]V_p'=24135[/tex] ,[tex]n=15.2\%[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=1500kg[/tex]
Velocity [tex]v=4.9m/s[/tex]
Coefficient of Rolling Friction [tex]\mu=0.06[/tex]
a)
Generally the equation for The Propulsion Force is mathematically given by
[tex]F_p=\mu*mg[/tex]
[tex]F_p=0.06*1500*9.81[/tex]
[tex]F_p=882N[/tex]
b)
Therefore Power Required at
[tex]V_p=5.0m/s[/tex]
[tex]P=F_p*V_p[/tex]
[tex]P=882*5[/tex]
[tex]P=4410W[/tex]
c)
[tex]V_p' =15mpg[/tex]
[tex]V_p'=15*\frac{1609}[/tex]
[tex]V_p'=24135[/tex]
Generally the equation for Work-done is mathematically given by
[tex]W=F_p*V_p'[/tex]
[tex]W=882*15*1609[/tex]
[tex]W=2.13*10^7[/tex]
Therefore
Efficiency
[tex]n=\frac{W}{E}*100\%[/tex]
Since
Energy in one gallon of gas is
[tex]E=1.4*10^8J[/tex]
Therefore
[tex]n=\frac{2.1*10^7}{1.4*10^8}*100\%[/tex]
[tex]n=15.2\%[/tex]
A pilot drops a bomb from a plane flying horizontally. Where will the plane be located when the bomb hits the ground
Answer:
The plane will be located directly above the bomb because they both have the same horizontal speed.