Explanation:
Earth or any planet are actually born from huge clouds of gas and dust. Their stellar mass are fairly distributed at a radius from the axis of rotation. Gravitational force cause the cloud to come together. Now the whole gathered in smaller area. Now, individual particles come close to the roational axis. Thus, decreasing the moment of inertia of the planet.
As
I=mr^2
reducing r reduces I. However, the angular moment of the system remains always conserved. So, to conserve the angular momentum the angular velocity of the planet increases and so did the otational kinetic energy
The additional rotational kinetic energy of the Earth compared to the initial cloud comes from the gravitational potential energy that was present in the cloud before it collapsed. The conservation of angular momentum dictates that as the cloud shrinks, its rotation speed must increase, leading to an increase in rotational kinetic energy.
The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed because of the conservation of angular momentum. As the cloud collapsed under its own gravity, the rotation rate increased to conserve angular momentum, which is the product of moment of inertia and rotational velocity. Since the moment of inertia decreases as the mass moves closer to the axis of rotation (due to the shrinking size of the collapsing cloud), the rotational velocity must increase to keep the angular momentum constant. This results in an increase in rotational kinetic energy, which is given by the equation [tex]\( KE_{rot} = \frac{1}{2} I \omega^2 \)[/tex], where [tex]\( I \)[/tex] is the moment of inertia and [tex]\( \omega \)[/tex] is the angular velocity.
The gravitational potential energy of the cloud is converted into kinetic energy as the cloud collapses. The initial potential energy is high because the particles in the cloud are far from the center of mass. As the cloud contracts, this potential energy is transformed into kinetic energy, both linear and rotational, due to the conservation of energy. The increase in rotational kinetic energy is a direct consequence of this conversion and the conservation of angular momentum.
In summary, the additional rotational kinetic energy of the Earth compared to the initial cloud comes from the gravitational potential energy that was present in the cloud before it collapsed. The conservation of angular momentum dictates that as the cloud shrinks, its rotation speed must increase, leading to an increase in rotational kinetic energy.
The complete question is:
The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. Where did this energy come from?
A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake. She hears the echo 2.9 s after shouting. The speed of sound in air is 343 m/s. Determine the length of the lake.
Answer:
497.35 m.
Explanation:
Echo: This is the sound heard after the reflection of sound wave on a plane surface.
v = 2x/t.......................................... Equation 1
Where v = speed of sound in air, x = length of the lake, t = time taken to hear the echo.
Making x the subject of the equation,
x = vt/2.......................................... Equation 2
Given: v = 343 m/s, t = 2.9 s.
Substitute into equation 2
x = 343×2.9/2
x = 497.35 m
Thus the length of the lake = 497.35 m.
In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. Find the minimum speed must the athlete leave the ground in order to lift his center of mass 1.65 m and cross the bar with a speed of 0.75 m/s.
Answer:
The answer to your question is v₁ = 5.74 m/s
Explanation:
Data
v₁ = ?
h₁ = 0 m
v₂ = 0.75 m/s
h₂ = 1.65 m
g = 9.81 m/s²
Formula
mgh₁ + 1/2mv₁² = mgh₂ + 1/2mv₂²
mass is not consider (if we factor mass, it is cancelled)
gh₁ + 1/2v₁² = gh₂ + 1/2v₂²
Substitution
(9.81)(0) + 1/2v₁² = (9.81)(1.65) + 1/2(0.75)²
Simplification
0 + 1/2v₁² = 16.19 + 0.28
Solve for v₁
1/2v₁² = 16.47
v₁² = 2(16.47)
v₁² = 32.94
Result
v₁ = 5.74 m/s
Minimum speed must 5.74 m/s in order to lift his center of mass 1.65 m and cross the bar with a speed of 0.75 m/s.
Given here,
v₁ - initial velocity = ?
h₁ - initial height = 0 m
v₂ - final velocity = 0.75 m/s
h₂ - final height = 1.65 m
g - gravitational acceleration = 9.81 m/s²
The speed can be calculated by using the formula,
[tex]\bold { mgh_1 + \dfrac 12 mv_1^2 = mgh_2 + \dfrac 12mv_2^2}[/tex]
factor the mass,
[tex]\bold { gh_1 + \dfrac 12 v_1^2 = gh_2 + \dfrac 12v_2^2}[/tex]
put the values in the formula, and solve it for V1
[tex]\bold { (9.81)(0) + \dfrac 12v_1^2 = (9.81)(1.65) + \dfrac 12(0.75)^2}\\\\\bold { \dfrac 12v_1^2 = 16.19 + 0.28}\\\\\bold { \dfrac 12v_1^2 = 16.47}\\\\ \bold {v_1^2 = 2(16.47)}\\\\\bold {v_1^2= 32.94}\\\\\bold { v_1 = 5.74\ m/s}[/tex]
Therefore, minimum speed must 5.74 m/s in order to lift his center of mass 1.65 m and cross the bar with a speed of 0.75 m/s.
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A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pushed 2.5 m straight backwards in 1.65 s. He breaks the tackle and runs straight forward another 24 m in 5.2 s.
Part (a) Calculate the wide receiver's average velocity in the horizontal direction during the first interval, in meters per second.
Part (b) Calculate the wide receiver's average velocity in the horizontal direction during the second interval, in meters per second.
Part (c) Calculate the wide receiver's average velocity in the horizontal direction during the third interval, in meters per second.
Part (d) Calculate the wide receiver's average velocity in the horizontal direction for the entire motion, in meters per second
Answer:
a) v1 = 5.52m/s
b) v2 = -1.52m/s
c) v3 = 4.62m/s
d) vt = 3.85m/s
Explanation:
The velocity of the football wide receiver is his displacement per unit time.
Velocity v = (displacement d)/time t
v = d/t .....1
For each of the cases, equation 1 would be used to calculate the velocity.
a) v1 = d1/t1
d1= 16m
t1 = 2.9s
v1 = 16m/2.9s
v1 = 5.52m/s
b) v2 = d2/t2
d2 = -2.5m
t2 = 1.65s
v2 = -2.5/1.65
v2 = -1.52m/s
c) v3 = d3/t3
d3 = 24m
t3 = 5.2s
v3 = 24/5.2
v3 = 4.62m/s
d) vt = dt/tt
dt = 16m - 2.5m + 24m = 37.5m
tt = 2.9 + 1.65 + 5.2 = 9.75s
vt = 37.5/9.75
vt = 3.85m/s
Brewed coffee is often too hot to drink right away. You can cool it with an ice cube, but this dilutes it. Or you can buy a device that will cool your coffee without dilution - a 200 aluminum cylinder that you take from your freezer and place in a mug of hot coffee.
q)If the cylinder is cooled to -20C, a typical freezer temperature, and then dropped into a large cup of coffee (essentially water, with a mass of 500g ) at 85C, what is the final temperature of the coffee?answer in 2 sig figs and answer in Celsius.
The final temperature of the coffee is approximately [tex]\( 57.21 \, ^\circ C \)[/tex] (rounded to 2 significant figures).
Given equations:
[tex]\[ Q_c = Q_a \][/tex]
[tex]\[ m_c \cdot c_w \cdot (T_{ic} - T_f) = m_a \cdot c_a \cdot (T_f - T_{ia}) \][/tex]
Substitute the given values:
[tex]\[ 0.5 \cdot 1000 \cdot (85 - T_f) = 0.2 \cdot 900 \cdot (T_f - (-20)) \][/tex]
Now, distribute and simplify:
[tex]\[ 42500 - 500 \cdot T_f = 180 \cdot T_f + 3600 \][/tex]
Combine like terms:
[tex]\[ 500 \cdot T_f + 180 \cdot T_f = 42500 - 3600 \][/tex]
[tex]\[ 680 \cdot T_f = 38900 \][/tex]
Now, solve for A [tex]\( T_f \):[/tex]
[tex]\[ T_f = \frac{38900}{680} \][/tex]
[tex]\[ T_f \approx 57.21 \, ^\circ C \][/tex]
So, the final temperature of the coffee is approximately [tex]\( 57.21 \, ^\circ C \)[/tex] (rounded to 2 significant figures).
a proton moving to the right at 3.8 enters a region where 56 electic field points ot the left, how far will the proton get and describe its motion
Answer: 0.07meter
Pointing to the left.
Explanation:
The Electric potential V of the proton is 3.8v. while the Electric Field strength is 56V/m. We first calculate the distance r the proton would travel
E= V/r
r = V/E
r = 3.8V/(56V/m)
r =0.07m.
The proton which is positively charged will move in the direction of the Electric Field that is to the left.
When a proton enters a region with an electric field pointing to the left, it will experience a force in the opposite direction. The force on the proton can be calculated using the formula F = q*E. The distance the proton will get before coming to a stop can be calculated using the formula d = (1/2)at^2.
Explanation:When a proton enters a region with an electric field pointing to the left, it will experience a force in the opposite direction. The magnitude of the force can be determined using the formula:
F = q*E
Where F is the force, q is the charge of the proton, and E is the magnitude of the electric field. In this case, the force on the proton is given by:
F = (1.6 x 10^-19 C)(56 N/C) = -8.96 x 10^-18 N
The negative sign indicates that the force is in the opposite direction of the electric field. Since the proton is moving to the right, it will slow down and eventually come to a stop. The distance the proton will get can be calculated using the formula:
d = (1/2)at^2
Where d is the distance, a is the acceleration, and t is the time. Since the proton starts from rest, its initial velocity is zero. The acceleration can be calculated using the formula:
a = F/m
Where a is the acceleration, F is the force, and m is the mass of the proton. The mass of a proton is approximately 1.67 x 10^-27 kg. Plugging in the values, we get:
a = (-8.96 x 10^-18 N) / (1.67 x 10^-27 kg) = -5.38 x 10^8 m/s^2
Substituting this value of acceleration and the given initial velocity into the distance formula, we can calculate the distance the proton will get:
d = (1/2)(-5.38 x 10^8 m/s^2)(t^2)
Where t is the time for which the proton travels. The time can be calculated using the formula:
t = v/a
Where t is the time, v is the initial velocity, and a is the acceleration. Plugging in the values, we get:
t = (3.8 m/s) / (-5.38 x 10^8 m/s^2) = -7.08 x 10^-9 s
Since the time cannot be negative, we take the magnitude of the time:
t = 7.08 x 10^-9 s
Substituting this value of time into the distance formula, we get:
d = (1/2)(-5.38 x 10^8 m/s^2)((7.08 x 10^-9 s)^2) = 1.05 x 10^-15 m
Therefore, the proton will get a distance of 1.05 x 10^-15 meters before coming to a stop.
A "free" electron and a "free" proton are placed in an identical electric field. Which of the following statements are true?
(Selectall that apply.)
a. Each particle is acted upon by the sameelectric force and has the same acceleration.
b. The magnitude of the acceleration of theelectron is greater than that of the proton.
c. The electric force on the proton isgreater in magnitude than the electric force on the electron, butin the opposite direction.
d. Both particles have the same acceleration.
e. The electric force on the proton is equalin magnitude to the electric force on the electron, but in the opposite direction.
Answer:b and e
Explanation:
When electron and proton are placed in the same Electric field then the force experienced by electron and proton is the same as the charge possessed by them is the same . The direction of force is different on them but the magnitude is the same.
The electric field force is the product of charge and strength of the electric field.
We know the mass of an electron is less than the mass of proton that is why the acceleration of electron is more as compared to proton for the same Electric force.
thus option b and e are correct.
The statements which are true of a "free" electron and a "free" proton are placed in an identical electric field are: Choice b and Choice e.
According to the question:
They are placed in an identical electric field.In essence, the magnitude of the Coulumb's force of attraction or repulsion is the same for the free electron and free proton.
However, the direction of the forces are opposite as the protons and electron have opposing signs.
Additionally, since the mass of an electron is relatively infinitesimal compared to a proton, The acceleration of the electron is greater than that of the proton.
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An expensive vacuum system can achieve a pressure as low as 1.83 ✕ 10−7 N/m2 at 28°C. How many atoms are there in a cubic centimeter at this pressure and temperature?
Answer:[tex]44.1\times 10^6[/tex] atoms
Explanation:
According to the ideal gas equation:
[tex]PV=nRT[/tex]
P = Pressure of the gas = [tex]1.83\times 10^{-7}N/m^2=1.81\times 10^{-12}atm[/tex] [tex]1N/m^2=9.87\times 10^{-6}atm[/tex]
V= Volume of the gas = [tex]1cm^3=1ml=0.001L[/tex] (1L=1000ml)
T= Temperature of the gas = 28°C = 301 K [tex]0^0C=273K[/tex]
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= ?
[tex]n=\frac{PV}{RT}=\frac{1.81\times 10^{-12}atm\times 0.001L}0.0821Latm/Kmol\times 301K}=7.32\times 10^{-17}moles[/tex]
Number of atoms =[tex]moles\times {\text {avogadro's number}}=7.32\times 10^{-17}mol\times 6.023\times 10^{23}mol^{-1}=44.1\times 10^6atoms[/tex]
Thus there are [tex]44.1\times 10^6[/tex] atoms in a cubic centimeter at this pressure and temperature.
Which of the following could be reasonable explanations for how a piece of invisible tape gets charged? Select all that apply.
(1) Protons are pulled out of nuclei in one tape and transferred to another tape.
(2) Charged molecular fragments are broken off one tape and transferred to another.
(3) Electrons are pulled out of molecules in one tape and transferred to another tape.
(4) Neutrons are pulled out of nuclei in one tape and transferred to another tape.
Answer:
2 and 3
Explanation:
The right answer is the option 2 and 3,
This is all about the electrons transfer from one material to the other material.
For example if the electrons in the valence shell of one material are loosely attached then the other material's atoms try to take those electron to complete their shells and that is how the charges transfers from one another.
And it could also be happen as in option 2.
As an example, a 3.80- kg aluminum ball has an apparent mass of 2.00 kg when submerged in a particular liquid: calculate the density of the liquid.
Answer:
The density of the liquid = 1278.95 kg/m³
Explanation:
Density: This can be defined as the ratio of the mass of a body to its volume. The S.I unit of density is kg/m³.
From Archimedes principle,
R.d = Density of object/Density of liquid = Weight of object in air/Upthrust in liquid.
D₁/D₂ = W/U .......................... Equation 1
D₂ = D₁(U/W)........................ Equation 2
Where D₁ = Density of aluminum, D₂ = Density of liquid, W = weight of aluminum, U = upthrust.
m₁ = 3.8 kg, m₂ = 2.00 kg, g = 9.8 m/s²
W = m₁g = 3.8(9.8) = 37.24 N.
U = lost in weight of the aluminum = (m₁ - m₂)g = (3.8-2.0)9.8
U = 1.8(9.8) = 17.64 N.
Constant: D₁ = 2700 kg/m³
Substituting these values into equation 2
D₂ = 2700(17.64)/37.24
D₂ = 1278.95 kg/m³
Thus the density of the liquid = 1278.95 kg/m³
A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no obstacles in its way. It is fired at an angle 33° with respect to the horizontal and with an initial speed of 49 m/s. Air resistance is negligible in this situation.
A) What is the horizontal distance that the projectile has traveled when it is at its maximum height?
B) The same projectile was then fired in the same way toward a wall that is a horizontal distance 55.8 m from where the projectile was fired.
What was the height of the projectile when it hit the wall?
C) Find the velocity of the projectile when it hit the wall.
D) What was the speed of the projectile when it hit the wall?
Answer:
(A) 111.77m
(B) 9.07m
(C)55.88m/s
(D)20.54m/s
Explanation:
step 1 " we have to calculate the time it took the projectile to get to its maximum height
(a) t = usinθ/g
= 49sin 33/9.81
49×0.5446/9.81
=2.72s
the horizontal distance = ucosθ×t , because the projectile horizontal motion is unaffected by the force of graavity
= 49cos33 ×2.72
=111.77m
(B) with the same projectile fired the same way , the horizontal distance = 55.8m
55.8 = ucosθ×t
55.8 = 49cos33 ×t
t = 55.8/49cos33
t= 1.36s
height of the projectile = 1/2 ×g×t²
=1/2 ×9.81×1.36²
= 9.07m
(c) Velocity of the projectile when it hits the wall
V₀ = ucosθ×t
49cos 33 × 1.36
=55.88 m/s
(D) speed = distance / time
distance /2×t ; total time of flight
= 55.88/ 2.72
=20.54m/s
The position of a particle moving along the x-axis varies with time according to x(t) = 5.0t2-4.0t3m. Find (a) the displacement, average velocity, average acceleration between 0.0 s and 1.0,
Answer:
[tex]\Delta x=1-0=1\ m[/tex]
[tex]\Delta v=-2-0=-2\ m.s^{-1}[/tex]
[tex]\Delta a=-14-10=-24\ m.s^{-1}[/tex]
Explanation:
The equation governing the position of the particle moving along x-axis is given as:
[tex]x=5\times t^2-4\times t^3[/tex]
we know that the time derivative of position gives us the velocity:
[tex]\frac{d}{dt} x=v[/tex]
[tex]v=10\ t-12\ t^2[/tex]
and the time derivative of of velocity gives us the acceleration:
[tex]\frac{d}{dt} v=a[/tex]
[tex]a=10-24\ t[/tex]
Now, when t = 0
[tex]x=0\ m[/tex]
[tex]v=0\ m.s^{-1}[/tex]
[tex]a=10\ m.s^{-2}[/tex]
When t=1 s
[tex]x_1=5\times 1^2-4\times 1^3=1\ m[/tex]
[tex]v_1=10\times 1-12\times 1^2=-2\ m.s^{-1}[/tex]
[tex]a_1=10-24\times 1=-14\ m.s^{-2}[/tex]
Hence,
Displacement between the stipulated time:
[tex]\Delta x=x_1-x[/tex]
[tex]\Delta x=1-0=1\ m[/tex]
Velocity between the stipulated time:
[tex]\Delta v=v_1-v[/tex]
[tex]\Delta v=-2-0=-2\ m.s^{-1}[/tex]
Acceleration between the stipulated time:
[tex]\Delta a=a_1-a[/tex]
[tex]\Delta a=-14-10=-24\ m.s^{-1}[/tex]
Here negative sign indicates that the vectors are in negative x direction.
The position of a particle is given by the expression x 5 4.00 cos (3.00pt 1 p), where x is in meters and t is in seconds. Determine
(a) the frequency
(b) period of the motion
(c) the amplitude of the motion
(d) the phase constant
(e) the position of the particle at t 5 0.250 s.
Answer:
1.5 Hz
0.67 s
4 m
[tex]\pi[/tex]
2.82842 m
Explanation:
The equation is
[tex]x=4cos(3\pi t+\pi)[/tex]
It is of the form
[tex]x=Acos(2\pi ft+\phi)[/tex]
Comparing the equations we get
[tex]3\pi=2\pi f\\\Rightarrow f=\dfrac{3}{2}\\\Rightarrow f=1.5\ Hz[/tex]
Frequency is 1.5 Hz
Time period is given by
[tex]T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1.5}\\\Rightarrow T=0.67\ s[/tex]
The period of the motion is 0.67 s
Amplitude
[tex]A=4\ m[/tex]
Amplitude is 4 m
Phase constant
[tex]\phi=\pi[/tex]
The phase constant is [tex]\pi[/tex]
At t = 0.25 s
[tex]x=4cos(3\pi t+\pi)\\\Rightarrow x=4cos(3\pi\times 0.25+\pi)\\\Rightarrow x=2.82842\ m[/tex]
The position of the particle is 2.82842 m
Galileo, in his Two New Sciences, states that "for elevations (angles of projection) which exceed or fall short of 45 degrees by equal amounts, the ranges are equal. Prove it.
Answer:
Please refer to the attachment below.
Explanation:
Please refer to the attachment below for explanation.
Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?
Answer:
[tex]L=6.21m[/tex]
Explanation:
For the simple pendulum problem we need to remember that:
[tex]\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0[/tex],
where [tex]\theta[/tex] is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:
[tex]\omega^{2}=\frac{g}{L}[/tex],
where [tex]\omega[/tex] is the angular frequency.
There is also an equation that relates the oscillation period and the angular frequeny:
[tex]\omega=\frac{2\pi}{T}[/tex],
where T is the oscillation period. Now, we can easily solve for L:
[tex](\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m[/tex]
A boat heads north across a river at a rate of 3 miles per hour. If the current is flowing east at a rate of 2 miles per hour, find the resultant velocity of the boat. (Assume that east lies in the direction of the positive x-axis and north in the direction of the positive y-axis.)
Answer:
The resultant velocity of the boat is 3.6 mil/h
Explanation:
given information:
flowing rate to the east (positive x-axis), v₁ = 2 mil/h (0.2) = 2i
boat's speed to the north (positive y-axis), v₂ = 3 mil/h (3,0) = 3j
the resultant velocity of the boat:
[tex]v_{R}[/tex] = √3²+2²
= √13
= 3.6 mil/h
The resultant velocity of the boat is approximately 3.61 mph, making an angle of 56.31° north of east. This is calculated using the Pythagorean theorem for magnitude and the inverse tangent function for direction.
Explanation:To find the resultant velocity of the boat when it heads north across a river at a rate of 3 miles per hour with an eastward current at 2 miles per hour, we can use vector addition. The velocity of the boat heading north (Vboat) is perpendicular to the velocity of the river's current (Vriver).
The northward velocity can be considered as the y-component (Vy = 3 mph), and the eastward current as the x-component (Vx = 2 mph). The resultant velocity (Vtot) is the vector sum of these two components and can be calculated using the Pythagorean theorem:
Vtot = √(Vx2 + Vy2)
Plugging in the values we get:
Vtot = √(22 + 32)
Vtot = √(4 + 9)
Vtot = √13
Vtot = 3.61 mph (approximately)
To find the direction of the resultant velocity, we can use the inverse tangent function (tan-1) to calculate the angle (θ) the resultant velocity vector makes with the positive x-axis (eastward direction).
θ = tan-1(Vy/Vx)
θ = tan-1(3/2)
θ = 56.31° (approximately)
Since the boat is heading north and the current flows east, the resultant velocity makes an angle of 56.31° north of east.
If the car is traveling at 83 km/h , will a friction force be required? If so, in what direction?
Answer:
a friction force that opposes its movement
All this force has a direction opposite to that of the car
Explanation:
A car traveling at 83 km / h with constant speed has a friction force that opposes its movement, this force comes from several sources:
.- The air resistance that is proportional to the speed
/ - the resistance between the rubbers and the pavement that is constant
.- The internal friction of the different engine components
.- The formation of eddies due to the lack of aerodynamic shape of the car
All these forces must be counteracted the motor force
All this force has a direction opposite to that of the car
Answer:
Yes, a frictional force would be required parallel to the bank in the direction downwards the bank to prevent the car from moving off the bank.
Question:
A 1050-kg car rounds a curve of radius 72 m banked at an angle of 14°. If the car is traveling at 83 kmh, will a friction force be required? If so, in what direction?
Explanation:
When a car is to move round a curve with a particular radius, the centrifugal force due to rounding the curve would act on it, which would try to push it off the bank.
As a result of the inclined bank, a force as a result of gravity will act on it to prevent it from moving off the bank.
Fc = mv^2/r
Fw = mgsin14
Where;
Fc is the centrifugal force.
m = mass
Fw = force as a result of weight.
v = velocity of car
r = radius of circular path
g = acceleration due to gravity.
Looking at the two forces, due to the high speed of the car, low angle of inclination of the bank and relatively low radius of round path, the centrifugal force will be high than the force resulting from the weight to keep the car on the track.
Fc > Fw
Fc is directed out of the bank
Fw is try to keep the car on track
Therefore, when Fc > Fw and without frictional force acting inward(down the bank) to neutralise the centrifugal force the car would move off the bank and out of the round curve track. So frictional force is needed in the direction downwards the bank to keep the car on track.
The terms intrusive and extrusive are used to describe which one of the three rock groups
Answer:
IGNEOUS ROCKS
Explanation: Igneous rocks are those rocks that solidify from magma.
Igneous rock is divided into two ,they are:
1. Intrusive
Igneous rocks crystallized belowearth"s crust. Its cooling material is called lava.
2 Extrusive igneous rock is also known as known as volcanic rocks
Two particles are located on the x axis of a Cartesian coordinate system Particle 1 carries a charge of 3.0 nC and is at x 40 mm and particle 2 carries a charge of 3.0 nC and is at x 40 mm Particle 3 which carries a charge of 9.0 μC is located on the positive y axis 90 mm from the origin What is the magnitude of the vector sum of the electric forces exerted on particle 3?
What is the direction angle of the vector sum of the electric forces exerted on particle 3 measured counterclockwise from the positive x axis?
The magnitude of the vector sum of the electric forces exerted on particle 3 is 0.598 N, with the direction angle being 90 degrees, as both forces act along the positive y-axis due to symmetry.
To solve the student's question, we'll have to calculate the forces exerted on particle 3 by particle 1 and particle 2, respectively, and then find the vector sum of these forces. Since both particle 1 and particle 2 are on the x-axis at the same distance from the origin and carry the same charge, the forces they exert on particle 3 will be equal in magnitude and opposite in direction along the y-axis. Therefore, the resulting force on particle 3 will be twice the force exerted by one particle along the y-axis.
Using the formula:
F = k × |q₁ × q₂| / r²
For particle 1 or 2 exerting force on particle 3:
F = (8.9875 × 10⁹) × (3.0 × 10⁻⁹ C × 9.0 × 10⁻⁶ C) / (0.09 m)²
F = 8.9875 × 10⁹ × 2.7 × 10⁻¹⁴ C / 0.0081 m²
F = 0.299 N (Force exerted by one particle)
The total force on particle 3 will be twice this value: 0.598 N.
The direction angle of the vector sum will be 90 degrees since both forces act along the y-axis and hence the resultant force is directed straight up along the positive y-axis when measured counterclockwise from the positive x-axis.
Magnitude is √2 N, Direction is 45° counterclockwise from positive x-axis.
Coulomb's Law: formula for electric force between point charges.
[tex]\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \][/tex]
[tex]\( q_1 = q_2 = 3.0 \, \text{nC} = 3.0 \times 10^{-9} \, \text{C} \),[/tex]
[tex]\( q_3 = 9.0 \, \text{μC} = 9.0 \times 10^{-6} \, \text{C} \),[/tex]
The distance between particle 3 and particles 1 and 2 is [tex]\( 90 \, \text{mm} = 0.09 \, \text{m} \)[/tex],
All charges are located on the x-axis.
The electric force exerted by particle 1 on particle 3:
[tex]\[ F_1 = \frac{k \cdot |q_1 \cdot q_3|}{r^2} \][/tex]
The electric force exerted by particle 2 on particle 3:
[tex]\[ F_2 = \frac{k \cdot |q_2 \cdot q_3|}{r^2} \][/tex]
Particles 1 and 2, equidistant from 3, have equal electric forces. The magnitude of each force:
[tex]\[ F_1 = \frac{(8.99 \times 10^9) \cdot |3.0 \times 10^{-9} \cdot 9.0 \times 10^{-6}|}{(0.09)^2} \][/tex]
[tex]\[ F_1 \approx \frac{8.09 \times 10^{-5}}{0.0081} \][/tex]
[tex]\[ F_1 \approx 1.00 \, \text{N} \][/tex]
Similarly, [tex]\( F_2 \approx 1.00 \, \text{N} \)[/tex].
Now, the total electric force [tex]\( F_{\text{total}} \)[/tex] exerted on particle 3 is the vector sum of [tex]\( F_1 \) and \( F_2 \)[/tex].
[tex]\[ F_{\text{total}} = \sqrt{F_1^2 + F_2^2} \][/tex]
[tex]\[ F_{\text{total}} = \sqrt{(1.00)^2 + (1.00)^2} \][/tex]
[tex]\[ F_{\text{total}} = \sqrt{2} \, \text{N} \][/tex]
Magnitude of total electric force on particle 3 is [tex]\( \sqrt{2} \, \text{N} \)[/tex].
[tex]\( \theta \)[/tex] is the angle from x-axis to [tex]\( F_{\text{total}} \)[/tex].
[tex]\[ \theta = \arctan\left(\frac{F_2}{F_1}\right) \][/tex]
[tex]\[ \theta = \arctan\left(\frac{1.00}{1.00}\right) \][/tex]
[tex]\[ \theta = \arctan(1) \][/tex]
[tex]\[ \theta \approx 45^\circ \][/tex]
Two 10-Hz, sine waves have a relative phase shift of 30 deg. What is the time difference between them? If the frequency of these sine waves doubles, but the time difference stays the same, what is the phase difference between them?
Answer:
The time difference is 8.33 ms.
The phase difference between them is 60°
Explanation:
Given that,
Frequency = 10 Hz
Angle = 30°
We need to calculate the time difference
Using formula of time difference
[tex]\Delta t=\dfrac{\phi}{360^{\circ}\times f}[/tex]
Put the value into the formula
[tex]\Delta t=\dfrac{30}{360\times10}[/tex]
[tex]\Delta t=8.33\ ms[/tex]
If the frequency of these sine waves doubles, but the time difference stays the same,
[tex]f=20\ Hz[/tex]
We need to calculate the phase difference between them
Using formula of phase difference
[tex]\Delta \phi=\Delta t\times360\times f[/tex]
Put the value in to the formula
[tex]\Delta=8.33\times10^{-3}\times360\times20[/tex]
[tex]\Delta \phi=60^{\circ}[/tex]
Hence, The time difference is 8.33 ms.
The phase difference between them is 60°
An object is undergoing SHM with period 0.300 s and amplitude 6.00 cm. At t = 0 the object is instantaneously at rest at x = 6.00 cm. Calculate the time it takes the object to go from x = 6.00 cm to x = -1.50 cm
Answer:
Time taken is 0.087 s
Solution:
As per the question:
Time period, T = 0.300 s
Amplitude, A = 6.00 cm
Now,
To calculate the time taken:
For SHM, we know that:
[tex]x = Acos\omega t[/tex] (1)
At x = 6.00 cm, the object comes to rest instantaneously at times t = 0.00 s
Thus from eqn (1), for x = 6.00 cm:
[tex]6.00 = 6.00cos\omega t[/tex]
[tex]cos\omega t = 1[/tex]
[tex]\omega t = cos^{- 1}(1)[/tex]
[tex]\omega t = 0[/tex]
Thus at t = 0.00 s, x = 6.00 cm
Now,
Using eqn (1) for x = - 1.50 cm:
[tex]- 1.50 = 6.00cos\omega t'[/tex]
[tex]cos\omega t' = -0.25[/tex]
We know that:
[tex]\omega = \frac{2\pi}{T}[/tex]
Thus
[tex]\frac{2\pi}{0.300} t' = cos^{- 1}(0.25)[/tex]
[tex]t' = 0.087\ s[/tex]
Time taken by the object in moving from x = 6.00 cm to x = 1.50 cm:
t' - t = 0.087 - 0.00 = 0.087 s
To find the time it takes for the object to go from x = 6.00 cm to x = -1.50 cm, we use the equation for simple harmonic motion. The time it takes is half of the period, which is 0.150 s.
Explanation:To find the time it takes for the object to go from x = 6.00 cm to x = -1.50 cm, we need to use the equation for simple harmonic motion (SHM). The equation is given by x = A * cos(2π/T * t + φ), where x is the position, A is the amplitude, T is the period, t is the time, and φ is the phase shift.
First, we need to determine the phase shift. At t = 0, the object is at rest at x = 6.00 cm. This means the phase shift is 0, because the cosine function is at a maximum at t = 0.
Next, we can plug in the values into the equation. The amplitude A is 6.00 cm and the period T is 0.300 s. We want to find the time it takes for the object to go from x = 6.00 cm to x = -1.50 cm. In SHM, the object goes from -A to A and back in one period, so the time it takes to go from x = 6.00 cm to x = -1.50 cm is half of the period. Therefore, the time is 0.300 s / 2 = 0.150 s.
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A block of 250-mm length and 54 × 40-mm cross section is to support a centric compressive load P. The material to be used is a bronze for which E = 95 GPa. Determine the largest load that can be applied, knowing that the normal stress must not exceed 80 MPa and that the decrease in length of the block should be at most 0.12% of its original length.
Answer:
P = 17.28*10⁶ N
Explanation:
Given
L = 250 mm = 0.25 m
a = 0.54 m
b = 0.40 m
E = 95 GPa = 95*10⁹ Pa
σmax = 80 MPa = 80*10⁶ Pa
ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m
We get A as follows:
A = a*b = (0.54 m)*(0.40 m) = 0.216 m²
then, we apply the formula
ΔL = P*L/(A*E) ⇒ P = ΔL*A*E/L
⇒ P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)
⇒ P = 24624000 N = 24.624*10⁶ N
Now we can use the equation
σ = P/A
⇒ σ = (24624000 N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa
So σ > σmax we use σmax
⇒ P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N
A charge of +2.00 x 10^-9 C is placed at the origin, and another charge of +4.50 x 10^-9 C is placed at x = 1.6 m. The Coulomb constant is 8.98755 x 10^9 N m2/C2. Find the point (coordinate) between these two charges where a charge of +3.70 x 10^-9 C should be placed so that the net electric force on it is zero.
Answer:
0.64 m from the first charge
Explanation:
Force is given by
[tex]F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F_1=\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}[/tex]
[tex]F_2=\dfrac{kq_2q_3}{r^2}\\\Rightarrow F_1=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}[/tex]
These forces are equal
[tex]\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{x^2}=\dfrac{4.5}{(1.6-x)^2}\\\Rightarrow \dfrac{2}{4.5}=\dfrac{x^2}{(1.6-x)^2}\\\Rightarrow \dfrac{4.5}{2}=\dfrac{(1.6-x)^2}{x^2}\\\Rightarrow \sqrt{\dfrac{4.5}{2}}=\dfrac{1.6-x}{x}\\\Rightarrow 1.5=\dfrac{1.6-x}{x}\\\Rightarrow 1.5x+x=1.6\\\Rightarrow x=\dfrac{1.6}{2.5}\\\Rightarrow x=0.64\ m[/tex]
The distance that charge should be placed is 0.64 m from the first charge
The charge +3.70 x [tex]10^{-9[/tex]C should be placed at approximately 0.96 meters from the origin for the net electric force to be zero. The solution involves balancing the Coulomb forces from two charges placed at different points. The calculations involve setting up the equation for the forces and solving it step-by-step.
Finding the Point Where the Net Electric Force is Zero
To determine the point where the charge +3.70 x [tex]10^{-9[/tex] C should be placed so that the net electric force on it is zero, we need to consider the forces exerted by both charges +2.00 x [tex]10^{-9[/tex] C (at the origin) and +4.50 x [tex]10^{-9[/tex] C (at x = 1.6 m).
Step-by-Step Explanation:
Let the position where the net force is zero be at distance x from the origin (charge +2.00 x [tex]10^{-9[/tex] C).
The distance from the charge +4.50 x [tex]10^{-9[/tex] C to this point will then be (1.6 - x) meters.
According to Coulomb's Law, the force due to a charge is given by F = k * |q1 * q2| / [tex]r^2[/tex], where k is the Coulomb constant (8.98755 x[tex]10^{9[/tex] N m²/C²).
For the net force on the charge +3.70 x [tex]10^{-9[/tex] C to be zero, the magnitudes of the forces due to the two other charges must be equal:
Force from +2.00 x [tex]10^{-9[/tex] C: F1 = k * |(2.00 x [tex]10^{-9[/tex]) * (3.70 x [tex]10^{-9[/tex])| / x²Force from +4.50 x [tex]10^{-9[/tex] C: F2 = k * |(4.50 x [tex]10^{-9[/tex] * (3.70 x [tex]10^{-9[/tex])| / (1.6 - x)²Set these forces equal to each other:
(8.98755 x[tex]10^{9[/tex] N m²/C²) * (2.00 x [tex]10^{-9[/tex]C) * (3.70 x [tex]10^{-9[/tex] C) / x² = (8.98755 x [tex]10^{9[/tex] N m²/C²) * (4.50 x [tex]10^{-9[/tex] C) * (3.70 x [tex]10^{-9[/tex] C) / (1.6 - x)²
Simplify and solve for x:
(2.00 x [tex]10^{-9[/tex]) / x² = (4.50 x [tex]10^{-9[/tex]) / (1.6 - x)²
2 / x² = 4.5 / (1.6 - x)²
x² / (1.6 - x)² = 2 / 4.5
x² / (1.6 - x)² = 0.444
To solve for x, take the square root of both sides:
x / (1.6 - x) = √0.444Finally, solve the equation x = √0.444 * (1.6 - x):
0.666x = 1.6 - xx(1 + 0.666) = 1.6x = 1.6 / 1.666x ≈ 0.96 metersThus, the charge +3.70 x [tex]10^{-9[/tex] C should be placed at approximately 0.96 meters from the origin to have the net electric force on it be zero.
After how many time constants has the voltage across a discharging capacitor decayed to 0.85{\rm \\%} of its initial value?
Answer:
After [tex]t=4.77\tau[/tex] voltage across the capacitor will be 0.85 % of the initial voltage across the capacitor
Explanation:
Let initially voltage across capacitor is [tex]v_0[/tex]
After discharging the voltage across the capacitor is .85% of its initial voltage
So final voltage [tex]v=0.0085v_0[/tex]
We know that voltage across capacitor is given by [tex]v=v_0e^{\frac{-t}{\tau }}[/tex]
So [tex]0.85v_0=v_0e^{\frac{-t}{\tau }}[/tex]
[tex]e^{\frac{-t}{\tau }}=0.0085[/tex]
[tex]{\frac{-t}{\tau }}=ln0.0085[/tex]
[tex]-t=-4.77\tau[/tex]
[tex]t=4.77\tau[/tex]
So after [tex]t=4.77\tau[/tex] voltage across the capacitor will be 0.85 % of the initial voltage across the capacitor
Time constant of capacitive circuit is equal to the ratio of resistance and capacitance
The number of time constants the voltage across a discharging cap has decayed to 0.85% of its' initial value is;
4.768τ
In RC time constants, the formula for voltage across capacitor when discharging is given by;
v = v₀e^(-t/τ)
where;
v₀ = initial voltage on the capacitor
v = the voltage after time t
t = time in seconds
τ = time constant
We are told that the capacitor decayed to 0.85% of its' initial vale. Thus;
v = 0.0085v₀
Thus;
0.0085v₀ = v₀e^(-t/τ)
v₀ will cancel out to give;
0.0085 = e^(-t/τ)
In 0.0085 = -t/τ
-4.768 = -t/τ
Since we want to find number of time constants, then let us make t the subject to get;
t = 4.768τ
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The legs of a weight lifter must ultimately support the weights he has lifted. A human tibia (shinbone) has a circular cross section of approximately 3.6 cm outer diameter and 2.30 cm inner diameter. (The hollow portion contains marrow.)If a 90.0 kg lifter stands on both legs, what is the heaviest weight he can lift without breaking his legs, assuming that the breaking stress of the bone is 150 MPa ?
To determine the heaviest weight a weight lifter can lift without breaking his legs, we need to calculate the stress on the bones. By using the dimensions of the tibia and the breaking stress of bone, we can calculate the maximum weight the lifter can lift.
Explanation:To determine the heaviest weight a weight lifter can lift without breaking his legs, we need to calculate the stress on the bones. Stress is defined as the force applied per unit area. In this case, we can calculate the stress on the tibia using the formula stress = force/area. The area of the cross section of the tibia can be found by subtracting the area of the inner circle from the area of the outer circle. Once we have the stress, we can use it to determine the maximum weight the lifter can lift without breaking his legs.
Using the given dimensions of the tibia, we can calculate the area and stress:
Outer radius = 3.6 cm / 2 = 1.8 cm = 0.018 m
Inner radius = 2.3 cm / 2 = 1.15 cm = 0.0115 m
Area of outer circle = π * (0.018 m)^2 = 0.001018 m^2
Area of inner circle = π * (0.0115 m)^2 = 0.000415 m^2
Area of tibia = Area of outer circle - Area of inner circle = 0.001018 m^2 - 0.000415 m^2 = 0.000603 m^2
Force on the tibia = weight lifted = mass * acceleration due to gravity = 90.0 kg * 9.8 m/s^2 = 882 N
Stress on the tibia = force/area = 882 N / 0.000603 m^2 = 1,460,070 Pa (or 1.46 MPa)
Therefore, the lifter can lift a maximum weight without breaking his legs if the stress on the tibia is less than or equal to the breaking stress of bone. In this case, the lifter can lift:
Maximum weight = breaking stress * area = 150 MPa * 0.000603 m^2 = 90.45 N
So, the lifter can lift a maximum weight of approximately 90.45 N without breaking his legs.
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A 1300 kgkg car drives around a flat 200-mm-diameter circular track at 40 m/sm/s . Part A What is the magnitude of the net force on the car? Express your answer to two significant figures and include the appropriate units. FnetFnet = nothingnothing
Answer:
[tex]F_C=20800\ N[/tex]
Explanation:
Given:
mass of the car, [tex]m=1300\ kg[/tex]diameter of the track, [tex]d=200\ m[/tex]speed of the car, [tex]v=40\ m.s^{-1}[/tex]During the motion of the car on a flat circular track there acts a force which pushes the car inwards to comply in the circular motion. This force is called centripetal force which is generated due to the frictional force between the road and the tyres of the car.
This centripetal force is given as:
[tex]F_C=m.\frac{v^2}{r}[/tex]
here: r = radius of the track = [tex]\frac{d}{2}[/tex]
[tex]F_C=1300\times \frac{40^2}{100}[/tex]
[tex]F_C=20800\ N[/tex] is the force which makes the car turn inwards going with the given uniform speed.
Final answer:
The net force on a 1300 kg car driving around a 200-m diameter track at 40 m/s is calculated using the centripetal force formula, resulting in a force of 20,800 N directed towards the center of the track.
Explanation:
The question asks to calculate the magnitude of the net force acting on a 1300 kg car driving around a 200-m diameter circular track at a speed of 40 m/s.
First, we convert the diameter to radius by dividing by 2, giving a radius (r) of 100 m. To find the net force (Fnet), we use the formula for centripetal force: Fnet = m * (v² / r), where m is mass, v is velocity, and r is radius.
Substituting the given values:
Mass (m) = 1300 kg
Velocity (v) = 40 m/s
Radius (r) = 100 m
This yields:
Fnet = 1300 kg * (40 m/s)² / 100 m
Calculating this, we get:
Fnet = 1300 kg * 1600 m²/s² / 100 m
Fnet = 1300 kg * 16 m/s²
The net force is 20,800 N directed towards the center of the circular track.
An electron is projected with an initial speed Vo = 5.35x10^6 m/s into the uniform field between the parallel plates. The direction of the field is vertically downward, and the field is zero except in the space between the two plates. The electron enters the field at a point midway between the plates.
Length of the plates= 2cm
Distance between the plates= 1cm
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Final answer:
To determine the magnitude of the electric field through which an electron narrowly misses the upper plate, apply principles of kinematics and the formula for electric force. Calculate the time of flight using the horizontal motion, and use this to establish the vertical acceleration due to electric force. From the electron's charge and acceleration, derive the electric field magnitude.
Explanation:
To calculate the magnitude of the electric field through which an electron passes and just misses the upper plate upon exit, we need to apply the concepts of kinematics and electric forces. Since the electrical force is the only vertical force acting on the electron, it will cause a vertical acceleration according to Newton's second law (F = ma), where the force F is the electric force (F = qE) and the acceleration a is due to this force.
The vertical distance traveled by the electron just before it exits the field is equal to half the distance between the plates, which is 0.5 cm (or 0.005 m). The time it takes for the electron to travel this vertical distance can be calculated by using the initial horizontal speed and the horizontal distance (2 cm or 0.02 m) it travels. Using the formula t = d/v (where d is the horizontal distance and v is the horizontal velocity), we find t = 0.02 m / 5.35×10¶6 m/s = 3.74×10¶8 seconds.
The vertical acceleration a can be found by using the vertical distance (s), initial vertical speed (u which is 0 in this case), and time (t), using the equation s = ut + 0.5at². From this, we find a = 2s/t². Knowing the charge of an electron (e = 1.6×10¶19 C) and its mass (m = 9.11×10¶31 kg), the electric field E can then be calculated as E = F/q = ma/q. Plugging in the values, we get the magnitude of the electric field.
Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsice force can be great enough to support the top pieces weight. Assuming equal point charges, calculate the magnitide of the charge if electrostatic force is great enough to support the weight of a 11.0 mg piece of tape held 1.00 cm above another.
Answer:
Q = 1.095 x 10^-9 C
Let the force experienced by the top piece of tape be F
F = kQ²/r²
r = distance between the two pieces tape = 1.00cm = 1.00 x 10^ -2 m
1/4(pi)*Eo = k = 8.99 x 10^9 Nm²/C²
The electric force of repulsion between the two charges and the weight of the top piece of tape are equal so
F = kQ²/r² = mg
Where m is the mass of the top piece of tape and g is the acceleration due to gravity
On re-arranging the equation above,
Q² = mgr²/k
Q² = ((11.0 x 10^-6) x 9.8 x (1.00x10^-2)²)/(8.99 x 10^9)
Q = 1.095x10^-9 C
Explanation:
The charge Q on both pieces of tape are equal and both act with a force of repulsion on each other.
The force of repulsion between both tapes pushes the top piece of tape upwards. The weight of the top piece of tape acts vertically downward. Since the top tape is in a position of equilibrium, the two forces acting on the top piece of tape must be equal to each other. This assumption is backed up by newton's first law of motion which states that the summation of all forces acting on a body at rest must be equal to zero. That is
Fe (electric force) - Fg (gravitational force) = 0
Fe = Fg
kQ²/r² = mg
On substituting the respective values for all variables except Q and rearranging the equation Q = 1.09 x 10^-9
Clearly cellulose is very abundant on earth, and it is a long-lasting stable substance. Many animals cannot digest cellulose. Given this, what prevents the bodies of dead plants from filling the earth?
Answer:The process that prevents the bodies of dead plant from filling the earth is decomposition of their remains.
Explanation:
1. Fungi and bacteria in the soil help decompose the remains of dead plant and uses them to enrich and nourish other plants that are alive.
2. Insects, worms and other invertebrates feed on the too
Mustang Sally just finished restoring her 1965 Ford Mustang car. To save money, she did not get a new battery. When she tries to start the car, she discovers that the battery is dead and she needs a jump start. While unhooking the jumper cables, the positive and negative cables almost touch and a spark jumps between the ends of the cables. This spark is caused by the movement of electrons through the air between the battery terminals. In what direction are the electrons traveling?
Answer:
the electron jump goes from the negative terminal to the positive terminal
Explanation:
In a battery the energy is produced by a chemical reaction, which accumulates electrons in a terminal, called a negative terminal, but by convention the current goes from the positive to the negative terminal.
We can see from this explanation that the negative wire is the one with an excess is electrons, so the electron jump goes from the negative terminal to the positive terminal
A charge of -3.50 nC is placed at the origin of an xy-coordinate system, and a charge of 1.95 nC is placed on the y axis at y = 3.75 cm .Find the distances to the center.
Answer:
3.72 cm
Explanation:
the coordiante of the negative charge =( 0,0)
the coordinate of the positive charge = (0,3.75)
the distance simply is 3.72-0 = 3.72 cm