Answer:
1) The mass of the continent is approximately [tex]1.608\times 10^{21}[/tex] kilograms.
2) The kinetic energy of the continent is approximately [tex]8.04\times 10^{16}[/tex] joules.
3) The speed of the 77 kg-jogger would be approximately [tex]45.698\times 10^{6}[/tex] meters per second.
Explanation:
1) The mass of the North American continent can be estimated by using the following formula under the assumption that rock has an uniform density:
[tex]m = \rho \cdot L^{2}\cdot h[/tex] (1)
Where:
[tex]m[/tex] - Mass of the continent, measured in kilograms.
[tex]\rho[/tex] - Average density of the rock, measured in kilograms per cubic meter.
[tex]L[/tex] - Side of the continent, measured in meters.
[tex]h[/tex] - Depth of the continent, measured in meters.
If we know that [tex]\rho = 2620\,\frac{kg}{m^{3}}[/tex], [tex]L = 4.450\times 10^{6}\,m[/tex] and [tex]h = 31\times 10^{3}\,m[/tex], then the mass of the continent is:
[tex]m = \left(2620\,\frac{kg}{m^{3}} \right)\cdot (4.450\times 10^{6}\,m)^{2}\cdot (31\times 10^{3}\,m)[/tex]
[tex]m = 1.608\times 10^{21}\,kg[/tex]
The mass of the continent is approximately [tex]1.608\times 10^{21}[/tex] kilograms.
2) By assuming that continent can be represented as a particle, we define its kinetic energy as:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
Where:
[tex]K[/tex] - Translational kinetic energy, measured in joules.
[tex]v[/tex] - Motion rate of the continent, measured in meters per second.
If we know that [tex]m = 1.608\times 10^{21}\,kg[/tex] and [tex]v = 1\times 10^{-2}\,\frac{m}{s}[/tex], then the kinetic energy of the continent is:
[tex]K = \frac{1}{2}\cdot (1.608\times 10^{21}\,kg)\cdot \left(1\times 10^{-2}\,\frac{m}{s} \right)^{2}[/tex]
[tex]K = 8.04\times 10^{16}\,J[/tex]
The kinetic energy of the continent is approximately [tex]8.04\times 10^{16}[/tex] joules.
3) The speed of the jogger is derived from the definition of translational kinetic energy:
[tex]v = \sqrt{\frac{2\cdot K}{m} }[/tex]
If we know that [tex]K = 8.04\times 10^{16}\,J[/tex] and [tex]m = 77\,kg[/tex], then the expected speed of the jogger is:
[tex]v = \sqrt{\frac{2\cdot (8.04\times 10^{16}\,J)}{77\,kg} }[/tex]
[tex]v\approx 45.698\times 10^{6}\,\frac{m}{s}[/tex]
The speed of the 77 kg-jogger would be approximately [tex]45.698\times 10^{6}[/tex] meters per second.
What is the main reason why the age of the oldest rocks can vary from one part of the continent to another?
Answer:
The options are
A. Older rocks are commonly remitted over huge regions
B. Older rocks have been uplifted and eroded away
C. Large parts of the continent are subducted deep within the mantle
D. Parts of the continent have been added by the accretion of tectonic terraces.
The answer is D. Parts of the continent have been added by the accretion of tectonic terraces.
The major reason why the age of the oldest rocks can vary from one part of the continent to another is that parts of continent have been added by the accretion of tectonic terraces.
The CEO, ellen misk, left her martian office but accidentally left a cylindricall can of coke (3.1 inches in diameter, 5.42 inches in height) on her desk. If the can exerts a pressure of 510 Pascals, what is the specific gravity of the can?
Answer:
Specific Gravity = 0.378
Explanation:
First, we will find the force exerted by the can on the table. This force will be equal to the weight of the can:
Pressure = Force/Area = Weight/Area
Weight = Pressure*Area
where,
Area = πdiameter²/4 = π[(3.1 in)(0.0254 m/1 in)]²/4 = 4.8 x 10⁻³ m²
Weight = (510 N/m²)(4.8 x 10⁻³ m²)
Weight = 2.48 N
Now, the weight is given as:
Weight = mg
2.48 N = m(9.8 m/s²)
m = (2.48 N)/(9.8 m/s²)
m = 0.25 kg
Now, we calculate volume of can:
Volume = (Area)(Height) = (4.8 x 10⁻³ m²)(5.42 in)(0.0254 m/1 in)
Volume = 6.6 x 10⁻⁴ m³
Hence, the density of can will be:
Density of Can = m/Volume = 0.25 kg/6.6 x 10⁻⁴ m³
Density of Can = 378.32 kg/m³
So, the specific gravity of Can will be:
Specific Gravity = Density of Can/Density of Water
Specific Gravity = (378.32 kg/m³)/(1000 kg/m³)
Specific Gravity = 0.378
What is Solar Energy used for? Use in your own words.
Answer:
TO POWER ELECTRIC GADGETS AND SAVE EARTH FROM POLLUTION
Explanation:
A wagon with a weight of 300 N is accelerated across a level surface at 0.5 m/s2. What net force acts on the wagon?
9.0 N
15 N
150 N
610 N
Answer:
F = 15.29[m/s²].
Explanation:
To solve this problem we must use Newton's second law which tells us that the force is equal to the product of mass by acceleration.
ΣF =m*a
where:
m = mass [kg]
a = acceleration = 0.5 [m/s²]
Therefore we must find first the mass.
[tex]W =m*g\\m=W/g\\m = 300/9.81\\m = 30.58 [kg][/tex]
Now replacing in Newton's second law.
[tex]F = m*a\\F = 30.58*0.5\\F=15.29[m/s^{2}][/tex]
Is a parked car potential or kinetic ?
Answer:
Potential energy is the energy that is stored in an object. ... When you park your car at the top of a hill, your car has potential energy because the gravity is pulling your car to move downward; if your car's parking brake fails, your vehicle may roll down the hill because of the force of gravity.
A 20.0-kg uniform plank (10.0 m long) is supported by the floor at one end and by a vertical rope at the other as shown in the figure. A 50.0-kg mass person stands on the plank a distance three-fourths of the length plank from the end on the floor.
Answer:
Tension= 475N
Force= 225N
Explanation:
The question is not complete, here is the complete question
Also, see attached a free body diagram for your reference
"A 20.0-kg uniform plank is supported by the floor at one end by a vertical rope at the other as shown in the figure. A 50.0-kg mass person stands on the plank a distance three-fourths of the length plank from the end on the floor.
a. What is the tension in the rope?
b. What is the magnitude of the force that the floor exerts on the plank?"
given data
mass of man=50kg
mass of plank=20kg
length of plank=10m
let us make the lenght of the rope be d
The torque about the floor
That is taking moment about the floor
[tex]N*0+T*d=20*10*d/2 + 50*10*3d/4\\\\T=100+375=475N\\\\[/tex]
Force will be also zero
[tex]N+T=20*10+50*10\\\\N+T=700 \\\\N=700-475=225 newtons\\\\N+T=20*10+50*10\\\\N+T=700\\\\N=700-475=225newtons[/tex]
Find the net work done by friction on the body of a snake slithering in a complete circle of 3.93 m radius. The coefficient of friction between the ground and the snake is 0.25, and the snake's weight is 54.0 N.
Answer:
The net work done by friction on the body of the snake is 333.35J
Explanation:
Work done is given by
W = F × s
Where W is the Work done
F is the force
and s is the distance covered
Since we are to determine the work done by friction, then we will determine the frictional force. The frictional force is given by
f = μN = μw
Where μ is the coefficient of friction
N is the normal reaction
and w is the weight
But, F = f
∴ W = μws
From the question
μ = 0.25
w = 54.0 N
Now, we will determine s
From the question,
We are to determine the work done by friction on the body of a snake slithering in a complete circle of 3.93 m radius.
The distance s here is given by the circumference of the circle. Circumference of a circle is given by 2πr
∴ s = 2πr
s = 2 × π × 3.93
s = 7.86π m
Hence,
W = 0.25 × 54.0 × 7.86π
W = 333.35 J
Hence, the net work done by friction on the body of the snake is 333.35J.
The net work done by friction on the body of the snake is :
-333.35J
FrictionFormulas:
Work done (W) = F × s
F = force
s = distance covered
f = μN = μw
μ = coefficient of friction
N = normal reaction
w = weight
Solution:
F = f
Weight is :
W = μws
μ = 0.25
w = 54.0 N
Distance covered:
s = 2πr
s = 2 × π × 3.93
s = 7.86π m
Therefore,
W = 0.25 × 54.0 × 7.86π
W = 333.35 J
The net work done by friction on the body of the snake is 333.35J.
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A circular conducting loop with a radius of 1.00 m and a small gap filled with a 10.0 Ω resistor is oriented in the xy-plane. If a magnetic field of 2.0 T, making an angle of 30º with the z-axis, increases to 11.0 T, in 2.5 s, what is the magnitude of the current that will be caused to flow in the conductor?
Answer:
ill get back to this question once i find the answer to it
SI unit differ from one country to another . true or false
Answer:
false ..........false
Answer:
FALSE
Explanation:
A car covers a distance of 200m. If its velocity is 20 m/s, calculate the time taken.
Answer:
10 seconds
Explanation:
velocity=displacement/timeA student kicks a soccer ball upward at a 30º angle with an initial speed of 20 m∕s. What expression should the student use to calculate the magnitude of the ball’s initial velocity in the horizontal direction?
Answer:
[tex]\displaystyle x=10\sqrt{3}\ m/s[/tex]
[tex]y=10\ m/s[/tex]
Explanation:
Rectangular coordinates of vectors in 2D
Given a vector with a magnitude v and angle θ with respect to the positive horizontal direction, the x and y components of the vector are given by:
[tex]x=v\cos\theta[/tex]
[tex]y=v\sin\theta[/tex]
The soccer ball is kicked upward at an angle θ = 30° and at a speed v=20 m/s.
The rectangular components of the vector are:
[tex]x=20\cos 30^\circ[/tex]
[tex]\displaystyle x=20\cdot \frac{\sqrt{3}}{2}[/tex]
Operating:
[tex]\mathbf{\displaystyle x=10\sqrt{3}\ m/s}[/tex]
[tex]y=20\sin 30^\circ[/tex]
[tex]\displaystyle y=20\cdot \frac{1}{2}[/tex]
Operating:
[tex]\mathbf{y=10\ m/s}[/tex]
If I travel 300 m east, then 400 m west, what is my distance &
displacement?
Answer:100m west
Explanation:
A ball is thrown vertically upward with an initial velocity of 23 m/s. What are its position and velocity after 2 s?
Answer:
The position of the ball after 2 s is 26.4 mThe velocity of the ball after 2 s is 3.4 m/sExplanation:
Given;
initial velocity of the ball, u = 23 m/s
time of motion, t = 2 s
The position of the ball after 2 s is given by;
h = ut - ¹/₂gt²
h = (23 x 2) - ¹/₂ x 9.8 x 2²
h = 46 - 19.6
h = 26.4 m
The velocity of the ball after 2 s is given by;
v² = u² + 2(-g)h
v² = u² - 2gh
v² = 23² - (2 x 9.8 x 26.4)
v² = 529 - 517.44
v² = 11.56
v = √11.56
v = 3.4 m/s
The earth's radius is 6.37×106m; it rotates once every 24 hours.What is the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator? (Hint: what is the radius of the circle in which the point moves?) Express your answer to two significant figures and include the appropriate units.
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
Match the term to its correct definition:
____ 1. Test variable (independent variable)
A. This is what is being measured during a scientific investigation.
____ 2. Outcome variable (dependent variable)
B. This serves as a reference for comparison during a scientific investigation.
____ 3. Control group
C. This is what is being purposefully changed during the scientific investigation.
Answer:
1. C
2. B
3. A
Explanation:
Hope this helped!
5.List the four goals of Psychology. Give your own example for each one using a behavior
Answer:
describe, explain, predict, and change/control behavior.
Explanation:
describe: What are they doing? -Pavlov noticed that dogs were salivating when they would see his lab assistant before food was presented to them. This observation acted as a description of what was happening to them.
explain: Why are they doing that?- Pavlov started to look into why they were doing it. There was a stimulus, the assistant giving them food in the past to where they started to salivate at the sight of the lab assistant
predict: What would happen if I responded in this way?- Pavlov predicted that he could get the same reaction if he used a bell as a stimulus. Using this he was able to condition dogs to salivate at the ring of the bell.
change/control: What can I do to get them to stop doing that? Because of this discovery we can use conditioning today. For example, in the classroom teachers can use conditioning with their students to make it easier, parents to teach their children right from wrong and to have good behavior. (you do this bad thing you get time out, do a good thing and I will praise you, etc) It can be used when training employees and many other places.
If you start at a speed of 4m/s and slow down to 2m/s in 4s what is your
acceleration?
Answer:
penis
Explanation:
student measures the weight of a bag of bananas with a spring balance.
Describe what is inside a spring balance and explain how it works.
A spring balance measures the weight of an object by opposing the force of gravity acting with force of an extending spring. May be used to determine mass as well as weight by recalibrating the scale. Some spring balances are available in gram or kilogram markings and are used to measure the mass of an object. Spring balances consist of a cylindrical tube with a spring inside. One end (at the top) is fixed to an adjuster which can be used to calibrate the device. The other end is attached to a hook on which you can hang masses etc.
The x and y coordinates of a particle at any time t are x = 5t - 3t2 and y = 5t respectively, where x and y are in meter and t in second. The speed of the particle at t = 1 second is
Answer:
[tex]v=\sqrt{26}~m/s[/tex]
Explanation:
Parametric Equation of the Velocity
Given the position of the particle at any time t as
[tex]r(t) = (x(t),y(t))[/tex]
The instantaneous velocity is the first derivative of the position:
[tex]v(t)=(v_x(t),v_y(t))=(x'(t),y'(t))[/tex]
The speed can be calculated as the magnitude of the velocity:
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
We are given the coordinates of the position of a particle as:
[tex]x=5t-3t^2[/tex]
[tex]y=5t[/tex]
The coordinates of the velocity are:
[tex]v_x(t)=(5t-3t^2)'=5-6t[/tex]
[tex]v_y(t)=(5t)'=5[/tex]
Evaluating at t=1 s:
[tex]v_x(1)=5-6(1)=-1[/tex]
[tex]v_y(1)=5[/tex]
The velocity is:
[tex]v=\sqrt{(-1)^2+5^2}[/tex]
[tex]v=\sqrt{1+25}[/tex]
[tex]\mathbf{v=\sqrt{26}~m/s}[/tex]
A pulley is in the form of a uniform solid cylinder of radius 7cm and mass 2kg. One end of a very light rope is fixed to wind the pulley and the other end is
free. When we pull the free end of the rope the pulley starts rotating from rest and accelerates uniformly. If the angular acceleration is 100rad/s2 so the
constant force that we exert on the pulley through the rope is:
Chọn một:
a.
200 N
b.
7N
C
0.49 N
d.
49 N
Answer:
correct is b 7N
Explanation:
The torque is
Στ = I α
torque
τ = F x R
bold, indicate vectors. The magnitude of torque is
τ = F R sin θ
in this case the angle is 90º so sin 90 = 1
τ = F R
The moment of inertia of a cylinder
I = ½ M R²
substitute
F R = ½ M R² α
F = ½ M R α
reduce to the SI system
R = 7 cm (1m / 100cm) = 0.07 m
calculate
F = ½ 2 0.07 100
F = 7 N
A jogger runs north 6 km, 5 km east, and 4 km north again. Time is 1.9 hr. What is average velocity?
Answer:
Average velocity = 5.9 km/ hr (or 1.64 m/s).
Explanation:
Velocity = [tex]\frac{displacement}{time}[/tex]
It is a vector quantity and has an SI unit of m/s.
Displacement = [tex]\sqrt{x^{2} + y^{2} }[/tex]
x = 5 km east
y = 6 km north + 4 km north
= 10 km north
So that,
Displacement = [tex]\sqrt{5^{2} + 10^{2} }[/tex]
= [tex]\sqrt{25 + 100}[/tex]
= [tex]\sqrt{125}[/tex]
= 11.1803
Displacement = 11.18 km = 11180 m
Time = 1.9 hr = 6840 seconds
Average velocity = [tex]\frac{11.180}{1.9}[/tex]
= 5.8842
Average velocity = 5.9 km/ hr
The average velocity is 5.9 km/ hr (or 1.64 m/s).
An ideal gas expands quasi-statically and isothermally from a state with pressurepand volumeVto a state with volume 4V. How much heat is added to the expanding gas?
Answer:
Q = PV(In 4)
Explanation:
We are told that the volume expands from V to a state with volume 4V.
Thus, initial volume is V and Final volume is 4V.
We want to find How much heat is added to the expanding gas.
For an isothermal process, the work done is calculated from;
W = nRT(In(V_f/V_i))
Where;
V_f is final volume
V_i is initial volume
Thus;
W = nRT(In(4V/V))
W = nRT(In 4)
Now, from ideal gas equation, we know that;
PV = nRT
Thus;
W = PV(In 4)
Now from first law of thermodynamics, we know that internal energy is zero and thus; Q = W
Where Q is quantity of heat
Thus;
Q = PV(In 4)
Which phrase desenbes an irregular galaxy ?
has a round shape
contains many young stars
has arms that extend from the center
Is larger than other types of galaxies
Answer:
contains many young stars
Explanation:
Irregular galaxies have no definite shape, which means that the first option is incorrect. They are definitely not round.
However, they contain many young stars because the degree of star formation is fast. They also contain old stars. Thus, the second choice is correct.
The "spiral galaxy" is the type of galaxy that has arms that extend from the center. These arms look "spiral," which influenced its name. This makes the last choice incorrect.
They are actually smaller than the other types of galaxies. This makes them prone to collisions. This makes the last choice incorrect.
Answer:
Contains many young stars
Explanation:
Convert 451 milliliters to fluid
ounces. Round your answer to 2
decimal places. **There are 29.57
milliliters in 1 fluid ounce***
Answer:
451 milliliters equals 15.25 fluid ounces
Explanation:
The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three.
To solve a direct rule of three, the following formula must be followed:
a ⇒ b
c ⇒ x
So: [tex]x=\frac{c*b}{a}[/tex]
The direct rule of three is the rule applied in this case where there is a change of units.
In this case, the rule of three can be applied in the following way: if there are 29.57 milliliters in 1 fluid ounce, in 451 milliliters how many fluid ounces are there?
[tex]fluid ounces=\frac{451 mL*1 fluid ounce}{29.57 mL}[/tex]
fluid ounces= 15.25
451 milliliters equals 15.25 fluid ounces
A sealed cubical container 28.0 cm on a side contains three times Avogadro's number of molecules at a temperature of 24.0°C. Find the force exerted by the gas on one of the walls of the container.
Answer:
3.32 atm
__________________________________________________________
We are given:
side of the cubical container = 28 cm
number of molecules in the container = 3 * Nₐ
[where Nₐ is the Avogadro's number]
Temperature = 24°C OR 297 K
We need to find the pressure exerted by the gas on the walls of the container
__________________________________________________________
Some Calculations:
Volume of the container
we are given the side of the cubical container = 28 cm
Volume of the cubical container = side³
Volume = 28³
Volume = 21952 cm³
We know that 1 cm³ = 1 mL
So,
Volume = 21952 mL
We also know that 1 L = 1000 mL
Volume = 21.952 L
Number of moles of Gas
We know that:
number of moles = number of molecules / Avogadro's number
number of moles = 3 * Nₐ / Nₐ [number of molecules = 3 * Nₐ]
number of moles = 3 moles
__________________________________________________________
Pressure Exerted by the Gas:
Using the ideal gas equation:
PV = nRT
Since the volume is in L, and Temperature is in K. R is equal to
0.082 L atm /mol K and the pressure will be in atm
P(21.952) = 3*(0.082)*(297)
P = 3.32 atm
Hence, the gas will exert a pressure of 3.32 atm on the walls of the container
How does increasing the width of a wire affect a circuit?
A. It restricts the flow of electrons.
B. It reduces the voltage
C. It allows electrons to flow more easily
D. It increases the resistance
Whoever gets this right I’ll give brainliest. Be sure that the answer is right. I’d love a explanation too if you could include one.
Answer:
The resistance of a wire decreases with increasing thickness.
Explanation:
Hope this helped!
Answer: C it allows electrons to flow more easily
Explanation:i got it right i hope this helps you
A cheetah can maintain a maximum constant velocity of 34.2 m/s for 8.70 s. What is
the displacement the cheetah covered at that velocity?
Answer:
297.54mExplanation:
step one:
given data
velocity v=34.2m/s
time t= 8.7s
Step two
Required is the distance the cheetah has covered on the condition
we know that speed= distance/time
make distance subject of formula we have
distance= velocity *time
distance= 34.2*8.7
distance = 297.54m
Therefore the displacement the cheetah covered at that velocity
is 297.54m
if on the average every man lives for 70 years, for how many microseconds is this life span
Answer:
3.1556926 × 10^13 microseconds
Explanation:
Hope this helped!
if on the average every man lives for 70 years, 3.1556926 × 10^13 microseconds is this life span
what is lifespan ?life span of an organism can be defined as the period of time between the birth and death of an organism, it is a common place that all organisms die.
Some of the organism die after a brief existence, for example mayfly, whose adult life burns out in a day, and the gnarled bristlecone pines, which have lived thousands of years.
The limits of the life span of each species appear to be determined by heredity, it can be locked within the code of the genetic material which are the instructions specify the age.
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An airtight box, having a lid of area 80.0 cm^2, is partially evacuated. Atmospheric pressure is 1.01 Times 10^5 Pa. A force of 108 lb is required to pull the lid off the box. The pressure in the box was:_________.
Answer:
5×10^4Pa
Explanation:
Given force of 108 lb is required to pull the lid off the box,
To convert "Ib"to Newton ,we use conversation rate below
1 pounds = 4.4482216282509 newtons
Then 108 lb=x Newton
Cross multiply we have
X= 480.41Newton
The force that is needed to open the lid is F and pressure P.
We know that Pressure= Force/Area
Area is given as 80.0 cm^2, we can convert to m^2 for unit consistency since 1cm^2= 0.001m^2 then
80.0 cm^2 = 80×10^-4m^2
Substitute to the equation of the pressure we have
P= 480.41Newton/(80×10^4m^2)
P=6×10^4 Pa
The pressure in the box will be difference between the initial pressure and final pressure
=( 1.01 ×10^5 Pa)-(6×10^4 Pa)
= 50100Pa
= 5×10^4Pa
Therefore, The pressure in the box was
5×10^4Pa
A force of 150 N is applied on an object at 60 degrees above the positive x-axis. Determine its
horizontal and vertical components.
Answer:
horizontal component=fcostita
=150cos60
use calculator to evaluate it
for vertical=fsintita
=150sin60