Answer:
The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.
Explanation:
We can determine the final velocity of the skier at the bottom of the ramp by Principle of Energy Conservation and Work-Energy Theorem, whose model is:
[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+W_{disp}[/tex] (1)
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.
[tex]W_{disp}[/tex] - Work dissipated by friction, measured in joules.
By definitions of gravitational potential and translational kinetic energy and work, we expand and simplify the model:
[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2}) =\mu_{k}\cdot N\cdot \Delta s[/tex] (2)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the skier, measured in meters.
[tex]N[/tex] - Normal force from the incline on the skier, measured in newtons.
[tex]\Delta s[/tex] - Distance covered by the skier, measured in meters.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.
The normal force exerted on the skier and the covered distance are, respectively:
[tex]N = m\cdot g\cdot \cos \theta[/tex] (3)
[tex]\Delta s = \frac{z_{1}-z_{2}}{\sin \theta}[/tex] (4)
Where [tex]\theta[/tex] is the angle of the incline above the horizontal, measured in sexagesimal degrees.
By applying (3) and (4) in (2), we get that:
[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot m\cdot g \cdot \cos \theta \cdot \left(\frac{z_{1}-z_{2}}{\sin \theta} \right)[/tex]
[tex]g\cdot (z_{1}-z_{2}) +\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})= \mu_{k}\cdot g \cdot \left(\frac{z_{1}-z_{2}}{\tan \theta} \right)[/tex] (5)
Then, we clear the velocity of the skier at the bottom of the ramp is: ([tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.1[/tex], [tex]\theta = 40^{\circ}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}-z_{2} = 40\,m[/tex])
[tex]\left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = \frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex]
[tex]2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = v_{1}^{2}-v_{2}^{2}[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}-2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2})}[/tex] (6)
[tex]v_{2} = \sqrt{\left(0\,\frac{m}{s} \right)^{2}-2\cdot \left(\frac{0.1}{\tan 40^{\circ}} -1\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m)}[/tex]
[tex]v_{2} \approx 26.288\,\frac{m}{s}[/tex]
The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.
An echo is sound that returns to you after being reflected from a distant surface (e.g., the side of a cliff). Assuming that the distances involved are the same, an echo under water and an echo in air return to you _____________________
a. at different times, the echo under water returning more slowly.
b. at different times, the echo under water returning more quickly.
c. at the same time
Answer:
The correct answer is B
An echo underwater and an echo in the air will return at different times. The echo underwater will return more quickly than the echo in the air.
Explanation:
The physics of this is simple.
Water and air are both made up of particles. The particles for water are more closely or densely arranged that those of the air molecules. Hence sound travels faster in water than in air. When measured, the speed actually differs by as much as 5 times with water being the fastest medium.
Think of it like this. Assume you have two stacks of dominoes, one closely packed than the other but exactly the same amount of dominos, you'd notice that the stack that is more tightly arranged will be the first to topple over because it takes less time for the kinetic energy from the first domino to reach the next and on and on like that until the last domino.
Cheers
Answer:
b. at different times, the echo under water returning more quickly.
a scuba diver in a pool looks at his instructor. the angle between the ray in the water and the perpendicular to the water is 25. find the height of the instructors head
Answer:
y= y' 0.67
Explanation:
This is an exercise in refraction of light,
n₁ sin θ₁ = n₂ sin θ₂
where subscript 1 is used for the incident medium and subscript 2 for the refracted medium
sin θ₁ = n2 /n1 sin θ₂
the incident medium is air with refractive index n1 = 1 and the medium where the ray is refracted is water with n = 1.33
let's calculate
sin θ₁ = 1.33 / 1 sin 25
θ₁ = sin⁻¹ (0.56208)
θ₁ = 34.2º
when the ray is refracted we can assume that the adjacent leg (water surface) is the same for the two media
let's use the trigonometry relationship
tan θ₁ = x / y
tan θ₂ = x / y '
tan θ₁ = y’ tan θ₂
y = y ’ tanθ₂ / tan θ₁
to finish exercise you must know the depth of the object
y =y' tan 25/ tan 34.2
y= y' 0.67
I went for a walk one day. I walked north 6.0 km at 6.0 km/h and then west 10 km at 5.0 km/hr.
Determine the total distance of the entire trip
Answer:
The total distance of the entire trip is 16 Km
Explanation:
Distance and Displacement
A moving object constantly travels for some distance at certain periods of time. The total distance is the sum of each individual distance the object traveled. It can be written as:
dtotal = d1 + d2 + d3 + ... + dn
This sum is obtained independently of the direction the object moves.
The displacement only takes into consideration the initial and final positions of the path defined by the object in its moving. The displacement, unlike distance, is a vectorial magnitude and can be even zero if the object starts and ends the movement at the same point.
The described movements are 6 Km North and 10 Km West. Regardless of the speed or time taken, the total distance is:
d = 6 Km + 10 Km = 16 Km
The total distance of the entire trip is 16 Km
If a rock is dropped from the top of a tower at the front of and it takes 3.6 seconds to hit the ground. Calculate the final velocity of the penny in m/s.
Answer:
36 m/s
Explanation:
t = 3.6s
u = 0m/s
a = +g = 10m/s²
v = ?
using,
v = u + at
v = 0 + 10(3.6)
v = 36 m/s
What is Solar Energy used for? Use in your own words.
Answer:
TO POWER ELECTRIC GADGETS AND SAVE EARTH FROM POLLUTION
Explanation:
A block slides down an inclined plane. Which force does zero work?
a. normal force
b. friction
c. the weight
d. all of the above
e. none of the above
Answer:
e
Explanation:
cause you have any attachments are not sure if you can get it
When a block slides down an inclined plane, normal force does zero work. Hence, option (a) is correct.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity.
An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.
The component of a contact force in mechanics known as the normal force is perpendicular to the surface that an item encounters.
As normal force always acts perpendicular to the motion, this force does zero work.
But friction force the weight acts in opposite direction of the motion, this force does work against the motion.
In an inclined plane, the parallel component of weight does work along the motion.
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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
Two kids are roller skating. Amy, with a mass of 55 kg, is traveling forward at 3 m/s. Jenny, who has a mass of 40 kg, is traveling in the opposite direction at 5 m/s. They crash into each other and hold onto each other so that they move as one mass. How fast are they traveling?
Answer:
-7/19
Explanation:
A Ferris wheel with 60 spokes has a diameter of 100 m. It makes one rotation every 60 seconds. Find the speed of the passengers when the Ferris wheel is rotating at this rate.
A. 5.24 m/s
B. 6.23 m/s
C. 7.45 m/s
D. 8.01 m/s
Answer:
v = 5.24 m/s
Explanation:
Given that,
No of spokes of a Ferris wheel = 60
The diameter of a wheel, d = 100 m
Radius, r = 50 m
It makes one rotation every 60 seconds.
We need to find the speed of the passengers when the Ferris wheel is rotating at this rate. Let it is v. It can be calculated as follows :
[tex]v=r\omega\\\\=\dfrac{2\pi r }{T}\\\\=\dfrac{2\pi \times 50 }{60}\\\\v=5.235\ m/s[/tex]
or
v = 5.24 m/s
So, the speed of the passengers is 5.24 m/s. Hence, the correct option is (A).
Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.0×10^4m/s when at a distance of 2.5×10^11m from the center of the sun, what is its speed when at a distance of 5.0×10^10?
hi ilove you jk it stoo stsfad
. A crane does 62,500 joules of work to lift a boulder a distance of 25.0 meters. How much did the boulder
weigh? (Hint: The weight of an object is considered to be a force in units of newtons.)
13
Answer:
2500 NExplanation:
The weight of the object can be found by using the formula
[tex]f = \frac{w}{d} \\ [/tex]
w is the workdone
d is the distance
From the question we have
[tex]f = \frac{62500}{25} \\ [/tex]
We have the final answer as
2500 NHope this helps you
Given F1: a force of magnitude 6 N at an angle of 30°
F2: a force of magnitude 8 N at an angle of 50°C
a. Find F1+ F2 analytically (using equations instead of graphing) and write it in the form Fr1i + Fr2 j
b. Find the magnitude FR and θ_resultant
Answer:
13.8 N
[tex]41.44^{\circ}[/tex]
Explanation:
[tex]F_1=6\ \text{N}[/tex]
[tex]F_2=8\ \text{N}[/tex]
[tex]F_1\cos\theta_1\hat{i}+F_1\sin\theta_1\hat{j}\\ =6\cos30^{\circ}+6\sin30^{\circ}\hat{j}\\ =5.2\hat{i}+3\hat{j}[/tex]
[tex]F_2\cos\theta_2\hat{i}+F_2\sin\theta_2\hat{j}\\ =8\cos50^{\circ}+8\sin50^{\circ}\hat{j}\\ =5.14\hat{i}+6.13\hat{j}[/tex]
[tex]F_R=F_1+F_2=10.34\hat{i}+9.13\hat{j}[/tex]
[tex]|F_R|=\sqrt{10.34^2+9.13^2}=13.8\ \text{N}[/tex]
The magnitude of the resultant is 13.8 N
Direction is given by
[tex]\tan^{-1}=\dfrac{y}{x}=\tan^{-1}\dfrac{9.13}{10.34}=41.44^{\circ}[/tex]
The angle of the resultant is [tex]41.44^{\circ}[/tex]
BRAINLIEST
Which statement is an assumption that is made about science?
Science is absolute.
The purpose of science is to advance technology.
The universe cannot be understood with certainty.
Science is democratic.
Answer: The purpose of science is to advance technology.
Explanation: My explanation is that science advances things, Such as science made it able to have Siri sand honestly creates what we are using now to type.
The statement "the universe cannot be understood with certainty" is an assumption that is made about science option (C) is correct.
What is science?Science is a rigorous process of acquiring and analyzing information about the earth in the form of testable understandings.
It is given that:
The statements are:
Science is absolute.The purpose of science is to advance technology.The universe cannot be understood with certainty.Science is democratic.As we know; from the definition of science; the rigorous process of acquiring and analyzing information about the earth in the form of testable understandings.
The assumption made about science is the universe cannot be understood with certainty.
Thus, the statement "the universe cannot be understood with certainty" is an assumption that is made about science option (C) is correct.
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At what distance on the axis of a current loop is the magneticfield half the strength of the field at the center of the loop? Give your answer as a multiple of R. z=?R
Answer:
x = 1.26 R
Explanation:
For this exercise let's find the magnetic field using the Biot-Savart law
B = μ₀ I/4π ∫ ds x r^ / r²
In the case of a loop or loop, the quantity ds is perpendicular to the distance r, therefore the vector product reduces to the algebraic product and the direction of the field is perpendicular to the current loop
suppose that the spiral eta in the yz plane, therefore the axis is in the x axis
B = μ₀ I/4π ∫ ds / (R² + x²)
The total magnetic field has two components, one parallel to the x axis and another perpendicular, this component is annual when integrating the entire loop, so the total field is
B = Bₓ i^
using trigonometry
Bₓ = B cos θ
we substitute
Bₓ = μ₀ I/4π ∫ ds cos θ / (x² + R²)
the cosine function is
cos θ = R /√(x² + R²)
The differential is
ds = R dθ
we substitute
Bₓ = μ₀ I/4π ∫ (R dθ) R /√( (x² + R²)³ )
we integrate from 0 to 2π
Bₓ =μ₀ I/4π R² / √(x² + R²)³ 2pi
therefore the final expression is
B = μ₀ I R²/ 2√(x² + R²)³ i^
In our case the distance is requested where B is half of B in the center of the bone loop x = 0
Spire center field x=0
B₀ = μ₀ I/2R
Field at the desired point (x)
B = B₀ / 2
we substitute
R² /√(x² + R²)³ = ½ 1 /R
2R³ =√(x² + R²)³
(x² + R²)³ = 4 (R²)³
(x²/R² + 1)³ = 4
The exact result is the solution of this equation, but it is quite laborious, we can find an approximate result assuming that the distance x is much greater than R (x »R)
B = μ₀ I/2x³
we substitute
R² / x³ = 1/2 1 / R
2R³ = x³
x = ∛2 R
x = 1.2599 R
The distance at which the magnetic field strength is half the strength of the field at the center of the loop in terms of R is 0.766 R
Suppose we consider a magnetic field located at point z on the axis of the current loop with radius R carrying a current (I), then the magnetic field can be represented as:
[tex]\mathbf{B = \dfrac{\mu_o}{2} \dfrac{IR^2}{(z^2+R^2)^{^{\dfrac{3}{2}}}}}[/tex]
And the field situated at the center of the loop is:
[tex]\mathbf{B_{z=0} =\dfrac{\mu_o}{2} \dfrac{I}{R} }[/tex]
Let consider a distance (z) on the axis of the loop, in which the magnetic field as a result of the loop is equal to half the strength of the magnetic field at the center of the loop;
Then;
[tex]\mathbf{B(z) = \dfrac{1}{2}B_{z=0}}[/tex]
[tex]\mathbf{\dfrac{\mu_o}{2} \dfrac{IR^2}{(z^2+R^2)^{\frac{3}{2}}} =\dfrac{1}{2} \Big (\dfrac{\mu_o}{2} \dfrac{I}{R} \Big )}[/tex]
Multiply both sides by (2);
[tex]\mathbf{\dfrac{R^2}{(z^2+R^2)^{\frac{3}{2}}} = \dfrac{1}{2R}}[/tex]
Cross multiply;
[tex]\mathbf{2R^3 = (z^2 +R^2)^{\dfrac{3}{2}}}[/tex]
[tex]\mathbf{4R^6 = (z^2 +R^2)^3}[/tex]
[tex]\mathbf{z = \sqrt{4^{1/3} R^2 -R^2 }}[/tex]
[tex]\mathbf{z = R\sqrt{4^{1/3} -1 }}[/tex]
[tex]\mathbf{z = R\sqrt{0.587401052 }}[/tex]z = 0.766 R
Therefore, we can conclude that the distance at which the magnetic field strength is half the strength of the field at the center of the loop in terms of R is 0.766 R
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A cable that weighs 4 lb/ft is used to lift 1000 lb of coal up a mine shaft 700 ft deep. Find the work done.
Answer:
980000ft-lbsExplanation:
Step one:
given data
mass of cable= 4lb/ft
mass of coal= 1000lb
dept of mine= 700ft
Step two:
Required
the work-done to lift the coal and the rope combined
Work-done to lift coal
Wc=1000*700= 700,000 lb-ft
Work-done to lift rope
[tex]Wr=\int\limits^{700} _0 {4(700-y)} \, dx \\\\Wr=4(700y-\frac{1}{2}y^2 )\limits^{700}_0[/tex]
substitute y=700 we have, since y=0 will result to 0
[tex]Wr=4(700*700-\frac{1}{2}*700^2 )\\\\Wr=4(490000-245000)\\\\Wr=4(245000)\\\\Wr=980000ft-lbs[/tex]
explain approach in volleyball
Answer:
This refers when a spiker quickly strides towards the net before they jump in the air for an attack.
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m^2. What is the maximum value of the magnetic field in the wave?
a. 77.9 nT
b. 55.1 nT
c. 38.9 nT
d. 108 nT
Answer:
a. 77.9 nT
Explanation:
Given;
average value of the electromagnetic wave, E = 0.724 W/m^2
The average value of a Poynting vector is given by;
[tex]S = \frac{1}{2\mu_o} *E_oB_o\\\\But, E_o = cB_o\\\\S = \frac{1}{2\mu_o} *cB_o^2\\\\cB_o ^2= 2\mu_o S\\\\B_o^2 = \frac{2\mu_o S}{c}\\\\B_o = \sqrt{ \frac{2\mu_o S}{c}} \\\\B_o = \sqrt{ \frac{2(4\pi*10^{-7}) (0.724)}{3*10^8}}\\\\B_o = 7.79*10^{-8} \ T\\\\B_o = 77.9*10^{-9} \ T\\\\B_o = 77.9 \ nT[/tex]
Therefore, the maximum value of the magnetic field in the wave is 77.9 nT.
Answer:
A I think : )
Explanation:
What preparation is usually needed for a career in physical science?
Answer:
It includes the sciences of chemistry and physics. Most careers in physical science require a 4-year college degree in one of these fields. Some careers require more advanced education as well. For example, an astronaut might have a master's degree or even a doctoral degree.
Explanation:
Answer: i have never seen two pretty best friends.
Explanation:
An object has an average acceleration of +6.07 m/s2 for 0.250 s . At the end of this time the object's velocity is +9.64 m/s .
What was the object's initial velocity?
Answer:
Explanation:
From the question we are given;
Acceleration a = 6.07m/s²
Time t= 0.25s
Final velocity v = 9.64m/s
Required
Initial velocity u
Using the equation of motion
v = u+at
9.64 = u+(6.07)(0.25)
9.64 = u+1.5175
u = 9.64-1.5175
u = 8.1225m/s
Hence the object's initial velocity is 8.1225m/s
What kind of acceleration occurs when an object speeds up? positive negative neutral zero
Answer:
POSITIVE
Explanation:
The acceleration of an object is given by the rate of change of velocity.
u and v are initial and final velocities
If the final velocity is more than that of the initial velocity, the acceleration of the object is positive. It means positive acceleration occurs when an object speeds up. Hence, the correct option is (a) "positive".
Answer:
Positive
Explanation:
Calculate the force of Earth's gravity on a spacecraft 2.00 Earth radii above the Earth's surface if its mass is 1650 kg .
Answer:
F = 1.8 KN
Explanation:
Given that the spacecraft is 2.00 Earth radii above the Earth's surface, the force of Earth's gravity on it can be determined by;
F = [tex]\frac{GMm}{(R + h)^{2} }[/tex]
Where: F is the force, G is the universal gravitation constant, M is the mass of the earth, m is the mass of the spacecraft, and R is the radius of the earth and h is the distance of the spacecraft to the surface of the earth.
G = 6.67 x [tex]10^{-11}[/tex] N[tex]m^{2}[/tex][tex]kg^{-2}[/tex], M = 5.972 x [tex]10^{24}[/tex] kg, m = 1650 kg, R = 6371 km, h = (2.0 x 6371 km) = 12742 km.
Thus,
F = [tex]\frac{6.67*10^{-11}*5.972*10^{24} *1650 }{(6371*10^{3}+12742*10^{3}) ^{2} }[/tex]
= [tex]\frac{6.5725*10^{17} }{(19113000)^{2} }[/tex]
= 1799.17
F = 1.8 KN
The force of the Earth's gravity on the spacecraft is 1.8 KN.
A bus is moving with
a velocity of lom/s. After 2 sec its velocity
becames 20/s find the distance from travelled
by a bus at that this time of interval
Answer:
30 m
__________________________________________________________
Explanation:
We are given:
Initial velocity (u) = 10 m/s
Final velocity (v) = 20 m/s
Time interval (t) = 2 seconds
Distance travelled by the bus (s) = s meters
Solving for the distance travelled:
Solving for the acceleration:
v = u + at [first equation of motion]
20 = 10 + a(2) [replacing the given values]
2a = 10 [subtracting 10 from both sides]
a = 5 m/s² [dividing both sides by 2]
Solving for the distance:
s = ut + 1/2 (at²) [second equation of motion]
s = 10(2) + 1/2(5)(2)² [replacing the given values]
s = 20 + 10
s = 30 m
Therefore, the bus travelled 30 m in the given time interval
If a runner travels 60 m in 12 s, his acceleration is *
Will be marked as brainiest!
Answer:
[tex]a=0.834\ m/s^2[/tex]
Explanation:
Given that,
Displacement, d = 60 m
Time, t = 12 s
We need to find the acceleration of the runner. Using second equation of motion as follows :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
Here, u = 0 (initial velocity = 0)
a is acceleration
[tex]a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 60}{(12)^2}\\\\a=0.834\ m/s^2[/tex]
So, the acceleration of the runner is [tex]0.834\ m/s^2[/tex].
The angle of incidence is the angle between the the normal of the surface. and A. point of incidence B. reflected ray C. incident ray D. refracted ray
Answer: c. incident ray
Explanation:
A uniform metal meter-stick is balanced with a 1.0 kg rock attached to the left end of the stick. If the support is located 0.25 m from the left end, what is the mass of the meter-stick? (Hint: meter-stick is 1 m long). a) 0.6 kg b) 2.0 kg c) 1.0 kg d) 3.0 kg
Answer:
c) 1.0 kg
Explanation:
The mass of the stick will be located at the centre of the metre rule. Since the rock is located 0.25m from the pivot, the mass of the meter rule is also 0.25m to the Right of the support
According to law of moment
Sum of clockwise moment = sum of anti clockwise moments
Clockwise moment = M×0.25(mass of metre rule is M)
CW moment = 0.25M
Anti clockwise moment = 0.25×1
ACW moments = 0.25kgm
Equate;
0.25M = 0.25
M = 0.25/0.25
M = 1.0kg
Hence the mass of the metre rule is 1.0kg
A good baseball pitcher can throw a baseball toward home plate at 97 mi/h with a spin of 1540 rev/min. How many revolutions does the baseball make on its way to home plate
Answer:
10778292789403987593790
Explanation:
I am a Cow'
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.
(A star if you answer this question) A school bus is traveling at 11.1 m/s and has a
momentum of 152,625 kgm/s. What is the mass of the bus?
[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Kinematics in real world.
So, as given here, we have to find the Mass of the bus from the given momentum, so we get as,
P = m * V
momentum = mass * velocity
here, P= 152625 kgm/s and v= 11.1 m/s
so substituting we get as,
m = 152625 ÷ 11.1 => 13,750 kg
hence,the mass of the bus is 13,750 kg.
two object, one of mass m and the other of mass 2m, are dropped from the top of a building. if there is no air resistance, when they hit the ground
Answer:
They hit the ground at the same time
Explanation:
Mass doesn't matter in free fall. As long as they were released with the same velocity and the same height, they'll hit the ground at the same time
1. Black Panther ran 567.5 ks to the east in 2.3 hours then tumed around and ran 2218 km to
the west in 1.2 hours. What was his average velocity during the trip in km/h?
Answer:Average Velocity of the trip=98.77km/h
Explanation:
Given that:
Black Panther started his journey with
Distance ran towards east = 567.5km
Time ran towards east = 2.3 hours
and then turned back,
Distance ran towards west, = 2218 km
Time taken towards west, = 1.2 hours
Let us treat the westward movement as negative movement of the east
So total displacement =567.5 km east + (-2218km east )= 1,650.5km
But it seems the value of the distance ran west is wrong , because i do not think he can run 2218km under 1.2hours considering he ran 567.5m due east at 2.3hours .
So Let we use 221.8 km as distance ran due west.
So that Total Displacement becomes = 567.5 km east + (-221.8km east )= 345.7km
Total time = 2.3 + 1.2 = 3.5hrs.
Average velocity = Total displacement / total time
.345.7km / 3.5hr = 98.77km/h