A long time ago on top of a mountain, the rocks were jagged and pointy. Over thousands of years, the rocks have become smooth.
What process explains this change?
pulling
weathering
depositing
eroding
Answer:
Eroding
Explanation:
A long time ago on top of a mountain, the rocks were jagged and pointy. Over thousands of years, the rocks have become smooth. The process explains this change is Eroding. Hence, Option D is the correct answer.
What is Erosion?Erosion is a process of eroding some solids by wind, water and other natural agent while weathering is breaking down of rocks, soil and minerals.
The different types of Erosion includes,
Gully erosionsheet erosionRill erosionsheet erosionThese factors that will contribute to the rate of water erosion includes the climatic factor (rainfall amount, intensity and frequency).
TopographySoil coverSoil propertiesMass wasting erosion if it is a type of erosion but not by water. It is a type of erosion that is generated mainly by the effect of gravity. It is also called mass movement, in which the material detaches from an inclined surface and falls as a result of gravity.
Here, A long time ago on top of a mountain, the rocks were jagged and pointy. Over thousands of years, the rocks have become smooth. The process explains this change is Eroding. Hence, Option D is the correct answer.
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19. What is shape of BF3
Explanation:
The geometry of the BF 3 molecule is called trigonal planar (see Figure 5). The fluorine atoms are positioned at the vertices of an equilateral triangle. The F-B-F angle is 120° and all four atoms lie in the same plane.
Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellite X has an orbital radius of 3R, and satellite Y has an orbital radius of 4R. The kinetic energy of satellite X is Kx . Satellite X is moved to the same orbit as satellite Y by a force doing work on the satellite. In terms of Kx , the work done on satellite X by the force is
Answer:
The work down on satellite X by the force in terms of Kx is [tex]\dfrac{-K_x}{4}[/tex].
Explanation:
The work done is given as in terms of
[tex]W=\Delta TE[/tex]
Where ΔTE is the change in total energy.
This is given as
[tex]W=\Delta TE\\W=TE_y-TE_x\\W=\dfrac{-GMm}{2(4R)}-\dfrac{-GMm}{2(3R)}\\W=\dfrac{-GMm}{8R}+\dfrac{GMm}{6R}\\W=\dfrac{-6GMm+8GMm}{48R}\\W=\dfrac{2GMm}{48R}\\W=\dfrac{GMm}{24R}[/tex]
Rearranging it in terms of K_x gives
[tex]W=\dfrac{GMm}{24R}\\W=\dfrac{GMm}{-4\times -6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{2(3R)}\\W=\dfrac{1}{-4}K_x\\W=\dfrac{-K_x}{4}[/tex]
Philosophy: The Big Picture Unit 8
How does pragmatism differ from the utilitarianism of the previous era?
A. Utilitarianism did not consider rights and pragmatism did.
B. Pragmatism is concerned with function, utilitarianism with happiness.
C. Utilitarianism was created in England and pragmatism came from Germany.
D. Pragmatism applies to everyone, but utilitarianism is concerned with the upper class.
How long will it take you to drive, at 80km/h a distance of 275km?
Answer:
3.44 s
Explanation:
speed=distance/time
=>t = s/v
=> t = 275/80
= 3.437.. s
has been proposed to use an array of infrared telescopes spread over thousands of kilometers of spaceto observe planets orbiting other stars. Consider such an array with an effective diameter of 6000 km, whichobserves infrared radiation at a wavelength of 10m. If it is used to observe a planet orbiting the star 70Virginis, which is 59 light years from our solar system, what is the size of the smallest details that the arraymight resolve on the planet
Answer:
[tex]\triangle x=1.135 *10^9km[/tex]
Explanation:
From the question we are told that:
Diameter of array[tex]d=6000km \approx 6000*10^3[/tex]
Infrared wavelength [tex]\lambda =10m[/tex]
Distance of planet [tex]d=59 light\ years \approx 55.82*10^{16}m[/tex]
Generally the equation for Diffraction is mathematically given by
[tex]sin\theta =1.22\frac{\lambda}{D}[/tex]
Given that
[tex]sin\theta=\frac{\triangle x}{R}[/tex]
Therefore
[tex]\frac{\triangle x}{R}=1.22\frac{\lambda}{D}[/tex]
[tex]\triangle x=1.22\frac{R \lambda}{D}[/tex]
[tex]\triangle x=1.22\frac{10 *55.82*10^{16}}{6000*10^3}[/tex]
[tex]\triangle x=1.135 *!0^1^2m[/tex]
[tex]\triangle x=1.135 *10^9km[/tex]
A 4 kg box is at rest on a table. The static friction coefficient u, between the box and table is 0.30, and
the kinetic friction coefficient Hi is 0.10. Then, a 10 N horizontal force is applied to the box.
Answer:
The box will not move from its position.
Explanation:
First, we will calculate the static frictional force that is stopping the box to move from its position:
[tex]f = \mu R = \mu W=\mu mg[/tex]
where,
f = static frictional force = ?
μ = coefficient of static friction = 0.3
m = mass of box = 4 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]f = (0.3)(4\ kg)(9.81\ m/s^2)\\f=11.77\ N[/tex]
Since the frictional force (11.77 N) is greater than the applied force (10 N).
Therefore, the box will not move from its position.
A body starts from rest and travels ‘s’ m in 2nd second, than acceleration is.
1) 2s m / s2 2) 3 s m / s2 3) 23 s m / s2 4) 32 s m / s2
Answer:
3m
Explanation:
A running Marites launched the egg she
stole as she was about to be caught with
a velocity of 25 m/s in a direction making
an angle of 20° upward with the
horizontal
a) What is the maximum height reached by
the egg?
b) What is the total flight time (between
launch and touching the ground) of the
egg?
c) What is the horizontal range (maximum
* above ground) of the egg?
d) What is the magnitude of the velocity
of the egg just before it hits the ground?
Answer:
a) y = 3.73 m, b) t = 1.74 s, c) R = 40.99 m,
d) vₓ = 23.49 m/s, v_y = -8.5 m / s
Explanation:
This is a projectile launching exercise, we start by breaking down the initial velocity
sin θ = v_{oy} / v₀
cos θ = v₀ₓ / v₀
v_{oy} = v₀ sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 25 sin 20 = 8.55 m / s
v₀ₓ = 25 cos 20 = 23.49 m / s
a) when the egg reaches the maximum height its vertical speed is zero
v_y² = v_{oy}² - 2 g y
0 = v_[oy}² - 2g y
y = v_{oy}² / 2g
y = [tex]\frac{8.55^2}{2 \ 9.8 }[/tex]
y = 3.73 m
b) flight time
y = v_{oy} t - ½ g t²
the time of flight occurs when the body reaches the ground y = 0
0 = (v_{oy} - ½ g t) t
The results are
t₁ = 0s this time is for using the body star
v_{oy} - ½ g t = 0
t = [tex]\frac{2v_{oy}^2}{g}[/tex]
t = 2 8.55 / 9.8
t = 1.74 s
c) the range
R = v₀² sin 2θ / g
R = 25² sin (2 20) / 9.8
R = 40.99 m
d) speed at the point of arrival
horizontal speed is constant
vₓ = v₀ₓ = 23.49 m/s
vertical speed is
v_y = Iv_{oy} - g t
v_y = 8.55 - 9.8 1.74
v_y = -8.5 m / s
(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is 3.85 x 108 m. Find the time it took for his voice to reach earth via radio waves. (b) Someday a person will walk on Mars, which is 5.6 x 1010 m from earth at the point of closest approach. Determine the minimum time that will be required for that person's voice to reach earth.
Answer:
a). 1.28333 seconds
b). 186.66 seconds
Explanation:
a). Given :
Distance between the earth and the moon, d = [tex]$3.85 \times 10^8$[/tex] m
Speed of the radio waves, c = [tex]$3 \times 10^8$[/tex] m/s
Therefore the time required for the voice of Neil Armstrong to reach the earth via radio waves is given by :
[tex]$t=\frac{d}{c}$[/tex]
[tex]$=\frac{3.85 \times 10^8}{3 \times 10^8}$[/tex]
= 1.28333 seconds
b). Distance between Mars and the earth, d = [tex]$5.6 \times 10^{10}$[/tex] m
Speed of the radio waves, c = [tex]$3 \times 10^8$[/tex] m/s
So, the time required for his voice to reach earth is :
[tex]$t=\frac{d}{c}$[/tex]
[tex]$=\frac{5.6 \times 10^{10}}{3 \times 10^8}$[/tex]
= 186.66 seconds
You are locked inside the train car and want to get it moving to draw attention to your plight. There is effectively no friction between the axle and the car, and the train is on horizontal tracks. To try and get the car moving with respect to the ground, you run and slam with all your force against the wall at the front. What happens with the car after you slammed against the wall of the car
Answer:
the car movves briefly as you ran, however, it stops again after you ran in to the wall
Explanation:
Our net total linear momentum was zero at the time both the train and the boy was at rest. As there is no pressure, the train and boy system's linear momentum can be conserved provided no external forces are working on it that might shift its momentum.
If the boy runs inside of the train with a velocity V1 in the forward direction, the train would have a velocity V2 in the reverse direction to V1 to conserve the system's momentum, resulting in Final momentum.
i.e
[tex]m \times V_1 + M \times V_2 = 0[/tex] --- (1)
here;
m = mass of the boy
M = mass of the train
Thus;
[tex]m \times V_1 =- M \times V_2[/tex] --- (2)
As the boy crashes into the train's wall, a pair of equal and opposing force F intervene on both the train and the boy. This force F causes the boy traveling with momentum m× V1 to come to a halt; its momentum remains zero. As the train moves with about the same momentum as the boy as seen in (2) and is subjected to the same force F, its momentum will be diminished to zero, and it will come to a halt.
Describing a Wave
What does a wave carry?
Answer:
Waves carry energy from one place to another.
Explanation:
Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals.
The diagram shows a charge moving into an electric field. The charge will most likely leave the electric field near which letter? OW OX OY OZ
you haven't attached the diagram, but i assume that this diagram is what you were talking about
Answer:
near Y
Explanation:
the electric field lines goes from a positive charge to a negative charge. This means that a positive charge would move in the same direction of the field lines, while a negative charge would move in the opposite direction of the field lines. the field lines are created from +vely charged plate to -vely charged plate so the negative charged particles moves towards the lower plate which is positively charged, and opposite to the direction of field lines.
The charge will most likely leave the electric field near Y letter. Hence option C is correct.
What is electric charge ?Electric charge is the physical property of matter that experiences force when it is placed in electric field. F = qE where q is amount of charge, E = electric field and F = is force experienced by the charge. there are two types of charges, positive charge and negative charge which are generally carried by proton and electron resp. like charges repel each other and unlike charges attract each other. the flow charges is called as current. Elementary charge is amount of charge a electron is having, whose value is 1.602 x 10⁻¹⁹ C
The charge on the electron is negative, and the force is directed in the opposite direction as the electric field. When an electron is projected perpendicular to a uniform electric field, it experiences an electric force in the opposite direction of the field.
the trajectory of the charged particle in the electric filed is parabolic and in magnetic field it is circular. when this electron moves perpendicular to the electric field, electron experience force in opposite direction to the electric field and due to parabolic nature, it will leave at Y.
Hence option C is correct.
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what is friction and the types withe examples.
Explanation:
The answer is In the picture. Thanks.
PLZ help 10 points!!! space question!
Answer:
3 i think
Explanation:
HURRY IM TIMED
How can you make people feel inspired?
By leading them on an emotional journey through various states to inspiration
By talking about something that interests you
By proving yourself to be a trustworthy speaker
By making them laugh and feel comfortable
Answer:
By talking about something that interesto you’
sorry if wrong
Explanation:
What is the term that describes waves that require a medium through which to travel?
Answer:
Mechanical waves are waves that require a medium. This means that they have to have some sort of matter to travel through. These waves travel when molecules in the medium collide with each other passing on energy. One example of a mechanical wave is sound.
What happens when Molecules collide?
Answer:
Collision theory is used to predict the rates of chemical reactions, particularly for gases. It is based on the assumption that for a reaction to occur for the reacting species (atoms or molecules) must come together or collide with one another. Not all collisions, however, bring about chemical change.
If two molecules collide with sufficient activation energy, there is no guarantee that the collision will be successful. In fact, the collision theory says that not every collision is successful, even if molecules are moving with enough energy. The reason for this is because molecules also need to collide with the right orientation so that the proper atoms line up with one another, and bonds can break and re-form necessarily.
How many times a minute does a boat bob up and down on ocean waves that have a wavelength of 35.6 m and a propagation speed of 4.68 m/s? (the answer may not be a whole number)
Answer:
It will bob 7.887640449 times a minute
Explanation:
I hope this is correct!!
⦁ Determining the magnetic flux, A rectangular piece of stiff paper measures 10 cm x 5 cm. You hold the piece of paper in a uniform magnetic field that has a magnitude of 4.0 x10-3 T For each situation below, - sketch a diagram showing the magnetic field and the paper, - determine the magnitude of the magnetic flux through the paper, when the magnitude of the flux is (a) maximized (b) minimized (c) halfway between its maximum and minimum value, when the angle between the magnetic field lines and the area normal vector is 300
Answer:
jguewjdofe
Explanation:
a student throws a 140 g snowball at 8.5 m/s at the side of a schoolhouse, where it hits and sticks. what is the magnitude of the average force on the wall if the duration of the collision is .16 s?
Answer:
7.4375N
Explanation:
According to Newtons second law
F = ma
m is the mass = 140g = 0.14kg
Get the acceleration
v = u + at
8.5 = 0+0.1a
a = 8.5/0.16
a = 53.125m/s²
Get the required force
F = 0.14 * 53.125
F = 7.4375N
Hence the magnitude of the average force on the wall is 7.4375N
A ray diagram is shown. A tree acts as the object further than 2 F away from a biconvex lens. The distance between 2 F and the object is labeled W. The distance between F and 2 F is labeled X. There I a light ray parallel to the principal axis is bent through F on the image side of the lens. There is a ray straight through the center of the lens. The rays intersect a point below the principle axis between F and 2 F on the image side of the lens and is closer to the principal axis than the object is tall. The intersect point is labeled Z and the distance between F and 2 F on the image side of the lens is labeled Y. Which letter represents the location of the image produced by the lens? W X Y Z
Answer:
Z
Explanation:
correct on edge
Answer: Z
good luck!
Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human is taken to be 90.0 years, a spaceship would need to achieve some minimum speed vmin to deliver a living human being to this galaxy. How close to the speed of light would this minimum speed be? Express your answer as the difference between vmin and the speed of light c.
Answer:
[tex]0.0018833\ \text{m/s}[/tex]
Explanation:
[tex]d[/tex] = Distance of Andromeda Galaxy from Earth = [tex]2.54\times 10^7\ \text{ly}[/tex]
[tex]t[/tex] = Time taken = [tex]90\ \text{years}[/tex]
[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]
We have the relation
[tex]t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}[/tex]
[tex]c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}[/tex]
The required answer is [tex]0.0018833\ \text{m/s}[/tex].
Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings and has a moment of inertia of 1850 kgm2. The turntable is at rest initially, but when the person begins running at a speed of 3.8 m/s (with respect to the turntable) around its edge, the turntable begins to rotate in the opposite direction. Calculate the angular velocity of the turntable. (Hint: use what you know about relative velocity to help solve the problem
Answer:
[tex]0.3165\ \text{rad/s}[/tex]
Explanation:
m = Mass of person = 65 kg
d = Diameter of round table = 6.5 m
r = Radius = [tex]\dfrac{d}{2}=3.25\ \text{m}[/tex]
v = Velocity of person running = 3.8 m/s
[tex]I_t[/tex] = Moment of inertia of turntable = [tex]1850\ \text{kg m}^2[/tex]
Moment of inertia of the system is
[tex]I=I_t+mr^2\\\Rightarrow I=1850+65\times 3.25^2\\\Rightarrow I=2536.5625\ \text{kg m}^2[/tex]
As the angular momentum of the system is conserved we have
[tex]L_i=L_f\\\Rightarrow mvr=I\omega_f\\\Rightarrow \omega_f=\dfrac{mvr}{I}\\\Rightarrow \omega_f=\dfrac{65\times 3.8\times 3.25}{2536.5625}\\\Rightarrow \omega_f=0.3165\ \text{rad/s}[/tex]
The angular velocity of the turntable is [tex]0.3165\ \text{rad/s}[/tex].
physics grade9 teacher guide
Answer:
huh
Explanation:
what is the relationship between Force, area and mass
Answer:
F=ma
Explanation:
Force is equal to mass multiplied by area
Force= mass × area
In turn the formula can be twisted so
Force/mass= area
Force/area= mass
Answer:
It states that the rate of change of velocity of an object is directly proportional to the force applied and takes place in the direction of the force. It is summarized by the equation: Force (N) = mass (kg) × acceleration (m/s²). Thus, an object of constant mass accelerates in proportion to the force applied.
Think of a hydropower dam . How is electrical energy produced from potential and kinetic energy ?
hydroelectric dam converts the potential energy stored in a water reservoir behind a dam to mechanical energy—mechanical energy is also known as kinetic energy. ... The generator converts the turbine's mechanical energy into electricity.
Hope this helps!
Answer:
Potential energy and kinetic energy are constituents of mechanical energy.
When a turbine is switched on, it rotates with mechanical energy.
Since a motor runs the turbine, it converts this mechanical energy to electrical energy.
A block and tackle of 6 pulleys is used to raise a load of 300 newton steadily through a height of 30 meters. If the work done against friction is 2000j calculate the work done by effort?
Answer:
W = F * S = 300 N * 30 m = 9000 Joules
This is the work out of the system regardless of how the pulleys are arranged
Conservation of energy tells that
Work Out = Work In - Work done by friction
So Work In = 9000 J + 2000 J = 11000 J The work input to the system
A 700-gram grinding wheel 22.0 cm in diameter is in the shape of a uniform solid disk. (We can ignore the small hole at the center.) When it is in use, it turns at a constant 215 rpm about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 50.0 s with constant angular acceleration due to friction at the axle.
What torque does friction exert while this wheel is slowing down?
Solution :
Given :
Mass of grinding wheel, m = 700 g
= 0.7 kg
Diameter of the grinding wheel, d = 22 cm
= 0.22 m
Radius of the grinding wheel, r = 0.11 m
Initial angular velocity of grinding wheel, [tex]$\omega_0$[/tex] = 215 rpm
[tex]$=215 \ rpm \times \frac{2 \pi \ rad}{1 \ rev}\times \frac{1 \ min}{60 \ s}$[/tex]
where, [tex]$\pi = \frac{22}{7}$[/tex]
Time taken to stop, t = 50 s
Final angular velocity is [tex]$\omega$[/tex] = 0
Angular acceleration of the grinding wheel is given by :
[tex]$\alpha = \frac{\omega-\omega_0}{t}$[/tex]
[tex]$=\frac{0-215 \ rpm \times \frac{2 \pi \ rad}{1 \ rev}\times \frac{1 \ min}{60 \ s}}{50 \ s}$[/tex]
[tex]$=-0.45 \ rad/s^2$[/tex]
Magnitude of the angular acceleration of grinding wheel [tex]$\alpha$[/tex] [tex]$=-0.45 \ rad/s^2$[/tex]
Moment of inertia of the grinding wheel (solid disk),
[tex]$I=\frac{1}{2}mR^2$[/tex]
[tex]$=\frac{1}{2} \times 0.7 \times 0.11^2$[/tex]
[tex]$=4.235 \times 10^{-3} \ kgm^2$[/tex]
Torque exerted by friction while the wheel is slowing down is
[tex]$\tau = I \alpha$[/tex]
[tex]$=4.235 \times 10^{-3} \times 0.45$[/tex]
[tex]$=1.90 \times 10^{-3} \ Nm$[/tex]
The half-life for a 400-gram sample of radioactive element X is 3 days. How much of element X remains after 15 days have passed?
A.
12.5 g
B.
25 g
C.
50 g
D.
100 g