explain an experiment of the phenomenon of rainfall
Unclear/incomplete question. However, I inferred you need an explanation of the phenomenon of rainfall.
Explanation:
Basically, the phenomenon of rainfall follows a natural cycle called the water cycle. What we call 'rainfall' occurs when water condensed (in liquid form) in the atmosphere is made to fall down on the ground as tiny droplets as a result of the forces of gravity.
The water cycle makes rainfall possible:
First, water on the earth's surface is evaporated (or is absorbed into) the atmosphere.Next, it then condensed into liquid form; which later falls to the surface to the ground again. And the process continues.Given f(x)=2x+7, which of the following is the value of x when f(x)=13?
Answer:
The value of x is 3
Explanation:
Equation Solving
We are given the equation
f(x) = 2x + 7
Let's find the value of such that f(x) = 13.
2x + 7 = 13
Subtracting 7:
2x = 13 - 7
2x = 6
Dividing by 2:
x = 6/2
x = 3
The value of x is 3
The compound formed from the elements calcium and chlorine is known as
Answer:
calcium chloride
Explanation:
an inorganic compound,a salt with the chemical formula CaCl2
You are interested in a new sports car that can go from 0 m/s to 120 m/s in 12 s. What is the acceleration of the car?
Formula
image
A.
0.1
m
s
2
B.
10
m
s
2
C.
60
m
s
2
D.
1440
m
s
2
Answer:
A
Explanation:
Option B) 10m/s² is the correct answer.
Hence, the acceleration of the new sports car in the given period of time is 10m/s².
What is Motion?Motion is simply the change in position of an object over time.
From the First Equation of Motion;
v = u + at
Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.
Given the data in the question;
Initial velocity of the sport car u = 0m/sFinal velocity of the sport car v = 120m/sElapsed time t = 12sAcceleration of the sports car a = ?To determine the acceleration of the new sports car, we substitute our given values into the expression above.
v = u + at
120m/s = 0m/s + (a × 12s)
a × 12s = 120m/s - 0m/s
a × 12s = 120m/s
a = 120m/s ÷ 12s
a = 10m/s²
Option B) 10m/s² is the correct answer.
Hence, the acceleration of the new sports car in the given period of time is 10m/s².
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A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 6.0 m/s in the opposite direction. If the object is in contact with the wall for 2.0 ms, what is the magnitude of the average force on the object by the wall?
a. 9.8 kN.
b. 8.4 kN.
c. 7.7 kN.
d. 9.1 kN.
e. 1.2 kN.
Given that,
Mass of the object, m = 1.2 kg
Initial speed of the object, u = 8 m/s
Final speed of the object, v = -6 m/s (in opposite direction)
Time, t = 2 ms
To find,
The average force on the object by the wall.
Solution,
Let F be the force. Using Newton's second law of motion,
F = ma, a is acceleration
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{1.2\times ((-6)-8)}{2\times 10^{-3}}\\\\=8400\ N[/tex]
or
F = 8.4 N
So, the magnitude of average force in the object by the wall is 8.4 N.
How would the mass and weight of an object on the Moon compare to the mass and weight of the same object on Earth? * Mass and weight would both be less on the Moon. Mass would be the same but its weight would be less on the Moon. Mass would be less on the Moon and its weight would be the same. Mass and weight would both be the same on the Moon.
Answer:
B. Mass would be the same but its weight would be less on the Moon.
Explanation:
The mass of a body can be expressed as the quantity of matter it contains. While the weight of a body is the extent of the gravitational force impressed on the body by a massive body.
Thus, the mass of a body is constant either on the Earth or on the Moon. But the weight would be less on the Moon because the gravitational force on the Moon is far less than that on the Earth. Therefore the weight would be less on the Moon.
The appropriate option is B.
The mass will remain same on both moon and Earth, but weight will be lesser on Moon than Earth. Hence, option (B) is correct.
The prime focus to solve this problem is the mass and weight of an object. The mass of a body can be expressed as the quantity of matter it contains. While the weight of a body is the extent of the gravitational force impressed on the body by a massive body.
So, the mass of a body is constant either on the Earth or on the Moon. But the weight of an object will depend on the mass and the gravitational acceleration.
W = mg
Here, W is weight, m is mass and g is gravitational acceleration.
Weight would be less on the Moon because the gravitational force on the Moon is far less (due to lower value of g) than that on the Earth. Therefore the weight would be less on the Moon.
Thus, we can conclude that the mass will remain same on both moon and Earth, but weight will be lesser on Moon than Earth. Hence, option (B) is correct.
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A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel before it stops?
Answer:
Δx = 39.1 m
Explanation:
Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:[tex]v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)[/tex]
where vf is the final velocity (0 in our case), v₀ is the initial velocity
(25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance
traveled since the brakes are applied.
Solving (1) for Δx, we have:[tex]\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)[/tex]
The car will travel a distance of 39.1 m before its stops.
To solve the problem above, use the equations of motion below.
Equation:
v² = u²+2as................... Equation 1Where:
v = final velocity of the automobileu = initial velocity of the automobilea = accelerations = distance coveredFrom the question,
Given:
v = 0 m/s (before its stops)u = 25 m/sa = -8 m/s² (decelerating)Substitute these values into equation 1
⇒ 0² = 25²+2(-8)(s)Solve for s
⇒ 0²-25² = -16s⇒ -16s = -625⇒ s = -625/16⇒ s = 39.1 mHence, The car will travel a distance of 39.1 m before its stops.
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A block of mass, m, sits on the ground. A student pulls up on
the block with a tension, T, but the block remains in contact
with the ground. What is the normal force on the block?
Answer a
Explanation: a
The speed of revolution of particle going around a circlr is doubled and its angular speed is havled. What happen to the centripetal acceleration?
a) unchanged
b) doubles
c) halves
d) becomes four times
Answer: The correct answer is C
Explanation:
Through what angle in degrees does a 33 rpm record turn in 0.32 s?
63°
35°
46°
74°
Answer:
1 rev = 2(pi) rad pi(rad) = 180 degrees
so 33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s = 63.36 degrees
Explanation:
63.36 estimated to 63 so 63
The angle in degrees where 33 rpm record turn in 0.32 s should be considered as the 63 degrees.
Calculation of the angle:Since we know that1 rev = 2(pi) rad
So here pi(rad) = 180 degrees
Now for 33 rpm it should be like
= 33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s
= 63.36 degrees
= 63 degrees
hence, The angle in degrees where 33 rpm record turn in 0.32 s should be considered as the 63 degrees.
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A bowling ball is 21.6 cm in diameter. What is the angular speed of these ball whenit is moving at 3.0 m/s?
Answer:
Angular speed = 27.78 rad/s (Approx)
Explanation:
Given:
Diameter = 21.6 cm
Speed = 3 m/s
Find:
Angular speed
Computation:
Radius = 21.6 / 2 = 10.8 cm = 0.108 m
Angular speed = v / r
Angular speed = 3 / 0.108
Angular speed = 27.78 rad/s (Approx)
g You heard the sound of a distant explosion (3.50 A/10) seconds after you saw it happen. If the temperature of the air is (15.0 B) oC, how far were you from the site of the explosion
Answer:
The answer is "1557 meters".
Explanation:
speed of sound in ([tex]\frac{m}{s}[/tex]) [tex]= 331.5 + 0.60 \ T^{\circ}\ C\\\\[/tex]
[tex]\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters[/tex]
A rigid tank contains an ideal gas at 300 kPa and 600 K. Now half of the gas is withdrawn from the tank and the gas is found at 100 kPa at the end of the process. Determine (a) the final temperature of the gas and (b) the final pressure if no mass was withdrawn from the tank and the same final temperature was reached at the end of the process.
20 pts.
Which of the following statements is true?
O Electromagnets use electrlcity and magnets.
O Magnetic fields are strongest around the poles of a magnet.
O The south pole of a magnet will repel the south pole of another magnet.
O all of the above
Answer:
all are true so d is right
Explanation:
Electromagnets use electrlcity and magnets is true.
Magnetic fields are strongest around the poles of a magnet is true.
The south pole of a magnet will repel the south pole of another magnet is true
and since all of them is true the answer is d all of the above
A gas cylinder holds 0.10 mol of O2 at 150 C and a pressure of 3.0 atm. The gas expands adiabatically until the pressure is halved
Part A
What is the final volume?
Part B
What is the final temperature?
Answer:
V2 = 1.899*10^-3 m^3
T2 = 347.125 K
Explanation:
Using gas law, we know that
PV = nRT,
Where
V1 = 0.00115743 m^3.
gamma = 1.4
Now, when we solve for final volume, V2 we get
V2 = V1/((P2/P1)^(1/gamma))
V2 = 1.899*10^-3 m^3
Using the same law and method, when we try to solve for the temperature, we find that the final temperature, T2 is
T2 = T1*((V1/V2)^(gamma-1))
T2 = 347.125 K
The final volume is 1.899*10^-3 m^3
And, the final temperature is 347.125 K
Gas law:here we used gas law,
we know that
PV = nRT,
Here
V1 = 0.00115743 m^3.
gamma = 1.4
Now final volume is
V2 = V1/((P2/P1)^(1/gamma))
V2 = 1.899*10^-3 m^3
Now the final temperature is
T2 = T1*((V1/V2)^(gamma-1))
T2 = 347.125 K
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Suppose a uniform solid sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The linear velocity of the sphere at the bottom of the incline depends on?
Answer:
None of the mass or the radius of the sphere
Explanation:
When a uniform solid sphere of any given mass, say M and any given radius, say R, rolls without slipping downwards an inclined plane that starts from rest. The linear velocity of the sphere at about the bottom of the inclined happens not to depend on either of its mass or that of the radius of its sphere.
A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 Hz. What driving frequency will set up a standing wave with five equal segments?
a) 360 Hz.b) 240 Hz.c) 600 Hz.d) 120 Hz.
Answer:
C) 600 Hz
Explanation:
The fundamental frequency can be related to the driving frequency by the expression below;
f(n) = n * f(1)
Where f(1)= fundamental frequency
f(n) = driving frequency
There are four equal segments in the standing wave , then our n= 4 and our f(n)=4, then we can get the fundamental frequency here
f(4) = 4× f(1)
480 = 4× f(1)
f(1) = 480/4
f(1)=120Hz
Hence, fundamental frequency is 120Hz
To calculate the driving frequency that will set up a standing wave with five equal segments?
n=5
f(n) = n× 120Hz
f(5) = 5×120Hz
= 600Hz.
Hence, the driving frequency that will set up a standing wave with five equal segments is 600Hz
two spheres A and B are projected off the edge of a 1.0 m high table with the same horizontal velocity . sphere A has a mass of 20.g and sphere B has a mass of 10.g.
If both spheres leave the edge of the table at the same instant, sphere A will land
a. at some time after sphere B.
b. at the same time as sphere B.
c. at some time before sphere B.
d. There is not enough information to decide.
A would land before since its heavier
What is the condition required of the phase difference (in radians) between two waves with the same wavelength if these waves interfere constructively?
a. (2m +1)π where m= 0, +1, +2, etc.
b. mπ where m = 0, +1, +2, etc.
c. 2mπ where m = 0, +1, +2, etc.
d. (m+1)π where m = 0, +1, +2, etc.
Answer:
c.
Explanation:
In order to two waves with the same wavelength can interfere constructively, their crests and valleys must coincide in space, so the phase difference must be equal to an integer number of wavelengths, i.e. m *(2 π rad), where m= 0, +1, +2, etc.This is equal to the stated by the answer c) , so c) it's the right answer.*
If a rock falls for 3 seconds off of a bridge, how far will the rock fall?
-30 m
-45m
-60m
-75m
what is mean by combination reaction ?
[tex] \underline{\purple{\large \sf Combination \: reaction :-}} [/tex]
Those reaction in which two or more substances combine to form a one new substance are called Combination reaction
In this reaction, We can add :
Two or more elements can combine to form a compound.Two or more compounds can combine to from a one new compound.An element and a compound can combine to form a new compound.[tex] \underline{\green{\large \sf For\: example :}} [/tex]
[tex] \sf 2H_{2} + O_{2} \: \underrightarrow{Combination} \: 2H_{2}o[/tex]
In this, Hydrogen is an element and Oxygen is another element. Both are combined to form compound 'Hydrogen oxide'. Hydrogen oxide is commonly known as water.
A horizontal force of 350 NN is exerted on a 2.0-kgkg ball as it rotates (at arm's length) uniformly in a horizontal circle of radius 0.90 mm. Part A Calculate the speed of the ball. Express your answer to two significant figures and include the appropriate units.
Answer:
The value is [tex]v = 12.6 \ m/s[/tex]
Explanation:
From the question we are told that
The horizontal force is [tex]F_h = 350\ N[/tex]
The mass of the ball is [tex]m = 2.0 \ kg[/tex]
The radius is [tex]r = 0.90 \ m[/tex]
Generally the speed of the ball is mathematically evaluated from the formula for centripetal force as
[tex]F_ h = \frac{mv^2}{ r }[/tex]
=> [tex]v = \sqrt{\frac{ F_h * r }{ m }}[/tex]
=> [tex]v = \sqrt{\frac{ 350 * 0.9 }{ 2 }}[/tex]
=> [tex]v = 12.6 \ m/s[/tex]
A radioactive nuclide of atomic number Z emits an alpha particle and the daughter nucleus then emits a beta particle. What is the atomic number of resulting nuclide?
A) Z-1
B) Z+1
C) Z-2
D) Z-3
Answer:
A) Z-1
Explanation:
when a radioactive element of atomic number Z emits an alpha particle, the mass of the new nucleus decreases by 2, i.e the new atomic number of the element = ( Z- 2).
Also, when the daughter nucleus emits a beta particle, the new nucleus increases by 1, that is the new atomic number of the element = (Z + 1).
Thus, the atomic number of resulting nuclide = Z ( - 2) + ( + 1).
= Z - 2 + 1
= Z - 1
Therefore, the atomic number of resulting nuclide is Z - 1
On Venus, the atmospheric temperature is a hot 720 K due to the greenhouse effect. It consists mostly of carbon dioxide (molar mass 44 g/mol) and the pressure is 92 atm. What is the total translational kinetic energy of 3 moles of carbon dioxide molecules?
Answer:
The value is [tex]E_t = 17958.2 \ J[/tex]
Explanation:
From the question we are told that
The atmospheric temperature is [tex]T_a = 720 \ K[/tex]
The molar mass of carbon dioxide is [tex]Z = 44 \ g/mol[/tex]
The pressure is [tex]P = 92 \ atm =[/tex]
The number of moles is [tex]n = 3 \ moles[/tex]
Generally the translational kinetic energy is mathematically represented as
[tex]E_t = \frac{f}{2} * n * R T[/tex]
Here R is the gas constant with value [tex]R = 8.314 J\cdot K^{-1}\cdot mol^{-1}[/tex]
Generally the degree of freedom of carbon dioxide in terms of translational motion is f = 3
So
[tex]E_t = \frac{ 3}{2} * 2 * 8.314 * 720[/tex]
=> [tex]E_t = 17958.2 \ J[/tex]
A balloon contains 0.075 m^3 of
gas. The pressure is reduced to
100kPa and fills a box of 0.45 m^3.
What is the initial pressure inside the
balloon if the temperature remains
constant?
Answer:
600 KPa.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V1) = 0.075 m³
Final volume (V2) = 0.45 m³
Final pressure (P2) = 100 KPa
Initial pressure (P1) =?
Temperature = constant
The initial pressure can be obtained by using the Boyle's law equation as shown below:
P1V1 = P2V2
P1 × 0.075 = 100 × 0.45
P1 × 0.075 = 45
Divide both side by 0.075
P1 = 45 / 0.075
P1 = 600 KPa.
Thus, the initial pressure in the balloon is 600 KPa.
Help me please,
A ball is thrown straight up in the air. What is the velocity and acceleration at the top of the path?
A) v 0m/s, = 0m/s/s
B) v = 0m/s, a 10m/s/s
C) v = 10m/s, a 10m/s/s
D) v = 10m/s, a = 0m/s/s
E) None of the above
Option B
Explanation:
no distance was given only the acceleration due to the fact that it went up (10m/s/s)
s0 it is
0 m/s and 10m/s/s (option B)