Given :
Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal.
To Find :
How long is it in the air.
Solution :
We know, the formula of time of flight is :
[tex]T = \dfrac{2usin\ \theta}{g}\\\\T = \dfrac{2\times 30\times sin\ 30^o}{9.8}\\\\T = 3.06\ seconds[/tex]
Therefore, the ball is in air for 3.06 seconds.
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.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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What happens to the molecules of water when it moves from a liquid to a gas?
A. Water molecules condense and move slower.
B. Water molecules spread out and move slower.
C. Water molecules spread out and move faster.
D. Water molecules condense and move faster.
its A or D but im not sure which one ik it moves fast
A heat pump has a coefficient of performance of 3.85 and operates with a power consumption of 7020 W. How much energy does it deliver into a home during 1 h of continuous operation?
Answer:
97.3 MJ
Explanation:
The formula for the coefficient of Perfomance is given as
COE = Q/W, where
COE is the coefficient of Perfomance
Q is the heat provided
W serves as the work input.
Dividing both sides of the equation by a factor of time t, we get the coefficient of Perfomance in terms of heating power and input power, so we say
COE = P / P(i),
making heating power, P the subject of formula, we have
P = COE * P(i)
P = 3.85 * 7020 * 1 * 3600
P = 97297200 J
P = 97.3 MJ
PLEASE HELP :(
I WILL GIVE EXTRA POINTS
1. When is the kinetic energy of an electron transformed into potential energy?
when it interacts with other electrons, decreasing its speed
when it interacts with neutrons without changing its speed
when it interacts with neutrons , increasing its speed
when it interacts with other electrons without changing its speed
2. Atoms bond to form molecules. Which structures or regions of the atoms interact in bonds ?
electric fields of particles with positive charge
electric fields of particles with no charge
electric fields of particles with negative charge
electric fields of particles with opposite charges
3. If two electrons that are apart get pushed toward each other, how does the repulsion between them change?
Initial repulsion is low and decreases as they approach .
Initial repulsion is high and decreases as they approach .
Initial repulsion is high and increases as they approach .
Initial repulsion is low and increases as they approach .
4. A positive charge of 5.0x10 ^ -5 C °is 0.040 m from a second positive charge of 2.0x10 ^ -6 C Calculate the force between the charges.
5.6x10^2 N
5.6x10^2 N
1.4X10^-2 N
2.3X10^1 N
(a) When the kinetic energy of an electron is transformed into potential energy is when it interacts with other electrons, decreasing its speed.
(b) The region of atoms that interact in bonds is electric fields of particles with negative charge.
(c) Initial repulsion is low and increases as they approach.
(d) The force between the charges is 562.5 N.
Kinetic theory of matter
This theory states that, the collision of particles (electrons) of matter is perfectly elastic. This implies that as the particles (electrons) collides with one another, kinetic energy is transferred from one electron to another.
ΔK.E = ΔP.E
Change in kinetic energy is equal to change in potential energy of the electrons.
Thus, when the kinetic energy of an electron is transformed into potential energy is when it interacts with other electrons, decreasing its speed. Decrease in speed implies decrease in kinetic energy and increase in potential energy.
Chemical bonds of moleculesChemical bond is formed from the transfer or sharing of electrons between atoms. (electrons between atoms implies negative charge to negative charge)
Thus, the region of atoms that interact in bonds is electric fields of particles with negative charge.
Coulomb's law
This law states that the force of attraction or repulsion between two charges is directly proportional to the product of the charges and inversely proportional to the distance between the charges.
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
When the distance between the electrons are large, the repulsive force is low and the distance is small, the repulsive force is high.
Force between the chargesThe force between the charges is determined by applying Coulomb's law,
[tex]F = \frac{kq_1q_2}{r^2} \\\\F = \frac{9\times 10^9\times (5 \times 10^{-5}) \times (2\times 10^{-6})}{0.04^2} \\\\F = 562.5 \ N[/tex]
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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
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 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)
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.
During an experiment, Ellie records a measurement of 25,000 m. How would she write her
measurement in scientific notation?
O 2.5 x 10^2 m
O 2.5 x 10^3 m
0 2.5 x 10^5 m
O 2.5 x 10^4.m
Answer:
D. 2.5 x 10^4
Explanation:
10 to the power of 4 is 10,000.
10,000 x 2.5 = 25,000
D. 2.5 x 10⁴ is the scientific notation of Ellie records.
What is scientific notation?Scientific Notation is the expression of a number n in the form a∗10ᵇ. where a is an integer such that 1≤|a|<10. and b is an integer too. It is a way of writing very large or very small numbers. A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10.
According to the question,
The given number is 25,000 m
i.e.
25 x 10³
Since we have to put the decimal between 2 and 5
It becomes,
2.5 x 10⁴
or
10 to the power of 4 is 10,000.
10,000 x 2.5 = 25,000
Therefore,
The answer is 2.25 x 10⁻⁶ M
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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:
*
If a rock falls for 3 seconds off of a bridge, how far will the rock fall?
-30 m
-45m
-60m
-75m
PLEASE HELP ME ASAP!! GUYSSS!! I AM IN CLASS AND DYING! LITERALLY
Billy and Ashley live in the same time zone. Billy lives in Brazil (blue smiley face on the image below). Ashley lives in Eastern Canada (yellow smiley face on the image below).
One day, Billy and Ashley are both outside at 1:32 pm. They are talking on the phone to each other. As they talk, Billy notices the sky in Brazil getting progressively darker. Eventually, it feels like it is nighttime, because it is so dark. Billy thinks the world is coming to an end. He asks Ashley if she is experiencing the same thing in Canada. Ashley has no idea what he is talking about. “It’s perfectly bright and sunny where I am,” she says.
Ashley and Billy conclude that the world is not coming to an end. They reach out to some 7th graders to figure out what is happening. The 7th graders tell Billy that he is experiencing a solar eclipse. To help Billy and Ashley understand, create a model to show why Billy is experiencing an eclipse in Brazil, but Ashley is not experiencing the eclipse in Canada.
Your model must include:
The sun, the earth, the moon, and solar energy (clearly labeled).
How accurate your scale is.
How solar energy interacts with both the moon and with Earth.
The tilt of the moon’s orbit relative to the Earth’s orbit
Why Billy is experiencing the solar eclipse and why Ashley is not.
Answer:
sounds like a you problem
Explanation:
yeah
Help me please..
When the slope of a velocity vs. time graph is negative and constant,
what type of motion is occurring?
A) No motion
B) Constant speed
C) Acceleration
D) Constant Velocity
E) Going in circles
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 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
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.
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
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 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.
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)
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.