Answer:
Mass was added to the water particles, resulting in deposition.
Explanation:
Answer: The energy was removed from the water particles resulting in condensation
Explanation Think about when you leave a cup of ice of and after a while you see water droplets thats what happened to the grass in the morning and thats called condensation.
MULTIPLE CHOICE QUESTION
What is the acceleration of the object?
(Remember a = Fnet/m and acceleration needs a
direction)
8 m/s/s
0.125 m/s/s to the right
8 m/s/s to the left
8 m/s/s to the right
Which object? More information is needed to answer this question
A rectangular loop with an area of 2 m2 is placed perpendicular to a uniform magnetic field of 1 Tesla. The field’s magnitude is increased to 6 Tesla in 4 seconds. The magnitude of the induced emf is equal to:
Answer:
Induced emf = 0
Explanation:
An emf can be induced due to the change in magnetic field. It can be given by :
[tex]\epsilon=\dfrac{d\phi}{dt}\\\\\because \phi=BA\cos\theta\\\\\epsilon=\dfrac{d(BA\cos\theta)}{dt}\\\\\epsilon=A\cos\theta\dfrac{dB}{dt}[/tex]
As the loop is placed perpendicular to a uniform magnetic field of 1 Tesla. It means that [tex]\theta=90^{\circ}[/tex] and cos(90) = 0. Hence, the induced emf is equal to 0.
One statement of the first law of thermodynamics is that:___________.
a. the amount of work done on a system is dependent of pathway.
b. the total work done on a system must equal the heat absorbed by the system.
c. the heat flow in or out of a system is independent of pathway.
d. the total energy flow in or out of a system is equal to the sum of the heat transferred to or from the system and the work done by or on the system.
e. in any chemical process the heat flow must equal the change in enthalpy.
Answer:
a. the amount of work done on a system is dependent of pathway
Explanation:
The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system.
ΔU = Q - W
Where;
Q, the net heat transfer into the system depends on the pathway
W, the net work done by the system also depends on the pathway
But, ΔU, the change in internal energy is independent of pathway
Therefore, the correct option is "A"
a. the amount of work done on a system is dependent of pathway
A coin is rolling across a table at 0.23 m/s. It rolls off the table and lands 0.15 meters away from the edge of the table. How high is the table? Show all your work.
Answer:
The table is 2.08 m high
Explanation:
Horizontal Motion
When an object is thrown horizontally with a speed vo from a height h, the range or maximum horizontal distance traveled by the object can be calculated as follows:
[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]
If we know the range d = 0.15 m and the speed v = 0.23 m/s, we can solve the above equation for h:
[tex]\displaystyle h=\frac{d^2\cdot g}{2v^2}[/tex]
[tex]\displaystyle h=\frac{0.15^2\cdot 9.8}{2\cdot 0.23^2}[/tex]
[tex]\boxed{h=2.08\ m}[/tex]
The table is 2.08 m high
A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen when the spectacles hit the ground? Neglect the air resistance.
Answer:
[tex]h_p = 30.46\ m[/tex]
Explanation:
Free Fall Motion
A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.
The speed vf of the object when a time t has passed is given by:
[tex]v_f=g\cdot t[/tex]
Where [tex]g = 9.8 m/s^2[/tex]
Similarly, the distance y the object has traveled is calculated as follows:
[tex]\displaystyle y=\frac{g\cdot t^2}{2}[/tex]
If we know the height h from which the object was dropped, we can solve the above equation for t:
[tex]\displaystyle t=\sqrt{\frac{2\cdot y}{g}}[/tex]
The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:
[tex]\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec[/tex]
The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:
[tex]t_2=2.56 - 2 = 0.56\ sec[/tex]
Therefore, it has traveled down a distance:
[tex]\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m[/tex]
Thus, the height of the pen is:
[tex]h_p = 32 - 1.54\Rightarrow h_p=30.46\ m[/tex]
The pen is 30.52 m above the ground.
Given that the height of the stadium is h = 32m
The initial velocity of the glasses will be 0.
[tex]h=\frac{1}{2}gt^{2} \\t=\sqrt{\frac{2h}{g} } \\t=\sqrt{\frac{2*32}{9.8} }\\t=2.55s[/tex]is the time taken for the glasses to hit the ground.
Now the pen is released 2 seconds later. So by the time the glasses hit the ground the pen has spent:
[tex]t^{'}=2.55-2\\t^{'}=0.55s[/tex]in the air
distance traveled by the pen:
[tex]d=\frac{1}{2}gt^{2}\\\\d=\frac{1}{2}*9.8*0.55*0.55\\\\d=1.48m[/tex]
So the pen is [tex]h-d=32-1.48=30.52m[/tex] above the ground.
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Consider a wire of length L =0.30 m that runs north-south on a horizontal surface. There is a current I = 0.50 A flowing north in tie wire. The Earth's magnetic field at this location has a magnitude of 0.50 gauss (or, in SI units. 0.5 x 10^—4 and points north and 38 degrees down from the horizontal, toward the ground. What is the size of the magnetic force on the wire due to the Earth's magnetic field?
Answer:
The size of the magnetic force on the wire due to the Earth's magnetic field is 4.62 × 10⁻⁶ N.
Explanation:
To determine the size of the magnetic force on the wire due to the Earth's magnetic field,
The magnetic force is given by the formula
F = ILB sinθ
Where F is the magnetic force on the wire
I is the electric current in Amperes (A)
L is is the length of wire in meters (m)
B is the magnetic field in Tesla (T)
and θ is the angle between current and magnetic field
From the question,
L = 0.30 m
I = 0.50 A
B = 0.50 gauss = 0.5 × 10⁻⁴ T (NOTE: 1 Gauss = 10⁻⁴ Tesla)
θ = 38°
Now, putting the values into the equation
F = ILB sinθ
F = 0.50 × 0.30 × 0.5 × 10⁻⁴ sin38°
F = 7.5 × 10⁻⁶ (0.61566)
F = 4.62 × 10⁻⁶ N
Hence, the size of the magnetic force on the wire due to the Earth's magnetic field is 4.62 × 10⁻⁶ N.
Are renewable resources always renewable, or can they become non-renewable
Answer:
Can become non-renewable but it depends on the object
Explanation:
Answer:
Renewable resources are so abundant or are replaced so rapidly that, for all practical purposes, they can't run out. Fossil fuels are the most commonly used non-renewable resources
Explanation:
Hope this helps (:
WHat is the correct definnition for recovery heart rate ?
Answer:
when your heart rate is not high than normal or lower than normally
Explanation:
Recovery heart rate is when your heart recovers from a certain heart rate eg high heart rate or even low heart rate this means you heart is now functioning normally .
Answer:
Number of beats per minute the heart drops after exercise.
Suppose a rocket-propelled motorcycle is fi red from rest horizontally across a canyon 1.00 km wide. (a) What minimum constant acceleration in the x-direction must be provided by the engines so the cycle crosses safely if the opposite side is 0.750 km lower than the starting point? (b) At what speed does the motorcycle land if it maintains this constant horizontal component of acceleration? Neglect air drag, but remember that gravity is still acting in the negative y-direction.
show answer No Attempt 50% Part (b) Calculate the non-relativistic speed of these electrons v in m/s.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is [tex]v = 9.18 *10^{6} \ m/s[/tex]
Explanation:
From question we are told that
The potential difference is [tex]\Delta V = 0.24 kV = 0.24 *10^{3} \ V[/tex]
Generally the the non-relativistic speed of these electrons is mathematically represented as
[tex]v = \sqrt{\frac{2 * e * \Delta V}{m} }[/tex]
Here m is the mass of an electron with value [tex]m = 9.11 *10^{-31 } \ kg[/tex]
e is the charge on an electron with value [tex]e = 1.60 *10^{-19} \ C[/tex]
So
[tex]v = \sqrt{\frac{2 * 1.60 *10^{-19} * 0.24 *10^{3} }{9.11*10^{-31}} }[/tex]
=> [tex]v = 9.18 *10^{6} \ m/s[/tex]
A 100 kg disc with radius 1.6 m is spinning horizontally at 25 rad/s. You place a 20 kg brick quickly and gently on the disc so that it sticks to the edge of the disc. Determine the final angular speed of the disc-brick system.
Answer:
20.83rad/s
Explanation:
Using the law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the mass of the disc and Brick respectively
u1 and u2 are the linear velocities respectively
v is their final velocity of the block-disc system
Given
m1 = 100kg
m2 = 20kg
u1 = wr =25×1.6
u1 = 40m/s
u2 = 0m/s (brick is stationary)
Substitute and the final linear velocity v
100(40)+20(0) = (100+20)v
4000 = 120v
v = 4000/120
v = 33.3m/s
From the formula;
v = wr
w = v/r
w= 33.3/1.6
w = 20.83rad/s.
Hence the final angular speed of the disc-brick system is 20.83rad/s
Can someone please answer this
Answer:
-) It is less than 8 [N]
See the explanation below.
Explanation:
To solve this problem we must analyze each of the answers, and in this way, we will come to the right solution. The most important thing is to keep in mind that the body is moving to the left.
-) It is equal to 8 [N]
This answer is not possible, since if the Force F was equal to 8 [N] the body would be in equilibrium and the body would not move.
-) It is equal to 10 [N]
This answer is not possible, since if Force F is greater than 8 [N] the body would move to the right.
-) It is greater than 10 [N]
This answer is not possible, since if Force F is greater than 8 [N] the body would move to the right.
-) It is less than 8 [N]
This is the correct option, as the force of 8 [N] will move the body to the left.
8. Using Newton's second law, solve the following problems using
GFFS.
a. A force of 35N is applied to a 25kg object, what is its
acceleration??
b. What force is required to accelerate a 14kg object at 4.5
m/s2?
c. Calculate the mass of an object that accelerated at a rate of
2.7 m/s2 by a force of 18N.
Answer:
a) 1.4 m/s2
b) 63N
c) 6.66kg
An underwater sound source emits waves of frequency 30 kHz in all directions. How does the intensity of the waves (in Watts/m2) vary with distance r from the source?
a) 1/r^3
b) 1/r^2
c) 1/r
d) None of above
When an underwater sound source emits waves of frequency 30 kHz in all directions, the intensity of the waves (in Watts/m2) vary with distance r from the source by the relation 1/r²
As the intensity mechanical sound wave is inversely proportional to the square of the distance from the source, therefore the correct option is B.
What is the Wavelength?Wavelength can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
It is the total length of the wave for which it completes one cycle.
The intensity of a mechanical wave is inversely proportional to the square of the distance from the source.
An underwater sound source emits waves of frequency of 30 kHz in all directions, the intensity of the waves (in Watts/m2) varies with distance r from the source by the relation 1/r², therefore the correct option is B.
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Help please!!!!!!!!!!!!!
Answer:
The second choice
Explanation:
I think the answer is the second choice because if the surface is smooth, there is less friction. With a boat, it is easier to pull it on water than on the sand, because water has less friction, and thus, the answer is the second choice because rough surfaces have more friction.
Hopefully the explanation and answer helps!
Question 4 of 10
A 6.2 kg bowling ball that weighs 60.76 N is accelerating at 1.8 m/s2. After 2
seconds, it reaches a speed of 3.6 m/s. What is its momentum at this point?
A. 11.2 kg-m/s
B. 1.7 kg.m/s
C. 22.3 kg-m/s
D. 16.9 kg.m/s
Answer:
22.3 kg•m/s
Explanation:
Apex;)
If the breaking strength of the string is 120 N, what is the minimum angle the string can make with the horizontal
Complete Question
A 940-g rock is whirled in a horizontal circle at the end of a 1.30-m-long string, If the breaking strength of the string is 120 N, what's the minimum angle the string can make with the horizontal?
Answer:
The value is [tex]\theta = 4.41^o[/tex]
Explanation:
From the question we are told that
The mass of the rock is [tex]m_r = 940 \ g = 0.94 \ kg[/tex]
The length of the string is [tex]l = 1.30 \ m[/tex]
The breaking strength(i.e the maximum tension) on the string is [tex]T = 120 \ N[/tex]
Gnerally the vertical component of the tension experienced by the string is mathematically represented as
[tex]T_v = T sin(\theta)[/tex]
Generally this vertical component of tension is equivalent to the weight of the rock
So
[tex]Tsin (\theta) = mg[/tex]
=> [tex]\theta = sin^{-1} [\frac{mg}{ T} ][/tex]
=> [tex]\theta = sin^{-1} [\frac{0.940 *9.8 }{ 120} ][/tex]
=> [tex]\theta = 4.41^o[/tex]
Jack is buying a subscription to a magazine club. The function c=45+3.75m
c
=
45
+
3
.
75
m
describes the cost of the yearly subscription, c
c
, in terms of the number of magazines, m
m
, he orders.
What is the appropriate domain for the function?
the set of integers
the set of whole numbers
the set of all real numbers
the set of rational numbers
Answer:
the set of all real numbers
Explanation:
Given the cost of the magazine expressed as:
c=45+3.75m where:
c describes the cost of the yearly subscription.
m is the amount of order.
You must understand that domain are the values of the input variable m that will make the function exist.
The value of m that will produce a positive output since we can only have a positive value of the cost. According to the option, we can say that the domain is on the set of all real numbers (some are negative but all positive real numbers)
A net force of 275 N accelerates a 18.0 kg mass. What is the resulting acceleration? (5 points)
Answer:
The answer is 15.28 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
m is the mass
f is the force
We have
[tex]a = \frac{275}{18} \\ = 15.277777...[/tex]
We have the final answer as
15.28 m/s²Hope this helps you
Calculate the maximum energy that a 3-MeV alpha particle can transfer to an electron in a single collision.
Answer:
The maximum energy that can be transferred to an electron in a single collision of alpha particle is 1.63 keV
Explanation:
let mass of electron, Me = 1
let mass of proton, Mp = 1836Me = 1836
let mass of alpha particle, Ma = 4Mp = 7344
The maximum energy that can be transferred to an electron in a single collision is given by;
[tex]Q_{max} = \frac{4mME}{(M+m)^2}\\\\Q_\alpha_{max} = \frac{4(1)(7344)(3\ MeV)}{(1+7344)^2} = 0.00163 \ MeV = 1.63 \ KeV[/tex]
Therefore, the maximum energy that can be transferred to an electron in a single collision of alpha particle is 1.63 keV
List some of the most common injuries that can occur from sports. Please hurry!
Which of the following does not discribe a mineral
Answer:
give us some further context to answer your question as well
Explanation:
Which example illustrates Newton's first law?
Answer: The motion of a ball falling down through the atmosphere, or a model rocket being launched up into the atmosphere are both examples of Newton's first law. The motion of a kite when the wind changes can also be described by the first law.
Explanation:
A cheetah can run at a maximum speed
91 km/h and a gazelle can run at a maximum speed of 72.7 km/h.
If both animals are running at full speed,
with the gazelle 87.5 m ahead, how long before
the cheetah catches its prey?
Answer in units of s.
Answer:
Approximately 17.21 seconds
Explanation:
With subtraction, we have the gazelle 18.3 km/h slower than the cheetah, which is about 5.08333 m/s. As the gazelle is 87.5 meters ahead of the cheetah, 87.5 divided by 5.083333333 is about 17.21 seconds.
1) Construct a graph showing the relationship between and independent variable and a dependent variable. 2) Demonstrate potential and kinetic energy. 3) Design a pendulum. 4) Investigate which variables influence the period of a pendulum. 5) Solve a problem using the scientific method. 6) Use graphs to characterize gow drop height responds to bounce height. (Next answer questions 1 - 6). 1) Write the goal of the lab or the question you tried to answer? 2) which observations, experiences, or lesson materials helped you form a hypothesis? 3) State your hypothesis? 4) Use the data from your experiment to complete these data tables? [Data Table 1: Period vs. Mass] Mass(g) / Time for 10 periods (s). [ Data Table 2: Period vs. String Length] String Length (cm) / Time for 10 periods (s). ( Question 5) Graph your data: A) Determine the values in seconds (s) for the y-axis based on your time data. B) Plot mass vs. time for 10 periods on the first graph. C) Plot string length vs. time for 10 periods on the second graph. ( question 6) Compare your data with your hypothesis. Explain how the data supports or does not support your hypothesis. If your data does not support your hypothesis, use the data you collected to answer the original question. { Truly Need Help and will mark brainlest} Please Please be very honest with your response.
A 1500 kg car traveling at 35 m/s hits its brakes and comes to rest in 5 seconds. Calculate the force applied by the brakes. (hint: solve for a first) Show all steps!!
Answer:
We are given:
m = 1500 kg
initial velocity (u) = 35 m/s
final velocity (v) = 0 m/s
time (t) = 5 seconds
acceleration = a m/s/s
Force (f) = F newtons
Solving for acceleration:
Using the first equation of motion:
v = u + at
replacing the variables
0 = 35 + a(5)
a = -35/5
a = -7 m/s/s
Force applied by the brakes:
From newton's second law of motion:
F = ma
replacing the variables
F = (1500)(-7)
F = -10500 N
Therefore, the force applied by the brakes is -10500 N, we have a negative sign since the force is being applied opposite to the direction of motion
A ball is dropped from a 19m high cliff. The acceleration on the ball was 9.8m/s². What was the ball's final velocity before hitting the ground?
Answer:
19.3 m/s
Explanation:
Take down to be positive. Given:
Δy = 19 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: v
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (9.8 m/s²) (19 m)
v = 19.3 m/s
Question 3 of 10
A 0.057 kg tennis ball and a tennis racket collide. The racket has an initial
momentum of -1.75 kg-m/s and a final momentum of -1.25 kg-m/s. The ball
has an initial momentum of 0.00684 kg-m/s. If you assume the collision is
elastic, what is the final velocity of the ball?
A. -0.50 m/s
B. -52.48 m/s
C. -2.99 m/s
d
D. -8.65 m/s
If we assume the collision to be elastic the final velocity of the ball will be -8.65 m/s, therefore the correct option is D.
What is elastic collision?It is a type of Collison for which the momentum, as well as the kinetic energy after and before the collision, is constant. there is no loss of energy in a perfectly elastic collision.
By using the law of conservation of momentum
momentum before collision = momentum after the collision
the initial momentum of rocket +initial momentum, of ball = final momentum of rocket + final momentum of the ball
By substituting the respective values
-1.75 + 0.00684 = -1.25 + m*v
-1.75 + 0.00684 = -1.25 + 0.057*v
v = -8.65 m/s
Thus, the final velocity of the ball is -8.65 m/s
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The number of
• in the atom of an element determines its chemical properties.
Answer:
Yes, the number of electrons determines the chemical properties of the atom.
Explanation:
A body is pulling a load of 50N with a string inclined at an angle of 30 degrees to the horizontal.
If the tension is 108N, calculate the force tending to lift the load off the ground.
Answer:
answer: 105 sin 30 = 52.5