1. The waterfall is 8.54 m high
2. The velocity with which the swimmer enters the water is 14.836 m/s
How to determine the time to reach the waterInitial velocity (u) = 1.9 m/sHorizontal distance (s) = 2.5 mTime (t) = ?s = ut
2.5 = 1.9 × t
Divide both sides by 1.9
t = 2.5 / 1.9
t = 1.32 s
1. How to determine the height of the waterfallTime (t) = 1.32 sAcceleration due to gravity (g) = 9.8 m/s²Height (t) =?h = ½gt²
h = ½ × 9.8 × 1.32²
h = 8.54 m
2. How to determine the velocityThe velocity with which the swimmer enters the water can be obtained as follow:
Initial velocity (u) = 1.9 m/sAcceleration due to gravity (g) = 9.8 m/s²Time (t) = 1.32 sFinal velocity (v) = ?v = u + gt
v = 1.9 + (9.8 × 1.32)
v = 1.9 + 12.936
v = 14.836 m/s
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How to find the angle in a projectile motion when displacement and heigh are the same
projectile's vertical speed v yv y v, start , y, drops prior to reaching its maximum height since its acceleration is in the opposite direction. Since the object's height is rising, the initial direction of the velocity is upward
.
At the projectile's highest point, vertical velocity zeroes out. After reaching the highest point, the vertical speed increases due of the same-direction acceleration. As the object's height lowers, the vertical velocity has a downward direction.
In projectile The beginning vertical velocity determines the maximum height. Increasing the launch angle raises the maximum height because steeper launch angles have a larger vertical velocity component.
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no sunlight
mineral-rich
extremely hot
chemosynthetic bacteria
These are all habitat features of __________________ within the oceans on Earth.
Responses
A estuariesestuaries
B benthic zonesbenthic zones
C kelp forestskelp forests
D hydrothermal ventshydrothermal vents
Answer: Hydrothermal vents is your answer.
Explanation:
it's very hot, there is no light, there are bacteria, and there it's mineral-rich.
what are the defining characteristics of matter
A Mass and time
B Mass and volume
C Mass and space
D volume and motion
At this rate, how long does it take to accelerate from 70 km/h to 110 km/h ?
Answer:
[tex]{ \tt{acceleration = \frac{ \triangle \: v}{t} }} \\ \\ { \tt{a = \frac{110 - 70}{t} }} \\ \\ { \tt{t = \frac{40}{a} } \: s}[/tex]
For a body under free fall; a = g = 9.8
[tex]{ \tt{t = \frac{40}{9.8} }} \\ \\{ \tt{t = 4.08 \: s}}[/tex]
in a hockey game, player A passes the puck to player B, who is standing on the blue line, as shown in the diagram. The puck travels distance d1=24.1m on its way to player B, making an angle = 52.5 degrees with the x axis. Player B then passes the puck so that it moves d2=7.9m on the blue line in the negative x direction. A) What is the magnitude of the total displacement in meters? B) What angle, in degrees, does the total displacement make with the x axis?
The magnitude of the total displacement is 20.28 m.
The angle of the total displacement with the x axis is 70.5⁰.
Total displacement
The total displacement of the player is calculated as follows.
Apply cosine rule as shown below;
d² = d₁² + d₂² - 2d₁d₂ cos(θ)
d² = (24.1)² + (7.9²) - (2 x 24.1 x 7.9) cos(52.5)
d² = 411.42
d = √411.42
d = 20.28 m
Angle of the displacement with horizontalApply sine rule as shown below;
d/sinD = d₂/sinD₂
20.28/sin(52.5) = 7.9/sinD₂
25.562 = 7.9/sinD₂
sinD₂ = 7.9/25.562
sinD₂ = 0.309
D₂ = sin⁻¹(0.309)
D₂ = 18⁰
angle with x axis = 18⁰ + θ
= 18⁰ + 52.5⁰
= 70.5⁰
Thus, the magnitude of the total displacement is 20.28 m.
The angle of the total displacement with the x axis is 70.5⁰.
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You jog at 9.5 km/h for 8.00 km; then you jump into a car and ride an additional 16.0 km. What average speed must the car have for the average speed for the entire 24.0-km trip to be 20 km/h
Average speed must the car have for the average speed for the entire 24.0 km trip 67.039 km/hr.
Equation :Average speed for entire journey (jogging + drive) = 20 km/hr
total distance traveled = 24 km
total time spent (on jog + drive) = total distance/average speed
total time spent (on jog + drive) = 24/20
total time spent (on jog + drive) = 1.2 hr
Jogging:
Speed = 9.5 km/hr
Distance = 8.0 km
Time spent jogging = Distance/Speed
= 8.0 / 9.5 = 0.842 hr
Driving car :
Speed = ?
Distance = 24 km
time spent driving = total time spent on journey - time spent jogging
time spent driving = 1.2 - 0.842 = 0.358 hr
average speed for driving = distance / time
average speed for driving = 24/0.358
average speed for driving = 67.039 km/hr
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Point of View 10:Question 5
Read the sentences below to answer the question.
Zoe glared at her soccer opponent from across the field. She had practiced all week, but still, she had to conceal the growing lump of fear inside. Was she good enough to defend Star, her greatest rival? Meanwhile, Star kept her focus on the soccer ball, visualizing her first play and identifying the path she would take to the goal.
The narration style in the writing above is..
Select one:
first person
third person omniscient
third person limited
second person
The narration style in the para is third person omniscient
As it narrates the whole incident which is going on.
What is third person omniscient?
The most flexible and open POV available to writers is third person omniscient. An omniscient narrator is one who has all knowledge and all sight, as the name suggests.
The narrator may occasionally access the consciousness of a few or many different characters while the narration is taking place outside of any one particular character.
Some authors employ this viewpoint to build a more "godlike" or purposefully "authorial" persona that enables them to offer dispassionate commentary on the action. Detailed location descriptions that contribute to the mood or atmosphere of a scene may be an example of this, as well as philosophical tangents that serve to develop concepts that only loosely link to the action of the story.
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Which of the following is a form of identity fraud?
A. Chipping
B. Skimming
C. Chirping
D. Skipping
Highway safety engineers build soft barriers so that cars hitting them will slow down at a safe rate. A person wearing a safety belt can withstand an acceleration of 300 m/s/s. How thick should barriers be to safely stop a car that hits the barriers at 76.86 km/h?
The thickness of the barrier that will stop a car travelling at 76.86 km/h is 0.76 m.
Thickness of the barriers
The thickness of the barriers can be calculated using a kinematic equation as shown below;
v² = u² + 2ad
where;
d is the thickness of the barrierv is the final velocityu is the initial velocity a is the acceleration of the carv is the final velocityThe given final velocity of the car;
v = 76.86 km/h = 21.35 m/s
Thickness of the barrier that will stop a car travelling at 76.86 km/h is calculated as follows;
d = v²/2a
d = (21.35²) / (2 x 300)
d = 0.76 m
Thus, the thickness of the barrier that will stop a car travelling at 76.86 km/h is 0.76 m.
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List the octal and hexadecimal numbers from 16 to 32. using A and B for the last two digits, list the numbers from 8 to 28 in base 12
The correct answer is :
In octal,
Decimal 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
Octal 20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 37 40
Hexadecimal 10 11 12 13 14 15 16 17 18 19 1A 1A 1C 1D 1E 1F 20
Decimal 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
Base13 8 9 A B C 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 20 21 22
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Write an experiment to show motion of particles in the matter
The experiment to show the random motion of the particles is the movement of particles in a solution is the simplest representation of motion of particles.
What is the experiment to show the motion of particle in a matter?Pick up two glass beakers. Fill one beaker with hot water and the second one with cold water. Add KMnO4 crystals in very small quantities to both. We thus see that KMnO4 particles move faster in hot water than in cold water.
What is the experiment to show evidence for attraction between the particles:Pick up some iron and a piece of chalk. We learn that it's quite difficult to work with iron. This leads us to believe that the Forces can affect matter particles. among them. The pressure kept between particles and the power of the attraction in materials is at its strongest between particles the least in gases.
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4. Invalid experiments sometimes result from scientists influencing them
unfairly. This is called:
O Bias
O Scientific Method
O Controlled Experiment
Hypothesis
Answer: Bias
Explanation:
If a liter of gasoline cost $0.58, and a car gets 11 km/L, what would it cost to travel 100km?
Answer: $638,00
Explanation: Hi Tania... I believe we're in the same class and I was just doing the Science homework when I saw your question... Take this as a gift for doing the Spanish homework last time... Thx! Oh, btw that's the procedure and the response.
X=(1100*0.58)/1= 638,00
which best describes savanna’s error
Answer:
The FM carrier waves should have different frequencies than the carrier waves, not different amplitudes.
Explanation:
The label at the top of the diagram should be “FM Radio,” not “Cell Phone.
1. Based on the mc free-body diagram, type an expression for the horizontal acceleration of the cart only in terms of T , mh , mc , μk , and g .
2.Based on the mh free-body diagram, type an expression for the vertical acceleration of the hanging mass.
3. As shown in the first figure, the horizontal acceleration of the cart to the right must be equal to the downward vertical acceleration of the hanging mass if the string does not stretch. Given that both the cart and the hanging mass have the same acceleration, eliminate the tension force ( T ) from your answers to Q1 and Q2 to find an expression for the acceleration only in terms of mh , mc , μk , and g .
(a) The expression for the horizontal acceleration of the cart is
a = (T - μN)/Mc.
(b) The expression for the vertical acceleration of the hanging mass is
a = (T - Mhg)/Mh.
(c) The expression for the acceleration of both cart and the hanging mass is a = (Mhg - μN) / (Mc - Mh).
Horizontal acceleration of the cart
The horizontal acceleration of the cart is calculated as follows;
The horizontal acceleration of the cart is determined from the net horizontal force acting on the system.
T - μN = Mca
where;
Mc is mass of the carta is the acceleration of the carta = (T - μN)/Mc
Vertical acceleration of the hanging massT - Mhg = Mha
where;
Mh is the mass of the hanging massa = (T - Mhg)/Mh
Acceleration of both objectsIf the acceleration of the cart and the hanging mass are equal, our final equation becomes;
From first equation;
a = (T - μN)/Mc
Mca = T - μN
Mca + μN = T
From second equation;
a = (T - Mhg)/Mh
Mha = T - Mhg
Mha + Mhg = T
Solve the two equation together;
Mca + μN = Mha + Mhg
Mca - Mha = Mhg - μN
a(Mc - Mh) = Mhg - μN
a = (Mhg - μN) / (Mc - Mh)
Thus, the expression for the acceleration of both cart and the hanging mass is a = (Mhg - μN) / (Mc - Mh).
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As you go up the y-axis, what happens to the number of sprouted bean seeds?
A. Sprouted bean seeds decrease.
B. Sprouted bean seeds increase.
C. Sprouted bean seeds remain constant.
D. None of the above
As we go up the y-axis, the number of sprouted bean seeds increase (option B).
What is a graph?Graph is a data chart intended to illustrate the relationship between a set (or sets) of numbers (quantities, measurements or indicative numbers) and a reference set.
In a graph, there are two axes as follows;
Y-axis or vertical axisX-axis or horizontal axisAccording to this question, a graph of number of sprouted bean seeds on the y-axis is plotted against temperature on the x-axis.
We can observe that as we go up the y-axis, the number of sprouted bean seeds increase.
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A bus travels from Houston, Texas to Dallas, Texas, in 4.7 h with an average velocity of 76 km/h to the north. What is the bus’s displacement?
The bus's displacement from Houston to Texas with average velocity = -16.170 km/h
Evaluation :
average velocity = ( initial velocity -- final velocity )
( v-u )
displacement = [tex]\frac{v-u}{t}[/tex]
displacement = [tex]\frac{0 - 76 }{4.7}[/tex]
= - 16.170 km/h
Average velocity :
Average velocity is determined as the change in position or displacement (∆x) divided by the time period (∆t) in which the displacement results . The average velocity could be positive or negative based upon the sign of the displacement. The SI unit of average velocity is defined as meters per second (m/s or ms-1).
Displacement :The displacement is commonly the difference in the position of the two marks and is independent of the path covered when traveling between the two of the marks. The distance covered , however, is the total length of the path chosen between the two marks. Displacement defined as a vector quantity that mean to "how far out of the place an object is"; it's the object overall changes in the position.
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and gravity is exerted on an object. DISTANCE VOLUME GRAVITY MASS both impact how much
Answer:
Explanation:
yes, it will.40
A 1.2kg stone is tied to string and swung in a vertical circle with a radius of 0.85 m. The string can withstand a tension of 40.0 N. At the maximum speed can the stone move at the bottom of its path without the string breaking
The maximum speed of the stone tied to the string is 5.32 m/s.
The mass of the stone is m = 1.2 kg.
The mass is tied to a string and it swung in a vertical radius of r = 0.85 cm.
The tension on the string is T = 40 N.
The centripetal force of a body moving in a circle of radius r is given as:
T = ( mv² / r )
Where T is the tension, m is the mass of the body and r is the radius of the circle.
Then,
v² = ( T × r ) / m
v = √ [ T × r / m ]
v = √ [ 40 × 0.85 / 1.2 ]
v = √ [ 34 / 1.2 ]
v = √28.3
v = 5.32 m/s
The maximum speed of the stone before the string breaks is 5.32 m/s.
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Can you please help me with this!?
To determine how classroom temperature affected test performance of students an experiment was conducted with each group of students in hot room, cold room and a room with normal temperature.
• Dependent variable- Test performance
• Independent variable- Room temperature
• Control variables/Constants - Group of students
• Control group – Students in the room with normal temperature
Test performance depends on the temperature of the room.
Room temperature is an independent variable that does not depend on any factor
Students were the constants in the experiment
Among the students, students in normal temperature room is the control group as they were placed just to check the correctness of the experiment.
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Layla jogs with an average velocity of 2.4 m/s east. What is her displacement after 46 seconds?
Answer:
[tex]{ \tt{displacement = velocity \times time }} \\ { \tt{ = 2.4 \times 46}} \\ { \tt{ = 110.4 \: m}}[/tex]
3) Two particles are travelling along a straight line AB of length 20m.
At the same instant one particle starts from rest at A and travels
towards B with a constant acceleration of 2 ms72 and the other
particle starts from rest at B and travels towards A with a constant
acceleration of 5 ms2, find how far from A the particles collide.
At distance of 40/7 m from A the particles will collide.
Let particles collide at C
And let distance AC = x and BC = y
then, x+y = 20
As the equation used for distance required here is: s= ut +1/2 a[tex]t^{2}[/tex] and here u=0
Then, 1/2 (2)[tex]t^{2}[/tex] +1/2 (5)[tex]t^{2}[/tex] =20
7/2[tex]t^{2}[/tex] =20
[tex]t^{2}[/tex] = 40/7
Now
x= 1/2 (2)[tex]t^{2}[/tex] = 40/7 m
Therefore, at distance of 40/7 m from A the particles will collide.
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Question 10 of 10 Why are experiments often performed in laboratories? A. You can eliminate variables like air resistance. B. Scientists only want to know what happens in labs. C. It is harder to control variables in a laboratory. D. They are better at representing the real world. Need answer asap.
Consider the five pairs of long, parallel wires shown. The arrows indicate the direction of the current in each wire, with small
arrows representing a current of 3 A and large arrows representing a current of 9 A. For each pair, determine in which one of the
regions the net magnetic field is zero someplace:
on the left of both wires
between the two wires
on the right of both wires
• none of the above (i.e., nowhere)
The Net magnetic field will be at the left in case I, right in Case II, In-between in Case III, and Nowhere in Case IV and V.
Case I
As the current in both wires is not equal, the net magnetic field will not be zero in the middle. It will be zero somewhere near the 3 A current carrying wire and farther from the 9A current carrying wire. Which is to the Left.
Case II
It is similar to the case I, as the current in both wires is not equal, and the net magnetic field will not be zero in the middle. It will be zero somewhere near the 3 A current carrying wire and farther from the 9A current carrying wire. Which in this case is to the Right.
Case III
As the current in both wires is the same and equal to 3 A, the net magnetic field will be zero in the middle. Which is in Between.
Case IV
As the current flowing in the wire is opposite in the direction, the magnetic field will always be in the same direction. So the magnetic field will be zero Nowhere.
Case V
It is similar to the above case IV, as the current flowing in the wire is opposite in the direction, and the magnetic field will always be in the same direction. So the magnetic field will be zero Nowhere.
Therefore, the Net magnetic field will be at left in case I, right in Case II, In-between in Case III, and Nowhere in Case IV and V.
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Abby throws a ball straight up and times it. She sees that the ball goes by the top of a flagpole after 0.50 s and reaches the level of the top of the pole after a total elapsed time of 4.1 s. What was the speed of the ball at launch? Neglect air resistance
The speed of the ball at launch is 23m/s.
After 0.5 seconds, the ball initially passed the pole.
Then it will again reach the top of the pole after 4.1 seconds.
Consequently, the ball's travel time above the pole is
Δt = 4.1 - 0.5 = 3.6sec
Now the time it will take to reach the highest point from the top of the pole is given as t = Δt/2 = 1.8sec.
Now the total time to reach the top from the bottom is given as 2.3sec.
After 0.5 seconds, the ball initially passed the pole.
Consequently, the ball's travel time above the pole is
[tex]v_{f} - v_{i} = at[/tex]
[tex]v_{i}=22.54m/s[/tex]
also the speed after 0.5 s is
[tex]v_{f} = 17.64m/s[/tex]
So, the speed of the ball at launch is 23m/s.
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A body, initially at rest, explodes into two masses M, and M₂ that
move apart with speeds v, and v₂ respectively. What is the ratio ?
V₂
A.
C.
M₁
M₂
(M₁2
1|2
B.
D.
M₂
M₁
M.
(M₂₁
Using conservation of momentum we will get that the ratio of velocity of mass M₁ and mass M₂ is ratio of M₂ to M₁
i.e. v₁/v₂=M₂/M₁
What is conservation of momentum ?
conservation of momentum states that in an isolated system the initial momentum will always be equal to final momentum.
given:
initial velocity of the object= 0m/s
initial momentum of the object = 0kg.m/s
now the mass M breaks into 2 pieces such that:
M₁ is the mass of first piece
M₂ is the mass of first piece
velocity of M₁ is v₁
velocity of M₂ is -v₂ as to conserve momentum
using conservation of momentum we get,
initial momentum = final momentum
0 = M₁×v₁ + M₂×(-v₂)
M₁×v₁= M₂×v₂
v₁/v₂ = M₂/M₁
using conservation of momentum we will get that the ratio of velocity of mass M₁ and mass M₂ is ratio of M₂ to M₁
i.e. v₁/v₂=M₂/M₁
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What precise meaning do you attach to the statement r = (24.0 ± 0.3) mm, where r is the radius of a tube.
The information provided above demonstrates that the range of the tube's radius within which we are certain is from 23.7 to 24.3.
Accurate findings are desired when any quantity is being measured. Closeness of the measurements to a particular value is referred to as accuracy (like the theoretical value). It is not insignificant that measurements can contain inaccuracies; as a result, the outcome frequently includes a margin error. The sentence above displays the radius's measured value along with its margin of error.
In other words, the aforementioned statement represents a quantity's numerical value along with its tolerance, or the only permissible (and conceivable) values to ensure that the measurement is accurate. The information provided above demonstrates that the range of the tube's radius within which we are certain is from 23.7 to 24.3.
These are all valid radius values, along with all others within this interval. By increasing the number of trials in a measurement, the value for the upper and lower bounds of the error can be decreased.
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For the vectors shown in the figure, determine
(Figure 1)
Figure
B (B=26.5)
56.0%
Ā (A = 44.0)
28.0°
C(C=31.0)
< 1 of 1 >
X
the magnitude of B-3A
Express your answer using three significant figures.
|B-3A| =
The magnitude of the given vectors for B - 3A expression is determined as 105.5 m.
What is magnitude of vectors?The magnitude of a vector is the length of the vector. The magnitude of the vector a is denoted as |a|.
The magnitude of a vector represents the absolute value of the given vector.
Magnitude of B - 3AThe magnitude of the expression B - 3A is calculated as follows;
B - 3A = 26.5 - 3(44)
B - 3A = 26.5 - 132
B - 3A = -105.5
|B - 3A| = 105.5
Thus, the magnitude of the given vectors for B - 3A expression is determined as 105.5 m.
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What two gases together make up about 99% of Earth's atmosphere? a. Carbon dioxide and nitrogen
A: Nitrogen and oxygen
B: Nitrogen and helium
C: Argon and oxygen
D: Trace gases and nitrogen
An arrow is shot vertically upward and then 1.87 s later passes the top of a tree 39.9 m high. How much longer will the arrow travel upward, how high will it go?
Time :
Height:
In order to determine the amount of time it would take the arrow to travel upward, we would have to first determine its initial velocity by using the second equation of motion.
Note: Since the arrow was projected upward, the acceleration due to gravity would be negative because it indicates deceleration.
From the second equation of motion, we have:
S = ut - ½gt²
39.9 = u(1.87) - ½ × (-10) × 1.87²
39.9 = 1.87u + 17.49
1.87u = 39.9 - 17.49
1.87u = 22.41
u = 22.41/1.87
Initial velocity, u = 11.98 m/s.
Now, we can calculate the amount of time it would take the arrow to travel upward by applying the first equation of motion:
V = u + at
0 = 11.98 + (-10)t
10t = 11.98
t = 11.98/10
Time, t = 1.198 seconds.
How to calculate the maximum height?In order to calculate the maximum height, we would apply the second equation of motion:
S = ut + ½gt²
S = 11.98(1.198) + ½ × (-10) × 1.198²
S = 14.35 - 7.18
Maximum height, S = 7.17 meters.
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