Under the low-power microscope, the kind of specimen Diatoms is Holotype.
What is Holotype Specimen?A solitary actual example of an organism known to have been utilized during the conventional portrayal of the species is known as a holotype.It is either the sole occurrence of this sort of actual model, however, it is explicitly recognized as the holotype.What are Diatoms?Diatom slides are often created by letting the suspension of diatoms settle out overnight on a coverslip, as explained below. It can take up to two days for this to disseminate the diatoms evenly on the coverslip, though.The frustule, a silica (hydrated silicon dioxide) cell wall, that surrounds diatoms is a distinctive aspect of their anatomy.To learn more about Diatoms refer to:
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A driver jumped from a top of a 15-meter-high cliff down to the water below. if he moved with an initial horizontal velocity of 10m/s
a.) Find his Vx or horizontal Velocity
b.) Calculate the time it took him to reach the water
c.) How many meters will his range or horizontal displacement
Given:
Formula:
Solution:
Final Answer:
.
Hi please help I need it bc I'm still confuse
(a) The initial horizontal velocity of the driver is 10 m/s.
(b) The time taken for him to reach the water is 1.75 s.
(c) The horizontal displacement is 17.5 m.
Horizontal velocity of the driver (Vx)
The initial horizontal velocity of the driver is given as 10 m/s.
Time of motion of the driverh = vy(t) + ¹/₂gt²
where;
vy is initial vertical velocityt is time of motionh is height of fallh = 0 + ¹/₂gt²
2h = gt²
t = √(2h/g)
t = √(2 x 15 / 9.8)
t = 1.75 s
Horizontal displacement of the driverX = Vx(t)
X = 10 x 1.75
X = 17.5 m
Thus, the initial horizontal velocity of the driver is 10 m/s, the time of motion is 1.75 s and the horizontal displacement is 17.5 m.
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what is forward and downward speed
Forward and downward speed are referred to as acceleration and deceleration respectively.
What is Acceleration?This is the rate at which an object gains speed with respect to time with its unit being meter per second square(m/s²).
Objects in constant speed are said to have a uniform motion. The downward speed is known as deceleration which involves the object losing speed to return to a state of rest.
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The _________ system produces electrical power while the engine is running to operate all the electrical components and recharge the battery
Answer:
kinetic energy and potential and electrical energy
Explanation:
it will be need to maintain the motion of the vehicle and produce electrical energy
meme on newtons law of motion (should me made ur self not from any searh engine)
For a pull box housing shielded conductors carrying over 1,000V and used in an angled pull, the distance between a conductor entry and exit must be________
Answer: For a pull box housing shielded conductors carrying over 1000V and used in an angled pull, the distance between a conductor entry and exit must be 36 times the outside diameter.
Explanation: To find the correct explanation to the question, we have to find more about the pull box housing shielded conductors.
What is pull box?Pull box is an underground electrical container, that made up of composite materials that used in traffic signal installation and also provide access to underground detectors as well as interconnected cables. These are used to simplify the wiring installation. What are the types of pull boxes?There we have two types of pull boxes, such as, one is for straight pulls and the other is for angle or U pulls.For straight pulls the length of the box shall not be less than 48 times the outside diameter of the cable.Angled pulls can be again classified as, Distance to opposite wall, Distance between entry and exit and removable sides.The distance between a conductor entry and exit from the box shall not be less than 36 times the outside diameter of the conductor.Thus, we can conclude that, for a pull box housing shielded conductors carrying over 1000V and used in an angled pull, the distance between a conductor entry and exit must be 36 times the outside diameter.
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4 . A negative charge -0.450 exerts an upward 0.150N force on an unknown charge 0.250m directly below it . ( a ) what is the unknown charge ( magnitude and sign ) ? ( b ) what are the magnitude and direction of the force that the unknown charge exert charge exerts on the -0.450 charge ?
(a) The unknown charge kept at distance 0.250m directly below -0.450 charge is 2.315 x 10⁻¹² C.
(b) The magnitude of the force is 29.16 x 10¹² N.
What is electrostatic force?The electrostatic force F between two charged objects placed distance apart is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.
F = kq₁q₂/r²
where k = 9 x 10⁹ N.m²/C²
Given is a negative charge -0.450 exerts an upward 0.150N force on an unknown charge 0.250m directly below it .
In equation of the magnitude of force, only the magnitude of charges are taken.
Substituting the values into the expression ,we get the charge
0.150 = (9 x 10⁹ x 0.450x q)/ (0.25)²
q = 2.315 x 10⁻¹² C
Thus, the magnitude of the unknown charge is 2.315 x 10⁻¹² C.
b)Let the unknown charge be 0.450C to exert on -450C.
Substituting the values into the expression ,we get the electrostatic force
F = kq₁q₂/r²
F = (9 x 10⁹ x 0.450x 0.450)/ (0.25)²
q = 2.315 x 10⁻¹² C
Thus, the magnitude of the force is 29.16 x 10¹² N.
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Please. Physics is so difficult.
A pendulum on a grandfather clock is supposed to oscillate once every 2.00 s, but actually oscillates once every 1.99 s. How much must you increase its length to correct its period to 2.00 s?
Answer:
0.010 m
Explanation:
So the equation for a pendulum period is: [tex]y=2\pi\sqrt{\frac{L}{g}}[/tex] where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:
[tex]1.99 s = 2(3.14)\sqrt{\frac{L}{9.8 m\backslash s^2}}\\[/tex]
Evaluate the multiplication in front
[tex]1.99 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}[/tex]
Divide both sides by 6.28
[tex]0.317 s= \sqrt{\frac{L}{9.8 m\backslash s^2}}[/tex]
Square both sides
[tex]0.100 s^2= \frac{L}{9.8 m\backslash s^2}[/tex]
Multiply both sides by m/s^2 (the s^2 will cancel out)
[tex]0.984 m = L[/tex]
Now now let's find the length when it's two seconds
[tex]2.00 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}}[/tex]
Divide both sides by 6.28
[tex]0.318 s = \sqrt{\frac{L}{9.8 m\backslash s^2}[/tex]
Square both sides
[tex]0.101 s^2 = \frac{L}{9.8 m\backslash s^2}[/tex]
Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)
[tex]0.994 m = L[/tex]
So to find the difference you simply subtract
0.984 - 0.994 = 0.010 m
the answer should be 0.00988 m
The deflection angle of the laser beam as it exits the prism is 22. 6º. If the prism had been made of glass instead of polystyrene plastic, what would the deflection angle have been?.
A wave traveling in a string has a wavelength of 35 cm, an amplitude of 8. 4 cm, and a period of 1. 2 s. What is the speed of this wave?.
0.29 m/s is the speed of this wave.
How can we determine a wave's speed?
Given: The wave's traveling wavelength is () = 35 cm
The wave's period (t) is 1.2 seconds.
The wave's frequency is 0.833 Hz.
We must ascertain a wave's wave speed.
Formula for wave speed is 35 x 0.833 (wavelength x frequency).
= 29.167 cm/sec
= 0.29 m/sec
As a result, the wave's speed is 0.29 m/s.
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The figure below shows a ball on a curved surface. The ball is released at point A. At which point does the ball have maximum gravitational potential energy?
Answer:
A
Explanation:
At point 'A' the ball is highest to the ground.As the formula of gravitational potential energy is Eg=MGH, in which 'M' is mass,'H' is height and 'G' is the gravitational pull of earth
Block X and Y are attached to each other by a light rope and can slide along a horizontal
surface. Mass of block X is 10 kg and that of block Y is 5 kg. The magnitude of force of
friction on blocks X and Y is 8.0 N and 4.0 N respectively. Find the action-reaction forces
that the blocks exert on each other if an applied force of 40 N [right] acts on block per the
illustration below
please answer asap
the action-reaction forces that the blocks exert on each other will 13 N and 22N
Force on a Horizontal PlaneForce is the product of mass and acceleration. There will be constant acceleration on both object X and Y.
Given that the Mass of block X is 10 kg and that of block Y is 5 kg. The magnitude of force of friction on blocks X and Y is 8.0 N and 4.0 N respectively. if an applied force of 40 N [right] acts on block,
We are given that:
Mass of block X = 10 KgMass of block Y = 5 KgApplied force = 40 NMagnitude of friction on X block = 8NMagnitude of friction on Y block = 4 NWith the use of the formula below,
F - [tex]F_{r}[/tex] = ma
Where
F = force applied[tex]F_{r}[/tex] = frictional forcem = massa = accelerationThe acceleration will be;
40 - (8 + 4) = (10 + 5)a
40 - 12 = 15 a
28 = 15a
a = 28/15
a = 1.8667 m/[tex]s^{2}[/tex]
The tension between the two blocks will be
Force applied-frictional force on X-tension = mass of X × acceleration
That is,
F - [tex]F_{r}[/tex] = ma
40-8-T = 10 × 1.8667
32 - T = 18.667
T = 32 - 18.667
T = 13.33 N
The opposite reaction will be
F - 8 = 10 x 1.8667
F - 4 = 18.67
F = 18.67 + 4
F = 22.67 N
Therefore, the action-reaction forces that the blocks exert on each other will 13 N and 22N
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write any three uses of solution in our daily life
Answer:
इअग्सिव्ह्ह्व्य्वुव्व्ह्विहेधेइएह
Explanation:
ह्य्स्स्येउ?स्य्ग्स्झ्स
A driver, traveling at 22.0 m/s, slows down her car and stops. What work is done by the friction force against the wheels if the mass of the car is 1500 kg? Group of answer choices -3.6 x 105 joules None of the above 3.6 x 106 joules -4.5 x 106 joules 4.5 x 105 joules
-3.6× 10^5J is the Work done by friction force
Work done = force × distance
v^2 = u^ +2as
u= 22m/s
a =10m/s^2
When the car stops the final velocity (v) =0
0= 22^2 +2×10×s
s = -484/20
s =-24.2m
Work done = force × distance
Force = mass × acceleration
Work done = 1500×10× -24.2
= -3.6×10^5J
Hence the Work done is -3.6×10^5J
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A sled having a certain initial speed on a horizontal surface comes to rest after traveling 20 m. If the coefficient of kinetic friction between the object and the surface is 0.20, what was the initial speed of the object
The initial speed of the object was 6.26 m/s.
Acceleration of the sled, a = -g*μk
= -9.8*0.2
= -1.96 m/s^2
Let u is the initial speed and v is the final speed.
Applying, v^2 - u^2 = 2*a*d
0^2 - u^2 = 2*(-1.96)*10
u = sqrt(2*1.96*10)
= 6.26 m/s
The kinetic friction coefficient μok is the ratio of the friction force to the normal pressure experienced by a body moving on a dry, non-easy floor. The determination of μok may be achieved quite appropriately through an alternative creative mechanism described with the aid of Timoshenko
The coefficient of friction (fr) is quite a number that is the ratio of the resistive force of friction (Fr) divided with the aid of the everyday or perpendicular force (N) pushing the items collectively. it is represented by the equation: fr = Fr/N.
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What is the amount of matter in a substance?
Answer:
mass
Explanation:
Mass (M) is the measure of the amount of matter in an object. Mass is measured in grams (g). Mass is measured on a balance by comparing the object against other objects with known masses.
Answer:
Mass is the amount of matter in a substance
A moving car has a kinetic energy of 0.3 MJ. If the car's speed decreases two times, its new kinetic energy will be
New kinetic energy:
If the initial kinetic energy of a car is 0.3 MJ and the speed of the car decreased two times then the new kinetic energy will be 0.075 MJ.
The calculation for kinetic energy:
Step-1:
It is given that the car has a kinetic energy of 0.3 MJ. It is required to find the new kinetic energy when the car's speed decreases two times from its initial speed.
Let the mass of the car is m and the initial speed of the car is v.
It is known that the kinetic energy is calculated as,
[tex]\text{K.E}=\frac{1}{2}\times \text{mass}\times\text{speed}^2[/tex]
Therefore the initial kinetic energy is the car is,
[tex]\text{K.E}_1=\frac{1}{2}\times \text{m}\times\text{v}^2[/tex]
Step-2:
The speed of the car is decreased two times. Thus the speed becomes v/2. Therefore the final kinetic energy is the car is,
[tex]\text{K.E}_2&=\frac{1}{2}\times \text{m}\times(\frac{v}{2})^2\\&=\frac{1}{4}\times\frac{1}{2}\times \text{m}\times\text{v}^2\\&=\frac{1}{4}\times\text{K.E}_1\\&=\frac{1}{4}\times {0.3}\text{ MJ}\\&=0.075 \text{MJ}[/tex]
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How do the accelerations of the rifle and bullet compare?
Answer:
The acceleration of the rifle equals the force that the bullet exerts on the rifle divided by the mass of the rifle. The two forces are equal, but since the mass of the rifle is much greater than the mass of the bullet, the acceleration of the rifle is much less than the bullet's acceleration.
Which phenomena support only the wave theory of light? select 2 options. reflection refraction diffraction interference photoelectric effect
Interference and diffraction are the phenomena that support only the wave theory of light. Options 2 and 3 are correct.
What is the interference of waves?The result of two or more wave trains flowing in opposite directions on a crossing or coinciding pathways. This phenomenon is known as the interference of waves.
The phenomenon of interference occurs when two wave pulses are traveling along a string toward each other.
The light wave hypothesis states that light behaves like a wave. Since light is an electromagnetic wave, it may be transmitted without a physical medium.
Light has magnetic and electric fields, much like electromagnetic waves do.
Transverse waves, such as those seen in light waves, oscillate in the same direction as the wave's path. A wave of light may experience interference as well as diffraction as a result of these properties.
All of the remaining options are the light phenomenon.
Hence, options 2 and 3 are correct.
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10) What is the total energy of a 450-g mass that is attached to a horizontal spring with a force constant of 260 N/m and oscillates along a frictionless horizontal surface with amplitude of 8.0 cm
Answer:
E = 1/2 * k x^2 is the energy stored in the spring
E = 1/2 * 260 N/m * .08^2 m = .832 Joules
When a second student joins the first, the height difference between the liquid levels in the right and left pistons is 35 cm . What is the second student's mass
Answer:
The mass of the second student is 60 kg.
Explanation:
Note: It is assumed that radius of piston is 0.23 m and density ρ is 850 kg/m^3 and mass of first student is 55 kg.
There occurs a change in pressure on a piston when some mass is added on the other piston.
This change of mass on the piston is given by the formula,
Δm=ρhA/g
where A is the area of piston.
Given h=35 cm or 0.35 m, change in mass is,
Δm=850*(0.35)*π*(0.23)^2 / 9.8
Δm=5.0 kg
Therefore, the mass of the second student is 55+5=60 Kg
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what are the types of waves
A 4.30 kg sign hangs from two wires. The
left wire exerts a 31.0 N force at 122°.
What is the magnitude of the force exerted
by the second wire?
F₁"
*W
magnitude (N)
F2
Enter
The magnitude of the force exerted by the second wire is 16.5 N.
Resultant of the forcesApply cosine rule to determine the force on the second wire;
W² = F₁² + F₂² - 2F₁F₂ cos(θ)
where;
W is the suspended weightθ is the angle between the two wires(4.3 x 9.8)² = 31² + F₂² - 2(31) F₂(cos 122)
1775.78 = 961 + F₂² + 32.85F₂
814.78 = F₂² + 32.85F₂
F₂² + 32.85F₂ - 814.78 = 0
Solve the quadratic equation using formula method;
F₂ = 16.5 N
Thus, the magnitude of the force exerted by the second wire is 16.5 N.
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Select the correct answer. You travel in a circle, whose circumference is 8 kilometers, at an average speed of 8 kilometers/hour. If you stop at the same point you started from, what is your average velocity? A. 0 kilometers/hour B. 2 kilometers/hour C. 4 kilometers/hour D. 8 kilometers/hour E. 16 kilometers/hour
Answer:
Since velocity is a vector quantity there is no net displacement and the average velocity is zero
(A) is correct
A person is using a rope to lower a 8.0-N bucket into a well with a constant speed of 2.0 m/s. What is the magnitude of the force exerted by the rope on the bucket
The magnitude of the force exerted by the rope on the bucket is 8N.
To find the answer, we need to know about the force.
What's force?According to Newton's second law, force is expressed as rate of change of momentum.It can be written as mass × acceleration.What's the force experienced by an object, when it moves with constant velocity?When an object moves with constant velocity, its acceleration is zero.Since acceleration is zero, so the force experienced by it is zero.Thus, we can conclude that there's no force on the bucket by the person as he is lowering it by constant velocity. But force on the bucket by the rope is 8N.
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A movement that keeps the distal end of the body segment fixed in one location describes what type of kinetic chain movement? Open kinetic chain movement Lateral kinetic chain movement Dynamic kinetic chain movement Closed kinetic chain movement
A movement which keeps the distal end of the body segment fixed in one location describes what type of kinetic chain movement is known as a closed kinetic chain movement.
What is a closed kinetic chain movement?A closed kinetic chain movement is defined as that position where the most distal aspects of a particular extremity are fixed to the earth or another solid object.
In conclusion; a movement which keeps the distal end of the body segment fixed in one location describes what type of kinetic chain movement is known as a closed kinetic chain movement.
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Anybody else play genshin??
I’m bored and don’t feel like doing hw (also my ar is 24 )
Answer:
no sorry
Explanation:
Answer:
yes i play genshin
Explanation:
ignore how this is a year late...what server r u on ?im on america and europe!!!!!!!!!!!!(≧∇≦)
A backpack has a mass of 8 kg. It is lifted and given 54.9 J of gravitational
potential energy. How high is it lifted? Acceleration due to gravity is g = 9.8
m/s2
The height is 0.70 meters
The formula for gravitational potential energy is
= mgh
m= mass, g= acceleration due to gravity, h= height
54.9= 8 × 9.8 × h
54.9= 78.4h
h= 54.9/78.4
= 0.70 meters
Hence the height at which the bacpack was lifted is 0.70 meters
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Answer:
its 0.7m
Explanation:
A sky diver, mass 90 kg, reaches a terminal velocity of 60 m/s. What is the approximate magnitude of the force of air resistance?.
The approximate magnitude of the force of air resistance is 882 N
Data obtained from the questionMass (m) = 90 KgTerminal velocity = 60 m/sForce (F) =?How to determine the forceThe magnitude of the force of air resistance can be obtained as illustrated below:
Mass (m) = 90 KgAcceleration due to gravity (g) = 9.8 m/s²Force (F) =?F = mg
F = 90 × 9.8
F = 882 N
Thus, approximate magnitude of the force of air resistance is 882 N
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A pendulum can be simply made of a suspended string tied to a weight. If the length of string is measured to be 1 m and the weight is measured to be 0.5 kg, determine its period of oscillation and its frequency of oscillation. Does its period depend on the weight tied to the string?
The time period of the pendulum is 2 sec, and the frequency will be 0.5 and will on depend on weight of the object.
The time period of a basic pendulum is defined as the time it takes to complete one full oscillation and is indicated by the letter "T".
This concept of frequency leads to the simplest frequency formula
Frequency = 1 / T.
T signifies the time it takes to complete one wave cycle in seconds, while f denotes frequency.
The formula for time period of simple pendulum is
Time Period = 2π√(l/g)
As we can see there is no mass term in the formula, Hence we can say that the time period of simple pendulum will not depend on mass or weight of the object tied to the string.
We have given length as 1 m
Time Period = 2π√(1/9.8)
Time Period = 2π×0.319
Time Period = 2 seconds
Time of the Simple Pendulum is 2 seconds.
Frequency = 1 / Time period
Frequency = 1 / 2
Frequency = 0.5 1/second
Frequency of the Simple Pendulum is 0.5
So we can conclude that the Time period of Simple pendulum is 2π√(l/g) and as we can see there is no term of mass, means the simple pendulum will not be depend on mass or weight. The time period came out to be 2 seconds and the frequency came out to be 0.5.
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Pls give full explanation!
1. The part that shows the body is accelerating is the first 2 hours (ABE). The acceleration is 20 Km/h²
2. The part that shows the body is coming to rest the last 2 hours (CDF)
3. The distance travelled in the first 2 hours is 40 Km
1. How to determine the accelerated motionAcceleration is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
a is the acceleration v is the final velocity u is the initial velocity t is the timeFrom the diagram (ABE), we can see that the body starts from rest and attain a final velocity (40 Km/h) within the first 2 hours. This clearly indicates that the body is accelerating.
The acceleration can be obtained as followed:
Initial velocity (u) = 0 Km/hFinal velocity (v) = 40 Km/hTime (t) = 2 hoursAcceleration (a) = ?a = (v – u) / t
a = (40 – 0) / 2
a = 20 Km/h²
2. How to determine the part showing the body is coming to restFrom the diagram (CDF), we can see that the object velocity changes from 40 to 0 within the last 2 hours. This clearly indicates that the object caming to rest.
3. How to determine the distance travelled in the first 2 hours Initial velocity (u) = 0 Km/hFinal velocity (v) = 40 Km/hTime (t) = 2 hoursAcceleration (a) = 20 Km/h²Distance (s) = ?v² = u² + 2as
v² = 2as
s = v² / 2a
s = 40² / (2 × 20)
s = 40 Km
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