Answer:
Explanation:
F = ma
F = 2.60(14.0)
F = 36.4 N
define heterotrophic
Answer:
Heterotrophic requiring complex organic compounds of nitrogen and carbon (such as that obtained from plant or animal matter) for metabolic synthesis.
Answer -:
⟹ It is a mode of nutrition in which organism are unable to synthesize organic substance by themselves and obtain part of whole of organic substance from external environment. the organism that obtain their food by this method are called heterotphs.
tha organism which lack green pigments chlorophyll are included in this group.all animal ,fungi and most of bacteria belong to this group.a large number of higher plant also Lack chlorophyll . they are also unable to synthesize their organic substance food .On the basis of types of food and feeding habit's nutrition classified into following group -:
(1) Holozoic Nutrition -:
nutrition and animal consumes a plant or an animal us whole or a part of it in solid or liquid form most of free living a cellular protist and all animal show Holozoic Nutrition
(2) Herbivores -:
animal eating grass or other plant material are called Herbivores.
Example - Grazers- horse , cow , goat etc .(3) Carnivorous
flesh eating animal are called carnivorous.
Example -: Lion , tiger , Wolf etc(4) Omnivorous
animal eating food or plant us well us animal origin are called omnivorous.
(5) Insectivorous
insect eating animal are calledinsectivorous .
(6) Frugivores
animal which mainly depend on fruit are called frugivores.
A domestic water heater holds 189 L of water at 608C, 1 atm. Determine the exergy of the hot water, in kJ. To what elevation, in m, would a 1000-kg mass have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water
A.
The energy of the hot water is 482630400 J
Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,
c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C
Substituting the values of the variables into the equation, we have
Q = mcΔT
Q = ρVcΔT
Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C
Q = 482630400 J
So, the energy of the hot water is 482630400 J
B.
The elevation the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water is 49248 m.
Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m
Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J
So, ΔU = mgΔh
ΔU = mg(h - h')
making h subject of the formula, we have
h = h' + ΔU/mg
Substituting the values of the variables into the equation, we have
h = h' + ΔU/mg
h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)
h = 0 m + 482630400 J/(9800 kgm/s²)
h = 0 m + 49248 m
h = 49248 m
So, the elevation the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water is 49248 m.
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calculate the mass of a block of ice having volume 5m³. (density of ice≈920 kg/m³)
Answer:
4600kg
Explanation:
Density=mass÷volume
920=m/5
m=920×5=4600kg
Which properties make a metal a good material to use for electrial wires
Answer:
Most importantly metals can pass an electric current without being affected and changed by the electricity. Electrical conductivity combined with ductility makes metals the most suitable materials for electrical transmission wires.
3) A 60. kg person is in an elevator. The elevator starts from rest and then accelerates upwards at 2.0 m/s^2 for 4.0 seconds. Calculate the work done by the normal force on the person. *
Answer:
WD = 960 J
Explanation:
WD = work done (J)
F = force (N)
s = displacement (m)
m = mass (kg) = 60
a = acceleration (m/s²) = 2
t = time (s) = 4
u = initial velocity (m/s) = 0
The formulas or equations that are relevant ate:
WD = F × s
F = m × a
s = u + at
We want to find WD, so we need to now the force and the displacement (or distance);
We calculate force, in Newtons, with the formula F = ma:
F = 60 × 2
F = 120 N
We also need displacement, which get with the formula s = u + at:
s = 0 + 2(4)
s = 8 m
Now we have F and s, we can calculate WD:
WD = 120 × 8
WD = 960 J
Methodology:
Starting with what you want to find, in this case WD, list the formula/s you could use;
Then, identify the information you need for the formula and whether or not you are given that information;
Next, list the formulas for the information you don't have and once again, identify whether the information you are given is sufficient to use those formulas;
Once you can calculate all necessary information, then proceed to calculate the values and finally, the answer;
I suggest also keeping a list of all the variables as I've done at the top of my working so it is clear for you to see and use.
Conservation of Energy Roller Coaster A roller coaster cart of mass 100kg travels on a track with one loop. Fill in blanks A-H. А. KE=OJ PE=120000J h= А. V= B B KE=___CE PE=60000J h= _D V= E KE=__F PE=40000J h=__G_ V= KE= PE= h=Om v= K D E F G H K
(a) The height of the roller coaster at 120,000 potential energy is 122.45 m.
(b) The velocity of the roller coaster at 0 J kinetic energy is 0.
(c) The height of the roller coaster at 60,000 potential energy is 61.23 m.
(d) The velocity of the roller coaster at 60,000 J kinetic energy is 34.64 m/s.
(e) The height of the roller coaster at 40,000 potential energy is 40.82 m.
(f) The velocity of the roller coaster at 80,000 J kinetic energy is 40 m/s.
The given parameters:
mass of the roller coaster, m = 100 kgWhen the kinetic energy = 0 and potential energy = 120,000 J
The height of the roller coaster is calculated as follows;
P.E = mgh
[tex]h = \frac{P.E}{mg}\\\\h = \frac{120,000}{100 \times 9.8} \\\\h = 122.45 \ m[/tex]
Since the kinetic energy = 0, the velocity of the roller coaster = 0
When the potential energy, P.E = 60,000 J, the kinetic energy, K.E is calculated as;
P.E + K.E = M.A
P.E + K.E = 120,000
60,000 + K.E = 120,000
K.E = 120,000 - 60,000
K.E = 60,000 J
The height of the roller coaster at 60,000 potential energy is calculated as follows;
[tex]h = \frac{P.E}{mg} \\\\h = \frac{60,000}{100 \times 9.8} \\\\h =61.23 \ m[/tex]
The velocity of the roller coaster at 60,000 J kinetic energy is calculated as follows;
[tex]K.E = \frac{1}{2} mv^2\\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{ \frac{2K.E}{m}} \\\\v = \sqrt{ \frac{2\times 60,000}{100}}\\\\v = 34.64 \ m/s[/tex]
When the potential energy, P.E = 40,000 J, the kinetic energy, K.E is calculated as;
P.E + K.E = M.A
40,000 + K.E = 120,000
K.E = 120,000 - 40,000
K.E = 80,000
The height of the roller coaster at 40,000 potential energy is calculated as follows;
[tex]h = \frac{P.E}{mg} \\\\h = \frac{40,000}{100 \times 9.8} \\\\h = 40.82 \ m[/tex]
The velocity of the roller coaster at 80,000 J kinetic energy is calculated as follows;
[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 80,000}{100} } \\\\v = 40 \ m/s[/tex]
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An object is moving with an initial velocity of 3.3 m/s. It is then subject to a constant acceleration of 3.7 m/s2 for 10 s. How far will it have traveled during the time of its acceleration?
I also need the complete Formula (Nothing left out)
Answer:
Explanation:
s = s₀ + v₀t + ½at²
ASSUMING the acceleration is in the direction of initial motion.
s = 0 + 3.3(10) = ½(3.7)(10²)
s = 218 m
The average normal body temperature measured in the mouth is 310 K. What would Celsius thermometers read for this temperature?
54.1°C
23.8°C
36.9°C
42.7°C
________
= 310 - 273
= 37°C
Actually,310 Kelvin is same with 37°C, and as you see, there is no 37°C
So, The Nearest Number To 37°C is 36,9°C
Answer:
36.9
Explanation:
Plato
A 115 kg hockey player, Adam, is skating east when he tackles a stationary 133 kg player, Bob. Afterward, they move at 1.35 m/s east. What was Adam's velocity before the collision? (Unit = m/s)
Answer:
Explanation:
Conservation of momentum
115v + 133(0) = (115 + 133)1.35
v = 2.911304...
v= 2.91 m/s east
Answer:
The velocity east is 2.91
Explanation:
Fill in the box
2.91
6) An object is released from rest at the top of a ramp inclined at 30. degrees up from the horizontal. Due to friction, the ramp is only 20. % efficient. What is the object's speed after it slides down ALONG the ramp for 2.0 m? *
Answer:
Explanation:
I've been doing these types of problems for many years and I don't think I've ever seen an "efficiency" rating on a ramp.
I'm going to ASSUME that 20% efficient means that 80% of the Potential energy that gets converted becomes system internal heat energy.
Potential energy at the start of a 2.0 m slide
PE = mgh = mg2sin30 = mg2(½) = mg J
0.8mg J gets converted to heat and 0.2mg converts to kinetic energy
0.2mg = ½mv²
v² = 0.4g
v = √(0.4(9.8)) = 1.979898... ≈ 2.0 m/s
numerical problems:
a.) convert 300K into the celsius scale.
b.) convert 220 centigrade scale into kelvin scale.
c.) convert 20 ventigrade scale into Fahrenheit scale.
d.) convert 260 Fahrenheit into centigrade. pls help me to solve this problems
plz answer the question.
Answer:
a
Explanation:
sana po makatulong <3♡♡
Which is the main gas that makes up the Earth's atmosphere?
Answer:
78 percent nitrogenExplanation:
I hope it's helpful for you
Help pls!
A 3 kg mass is raised a distance of 14 m above the earth by a vertical force of 93 N.
The final kinetic energy of the mass, to 3 significant figures, if it was originally at rest is:
[tex] \large★·.·´¯`·.·★ {Answer}★·.·´¯`·.·★[/tex]
As we know that Kinetic Energy is the Energy that is possessed by a moving object. and if the object is at rest then it doesn't have velocity therefore there is no kinetic Energy.
In the numerical terms we can express it as : -
[tex] \sf0.00 \: \: joules[/tex][tex]꧁ \: \large \frak{Eternal \: Being } \: ꧂[/tex]
If the penny is thrown horizontally at 25 m/s from the 170 meter building, how long will it take for the penny to hit the ground?
9514 1404 393
Answer:
about 5.89 seconds
Explanation:
The penny will hit the ground at the same time it would if it were simply dropped. The equation for the vertical motion is ...
h(t) = -4.9t^2 +170 . . . . . where 170 is the initial height in meters
h(t) = 0 when ...
4.9t^2 = 170
t = √(170/4.9) ≈ 5.89
The penny will hit the ground in about 5.89 seconds.
If an electron moves in a direction perpendicular to the same magnetic field with this same linear speed,
what is the radius of its circular orbit?
Answer:
An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.
Explanation:
Answer:
An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.
Explanation:
a box has a mass of 4 kg and surface area 4 metre square calculate the pressure exerted by the box on the floor
- BRAINLIEST answerer ❤️✌
Answer:
9.8
Explanation:
we know ,
f=m.g(g=9.8m/s^2)
now,
p=force/area
p=m.g/a
p=4×9.8/4
p=4 pascal.
if u put g as 10 then ypu will get 10 pascal
Is electrical energy the same or different from energy it takes to play a soccer game
Answer:
Diffrent
Explanation:
Electrical energy is electrical charges moving. When you play soccer kinetic energy is used. Kinetic energy is the movement of atoms, objects, and electrons.
A racing car on the straight accelerates from 100 km/h to 316 km/h in three seconds.
What is its acceleration?
40m/s2
30m/s2
20m/s2
72m/s2
Answer:
[tex]20m/s^2[/tex]
Explanation:
Solution is attached. I apologize if it is a little messy.
Need help with dot product
[tex]\textbf{A}\cdot\textbf{B} = 11.5[/tex]
Explanation:
The dot product between two vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B}[/tex] is defined as
[tex]\textbf{A}\cdot\textbf{B} = AB\cos{\theta}[/tex]
where A and B are the magnitudes of the vectors [tex]\textbf{A}[/tex] and [tex]\textbf{B},[/tex] respectively and [tex]\theta[/tex] is the angle between the two. Since A = 3, B = 5 and [tex]\theta = 40°,[/tex] the dot product [tex]\textbf{A}\cdot\textbf{B}[/tex] is
[tex]\textbf{A}\cdot\textbf{B} = (3)(5)(0.766) = 11.5[/tex]
please answer this as fast as you can i need it
Answer:
it says pdf only i dont knowwhat u want me to do
A car travelling at 79.3 Km/h on a highway has 4.22x10 5 J of kinetic energy.
a. What is the mass of the car?
b. If brakes are applied with a force of 2100 N, what distance will it take for the car to slow down to a speed of 56 Km/h?
Answer:
[tex]1.74\times10^3 kg; 100m[/tex]
Explanation:
Step a: mass of the car. Let's grab the definition of kinetic energy: [tex]K= \frac12 mv^2[/tex]. We have K, we have v (which we should convert in meters per second, dividing by 3.6) to get:[tex]4.22\times10^5 = \frac12m(22.03)^2 \rightarrow m= 2\times4.22 / 495.22 \times 10^5 = 1.74 \times 10^3 kg[/tex]
Point a is done.
Now for the (b)reaking part. (I'm sorry, it's an horrible joke, but I couldn't resist)
In theory we have the mass, we have the force, so we could find the acceleration, find how long it takes to slow down, and then find the distance traveled. Too long. Let's do things more easily: when the car slows down to 56 km/h it will have a different kinetic energy. The difference in kinetic energy is the work done by the breaking force ofer the slowing distance.
[tex]K_f-K_i=W[/tex] A quick note on signs: if you look carefully the final kinetic energy will be less than the initial value, thus the work will be negative: it means it's correct, since the work is against the motion, slowing it down. Let's get calculating, first by converting 56 kmh in m/s (15,56 m/s), then finding the final kinetic energy:
[tex]K_f =\frac12 (1.74\times10^3) (15.56)^2 =2.11 \times 10^5 J[/tex]
The difference will be the work done by the force, or
[tex](2.11 - 4.22) \times 10^5 = \vec F\cdot \vec x=Fx[/tex] where we are assuming that force and displacement have the same line of actions to simplify the dot product.
[tex]2.11\times 10^5 = 2100x = 1.00\times 10^2 m[/tex]
given two vector
p= 2i + 2j + 4k
q = i - 4j + 4k
find p+ q
Answer:
3i - 2j + 8k
Explanation:
p + q = (2i + i) + (2j - 4j ) + (4k + 4k )
= 3i -2j + 8k
An object with an initial velocity of 10 m/s accelerates at a rate of 3.5 m/s2 for 8 seconds. How far will it have traveled during that time?
Free-fall Acceleration is -10 m/s^2
I also need the formula
Answer:
Explanation:
s = s₀ + v₀t + ½at²
assuming that the acceleration is in the direction of initial velocity.
(it would not have to be so)
s = 0 + 10(8) + ½(3.5)(8²)
s = 192 m
Two blocks are set in a pully system as shown in fig below. Block A sits on the frictionless table while block B hags freely. The pully is light and frictionless towards the light string that runs over it. If the Block A has mass of 3.4 kg and Block has 3.5 kg, what would be the magnitude of the acceleration (in ms-2) of the blocks? [g = 9.8 ms=2]
Answer:
Explanation:
F = ma
a = F/m
a = mBg / (mB + mA)
a = 3.5(9.8)/(3.5 + 3.4)
a = 4.971014...
a = 5.0 m/s²
If you want to use individual Free Body Diagrams
mass A will have downward weight and upward normal forces equal at mAg
and a horizontal force of string tension T
F = ma
T = mAa
mass B will have a downward force of mBg and an upward force of T
mBg - T = mBa
substitute for T
mBg - mAa = mBa
mBg = a(mB + mA)
a = mBg / (mB + mA) which is identical to the above answer.
An object moving at a constant velocity of 5.4 m/s travels for 12 s. How far will it move during that time?
Free-fall Acceleration is -10 m/s^2
I also need the formula
Answer:
Explanation:
s = s₀ + v₀t + ½at²
s = 0 + 5.4(12) + ½(0)12²
s = 64.8 m
describe the motion of objects that are viewed from your reference frame both inside and outside while you travel inside a moving vehicle
Answer:
The objects outside the reference frame aren't moving. It appears this way since the vehicle you are inside is moving, but unless the objects are people, animals, or other vehicles, the objects aren't moving.
How much did the pressure drop in the storm's center from November 9, 1200z, until November 11, 0000z
The pressure dropped by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.
Using the attached map below, in the morning of November 9, the pressure situated at the storm's center = 1000 MB isobar located at the center.
Meanwhile, on November 11, 0000z the pressure situated at the storm's center = 976 MB isobar
The difference in this pressure is regarded as the pressure drop in the storm's center and it is determined as follows;
= 1000 MB - 976 MB
= 24 MB
Therefore, we can conclude that the pressure drop by 24 MB in the storm's center from November 9, 1200z, until November 11, 0000z.
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Objects 1 and 2 attract each other with a gravitational force of 16 units. If the mass of object 1 is one-third the original value AND the mass of object 2 is doubled AND the distance separating objects 1 and 2 is doubled, then the new gravitational force will be _____ units.
Explanation:
Fgravity = G*(mass1*mass2)/D²
G is the gravitational constant, which has the same value throughout our universe.
D is the distance between both objects.
so, now some numbers change
Fgravitynew = G*((1/3)*mass1*2*mass2)/(2D)² =
= G*((2/3)*mass1*mass2)/(4D²) =
= (2/3)* (G*(mass1*mass2)/D²) / 4 =
= ((2/3)/4) * G*(mass1*mass2)/D² =
= (2/12) * Fgravity = Fgravity/6
the new gravitational force will be 16/6 = 8/3 units.
How do I resolve moments about the point P?
Answer:
By applying the definition of torques ( [tex]\vec \tau = \vec r \times \vec F[/tex] ) and them remembering a few tricks.
Namely: if you wrap your RIGHT hand fingers around something and stick your thumb out, the direction your finger wraps gives you the verse of rotation and the thumb the orientation of the torque. Bottom force (4N) will give a counterclockwise rotation, torque is pointing up; top force (3N) will give a clockwise rotation and its torque its pointing down (read up and down as if the sheet the image is printed on is on your table).
In terms of magnitude the trick is easy: You want to multiply the intensity of the force (3N and 4N) by the distance between the point and the line the force it is applied to (that is, you don't care about the length of r itself, but the distance at a right angle, which is 0.9 and 0.8m respectively.
At this point, assuming "upwards" (relative to the plane of the sheet that is) torques positive, the 3N force gives you a torque of [tex]- 3N \times 0.9m = - 2.7N\cdot m[/tex] and the 4N force provides [tex]+4N\times 0.8 m = +3.2 N\cdot m[/tex]