The simplest definition of a black hole is an object that is so dense that not even light can escape its surface. If we squished the Earth's mass into a sphere with a radius of 9 mm, the escape velocity would be the speed of light. Just a wee-bit smaller, and the escape velocity is greater than the speed of light.
HopeItHelps!
A dog finds a toy at rest on the floor. The dog pushes the toy horizontally on a frictionless floor with a net force of 2.0 Newtons for 3.0 meters. How much kinetic energy does the toy gain? Round your answer to the nearest whole number.
Answer:
the kinetic energy gained by the toy is 6J.
Explanation:
Given;
net applied to the toy by dog, F = 2 N
distance moved by the toy, d = 3 m
Apply the principle of work-energy theorem to determine the kinetic energy gained by the toy.
ΔK.E = W
= F x d
= 2 x 3
= 6 J
Therefore, the kinetic energy gained by the toy is 6J.
1. A block with mass 20 kg is
sliding up a plane (Ukinetic=0.3,
inclined at 10°) at a speed of
2 m/s to the right (positive
X-direction). How far does it
go up along the plane before
it comes to rest momentarily?
Answer: 0.435 m
Explanation:
Given
mass m=20 kg
initial speed u=2 m/s
coefficient of kinetic friction [tex]\mu_k=0.3[/tex]
deceleration which opposes the motion is given by
[tex]\Rightarrow a=g\sin \theta+\mu_kg\cos \theta\\\Rightarrow a=g(\sin \theta +\mu_k\cos \theta)[/tex]
[tex]\Rightarrow a=9.8(\sin 10^{\circ}+0.3\times \cos 10^{\circ})\\\Rightarrow a=4.59\ m/s^2[/tex]
using [tex]v^2-u^2=2as[/tex]
[tex]\Rightarrow s=\dfrac{2^2}{2\times 4.59}=0.435\ m[/tex]
State three factors affecting pressure in liquids
Answer:
Density of liquid
Depth of liquid
Acceleration due to gravity
help please i will mark brainlist!!!
Answer:
.50 M
Explanation:
5*.50=2.5 + 2*.25=.5 = 3n
6*.50= 3N
Final answer is .50M
A 60 kg student weighs 600 N.
He does a bungee jump.
Calculate the change in gravitational potential energy as the student falls 50 m.
Give the unit.
Will give brainliest!
Answer:
30 000 J/Nm
Explanation:
60 x 10 x 50
=600 x 50
=30000 J/Nm
The change in gravitational potential energy of the student is 18000 Joule.
What is gravitational potential energy?
The energy that an item has or acquires when its location changes as a result of being in a gravitational field is known as gravitational potential energy. Gravitational potential energy can be defined as an energy that has a connection to gravitational force or gravity.
Given parameters:
Mass of the student: m = 60 kg.
Weighs of the student: W = 600 N.
the student falls h =50 m.
Hence, change in gravitational potential energy of the student =
Weighs of the student × change in height
= 600 N × 30 m
= 18000 Joule.
So, the change in gravitational potential energy of the student is 18000 Joule.
Learn more about potential energy here:
https://brainly.com/question/24284560
#SPJ2
If the magnitude of vector A⃗ is less than the magnitude of vectorB⃗ , then the x component of A⃗ is less than the x component ofB⃗ . If the magnitude of vector is less than the magnitude of vector, then the component of is less than the component of.
a. True
b. False
A spring with a constant of 76 N/m is extended by 0.9 m. How much energy is stored in the extended spring?
Answer:
[tex]E=30.78\ J[/tex]
Explanation:
The force constant of the spring, k = 76 N/m
The extension in the spring, x = 0.9 m
We need to find the energy is stored in the extended spring. The energy stored in the spring is given by :
[tex]E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 76\times (0.9)^2\\\\E=30.78\ J[/tex]
So, 30.78 J of energy is stored in the spring.
A spring is stretched 5 mm by a force of 125 N. How much will the spring stretch
when 250 N force is applied?
Answer:
10 mm
Explanation:
We'll begin by calculating the spring constant of the spring. This can be obtained as follow:
Extention (e) = 5 mm
Force (F) = 125 N
Spring constant (K) =?
F = Ke
125 = K × 5
Divide both side by 5
K = 125 / 5
K = 25 N/mm
Finally, we shall determine how much the spring will stretch when a 250 N force is applied. This can be obtained as follow:
Force (F) = 250 N
Spring constant (K) = 25 N/mm
Extention (e) =?
F = Ke
250 = 25 × e
Divide both side by 25
e = 250 / 25
e = 10 mm
Thus, the spring will stretch 10 mm when a 250 N force is applied.
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8 in. to the final position x2 = 5 in. determine (a) the work done on the cart by the spring and (b) the work done on the cart by its weight.
This question is incomplete, the missing diagram is uploaded along this Answer below.
Answer:
a) the work done on the cart by the spring is 4.875 lb-ft
b) the work done on the cart by its weight is - 3.935 lb-ft
Explanation:
Given the data in the question;
(a) determine the work done on the cart by the spring
we calculate the work done on the cart by the spring as follows;
[tex]W_{spring}[/tex] = 1/2×k( [tex]x^{2} _{1}[/tex] - [tex]x^{2} _{2}[/tex] )
where k is spring constant ( 3 lb/in )
we substitute
[tex]W_{spring}[/tex] = 1/2 × 3( (-8)² - (5)² )
[tex]W_{spring}[/tex] = 1/2 × 3( 64 - 25 )
[tex]W_{spring}[/tex] = 1/2 × 3( 39 )
[tex]W_{spring}[/tex] = 58.5 lb-in
we convert to pound force-foot
[tex]W_{spring}[/tex] = 58.5 × 0.0833333 lb-ft
[tex]W_{spring}[/tex] = 4.875 lb-ft
Therefore, the work done on the cart by the spring is 4.875 lb-ft
b) the work done on the cart by its weight
work done by its weight;
[tex]W_{gravity}[/tex] = -mgsin∅( x₂ - x₁ )
we substitute in of values from the image below;
[tex]W_{gravity}[/tex] = -14 × sin(15°)( 5 - (-8) )
[tex]W_{gravity}[/tex] = -14 × 0.2588 × 13
[tex]W_{gravity}[/tex] = -47.1 lb-in
we convert to pound force-foot
[tex]W_{gravity}[/tex] = -47.1 × 0.0833333 lb-ft
[tex]W_{gravity}[/tex] = - 3.935 lb-ft
Therefore, the work done on the cart by its weight is - 3.935 lb-ft
a) the work done on the cart by the spring is 4.875 lb-ft.
b) the work done on the cart by its weight is - 3.935 lb-ft.
Calculation of the work done:a. The work done on the cart by the spring is
= 1/2 × 3( (-8)² - (5)² )
= 1/2 × 3( 64 - 25 )
= 1/2 × 3( 39 )
= 58.5 lb-in
Now we have to convert to pound force-foot
So,
= 58.5 × 0.0833333 lb-ft
= 4.875 lb-ft
b) Now
work done by its weight;
= -mgsin∅( x₂ - x₁ )
So,
= -14 × sin(15°)( 5 - (-8) )
= -14 × 0.2588 × 13
= -47.1 lb-in
Now we convert to pound force-foot
= -47.1 × 0.0833333 lb-ft
= - 3.935 lb-ft
Learn more about spring here: https://brainly.com/question/16060099
You are standing on the bottom of a lake with your torso above water. Which statement is correct?
a. You feel a buoyant force only when you momentarily jump up from the bottom of the lake.
b. There is a buoyant force that is proportional to the weight of your body below the water level.
c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.
d. There is no buoyant force on you since you are supported by the lake bottom.
Answer:
c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.
Explanation:
Buoyancy can be defined as a force which is created by the water displaced by an object.
Simply stated, buoyancy is directly proportional to the amount of water that is being displaced by an object.
Hence, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up.
The buoyancy of an object is given by the formula;
[tex] Fb = pgV [/tex]
[tex] But, \; V = Ah [/tex]
[tex] Hence, \; Fb = pgAh [/tex]
Where;
Fb = buoyant force of a liquid acting on an object.
g = acceleration due to gravity.
p = density of the liquid.
v = volume of the liquid displaced.
h = height of liquid (water) displaced by an object.
A = surface area of the floating object.
The unit of measurement for buoyancy is Newton (N).
In this scenario, you are standing on the bottom of a lake with your torso above water. Thus, there is a buoyant force that is proportional to the volume of your body that is below the level of the water.
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25 cm and the point-like object with charge q2 = −2.14 µC is located at x2 = −1.80 cm.
A) Determine the total electric potential (in V) at the origin.
B) Determine the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
Answer:
a) the total electric potential is 2282000 V
b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
Explanation:
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
so
Electric potential at p in the diagram 1 below is;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we know that; Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)
the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
r1² = 0.015² + 0.0125²
r1 = √[ 0.015² + 0.0125² ]
r1 = √0.00038125
r1 = 0.0195
Also
r2² = 0.015² + 0.018²
r2 = √[ 0.015² + 0.018² ]
r2 = √0.000549
r2 = 0.0234
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
a) The total electric potential is 2282000 V
b) The total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
What is electric potential?The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to a specific point in an electric field.
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
Electric potential at p in diagram 1 below is;
[tex]V_P=V_1+V_2[/tex]
[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]
we know that; the Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)The total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
[tex]r_1^2=0.015^2+0.0125^2[/tex]
[tex]r_1 = \sqrt{[ 0.015^2 + 0.0125^2 ][/tex]
[tex]r_1 = \sqrt{0.00038125}[/tex]
[tex]r_1 = 0.0195[/tex]
Also
[tex]r_2^2 = 0.015^2 + 0.018^2[/tex]
[tex]r_2 = \sqrt{0.015^2 + 0.018^2}[/tex]
[tex]r_2 = \sqrt{0.000549[/tex]
[tex]r_2 = 0.0234[/tex]
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
To know more about electric potential follow
https://brainly.com/question/25923373
Standing at a crosswalk, you hear a frequency of 530 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 424 Hz. Determine the ambulance's speed from these observations.
Answer:
_s = 37.77 m / s
Explanation:
This is an exercise of the Doppler effect that the change in the frequency of the sound due to the relative speed of the source and the observer, in this case the observer is still and the source is the one that moves closer to the observer, for which relation that describes the process is
f ’= f₀ [tex]\frac{v}{v - v_s}[/tex]
where d ’= 530 Make
when the ambulance passes away from the observer the relationship is
f ’’ = f₀ [tex]\frac{v}{v + v_s}[/tex]
where d ’’ = 424 beam
let's write the two expressions
f ’ (v-v_s) = fo v
f ’’ (v + v_s) = fo v
let's solve the system, subtract the two equations
v (f ’- f’ ’) - v_s (f’ + f ’’) = 0
v_s = v [tex]\frac{ f' - f''}{ f' + f''}[/tex]
the speed of sound is v = 340 m / s
let's calculate
v_s = 340 [tex](\frac{ 530 -424}{530+424} )[/tex]
v_s = 340 [tex](\frac{106}{954}[/tex])
v_s = 37.77 m / s
what are the types of energy sources based on
time of replacement ? write down their names
Answer:
solar energy
wind power
geothermal energy
hydraulic power
biomass energy
energy storage
(That's all I know).
1. With the exception to water, matter (expands, contracts) when it gets
hotter. *
A)Expands
B)Contracts
what is the average velocity of a van that moves from 0 to 60 m east and 20 seconds
Explanation:
I have a lot to say it was very nice to meet my parents are u doing well I dont want too its been so much I love you so I was like u know I am not a man but you are the auditions I have been in a long long long life is a triangle and a chair for me and my parents think about the way I
Captain Jack Sparrow has been marooned on an island in the Atlantic by his crew, and decides to builda raft to escape. The wind seems quite steady, and first blows him due east for 11km, and then 6km ina direction 6degrees north of east. Confident that he will eventually find himself in safety, he fallsasleep. When he wakes up, he notices the wind is now blowing him gently 11degrees south of east -but after traveling for 21km, he finds himself back on the island.
Variable Name Min Max Step Sample Value
thetab 5 10 1 6
a 10 20 11 1
b 5 15 1 6
c 20 30 1 21
thetac 10 15 11 1
Required:
How far (in km) did the wind blow him while he was sleeping?
Answer:
d₃ = 37,729 km, θ= 5.1º North of West
Explanation:
This is a velocity addition problem, the easiest way to solve it is to decompose the velocities in a Cartesian system, the x-axis coincides with the West-East direction and the y-axis with the South-North direction
* first displacement is
d₁ₓ = 11 km
* second offset is
cos 6 = d₂ₓ / d₂
sin 6 = d_{2y} / d₂
d₂ₓ = d₂ cos 6
d_{2y} = d₂ sin 6
d₂ₓ = 6 cos 6 = 5.967 km
d_{2y} = 6 sin 6 = 0.6272 km
* third displacement is unknown
* fourth and last displacement
cos (-11) = d₄ₓ / d₄
sin (-11) = d_{4y} / d₄
d₄ₓ = d₄ cos (-11)
d_{4y} = d₄ sin (-11)
d₄ₓ = 21 cos (-11) = 20.61 km
d_{4y} = 21 sin (-11) = -4.007 km
They tell us that at the end of the tour you are back on the island, so the displacement must be zero
X axis
x = d₁ₓ + d₂ₓ + d₃ₓ + d₄ₓ
0 = 11 +5.967 + d₃ₓ + 20.61
d₃ₓ = -11 - 5.967 - 20.61
d₃ₓ = -37.577 km
Y axis
y = d_{1y} + d_{2y} + d_{3y} + d_{4y}
0 = 0 + 0.6272 + d_{3y} -4.007
d_{3y} = 4.007 - 0.6272
d_{3y} = 3.3798 km
This distance can be given in the form of module and angle
Let's use the Pythagorean theorem for the module
d₃ = [tex]\sqrt{d_{3x}^2 + d_{3y}^2}[/tex]
d₃ = [tex]\sqrt{37.577^2 + 3.3798^2}[/tex]
d₃ = 37,729 km
Let's use trigonometry for the angle
tan θ = d_{3y} / d₃ₓ
θ = tan⁻¹ [tex]\frac{d_{3y}}{d_{3x}}[/tex]
θ = tan-1 (-3.3798 / 37.577)
θ = 5.1º
Since the y coordinate is positive and the x coordinate is negative, this angle is in the second quadrant, so the direction given in the form of cardinal coordinates is
θ= 5.1º North of West
a. Using the ideas of electric field and force, explain what would happen to an electron if released from rest at r=2.0m?
b. Would the electron released from rest move to a region of higher electrical potential or lower electrical potential?
c. Would the electron released from rest move such that the system would have higher potential energy or lower potential energy?
If the diameter of a moose eye is 40 mm, what is the total refractive power of the anterior portion of the eye?
Answer:
-the ratio of the speed of light
in air to the speed of light in the substance.
-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.
-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5
Explanation:
A bowling ball and a small marble will fall downward to the surface of the Moon at the same rate because ____________.a. the force of gravity is the same for each object.
b. the force of gravity on an object in a vacuum is zero.
c. the ratio of the force of gravity exerted on an object to the object's mass is the same.
Answer:
The answer is C
Explanation:
Candice is examining a cell under a microscope. She has identified a cell wall, a nucleus, and a chloroplast. What type of organism does this most likely belong to?
A. A plant B. An animal C. A fungus D. A bacterium
Answer:
A plant
Explanation:
because animals don't have cell walls, and fungus and bacteria dont have chloroplasts
A potter’s wheel moves from rest to an angular speed of 0.10 rev/s in 36.5 s.
Assuming constant angular acceleration,
what is its angular acceleration in rad/s2?
Answer in units of rad/s2
.
Answer:
please find attached pdf
Explanation:
How does Physics help you as a student?
Answer:
The goal of physics is to understand how things work from first principles. ... Courses in physics reveal the mathematical beauty of the universe at scales ranging from subatomic to cosmological. Studying physics strengthens quantitative reasoning and problem solving skills that are valuable in areas beyond physics
Answer:
you get to understand why things happen this way
Explanation:
for example, are you not curious about why when standing in the bus and when the bus stops, you will might feel like you are going to fall ,
why does this happen because....
newton's laws explains it,
inertia causes you to be reluctant to change your initial state of motion due to your mass so you fall because you are still moving at the 'speed of the bus ' , something in like that
hope this helps,
please mark also
A bicyclist rides 5.0 km due east, while the resistive forcefrom the air has a magnitude of 3.0 N and points due west. Therider then turns around and rides 5.0 km due west, back to herstarting point. The resistive force from the air on the return triphas a magnitude of 3.0 N and points due east.
a) Find the work done by the resistive force during the roundtrip.
Based on answer in part A.
b) Is the resistive force a conservative force? explain.
Answer:
a) Find the work done by the resistive force during the roundtrip.
W=-30kJ
b) Is the resistive force a conservative force? explain.
The resistive force is not a conservative force since the work done during the round trip is not zero
Explanation:
The worf done on object y a constant force F is given by:
W= (F cos ∅)S
Where S is the displacement and ∅ is the angle between the force and the displacement.
The displacement of the bicycle during each part of the trip is s=5000m and teh magnitude of teh resistance force is F=3.0N
∅1=180° he angle between the displacement and the force
W1=W2
W1 = (3.0 cos180) 5000m
W1=-15.O kJ
W=W1+W2
W=-30kJ
The resistive force is not a conservative force since the work done during the round trip is not zero
(a) The work done by the resistive force is 15,000 J
(b) The work done the resistive force is non-conservative since the resultant resistive force in not zero.
Work doneWork is said to be when an applied force displaces an object from its initial position.
Work done by resistive forceThe work done by the resistive force is calculated as follows;
W = FΔr
W = 3 x (5,000 - 0)
W = 15,000 J
Thus, the work done the resistive force is non-conservative since the resultant resistive force in not zero.
Learn more about conservative force here: https://brainly.com/question/15357875
Messages from the Perseverance Rover on Mars reach Earth in 11 min. The speed of light is 3.00 x 108 m/s. Using this information, how far is Mars from Earth?
Answer:
[tex]d=1.98\times 10^{11}\ m[/tex]
Explanation:
Messages from the Perseverance Rover on Mars reach Earth in 11 min i.e. time is 660 s
The speed of light is [tex]3\times 10^8\ m/s[/tex]
We need to find the distance between Mars and Earth. Let the distance be d.
We know that,
Distance = speed × time
So,
[tex]d=660\times 3\times 10^8\\\\d=1.98\times 10^{11}\ m[/tex]
So, Mars is [tex]1.98\times 10^{11}\ m[/tex] from the Earth.
Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1, the ball bounces off a cement floor, and in Case 2, the ball bounces off a piece of stretchy rubber. Two cases of a ball dropping and bouncing off of a surface. In case one, the ball approaches a cement floor, comes in contact, momentarily comes to rest, and then bounces off with a velocity smaller than the approach velocity. In case 2, the ball approaches a piece of stretchy rubber. When the ball comes in contact with the rubber, it deforms the rubber and comes momentarily to rest. The ball bounces off again with a velocity smaller than the approach velocity, and the rubber regains its original shape. In both cases, the balls are dropped from the same initial height and reach the same final height. In which case is the magnitude of the ball's change in momentum the greatest
Answer:
the impulse must be the same in these two cases F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
Explanation:
For this exercise we use the relationship between momentum and momentum
I = Δp
F t = m v_f - m v₀
To know the speed we use the conservation of energy
starting point. Highest point
Em₀ = U = m g h
fincla point. Just before the crash
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
we substitute in the impulse relation
F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases
A 50kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3m/s. What was the force acting on the mass?
Answer:
75N
Explanation:
a = v/t = 3/2
F = ma = 50(3/2) = 75
The resistivity of blood is related to its hematocrit, the volume fraction of red blood cells in the blood. A commonly used equation relating the hematocrit h to the blood resistivity rho (in Ω⋅m) is rho=1.32/(1−h)−0.79. In one experiment, blood filled a graduated cylinder with an inner diameter of 0.90 cm. The resistance of the blood between the 1.0 cm and 2.0 cm marks of the cylinder was measured to be 198 Ω.
Required:
What was the hematocrit for this blood?
Answer:
[tex]0.35598[/tex]
Explanation:
r = Radius = [tex]\dfrac{0.9}{2}=0.45\ \text{cm}[/tex]
R = Resistance = [tex]198\ \Omega[/tex]
A = Area = [tex]\pi r^2[/tex]
l = Length of blood in cylinder = 1 cm
h = Hematocrit of the blood
Resistivity is given by
[tex]\rho=\dfrac{1.32}{1-h}-0.79[/tex]
Resistance is given by
[tex]R=(\dfrac{1.32}{1-h}-0.79)\dfrac{l}{\pi r^2}\\\Rightarrow h=1-\dfrac{1.32}{\dfrac{R\pi r^2}{l}+0.79}\\\Rightarrow h=1-\dfrac{1.32}{\dfrac{198\times \pi\times (0.45\times 10^{-2})^2}{0.01}+0.79}\\\Rightarrow h=0.35598[/tex]
The hematocrit of the blood is [tex]0.35598[/tex].
List down the types of centripetal force?
Answer: Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force.
Answer:
roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge
Explanation:
4. Which of these is exerted by an engine?
(1 Point)
Thrust
Water resistance
Friction
Air resistance
Air resistance is exerted by an engine
A liquid fueled rocket is red on a test stand. The rocket nozzle has an exit diameter of 30 cm and the combustion gases leave the nozzle at a velocity of 3800 m/s and a pressure of 100 kPa, which is the same as the ambient pressure. The temperature of the gases in the combustion area is 2400 C. Find (a) the temperature of the gases at the nozzle exit plane, (b) the pressure in the combustion area, and (c) the thrust developed. Assume that the gases have a speci c heat ratio of 1.3, and a molar mass of 9. Assume that the ow in the nozzle is isentropic.
Answer:
1. Temperature= 869.35 K
2. Pressure of combustion = 12994.043 kpa
3. Thrust = 127x10⁶N
Explanation:
this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.
1.
The temperature = (273+2400k) - (3800)²/2(4003)
= 2673 - 14440000/8006
= 2673 - 1803.65
= 869.35 K
Approximately 869.4K
2. We first get mach number
= 3800/√1.3(923.8)(869.35)
= 3800/1021.78
= 3.719
Pressure = 100kpa[1+2.07464415]^1.3/0.3
= 12995.043kpa
C. Thrust
Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)
= 12678.621
= 126.781 kN
Thrust is approximately 127kN = 127x10⁶N