A recipe gives the instructions below
After browning the meat pour off fat from the pan to further reduce fat use a strainer.
what type lf separation methods are described in the recipe
A decantation and screening
B distillation and screening
C decantation and centrifugation
D distillation and filtration
Answer:
A. decantation and screening
Explanation:
Decantation is the one of the process of separating the mixture. In this process the precipitated liquid is separated from the solid. According to the given instruction for the recipe, the fat which is in liquid state is separated from meat. In the process of screening, more liquid is separated by placing the mixture on the screen. Here, the gravity plays an important role for the process of separation.
Answer:
a
Explanation:
Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The expression is [tex]W_c = P_o V_o ln (R_v)[/tex]
Explanation:
Generally smallest workdone done by a gas is mathematically represented as
[tex]dW = PdV[/tex]
Generally for an isothermal process
[tex]PV = nRT = constant [/tex]
=> [tex]P = \frac{nRT}{V}[/tex]
Generally the total workdone is mathematically represented as
[tex]W_c = \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV[/tex]
=> [tex]W_c = nRT \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV[/tex]
=> [tex]nRT [lnV] | \left \ {V_f}} \atop {V_o}} \right.[/tex]
=> [tex]W_c = nRT [ln(V_f) - ln(V_o)][/tex]
=> [tex]W_c = nRT ln \frac{V_f}{V_o}[/tex]
From the question [tex]\frac{V_f}{V_o } = R_v[/tex]
=> [tex]W_c = P Vln (R_v)[/tex]
at initial state
[tex]W_c = P_o V_o ln (R_v)[/tex]
distance is constant and time increseas
Will Speed increase or decrease?
Answer:
The speed will be decrease
Which best explains a difference between Einstein’s general theory of relativity and his special theory of relativity?
His general theory includes uniform and accelerated motion, but his special theory applies only to uniform motion.
His general theory includes uniform and accelerated motion, but his special theory applies only to accelerated motion.
His general theory applies only to accelerated motion, but his special theory includes uniform and accelerated motion.
His general theory applies only to uniform motion, but his special theory includes uniform and accelerated motion.
Answer:
His general theory includes uniform and accelerated motion, but his special theory applies only to uniform motion.
Explanation:
According to Einstein's 1915 general theory of relativity, the force of gravity arises from the curvature of space and time.
According to theory of special relativity:
1. The laws of physics are the same for all non-accelerating observers
2. The speed of light in a vacuum was independent of the motion of all observers.
His general theory includes uniform and accelerated motion, but his special theory applies only to uniform motion.
Answer:
for those who dont like to read
the answer is A.
hope i helped
Explanation:
calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun
Answer:
Hello your question is incomplete below is the complete question
Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000
answer : V = 1.624* 10^-5 m/s
Explanation:
First we have to calculate the value of a
a = 93 * 10^6 mile/m * 1609.344 m
= 149.668 * 10^8 m
next we will express the distance between the earth and the sun
[tex]r = \frac{a(1-E^2)}{1+Ecos\beta }[/tex] --------- (1)
a = 149.668 * 10^8
E (eccentricity ) = ( 1/60 )^2
[tex]\beta[/tex] = 90°
input the given values into equation 1 above
r = 149.626 * 10^9 m
next calculate the Earths velocity of approach towards the sun using this equation
[tex]v^2 = \frac{4\pi^2 }{r_{c} }[/tex] ------ (2)
Note :
Rc = 149.626 * 10^9 m
equation 2 becomes
([tex]V^2 = (\frac{4\pi^{2} }{149.626*10^9})[/tex]
therefore : V = 1.624* 10^-5 m/s
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
Answer:
303 Ω
Explanation:
Given
Represent the resistors with R1, R2 and RT
R1 = 633
RT = 205
Required
Determine R2
Since it's a parallel connection, it can be solved using.
1/Rt = 1/R1 + 1/R2
Substitute values for R1 and RT
1/205 = 1/633 + 1/R2
Collect Like Terms
1/R2 = 1/205 - 1/633
Take LCM
1/R2 = (633 - 205)/(205 * 633)
1/R2 = 428/129765
Take reciprocal of both sides
R2 = 129765/428
R2 = 303 --- approximated
What type of force holds atoms together in a crystal?
Answer:
Covalent Bond
Explanation:
i took the test , mark me brainliest.
Answer: Electrical
Explanation: Atoms are tied together by electrical bonding forces.
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
This question is incomplete, the complete question is;
The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
Answer:
the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Explanation:
Given that;
P₁ = 1.00 atm
P₂ = ?
V₁ = 1 L
V₂ = 1.60 L
the temperature of the gas is kept constant
we know that;
P₁V₁ = P₂V₂
so we substitute
1 × 1 = P₂ × 1.60
P₂ = 1 / 1.60
P₂ = 0.625 atm
Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
What does g stand for
Group of answer choices
gravity
The acceleration of gravity
The force of gravity
Answer:
the acceleration of gravity.
Answer:
g stand for the acceleration of gravity .
Explanation:
In a lightning bolt, a large amount of charge flows during a time of 1.2 x 10-3 s. Assume that the bolt can be represented as a long, straight line of current. At a perpendicular distance of 21 m from the bolt, a magnetic field of 8.4 x 10-5 T is measured. How much charged has flowed during the lightning bolt?
Answer: 10.58 C has flowed during the lightning bolt
Explanation:
Given that;
Time of flow t = 1.2 × 10⁻³
perpendicular distance r = 21 m
Magnetic field B = 8.4 x 10⁻⁵ T
Now lets consider the expression for magnetic field;
B = u₀I / 2πr
the current flow is;
I = ( B × 2πr ) / u₀
so we substitute
I = ( (8.4 x 10⁻⁵) × 2 × 3.14 × 21 ) / 4π ×10⁻⁷
= 0.01107792 / 0.000001256
= 8820 A
Hence the charge flows during lightning bolt will be;
q = It
so we substitute
q = 8820 × 1.2 × 10⁻³
q = 10.58 C
therefore 10.58 C has flowed during the lightning bolt
A car starts from rest and accelerates uniformly over a time of 18 seconds for a distance of 390 m. Determine the acceleration of the car.
Answer:
[tex]a=2.4\ m/s^2[/tex]
Explanation:
Given that,
The initial speed of a car, u = 0
Time, t = 18 s
Distance, d = 390 m
We need to find the acceleration of the car. Let it is a. Using the second equation of motion to find it.
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
or
[tex]d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 390}{(18)^2}\\\\a=2.4\ m/s^2[/tex]
So, the acceleration of the car is [tex]2.4\ m/s^2[/tex].
describe the energy conversion that occurs in a diesel engine
PLEASE PROVIDE AN EXPLANATION!!
THANK YOU.
Answer:
(a) 3.43 m/s
(b) 3.43 m/s
(c) 95.8 kPa
Explanation:
Use Bernoulli equation:
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
where P is pressure (either absolute or gauge), ρ is density, v is velocity, g is acceleration due to gravity, and h is elevation.
(a) Let's choose point 1 at the surface of the fluid in the container, and point 2 at point Z at the exit of the tube. I'll say 0 elevation is at point Z, and I'll use gauge pressure.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
0 Pa + ½ ρ (0 m/s)² + ρ (9.8 m/s²) (0.60 m) = 0 Pa + ½ ρ v² + 0
ρ (9.8 m/s²) (0.60 m) = ½ ρ v²
5.88 m²/s² = ½ v²
v = 3.43 m/s
(b) The tube's cross section is constant, so the fluid's speed is the same at all points in the tube. v = 3.43 m/s.
(c) Use Bernoulli equation again, choosing point 2 to be at Y. I'll say 0 elevation is at the surface of the fluid, and again use gauge pressure.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
0 + 0 + 0 = P + ½ (700 kg/m³) (3.43 m/s)² + (700 kg/m³) (9.8 m/s²) (0.20 m)
0 = P + 4116 Pa + 1372 Pa
P = -5488 Pa
The gauge pressure is -5488 Pa, so the absolute pressure is 101,300 Pa + -5488 Pa = 95812 Pa, or approximately 95.8 kPa.
what is the meaning of the word physics
Answer:
the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.
Explanation:
mark as brainliest
Calculate the effective charges on the H and F atoms of the HF molecule in units of the electronic charge, e.
Answer:
Explanation:
Hydrogen fluoride (HF) is an ionic/electrovalent compound that dissociates into ions when dissolved in water. It's dissociation is as seen below
HF ⇄ H⁺ + F⁻
There is a transfer of electron from the hydrogen atom which produces the hydrogen ion (H⁺), while the fluorine atom receives the donated ion to become negatively charged (F⁻). The amount of charge in one electron is generally given as 1.602 × 10⁻¹⁹ coloumbs.
The required value of effective charge on HF molecule, due to H and F is 1.602 × 10⁻¹⁹ Coulombs.
The given problem is based on the concept of effective charges. The net positive charge carried out by the electrons of atomic species, after forming a polyelectronic atom is known as Effective charge.
As per the given problem, the Hydrogen fluoride (HF) is an ionic/electrovalent compound that dissociates into ions when dissolved in water. It's dissociation is given as,
HF ⇄ H⁺ + F⁻
There is a transfer of electron from the hydrogen atom which produces the hydrogen ion (H⁺), while the fluorine atom receives the donated ion to become negatively charged (F⁻). The amount of charge in one electron is generally given as 1.602 × 10⁻¹⁹ Coulombs.
Thus, we can conclude that the required value of effective charge on HF molecule, due to H and F is 1.602 × 10⁻¹⁹ Coulombs.
Learn more about the effective charge here:
https://brainly.com/question/25002720
Before digital filmmaking, what tool was used to control the speed of movement on the screen after filming?
Answer:
The appropriate response is "Optical printer ".
Explanation:
A photographic printer used mostly for optical aberrations, comprised simply of either a camera that captures the frame to expand, minimize, deform, respectively. through magnifying lenses. A projector that always, as distinct from some kind of touch printer, transferred the image to something like the printing supply.what causes chest pain is it by eating peppery food?
Answer:
Yes that is almost always the problem!
Explanation:
Answer:
the answer is Acid reflux hope this helped!
A studious physics student is interrupted by a swarm of bees and chased off a cliff. Since she has her calculator in hand she quickly punches in numbers to figure out the initial velocity she needs to make it into the lake below. The cliff is 10 m high and the lake is 15 m away from the edge of the cliff. Find the time it takes her to drop. Find her initial velocity,
Answer:
The time is 1.4 sec
The initial velocity is 10.7 m/s.
Explanation:
Given that,
Height = 10 m
Distance = 15 m
We need to calculate the time
Using equation of motion
[tex]s=ut-\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]10=0+\dfrac{1}{2}\times9.8\times t^2[/tex]
[tex]t^2=\dfrac{2\times10}{9.8}[/tex]
[tex]t=\sqrt{\dfrac{2\times10}{9.8}}[/tex]
[tex]t=1.4\ sec[/tex]
We need to calculate the initial velocity
Using formula of velocity
[tex]v=\dfrac{d}{t}[/tex]
Put the value into the formula
[tex]v=\dfrac{15}{1.4}[/tex]
[tex]v=10.7\ m/s[/tex]
Hence, The time is 1.4 sec
The initial velocity is 10.7 m/s.
Determine the mass m1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration).
This question is incomplete, the complete question is;
Two blocks, M1M1 and M2,M2, are connected by a massless string that passes over a massless pulley as shown in the figure. M2,M2, which has a mass of 25.0 kg,25.0 kg, rests on a long ramp of angle =25.0∘.θ=25.0∘. Friction can be ignored in this problem.
Determine the mass 1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration).
Answer:
the mass m1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration) is 10.57 kg
Explanation:
Given that;
m₂ = 25 kg
θ = 25°
Now at equilibrium, T = m₁g ------------------lets say equ 1
and also T = m₂gsinθ
therefore
m₁g = m₂gsinθ
m₁ = m₂sinθ
so we substitute
m₁ = 25 × sin(25)
m₁ = 25 × 0.4226
m₁ = 10.565 ≈ 10.57 kg
therefore the mass m1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration) is 10.57 kg
HELP PLS7. A steel ball is dropped from a height of 100 meters. Which velocity-time graph best describes the
motion of the ball?
Answer:
Option C.
Explanation:
To know which velocity-time graph best describes the motion of the ball, let us calculate the velocity of the ball and the time taken for the ball to get the ground. This can be obtained as follow:
1. Determination of the velocity.
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 100 m
Final velocity (v) =.?
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 100)
v² = 0 + 1960
v² = 1960
Take the square root of both side.
v = √(1960)
v = 44.27 m/s
2. Determination of the time taken.
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 100 m
Time (t) =.?
h = ½gt²
100 = ½ × 9.8 × t²
100 = 4.9 × t²
Divide both side by 4.9
t² = 100 / 4.9
Take the square root of both side
t = √(100 / 4.9)
t = 4.52 s
From the above illustration,
Initial time (t1) = 0 s
Final time (t2) = 4.52 s
Initial velocity (u) = 0 m/s
Final velocity (v) = 44.27 m/s
Thus, we can see that as the time increase, the velocity also increase. Therefore, option C gives the correct answer to the question.
A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping distance of the car?
v² - u² = 2 a ∆x
where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.
So
0² - (27 m/s)² = 2 (-8 m/s²) ∆x
∆x = (27 m/s)² / (16 m/s²)
∆x ≈ 45.6 m
The stopping distance of car achieved during the braking is of 45.56 m.
Given data:
The initial speed of car is, u = 27 m/s.
The final speed of car is, v = 0 m/s. (Because car comes to stop finally)
The magnitude of deacceleration is, [tex]a = 8\;\rm m/s^{2}[/tex].
In order to find the stopping distance of the car, we need to use the third kinematic equation of motion. Third kinematic equation of motion is the relation between the initial speed, final speed, acceleration and distance covered.
Therefore,
[tex]v^{2}=u^{2}+2(-a)s[/tex]
Here, s is the stopping distance.
Solving as,
[tex]0^{2}=27^{2}+2(-8)s\\\\s = 45.56 \;\rm m[/tex]
Thus, we can conclude that the stopping distance of car achieved during the braking is of 45.56 m.
Learn more about the kinematic equation of motion here:
https://brainly.com/question/11298125
what is the summary for Electrons and protons
Explanation:
the link enjoy
A truck with a mass of 1330 kg and moving with a speed of 15.0 m/s rear-ends a 805 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.
Answer:
The Speed of the vehicles is 9.34m/s
Explanation:
For an elastic collision the two bodies move with similar velocities after collision
Given
M1=1330kg
V1=15m/s
M2=805kg
V2=0(the car is parked on neutral)
The formula is
M1V1+M2V2=(M1+M2)V
1330*15+805*0=(1330+805)V
19950+0=2135V
2135V=19950
divide both sides by 2135
V=19950/2135
V=9.34m/s
If car A is at 40km/h and car B is at 10km/h in the opposite direction, what is the velocity of the car A relative to the car B?.
Explanation:
The velocity of car A relative to car B is (10km/h+40km/h)=50km/h
How much energy is used by a 1000 W microwave that operates for 5
minutes?
Answer:
300000 Joules
Explanation:
Recall that the unit Watts is Joules per second, derived from the quotient of energy per unite of time. Therefore, to calculate the energy used on the given time, we need to multiply the power used (1000 W) times the time used. Then we need to express the 5 minutes in seconds to get our answer in Joules:
5 minutes = 5 * 60 seconds = 300 seconds
Final energy calculation gives:
E = 1000 W * 300 sec= 300000 Joules
A person drops a marble from the top of a skyscraper. After falling four floors the marble has gained a certain speed. How many more floors will the marble have to fall to triple this speed?
a. 8
b. 12
c. 32
d. 48
Answer:
B. 12
Explanation:
4 x 3 = 12
Find the gravitational potential energy of an 84 kg person standing atop Mt. Everest at an altitude of 8848 m. Use sea level as the location for y
Answer:
[tex]E=7.28\times 10^6\ J[/tex]
Explanation:
Given that,
Mass of a person, m = 84 kg
The person is standing at a top of Mt. Everest at an altitude of 8848 m
We need to find the gravitational potential energy of the person. We know that the gravitational potential energy is possessed due to the position of an object. It is given by :
E = mgh, g is the acceleration due to gravity
[tex]E=84\ kg\times 9.8\ m/s^2\times 8848\ m\\\\E=7283673.6\ J\\\\E=7.28\times 10^6\ J[/tex]
So, the gravitational potential energy of the person is [tex]7.28\times 10^6\ J[/tex]
What type of research based on approach that used to describe variables rather than to test a predicted relationship between variables?
Answer:
Correlational research can be used to see if two variables are related and to make predictions based on this relationship.
Find the linear velocity of a point moving with uniform circular motion, if the point covers a distance s in the given amount of time t. s
Answer:
The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]
Explanation:
From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:
[tex]v = r\cdot \omega[/tex] (Eq. 1)
Where:
[tex]r[/tex] - Radius of rotation of the particle, measured in meters.
[tex]\omega[/tex] - Angular velocity, measured in radians per second.
[tex]v[/tex] - Linear velocity of the point, measured in meters per second.
But we know that angular velocity is also equal to:
[tex]\omega = \frac{\theta}{t}[/tex] (Eq. 2)
Where:
[tex]\theta[/tex] - Angular displacement, measured in radians.
[tex]t[/tex] - Time, measured in seconds.
By applying (Eq. 2) in (Eq. 1) we get that:
[tex]v = \frac{r\cdot \theta}{t}[/tex] (Eq. 3)
From Geometry we must remember that circular arc ([tex]s[/tex]), measured in meters, is represented by:
[tex]s = r\cdot \theta[/tex]
[tex]v = \frac{s}{t}[/tex]
The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]
A designer is creating an obstacle for an obstacle course where a person starts on a moveable platform of height H from the ground. The person grabs a rope of length L and swings downward. At the instant the rope is vertical, the person lets go of the rope and attempts to reach the far side of a water-filled moat. The left side of the moat is directly below the position where the person will let go of the rope. The designer runs several tests in which the rope has different lengths and moves the platform such that the rope is always initially horizontal. The designer notices that the person cannot land on the other side if the length L is very short. The designer also notices that the person also cannot land on the other side if the length L is very close to the height H.
Assume the size of the person is much smaller than the lengths L and H. Let D represent the horizontal distance from below the release point to where the person lands.
Required:
a. Why does the person land in the moat if the rope's length is very short?
b. Why does the person land in the moat if the length is nearly the same as the height of the platform?
Answer:
* when L → H chord too long
in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross
* when L → 0 very short string
the speed of the platform is very small, so we do not have the minimum required value
vox = √ (g / (2 (H)) D
Explanation:
For this exercise we are going to solve it using conservation of energy to find the velocity of the body and the launch of projectiles to find the velocity to cross the well.
Let's start with the projectile launch
as the body leaves the vertical its velocity must be horizontal
x = v₀ₓ t
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
when reaching the ground its height of zero (y = 0) and the initial vertical velocity is zero
t = √ 2 y₀ / g
we substitute
x = vox √2y₀ / g
v₀ₓ = √(g / 2y₀) x
In the exercise, it tells us that the width of the well is D (x = D) and the initial height is the height of the platform minus the length of the rope (I = H - L)
v₀ₓ = √(g /(2 (H -L)) D
this is the minimum speed to cross the well.
Now let's use conservation of energy
starting point. On the platform
[tex]Em_{o}[/tex] = U = m g H
final point. At the bottom of the swing
Em_{f} = K + U = 1 / 2m v² + m g (H -L)
as there is no friction the mechanical energy is conserved
Em_{o} = Em_{f}
m g H = 1 / 2m v² + m g (H -L)
v = √ (2gL)
let's write our two equations
the minimum speed to cross the well
v₀ₓ = √ (g /(2 (H -L)) D
the speed at the bottom of the oscillatory motion
v = √ (2g L)
we analyze the extreme cases
* when L → H chord too long
in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross
* when L → 0 very short string
the speed of the platform is very small, so we do not have the minimum required value
vox = √ (g / (2 (H)) D
From this analysis we see that there is a range of lengths that allows us to have the necessary speeds to cross the well
V₀ₓ = v
g / (2 (H -L) D² = 2g L
4 L (H- L) = D²
4 H L - 4 L2 - D² = 0
L² - H L - D² / 4 = 0
let's solve the quadratic equation
L = [H ± √ (H2-D2)] / 2
we assume that H> D
L = ½ H [1 + - RA (1 - (D / H) 2)]
The two values of La give the range of values for which the two speeds are equal
A) The person lands in the moat if the rope's length is very short because :
The speed of the platform is less than the required minimum speedB) The person lands in the moat if the rope length is similar to the height of the platform because :
The speed required to cross the moat approaches infinityFollowing the assumptions;
size of the person is much smaller than L and H
D = horizontal distance
The conditions that will cause the person to land on the moatThe person will land in the moat when the rope's length is very short because as the rope reduces in length the speed reduces as well such that the speed of the platform goes below the required minimum speed which will enable the person cross over. while As the magnitude of the length tends towards the same magnitude of the height the speed required to cross the moat increases towards infinity and this speed cannot be attained by the person hence he will land in the moat.Hence we can conclude that The person lands in the moat if the rope's length is very short because The speed of the platform is less than the required minimum speed and The person lands in the moat if the rope length is similar to the height of the platform because,the speed required to cross the moat approaches infinity.
Learn more about obstacle course : https://brainly.com/question/241926