Answer:
Explanation:
From the given information:
At state 1:
Initial Quality [tex]= x_1 = 0.85[/tex]
mass = 10.0 kg
At state 2:
Temperature [tex]T_2 = 320^0[/tex]
mass of the piston [tex]m_p = 204 \ kg[/tex]
area of the piston [tex]A_p = 0.00 5 \ m^2[/tex]
Atmospheric pressure [tex]P_{atm}= 100 \ kPa = 100 \times 10^3 \ Pa[/tex]
Gravitational acceleration = 9.81 m/s²
[tex]\mathbf{P= P_1=P_2}[/tex], This is because there exists no restriction to the movement of the piston and provided the process is frictionless. So, the process 1-2 is regarded as constant.
To calculate the applying force balance over the piston by using force balance in the vertical direction:
[tex]\mathbf{P_{AP} = P_{atmA_p} + m_pg}[/tex]
∴
(100 × 10³)×0.005 + 204 × 9.31 = P × 0.05
P = 500248 Pa
P = 500.25 kPa
At state 1:
[tex]\mathbf{P_1 = P = 500.25 \ kPa}[/tex]
[tex]x_1 = 0.85[/tex]
Hence, this is a saturated mixture of liquid and vapor
Using the steam tables at 500.25 kPa
[tex]V_f = 1.093 \times 10^{-3} \ m^3/kg \\ \\ V_g = 0.375 \ m^3/kg \\ \\ U_f = 639.72 \ kJ/kg \\ \\ U_g = 2560.72 \ kJ/kg[/tex]
∴
Specific volume at state 1 is given as:
[tex]V_1 = [ V_f +x_1(v_g -v_f) ] \ at \ 500.25 \ kPa \\ \\ V_1 = 0.319 \ m^3/kg[/tex]
volume at state 1 is given by:
[tex]V_1 = mV_1 = 10 \times 0.319 \\ \\ V_1 = 3.19 \ m^3[/tex]
Similarly, the specific internal energy is:
[tex]U_1 = [U_f +x_1 (U_o-Uf)] \ at \ 500.25 \ kPa[/tex]
[tex]U_1 = 639.72 +0.82 (2560.72 -639.72)[/tex]
[tex]U_1 = 2272.57 \ kJ/kg[/tex]
At state 2:
[tex]P = P_1 = P_2 = 500.25 \ kPa \\ \\ T_2 = 320^0 \ C[/tex]
Using steam tables at P = 500.25 kPa and T = 320° C
[tex]V_2 = 0.541 \ m^3/kg \\ \\ U_2 = 2835.08 \ kJ/kg[/tex]
∴
[tex]V_2 = mV-2 = 10 \times V_2 = 5.41 \ m^3[/tex]
[tex]\text{Now; Applying the 1st law of thermodynamics to the system}[/tex]
[tex]_1Q_2 -_1W_2 = \Delta V =m(u_2-u_1) \\ \\ where;\ _1W_2 = P(V_2-V_1) \\ \\ _1Q_2 -P(V_2-V_1) = m(u_2-u_1) \\ \\ _1Q_2 - 500.25(5.91 -3.19) = 10( 2835.08 -2272.57) \\ \\ \mathbf{ _1Q_2 = 6735.66 \ kJ}[/tex]
What is The first stage in cellular respiration, what cycle?
ANSWER:
What is The first stage in cellular respiration, what cycle?- GLYCOLYSIS
In glycolysis, the beginning process of all types of cellular respiration, two molecules of ATP are used to attach 2 phosphate groups to a glucose molecule, which is broken down into 2 separate 3-carbon PGAL molecules. PGAL releases electrons and hydrogen ions to the electron carrier molecule NADP+.What are the 4 stages of cellular respiration?It has four stages known as glycolysis, Link reaction, the Krebs cycle, and the electron transport chain. This produces ATP which supplies the energy that cells need to do work.What are characteristics of wide-angle lenses?
A tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J
Answer:
3.9 × 10^7 J
Explanation:
Given that a tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J
Solution
Since the tank is half full, the height = 2.5m
Pressure = density × gravity × height
Pressure = 900 × 9.8 × 2.5
Pressure = 22050 Pascal
The cross sectional area of the pump will be area of a circle.
A = πr^2
A = π × 15^2
A = 706.858 m^2
Using the formula
Density = mass/volume
Mass = density × volume
Mass = 900 × 706.86 × 2.5
Mass = 1590.435
Energy = mgh
Energy = 1590.435 × 9.8 × 2.5
Energy = 38965657.8 J
Since the work done = energy
Therefore, the work done = 3.9 × 10^7 J
Velocity tells us not only how fast something is going, but in what direction it is traveling.
True
False
Answer:
true
Explanation:
because I tride it
A dump truck, whose bed is made of steel, holds an old steel watering trough. The bed of the truck is slowly raised until the trough begins to slide. For dry steel to steel μs= 0.80, μk= 0.60. What is the acceleration of the trough as it slides down the truck bed? Express your answer with the appropriate units.
Answer:
a = 1,538 m / s²
Explanation:
Let's use Newton's second law, let's set a reference system where the x-axis is parallel to the sloping floor of the truck and the positive direction is in the direction of movement of the trough, for this case the weight is the only force to decompose
sin θ = Wₓx / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Y axis
N -W_y = 0
N = mg cos θ
X axis
Wₓ - fr = m a
the friction force has the expression
fr = μ N
There are values of the friction coefficient (μ_s) one for when the movement has not started and it takes a smaller value for when the bodies are moving.
In this case we first find the angle for which the movement begins, in this part we use the static coefficient and the acceleration is zero
Wₓ - μ_s N = 0
m g sin θ = μ_s mg cos θ
tan θ = μ_s
θ = tan⁻¹ μ_s
we calculate
θ = tan⁻¹ 0.8
θ = 38.7º
For this angle, how the trough begins to move, the coefficient is reduced to the dynamics coefficient (μ_k) and the acceleration is different from zero.
we substitute
mg sin θ - μ_k mg cos θ = m a
a = g (sin θ - μ_k cos θ)
let's calculate
a = 9.8 (sin 38.7 - 0.6 cos 38.7)
a = 1,538 m / s²
The rate at which work is done is known as which of the following?
A. Power,
B. Energy,
C. Momentum,
Answer:
A , power
Explanation:
Hope this is useful. Have a lovely rest of your day! God bless you.
the motion of a body with respect to another body is?
Answer:
Motion that changes the orientation of a body is called rotation. ... In both cases all points in the body have the same velocity (directed speed) and the same acceleration (time rate of change of velocity).
Answer:
its called relative speed
What x rays travel at the speed of
HELP
HELP
SORRY TO BEG :/
HELP ASAP :) PLZ
Answer:
It's b because he discovered galaxies .
Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for slinging supplies to affected areas, similar to the electromagnetic lift catapults used to launch planes from aircraft carriers. This catapult is located at a fixed point 400 meters away and 50 meters below the target site. The catapult is capable of launching the payload at 67 meters per second and an initial launch angle of 50 degrees. Using your knowledge of kinematics equations, determine whether this would be sufficient to deliver the payload to the drop site.
Answer:
Please see below as the answer is self-explanatory.
Explanation:
We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).Both movements are independent each other, due to they are perpendicular.In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:[tex]v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)[/tex]
Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:[tex]t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)[/tex]
In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:[tex]\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)[/tex]
In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:[tex]v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)[/tex]
Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:[tex]\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)[/tex]
Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.A friend pushes a sled across horizontal snow and when it gets up to speed the friend jumps on. After the friend jumps on, the sled gradually slows down. Which forces act on the combined sled plus friend after the friend jumps on
Answer:
v’ =( [tex]\frac{1}{1+ \frac{M}{m} }[/tex] ) v
we see that the greater the difference, the more the sled slows down.
friction force
Explanation:
When the man pushes the sled he does work and the sled acquires a speed and as long as it is supplied with an energy equal to the work of the chipping force with the snow, the speed is maintained.
When he jumps on the sled, a collision occurs and the initial moment
p₀ = mv
is increased by the increase in mass
m_f= (m + M_{man} ) v '
In this case there is no longer any external force applied and the only external force is friction, which causes the sled to stop, even when it is small, but the significant reduction in speed is due to the increase in masses.
p₀ = p_f
mv = (m + M_{man}) v '
v ’= [tex]\frac{m}{m+M}[/tex] v
v’ =( [tex]\frac{1}{1+ \frac{M}{m} }[/tex] ) v
Therefore, we see that the greater the difference, the more the sled slows down.
The only forces that act on the sled with the man are the friction that is responsible for the decrease in speed and weight with the normal
what is the potential energy the greatest in a roller coaster
Answer:
Gravitational potential energy is the greatest
Explanation
It is the highest point of a roller coaster.
A woman pushes a 35.0 kg object at a constant speed for 10.8 m along a level floor, doing 280 J of work by applying a constant horizontal force of magnitude F on the object. (a) Determine the value of F (in N). (Enter the magnitude.) N (b) If the worker now applies a force greater than F, describe the subsequent motion of the object. The object's speed would increase with time. The object's speed would remain constant over time. The object would slow and come to rest. (c) Describe what would happen to the object if the applied force is less than F. The object's speed would increase with time. The object's speed would remain constant over time. The object would slow and come to rest.
Answer:
a) F= 25.9 N
b) The object's speed would increase with time.
c) The object would slow and come to rest.
Explanation:
a)
By definition of work, this is the process through which a force applied on an object, produces a displacement of the object.If the force is constant, and the displacement is parallel to the direction of the force, this work is just the product of the applied force times the distance, as follows:[tex]W = F_{app} * d (1)[/tex]
We can solve for Fapp, replacing W and d by their values:[tex]F_{app} =\frac{W}{d} = \frac{280J}{10.8m} = 25.9 N (2)[/tex]
b)
If the object is moving at constant speed when it is applied a force F, this means that this force must be compensated by an equal opposite force, in this case, the kinetic friction force.Since this force is constant while the object is moving, if we increase the force F making it larger, there will be a net force in the direction of the displacement, which will cause an acceleration that will increase the speed with time.c)
We have already said in b) that if the object is moving at constant speed, there must be an equal and opposite force to the applied force F, the kinetic friction force, which is constant, acting on the object.If we apply a force less than F, there will be a net force in the direction opposite to the displacement, that will cause a acceleration opposite to the displacement, which will make the object to slow down and eventually come to rest.1. Describe how unequal heating and rotation of the Earth causes atmospheric and oceanic circulation patterns that determine the regional and global climate.
Which type of electricity is a one-time event, caused by unbalanced charges trying to become neutral again?
Answer: yes
Explanation:
A 65.0 kg skier slides down a 37.20 slope with mu = 0.107.
What is the friction force?
Answer:
54.3N
Explanation:
The normal force is perpendicular to the slope, so:
Normal Force = cos(37.2)(9.8*65).......507.39N
F(friction)=mu*F(normal)
F(friction)=(0.107)(507.39)
F(friction)=54.3N
Answer:
magnitude of friction force- 54.3
friction force- -54.2
Explanation:
If 155 g of sugar can be dissolve in 100g of water at 20 degree than how much sugar will be dissolved in 300 g of water at same temperature if someone do I will make you brainly
Answer:
465 grams of sugar.
Explanation:
I'm not sure this is true: is the relationship linear?
If it is then 300 grams of water should be able told 3*155 grams of sugar
3 * 155 = 465 grams of sugar in 300 grams of water.
a ball is spun around in circular motion such that it completes 50 rotations in 25 s.
1) What is the period of its rotation?
2) what is the frequency of its rotation?
Answer:
(A) The period of its rotation is 0.5 s (2) The frequency of its rotation is 2 Hz.
Explanation:
Given that,
a ball is spun around in circular motion such that it completes 50 rotations in 25 s.
(1). Let T be the period of its rotation. It can be calculated as follows :
[tex]T=\dfrac{25}{50}\\\\T=0.5\ s[/tex]
(2). Let f be the frequency of its rotation. It can be defined as the number of rotations per unit time. So,
[tex]f=\dfrac{50}{25}\\\\f=2\ Hz[/tex]
Hence, this is the required solution.
An object which is dropped from a certain height has zero (0) initial velocity.
Answer:
0 m/s
Explanation:
if an object is dropped we know the initial velocity is zero when in free fall
What is the magnitude of the force that is exerted on a 20 kg mass to give it an acceleration of 10.0
m/s2?
Answer:Mass of the body = 20 kg.
Final Velocity = 5.8 m/s.
Initial velocity = 0
Time = 3 seconds.
Using the Formula,
Acceleration = (v - u)/ t
= (5.8 - 0)/ 3
= 1.6 m /s².
Now, Using the Formula,
Force = mass × acceleration
= 20 × 1.6
=
Explanation: I REALLY HOPE THIS HELPS I'M KINDA NEW AT THIS :] :]
The magnitude of requires force, that is exerted on a 20 kg mass to give it an acceleration of 10.0 m/s^2 is 200 Newton.
What is force?The definition of force in physics is: The push or pull on a mass-containing item changes its velocity.
An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force.
Given parameters:
Mass of the object: m = 20 kg.
Acceleration of the object: a = 10.0 m/s^2.
Hence, according to Newton's 2nd law of motion:
the magnitude of requires force = mass ×acceleration
= 20 × 10 Newton
= 200 Newton.
Hence, the magnitude of requires force is 200 Newton.
Learn more about force here:
https://brainly.com/question/13191643
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What is the magnitude of the electric field at the point (8.70 - 9.10 7.20 ) m if the electric potential is given by V
This question is incomplete, the complete question is;
What is the magnitude of the electric field at the point (8.70i - 9.10j + 7.20k ) m if the electric potential is given by V = 4.30xyz², where V is in volts and x, y and z are in meters.
Answer:
the magnitude of the electric field is 5648.67 N/C
Explanation:
Given the data in the question;
Suppose electric field at this point is;
E = Exi + Eyj + Ezk
Now, electric field is given by;
E = -dV/dr
so,
Ex = -d( 4.30xyz² )/dx = -4.30yz²
Ey = -d( 4.30xyz² )/dy = -4.30xz²
Ez = -d( 4.30xyz² )/dz = -8.60xyz
so
E = -4.30yz² i - 4.30xz² j - 8.60xyz k
now, at the point (8.70i - 9.10j + 7.20k )
E = (-4.30(-9.10)(7.20)²) i + (-4.30(8.70)(7.20)²) j + (-8.60(8.70)(-9.10)(7.20) k
E = 2028.499 i - 1939.334 j + 4902.206 k
so, Magnitude of electric field will be;
|E| = √( Ex² + Ey² + Ez² )
we substitute
|E| = √( (2028.499)² + (-1939.334)² + (4902.206)² )
|E| = √( 31907448.222993 )
|E| = 5648.67 N/C
Therefore, the magnitude of the electric field is 5648.67 N/C
If a person walks 3 m north and 5 meters east, how would you find the displacement for that person? what would the displacement be?
Answer:
AC)=(AB)2+(BC)2−−−−−−−−−−−−√=42+32−−−−−−√
⇒displacement=16+9−−−−−√=25−−√=5m
14. Which of the following is not an example of work being done?
A. pushing a basketball away from your body
B. holding a coffee mug
C. carrying boxes across a warehouse floor
Answer:
B. holding a coffee mug
Explanation:
Something must move a distance for work to be done.
6)
A 50 kg bike and rider accelerate from 4 to 8 m Determine how much kinetic energy must be added to make this occur?
S
S
A)
200
B)
400)
800)
D)
1200
Answer:
A.) 200 in my opinion so I hope this helps
Answer: 1200 j
Explanation:
I just took it
Help please, i am lost.
Answer:
nas wr hdrtuendnetyje because
Explanation:
list the 5 components of fitness
Pleaseee I need help!!
Answer:
R1=3.333 Ohms
R2=10 Ohms
R3=16.666 Ohms
Explanation:
30 total
30=R1+R2R3
30=1(x):3(x):5(x)
x=3.3333
R1=3.333 Ohms
R2=10 Ohms
R3=16.666 Ohms
if you ride a bike at 2 km/h and travel a total distance of 20km, how long willnit takr ( in second) you to teach your destination
Time = (20 km) x (1 hr/2 km) x (3,600 sec/hr)
Time = (20 x 1 x 3,600 km-hr-sec) / (2 km-hr)
Time = (20 x 1 x 3,600 / 2) (km-hr-sec / km-hr)
Time = 36,000 seconds
(That's 10 hours.)
HELP me please cause I don't understand it.
Answer:
Force = 0.49 N (Approx)
Explanation:
Given:
Mass = 50 grams = 0.05 Kg
Acceleration = 9.81 m/s²
Find:
Force
Computation:
Force = Mass x Acceleration
Force = 0.05 x 9.81
Force = 0.4905
Force = 0.49 N (Approx)
A ray of light strikes a flat mirror at an angle of 40° from the perpendicular. The light is reflected at
what opposite angle to the same perpendicular line?
a) 40°
b) 50°
c) 30°
d) 60°
Answer:
a
Explanation:
angle of incidence equals angle of reflection on flat surface
Answer:
40 degress
Explanation: