Answer:
a) [tex]V_2=8m/s[/tex]
b) [tex]P_2=9.54*10^4 Pa[/tex]
Explanation
From the question we are told that:
Initial Area of pipe [tex]A_1=8.00 cm^2[/tex]
Initial Fluid flow speed [tex]r_1 =320 cm/s,\approx 320*10^{-2}[/tex]
Initial Pressure of [tex]\rho_1 =1.40*10^5 Pa[/tex]
Final area of pipe [tex]A_2 =3.70 *10^{-2} cm^2[/tex]
Density of acid [tex]\rho=1660kg/m^3[/tex]
a)
Generally the equation for continuity is mathematically given by
[tex]A_1V_1=A_2V_2\\\\V_2=\frac{A_1*V_1}{A_2}[/tex]
Since volume is directly proportional to rate of flow
[tex]V_2=\frac{8*320}{3.20} *10^{-2}[/tex]
[tex]V_2=8m/s[/tex]
b)
Generally the Bernoulli's equation is mathematically given by
[tex]p_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_2+\frac{1}{2}\rho v_2^2+\rho gh_2\\\\with\ h_1=h_2\\\\p_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2[/tex]
Therefore
[tex]P_2=P_1+\frac{1}{2}\rho(v_1^2-V_2^2)\\\\P_2=(1.40*10^5)+\frac{1}{2}(1660)(v_1^2-V_2^2)[/tex]
[tex]P_2=9.54*10^4 Pa[/tex]
A 4500-kg helicopter accelerates upward at 2.0m/s2 what lift force is exerted by the air on the
helicopter?
Answer:
9000
Explanation:
f=ma
f= force
m=mass
a=acceleration
f=ma
4500×2.0
9000N
If the centripetal force is of the form m^a v^b r^c, find the values of a, b and c.
Answer with explanation:
20/05/2021
BP102T
- Which one of the following statements about reaction rate is false?
3253505030
Reaction rates are not sensitive to temperature.
Reaction rate is governed by the energy barrier between reactants and products.
Reaction rate is the speed at which the reaction proceeds.
Enzymes can accelerate the rate of a reaction.
Answer:
Reaction rate are not sensitive to temperature.
A convex lens has a focal length of 0.33 m. The object distance is 0.7 m. What is the image distance?
Answer:
Explanation:
1/v - 1/u = 1/f
given, f = 0.3 m, u = -0.4m
so, 1/v - 1/-0.4 = 1/0.3
or, 1/v = 1/0.3 - 1/0.4 = 1/1.2
v = 1.2 m
now, differentiating 1/v - 1/u = 1/f with respect to t,
-1/v² dv/dt + 1/u² du/dt = 0
or, dv/dt = (v/u)² du/dt
putting, du/dt = 0.01 m/s , v = 1.2 m and u = -0.4 m
so, dv/dt = (1.2/-0.4)² × 0.01
= 0.09 m/s
hence, speed of image with respect to lens is 0.09 m/s .
from formula of magnification
magnification, m = v/u
differentiating with respect to time both sides,
dm/dt = (u dv/dt - vdu/dt)/u²
= (-0.4 × 0.09 - 1.2 × 0.01)/(-0.4)²
= (-0.036 - 0.012)/0.16
= -0.048/0.16
= -0.3 m/s
hence, magnitude of rate of change of lateral magnification is 0.3 m/s
Solid pressure depends on?
Answer:
The pressure of the solid on the surface depends on the area of contact. The area of contact between the two surfaces. The greater the force or the smaller the area the greater the pressure.
(credits to the rightful owner for these answers :)
State the
the properties of magnets.
Answer:
All magnets have two poles: the North Pole and the South Pole.
Magnets attract ferromagnetic materials such as iron, nickel, and cobalt.
The magnetic force of a magnet is stronger at its poles than in the middle.
A freely suspended magnet always points in North-South direction.
Hope this helps
If the frequency of a wave is doubled, what happens to that waves energy
Answer:
If the frequency of a wave is doubled, what happens to its wavelength? If the frequency is doubled, the wavelength is only half as long. 3. ... As the frequency slows down, the wavelength increases.
Explanation:
oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student observes that a particular spring-mass system has a frequency of oscillation of 10 Hz. the spring constant of the spring is 250 N/m. what is the mass?
Answer:
Mass, m = 6.18 kg
Explanation:
Given the following data;
Frequency, F = 10 Hz
Spring constant, k = 250 N/m
We know that pie, π = 22/7
To find the mass, we would use the following formula;
F = 1/2π√(k/m)
Where;
F is the frequency of oscillation.
k is the spring constant.
m is the mass of the spring.
Substituting into the formula, we have;
10 = 1/2 * 22/7 * √250/m
10 = 22/14 * √250/m
Cross-multiplying, we have;
140 = 22 * √250/m
Dividing both sides by 22, we have;
140/22 = √250/m
6.36 = √250/m
Taking the square of both sides, we have;
6.36² = (√250/m)²
40.45 = 250/m
Cross-multiplying, we have;
40.45m = 250
Mass, m = 250/40.45
Mass, m = 6.18 kg
a 4 kg block is moving at 12 m/s on a horizontal frictionless surface. a constant force is applied such that the block slows with an acceleration of 3 m/s^2. how much work must this force do to stop the block?
a. -576 J
b. -360 J
c. -288 J
d. 360 J
e. 576 J
Answer:
c. -288 J
Explanation:
Given;
mass of the block, m = 4 kg
velocity of the block, v = 12 m/s
deceleration of the block, a = 3 m/s²
Apply work-energy theorem;
work done in stopping the block = kinetic energy of the block
W = ¹/₂mv²
where;
W is the magnitude of the work done by the force
W = ¹/₂ x 4 x 12²
W = 288 J
Since the work done by the force is in opposite direction, then the value of the work done by the force is -288 J
Four point masses are connected by rods of negligible mass and form a square with sides of length 32.2 cm. Three of the masses are 1.5 kg and one is 3.0 kg. How far from the 3.0 kg mass is the center of mass of the system
Solution :
Placing the[tex]$3 \ kg$[/tex] mass at the [tex]$\text{origin}$[/tex] and line up the square up with the axes.
[tex]$x_{cm} = \frac{\sum_i x_i m_i}{\sum_i m_i }$[/tex]
[tex]$=\frac{1.5 a + 1.5a +0 +0}{7.5}$[/tex]
[tex]$=\frac{3a}{7.5}$[/tex]
[tex]$=\frac{2a}{5}$[/tex]
[tex]$y_{cm} = \frac{\sum_i y_i m_i}{\sum_i m_i }$[/tex]
[tex]$=\frac{1.5 a + 1.5a +0 +0}{7.5}$[/tex]
[tex]$=\frac{3a}{7.5}$[/tex]
[tex]$=\frac{2a}{5}$[/tex]
Therefore, r = [tex]$\sqrt2 \left(\frac{2a}{5}\right)$[/tex]
[tex]$=\frac{2 \sqrt2}{5}a$[/tex]
It s given that the side of the square is a = 32.2 cm
So, r [tex]$=\frac{2 \sqrt2}{5}a$[/tex]
[tex]$=\frac{2 \sqrt2}{5}\times 32.2$[/tex]
= 18.21 cm
So the distance of the 3 kg mass from the center of mass, r= 18.21 cm
When a massive star greater than 8 solar masses has its core filled with iron, the core can no longer continue fusion, and it becomes supported by electron degeneracy pressure for a while. When the mass of the accumulating iron becomes greater than 1.4 solar masses, what happens
If the hairdryer was put on full power, what would happen? Why?
Most hair dryers are in the 1800 watt range and lets assume you turned it to the highest speed and heat setting.
PLEASE MARK AS A BRAINLISTIt will heat up the air in the box, failry quickly. Rules of thermodynamics says that the heat will increase the temperature according to the rate of heat input minus the rate of heat loss from the box… due to conduction, radiation and convection (in this case none of the latter because the box is sealed).
HOPE IT WILL HELP YOUThe loss rate depends in the temperature difference and the thermal resistance. THicker and better insulation increases the resistance; larger area decreases the resistance. Eventually the heat (temperature) will rise and the difference between inside and outside will be so large that the temperature will rise no more. But in the case of a 1800 W hairdryer the temperature will exceed the melting point of plastics and wire insulation and if allowed to come to heat equilibrium will probably short out and catch fire or blow an external fuse.
Most modern hair dryers have a internal thermal fuse that cuts out at temperatures below the melting point and probably this will cut off the dryer before catastrophic meltdown. Its a one time fuse and not readily available, mostly you toss the dryer when the fuse goes
8. Aunt Barbara sets a full 2.0 kg milk carton on the kitchen table for breakfast. After the family has eaten, she pours herself a cup of coffee, sits down, and picks up the milk carton, only to find that the 20 N force she exerts accelerates the carton upward at a surprising 50 m/s². Calculate and describe why the milk carton accelerates up faster than Aunt Barbara expect
If 03.4 kg of water at 69°C is changed into steam at 100°C, how will the energy be used? specific heat capacity of water is 4200J/kg C and latent heat is 2,260 000J/Kg.
A. 7864513 J
B. 8126680 J
C. 329758 J
D. 742976 J
With explanation please
Answer:
The answer is B.
Explanation:
مافيني اشرح
You are standing outside with two speakers. The temperature is 0 degrees C. The two speakers are playing sound of the same frequency: a pure sinusoidal sound of the same frequency at the same phase. The speakers are playing at different amplitudes. One speaker is 1.230 meters from you and one is 1.425 meters from you. You hear no sound. Which of the following is a possible frequency of the sound being played?
a. 993 Hz
b. 331 Hz
c. 4965 Hz
d. 1655 Hz
e. 662 Hz
Answer:
Explanation:
No sound is heard , that means there is destructive interference at the place where sound is heard .
Path difference of the source of sound = 1.425 m - 1.230 m
= 0.195 m
Let frequency of sound be n .
wave length = velocity of sound at 0°C / n
λ = 330 / n
for destructive interference ,
path difference = ( 2m+1) λ /2 , where λ is wave length of sound.
0.195 m = ( 2m+1) λ /2
0.195 m = ( 2m+1)330 / 2n
2n = 1692.3 (2m+1)
If m = 0
n = 846 . which is nearest to given frequency of 993 Hz
So 993 Hz is the answer.
The proper time between two events is measured by clocks at rest in a reference frame in which the two events: The proper time between two events is measured by clocks at rest in a reference frame in which the two events:
a. are separated by the distance a light signal can travel during the time interval
b. occur at the same time
c. satisfy none of the above
d. occur in Nashville occur at the same coordinates
The proper time between two events is measured by clocks at rest in a reference frame in which the two events occur in Nashville occur at the same coordinates. So, option d.
What is meant by frame of reference ?The perspective from which you perceive and measure things is known as a reference frame. It's applied to describe how an object is moving or where it is.
Here,
Depending on the frame of reference, the time elapsed between two occurrences may vary, according to the Theory of Special Relativity.
The duration between two events occurring in the rest frame, as measured from a reference frame in motion relative to the rest frame, will always be longer than the correct time. Time dilation is a phenomenon that develops in accordance with the relativity of space and time.
A single clock that is present for both occurrences can be used to determine the right time difference between them.
The appropriate time between two events is the amount of time that elapses within a frame in which the two events take place at the same location, to put it another way.
Hence,
The proper time between two events is measured by clocks at rest in a reference frame in which the two events occur in Nashville occur at the same coordinates. So, option d.
To learn more about frame of reference, click:
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We know that the frequency and wavelength of an oscillation are related by the velocity of the wave . In standing waves, the wave velocities of the different harmonics are the same.Think about why this may be and explain that in your notebook. (Hint: The wave velocity is dependent on the static properties of the oscillating object like mass per unit length and tension)
Answer
the medium does not change the speed of the on they should change.
Explanation:
The speed of the waves is constant for a given medium, depending on the physical properties of the medium,
When a wave is strapped on a wall of a medium it does not change the properties of the medium, the wave changes direction, but since the medium does not change the speed of the on they should change.
A radio antenna broadcasts a 1.0 MHz radio wave with 20.0 kW of power. Assume that the radiation is emitted uniformly in all directions.
1) What is the wave's intensity 30 km from the antenna
2) What is the electric field amplitude at this distance?
Answer:
1) [tex]I=1.8*10^{-6}\: W/m^{2}[/tex]
2) [tex]E=0.037 \: V/m[/tex]
Explanation:
1)
The intensity equation is given by:
[tex]I=\frac{P}{4\pi r^{2}}[/tex]
Where:
P is the power of the radio waver is the distance from the source[tex]I=\frac{20000}{4\pi (30000)^{2}}[/tex]
[tex]I=1.8*10^{-6}\: W/m^{2}[/tex]
2)
Now, the intensity and the electric field are related as:
[tex] I=0.5c\epsilon_{0}E^{2}[/tex]
Here:
c is the speed of lightε₀ is the electric permittivityE is the electric fieldWe need to solve it for E.
[tex]E= \sqrt{\frac{2I}{c\epsilon_{0}}}[/tex]
[tex]E= \sqrt{\frac{2(1.8*10^{-6})}{(3*10^{8})(8.85*10^{-12})}}[/tex]
[tex]E=0.037 \: V/m[/tex]
I hope it helps you!
Which of the following rays is NOT possible for a converging lens?
Ray B. isn't possible for a converging lens, because after passing through focus ot would go parallel to the principal axis,
So, Correct option is :
=》Option B.)
A hydraulic system is lifting a 11760 N car using a cylinder with an area of 0.25 m
squared. What force is applied to the small cylinder if it has an area of 0.0125 m
squared?
Answer: 588 N
Explanation: pressure = force/area, or p = F/A. Then p1=p2
And F1//A1 = F2/A2 . F2 = F1·A2/A1 = 11760 N·0.0125 m²/ 0.25 m²
Which has a greater momentum: a 0.0010 kg bullet going
250 m/s OR a 80 kg student walking at 4 m/s? Which has
more inertia? Which has more kinetic energy?
Answer:
Momentum is Mass x Velocity.
Its pretty obvious that the 80kg student Moving at 4ms-¹ has more momentum.
80x4 = 320kgms-1
0.001x250= 0.25kgms-¹
The Second student also Has More Inertia. Inertia is the resistance to motion offered by a Body. An Object with greater mass has more tendency to resist Motion .
So
The 80Kg student wins all.
Answer:
i think it is 80 kg
Explanation:
Plz help The momentum of a baseball changes dramatically when struck by a bat.Momentum of the ball is not conserved. The best explanation for this is that
Answer:
The answer is C the ball is not in a closed system and experience an impulse.
Explanation:
Claire pushes a box to the left while Jamie pushes the same box to the right. The box stays in
between them in the same position.
Which best explains why Claire and Jamie are unable to move the box?
A. They are applying an unbalanced force.
B. They are applying a balanced force.
C. They are applying gravity.
D. They are applying friction.
Particle A has less mass than particle B. Both are pushed forward across a frictionless surface by equals forces for 1 s. Both start from rest.
a. Compare the amount of work done on each particle. That is, is the work done on A greater thane, less than, or equal to the work done on B? Explain.
b. Compare the impulses delivered to particles A and B. Explain.
c. Compare the final speeds of particles A and B. Explain.
An Particle a has Weston practical.
What is Friction?
The resistance to motion of one object moving in relation to another is known as friction. It is not regarded as a fundamental force like gravity or electromagnetic, according to the International Journal of Parallel, Emergent and Distributed Systems(opens in new tab).
According to the book Soil Mechanics(opens in new tab), scientists started putting together the laws governing friction in the 1400s.
However, because the interactions are so complex, characterizing the force of friction in various circumstances typically requires experiments and can't be derived from equations or laws alone. There are numerous exceptions to every frictional general rule.
Therefore, An Particle a has Weston practical.
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Pre-laboratory Assignment: Experiment 20 Reflection and Refraction of Light 1. When light is incident on a reflective surface, what can be said about the angle and speed at which the light is reflected? (Information is in your ‘General Physics Laboratory Manual’ Chapt. 20) 2. At what angle is the normal drawn to the reflective surface or air-medium interface? 3. How are angles of incidence, angles of reflection and of refraction measured? 4. Describe what happens to a light ray as it enters from a medium of greater refractive index to a medium of lesser refractive index
Answer:
1) ngle of incidence and reflection are equal, light carries does not change
2) the angle of this line with respect to the surface is 90º
3) protractor
4) n₂ sin θ₂ = n_1 sin θ₁, light ray must have a greater angle than the incident ray ,
Explanation:
1) When light falls on a reflective surface, the angle of incidence and reflection are equal and as it travels in the same medium, the speed that the light carries does not change
2) The normal is a line perpendicular to the point of incidence of light, so the angle of this line with respect to the surface is 90º
3) Angles are measured with a protractor
4) When light passes from one medium to another, the speed of the ray changes due to the difference in the refractive index in each medium, due to this change in speed the transmitted light ray must have a greater angle than the incident ray , since the speed increases as the density of the medium decreases
[tex]\frac{sin \theta _2}{ sin \theta_1} = \frac{v_2}{v_1}[/tex]
[tex]\frac{c}{v_2} \ sin \theta_2 = \frac{c}{v_1} \ sin \theta_1[/tex]
n₂ sin θ₂ = n_1 sin θ₁
name the basic principle on which generators work.
electromagnetic induction?
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 48.0 cm . The explorer finds that the pendulum completes 93.0 full swing cycles in a time of 141 s.
Required:
What is the value of g on this planet?
Answer:
8.24 m/s²
Explanation:
Applying,
T = 2π√(L/g).................... Equation 1
Where T = period of the pendulum, L = Length of the pendulum, g = acceleration due to gravity of the planet, π = pi
From the question, we were asked to find the value of g,
There we make g the subject of the equation
g = 4π²L/T²..................... Equation 2
Given: L= 48 cm = 0.48 m, T = (141/93) s = 1.516 s
Constant: π = 3.14
Substitute these values into equation 2
g = 4(3.14²)(0.48)/(1.516²)
g = 18.93/2.298
g = 8.24 m/s².
Hence the acceleration of the planet is 8.24 m/s²
A soccer player with a mass of 80.0 kg kicks a ball by applying a 20.0 N force. What force does the ball exert on the player?
A. 160. N
C. 2.50 N
B. 0.400 N
D. 20.0N
Answer:
F = 160.0 N
Explanation:
Given: Soccer payer with a mass = 80 kg, force = 20 N
To find: force
Formula: [tex]F=ma[/tex]
Solution: It is summarized by the equation: Force (N) = mass (kg) × acceleration (m/s²). Thus, an object of constant mass accelerates in proportion to the force applied.
F = m × a
F = 20 kg - 10 = 2
F = 80 × 2 = 160
F = 160.0 N
Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.
Aluminum has an alpha value of 25 x 10^-6 1/degrees C. What is the change in
length of a 4850 cm wire at 2 degrees C when the temp increase to 55 degrees
C?
Answer:
6.4 cm.
Explanation:
From the question given above, the following data were obtained:
Coefficient of linear expansion (α) = 25×10¯⁶ / °C
Original length (L) = 4850 cm
Initial temperature (T₁) = 2 °C
Final temperature (T₂) = 55 °C
Change in length (ΔL) =?
The change in the length of the wire can be obtained as follow:
α = ΔL / L(T₂ – T₁)
25×10¯⁶ = ΔL / 4850(55 – 2)
25×10¯⁶ = ΔL / 4850 × 53
25×10¯⁶ = ΔL / 257050
Cross multiply
ΔL = 25×10¯⁶ × 257050
ΔL = 6.4 cm
Therefore, the change in the length of the wire is 6.4 cm
A +26.3 uC charge qy is repelled by a force
of 0.615 N from a second charge 92 that is
0.750 m away. What is the value of 92?
Include the sign of the charge (+ or -).
(u stands for micro.)
[?] x 10-6 C
Answer:
+1.46×10¯⁶ C
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C
Force (F) = 0.615 N
Distance apart (r) = 0.750 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Charge 2 (q₂) =?
The value of the second charge can be obtained as follow:
F = Kq₁q₂ / r²
0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²
0.615 = 236700 × q₂ / 0.5625
Cross multiply
236700 × q₂ = 0.615 × 0.5625
Divide both side by 236700
q₂ = (0.615 × 0.5625) / 236700
q₂ = +1.46×10¯⁶ C
NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C
Answer:
+1.46
Explanation:
acellus