Answer:
Speed=28.1m/s(to 3s.f.) , Time=2.19s(to 3s.f.)
Explanation:
Time=Distance/Speed
=14.5/6.63
=2.19s(to 3s.f.)
Acceleration=Final Velocity(v)-Initial Velocity(u)/Time
9.81=v-6.63/2.19
v-6.63=21.5
v=28.1m/s
why did Rita's hands get hot when she rubbed them ?
Answer:
due to production of heat through friction
Explanation:
because of the friction produce between her hands
how can you convert galvanometer into ammeter?
Answer:
A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.
Explanation:
This low resistance is called shunt resistance S. The scale is now calibrated in ampere and the range of the ammeter depends on the values of the shunt resistance.
Two masses of 3 kg and 5 kg are connected by a light string that passes over a smooth polley as shown in the Figure.
QL
Determine:
i. the tension in the string,
ii. the acceleration of each mass, and
iii. the distance each mass moves in the first second of motion if they start from rest
i. [tex]T = 36.8\:\text{N}[/tex]
ii. [tex]a = 2.45\:\text{m/s}^2[/tex]
iii. [tex]x = 1.23\:\text{m}[/tex]
Explanation:
Let's write Newton's 2nd law for each object. We will use the sign convention assigned for each as indicated in the figure. Let T be the tension on the string and assume that the string is inextensible so that the two tensions on the strings are equal. Also, let a be the acceleration of the two masses. And [tex]m_1 = 3\:\text{kg}[/tex] and [tex]m_2 = 5\:\text{kg}[/tex]
Forces acting on m1:
[tex]T - m_1g = m_1a\:\:\:\:\:\:\:(1)[/tex]
Forces acting on m2:
[tex]m_2g - T = m_2a\:\:\:\:\:\:\:(2)[/tex]
Combining Eqn(1) and Eqn(2) together, the tensions will cancel out, giving us
[tex]m_2g - m_1g = m_2a + m_1a[/tex]
or
[tex](m_2 - m_1)g = (m2 + m_1)a[/tex]
Solving for a,
[tex]a = \left(\dfrac{m_2 - m_1}{m_2 + m_1}\right)g[/tex]
[tex]\:\:\:\:= \left(\dfrac{5\:\text{kg} - 3\:\text{kg}}{5\:\text{kg} + 3\:\text{kg}}\right)(9.8\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:= 2.45\:\text{m/s}^2[/tex]
We can solve for the tension by using this value of acceleration on either Eqn(1) or Eqn(2). Let's use Eqn(1).
[tex]T - (3\:\text{kg})(9.8\:\text{m/s}^2) = (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]
[tex]T = (3\:\text{kg})(9.8\:\text{m/s}^2) + (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:= 29.4\:\text{m/s}^2 + 7.35\:\text{m/s}^2 = 36.8\:\text{N}[/tex]
Assuming that the two objects start from rest, the distance that they travel after one second is given by
[tex]x = \frac{1}{2}at^2 = \frac{1}{2}(2.45\:\text{m/s}^2)(1\:\text{s})^2 = 1.23\:\text{m}[/tex]
The exponent of the exponential function contains RC for the given circuit, which is called the time constant. Use the units of R and C to find units of RC. Write ohms in terms of volts and amps and write farads in terms of volts and coulombs. Simplify until you get something simple. Show your work below.
Answer:
The unit of the time constant RC is the second
Explanation:
The unit of resistance, R is the Ohm, Ω and resistance, R = V/I where V = voltage and I = current. The unit of voltage is the volt, V while the unit of current is the ampere. A.
Since R = V/I
Unit of R = unit of V/unit of I
Unit of R = V/A
Ω = V/A
Also, The unit of capacitance, C is the Farad, F and capacitance, F = Q/V where Q = charge and V = voltage. the unit of charge is the coulomb, C while the unit of voltage is the volt, V
Since C = Q/V
Unit of C = unit of Q/unit of V
Unit of C = C/V
F = C/V
Now the time constant equals RC.
So, the unit of the time constant = unit of R × unit of C = Ω × F = V/A × C/V = C/A
Also. we know that the 1 Ampere = 1 Coulomb per second
1 A = 1 C/s
So, substituting 1 A in the denominator, we have
unit of RC = C/A = C ÷ C/s = s
So, the unit of RC = s = second
So, the unit of the time constant RC is the second
if 6000j of energy is supplid to a machine to lift a load of 300N through a vvertical height of 1M calculatework out put
Answer:
300J
Explanation:
Work done = Force x the distance travelled in the direction of the force
=300 x 1
=300J
1. A 63 kg driver gets into an empty taptap to start the day's work. The springs compress 1.5x10-2 m. What is the effective spring constant of the spring system in the taptap?
2. After driving a portion of the route, the taptap is fully loaded with a total of 24 people including the driver, with an average mass of 68 kg per person. In addition, there are three 15-kg goats, five 3-kg chickens, and a total of 25 kg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed?
(1) When the driver is at rest, the restoring force exerted by spring is equal in magnitude to the driver's weight, so that
∑ F = s - mg = 0 ==> s = mg = 617.4 N
If the spring is compressed 0.015 m, then the spring constant k is such that
617.4 N = k (0.015 m) ==> k = 41,160 N/m ≈ 41 kN/m
(2) The total mass of the passengers is
24 (68 kg) + 3 (15 kg) + 5 (3 kg) + 25 kg = 1717 kg
so that if everyone is at rest, the spring is compressed a distance x such that
kx = (1717 kg) g ==> x ≈ 0.41 m
Trình bày những hiểu biết của em về đại lượng vận tốc dài, vận tốc góc(định nghĩa, công thức, ý nghĩa, đơn vị, loại đại lượng).
A 100 kg man is one fourth of the way up a 4.0 m ladder that is resting against a smooth, frictionless wall. The ladder has mass 25 kg and makes an angle of 56 degrees with the ground. What is the magnitude of the force of the wall on the ladder at the point of contact, if this force acts perpendicular to the wall and points away from the wall
Answer:
[tex]N_f=248N[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=100kg[/tex]
Ladder Length [tex]l=4.0m[/tex]
Mass of Ladder [tex]m_l=25kg[/tex]
Angle [tex]\theta=56 \textdegree[/tex]
Generally the equation for Co planar forces is mathematically given by
[tex]mgcos \theta *2+Mgcos\theta*1 -N_fsin \theta*4=0[/tex]
Therefore
[tex]25*9.81cos 56 *2+100*9.81cos56*1 -N_fsin 56*4=0[/tex]
[tex]N_f=248N[/tex]
A planet of mass m = 4.25 x 1024 kg orbits a star of mass M = 6.75 x 1029 kg in a circular path. The radius of the orbits R = 8.85 x 107 km. What is the orbital period Tplanet of the planet in Earth days?
285.3 days
Explanation:
The centripetal force [tex]F_c[/tex] experienced by the planet is the same as the gravitational force [tex]F_G[/tex] so we can write
[tex]F_c = F_G[/tex]
or
[tex]m\dfrac{v^2}{R} = G\dfrac{mM}{R^2}[/tex]
where M is the mass of the star and R is the orbital radius around the star. We know that
[tex]v = \dfrac{C}{T} = \dfrac{2\pi R}{T}[/tex]
where C is the orbital circumference and T is orbital period. We can then write
[tex]\dfrac{4\pi^2R}{T^2} = G\dfrac{M}{R^2}[/tex]
Isolating [tex]T^2[/tex], we get
[tex]T^2 = \dfrac{4\pi^2R^3}{GM}[/tex]
Taking the square root of the expression above, we get
[tex]T = 2\pi \sqrt{\dfrac{R^3}{GM}}[/tex]
which turns out to be [tex]T = 2.47×10^7\:\text{s}[/tex]. We can convert this into earth days as
[tex]T = 2.47×10^7\:\text{s}×\dfrac{1\:\text{hr}}{3600\:\text{s}}×\dfrac{1\:\text{day}}{24\:\text{hr}}[/tex]
[tex]\:\:\:\:\:= 285.3\:\text{days}[/tex]
It was recorded that the temperature of a body was 320 degree F determine the value of the temperature in kelvin
Answer:
433.15K
Explanation:
(320°F − 32) × 5/9 + 273.15 = 433.15K
Identify each action as a wave erosion war wind erosion
Answer:Lesson Objectives
Describe how the action of waves produces different shoreline features.
Discuss how areas of quiet water produce deposits of sand and sediment.
Discuss some of the structures humans build to help defend against wave erosion.
Vocabulary
arch
barrier island
beach
breakwater
groin
refraction
sea stack
sea wall
spit
wave-cut cliff
wave-cut platform
Introduction
Waves are important for building up and breaking down shorelines. Waves transport sand onto and off of beaches. They transport sand along beaches. Waves carve structures at the shore.
Wave Action and Erosion
All waves are energy traveling through some type of material, such as water (Figure below). Ocean waves form from wind blowing over the water.
Ocean waves are energy traveling through water.
The largest waves form when the wind is very strong, blows steadily for a long time, and blows over a long distance.
The wind could be strong, but if it gusts for just a short time, large waves won’t form. Wave energy does the work of erosion at the shore. Waves approach the shore at some angle so the inshore part of the wave reaches shallow water sooner than the part that is further out. The shallow part of the wave ‘feels’ the bottom first. This slows down the inshore part of the wave and makes the wave ‘bend.’ This bending is called refraction.
Wave refraction either concentrates wave energy or disperses it. In quiet water areas, such as bays, wave energy is dispersed, so sand is deposited. Areas that stick out into the water are eroded by the strong wave energy that concentrates its power on the wave-cut cliff (Figure below).
The wave erodes the bottom of the cliff, eventually causing the cliff to collapse.
Other features of wave erosion are pictured and named in Figure below. A wave-cut platform is the level area formed by wave erosion as the waves undercut a cliff. An arch is produced when waves erode through a cliff. When a sea arch collapses, the isolated towers of rocks that remain are known as sea stacks.
(a) The high ground is a large wave-cut platform formed from years of wave erosion. (b) A cliff eroded from two sides produces an arch. (c) The top of an arch erodes away, leaving behind a tall sea stack.
Wave Deposition
Rivers carry sediments from the land to the sea. If wave action is high, a delta will not form. Waves will spread the sediments along the coastline to create a beach (Figure below). Waves also erode sediments from cliffs and shorelines and transport them onto beaches.
Sand deposits in quiet areas along a shoreline to form a beach.
Beaches can be made of mineral grains, like quartz, rock fragments, and also pieces of shell or coral (Figure below).
Quartz, rock fragments, and shell make up the sand along a beach.
Waves continually move sand along the shore. Waves also move sand from the beaches on shore to bars of sand offshore as the seasons change. In the summer, waves have lower energy so they bring sand up onto the beach. In the winter, higher energy waves bring the sand back offshore.
Some of the features formed by wave-deposited sand are in Figure below. These features include barrier islands and spits. A spit is sand connected to land and extending into the water. A spit may hook to form a tombolo.
Examples of features formed by wave-deposited sand.
Shores that are relatively flat and gently sloping may be lined with long narrow barrier islands (Figure below). Most barrier islands are a few kilometers wide and tens of kilometers long.
(a) Barrier islands off of Alabama. A lagoon lies on the inland side. (b) Barrier islands, such as Padre Island off the coast of Texas, are made entirely of sand. (c) Barrier islands are some of the most urbanized areas of our coastlines, such as Miami Beach.
In its natural state, a barrier island acts as the first line of defense against storms such as hurricanes. When barrier islands are urbanized (Figure above), hurricanes damage houses and businesses rather than vegetated sandy areas in which sand can move. A large hurricane brings massive problems to the urbanized area.
Protecting Shorelines
Intact shore areas protect inland areas from storms that come off the ocean (Figure below).
Dunes and mangroves along Baja California protect the villages that are found inland.
Explanation:
Answer: Below
Explanation: Correct on Edmentum
Three wires are connected at a branch point. One wire carries a positive current of 18 A into the branch point, and a second wire carries a positive current of 7 A away from the branch point. Find the current carried by the third wire into the branch point.
Answer:
The current in third branch is 11 A.
Explanation:
incoming current in one branch = 18 A
outgoing current in the other branch = 7 A
let the current in the third branch is i.
According to the Kirchoff's fist law in electricity
incoming current = out going current
18 = 7 + i
i = 11 A
The current in third branch is 11 A.
A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.
Answer:
Explanation:
An impulse results in a change of momentum.
The impulse is the product of a force and a distance. This will be represented by the area under the curve
a) W = ½(4.00)(3.00) = 6.00 J
b) W = (11.0 - 4.00)(3.00) = 21.0 J
c) W = ½(17.0 - 11.0)(3.00) = 9.00 J
d) ASSUMING the speed at x = 0 is in the direction of applied force
½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00
v₄ = 2.05 m/s
½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00
v₁₇ = 4.92 m/s
If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.
If a body travels 6km in 30 minutes in a fixed direction, calculate it's velocity.
Plz show me the process too.
We know
[tex]\boxed{\large{\sf Velocity=\dfrac{Distance}{Time}}}[/tex]
[tex]\\ \Large\sf\longmapsto Velocity=\dfrac{6}{\dfrac{1}{2}}[/tex]
[tex]\\ \Large\sf\longmapsto Velocity=6\times 2[/tex]
[tex]\\ \Large\sf\longmapsto Velocity=12km/h[/tex]
Is this the right answer??
We should keep km and min in smallest SI unit
Michelson and Morley concluded from the results of their experiment that Group of answer choices the experiment was successful in not detecting a shift in the interference pattern. the experiment was a failure since they detected a shift in the interference pattern. the experiment was a failure since there was no detectable shift in the interference pattern. the experiment was successful in detecting a shift in the interference pattern.
Answer:
The results of the experiment indicated a shift consistent with zero, and certainly less than a twentieth of the shift expected if the Earth's velocity in orbit around the sun was the same as its velocity through the ether.
Explanation:
Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.0 cm and a current of 12 A. The bigger loop has a current of 20 A. The magnetic field at the center of the loops is found to be zero.
Required:
What is the radius of the bigger loop?
Answer:
the radius of the bigger loop is 5 cm.
Explanation:
Given;
current in the smaller loop, I₁ = 12 A
current in the larger loop, I₂ = 20 A
radius of the smaller loop, r₁ = 3 cm
let the radius of the larger loop, = r₂
Apply Biot-Savart's law to determine the magnetic field at the center of the circular loops.
[tex]B= \frac{\mu_0 I}{2r}[/tex]
The magnetic field at the center of the smaller loop;
[tex]B_1 = \frac{\mu_0 I_1}{2 r_1}[/tex]
The magnetic field at the center of the bigger loop;
[tex]B_2 = \frac{\mu_0 I_2}{2 r_2}[/tex]
If the magnetic field at the center is zero, then B₁ = B₂
[tex]B_1 = B_2 = \frac{\mu_0 I_1}{2 r_1} = \frac{\mu_0 I_2}{2 r_2} \\\\\frac{I_1}{ r_1} = \frac{ I_2}{r_2} \\\\r_2 = \frac{I_2 r_1}{ I_1} = \frac{(20 \ A) \times (3.0 \ cm)}{12 \ A} = 5 \ cm[/tex]
Therefore, the radius of the bigger loop is 5 cm.
Accommodation of the eye refers to its ability to __________. see on both the brightest days and in the dimmest light see both in air and while under water move in the eye socket to look in different directions focus on both nearby and distant objects
Answer:
to adjust from distant to the near objects
Explanation:
The process of accommodation is achieved by changing in the shape and position of the eye ball. Just like adjusting the lens of the camera.Answer:
The ability of eye lens to change the focal length of eye lens is called accommodation power of eye.
Explanation:
The human eye is the optical instrument which works on the refraction of light.
The ability of eye lens to change its focal length is called accommodation power of eye.
The focal length of eye lens is changed by the action of ciliary muscles.
When the ciliary muscles are relaxed then the thickness of lens is more and thus the focal length is small. When the ciliary muscles is stretched, the lens is thin and then the focal length is large.
For a spring-mass oscillator if you double the mass but keep the stiffness the same, by what numerical factor does the pena original period was and the new period is DT, what is b7 It is useful to write out the expression for the period and ask yours you doubled the mass.
b = _____
If, instead, you double the spring stiffness but keep the mass the same, what is the factor b?
b = _____
If, instead, you double the mass and also double the spring stiffness, what is the factor b?
b = _____
If, instead, you double the amplitude (keeping the original mass and spring stiffness), what is the factor b?
b = _____
Answer:
ygguguguhhihhihijijijjojojinjbgy
Imagine you’re driving along a road and you approach a bridge. You notice a sign that reads, “Bridge freezes before road.” Why do bridges become covered with ice before roads do? Research this question and respond in depth, writing a full paragraph. Be sure to include examples. At the end of your response, provide at least two authoritative sources that you used in your research.
Answer:
During wet and freezing temperatures, ice is able to form at a faster pace on bridges because freezing winds blow from above and below and both sides of the bridge, causing heat to quickly escape. The road freezes slower because it is merely losing heat through its surface.Sources:
-- https://intblog.onspot.com/en-us/why-do-bridges-become-icy-before-roads
and
-- https://www.accuweather.com/en/accuweather-ready/why-bridges-freeze-before-roads/687262
I hope this helps you! ^^
Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.
Use the pressure meter to read the pressure in Fluid A at the bottom of the tank. Do not move the pressure meter. Switch to Fluid B and read the pressure in fluid B. Based on the two readings, compare the density of fluid B to the density of fluid A. Which statement is correct?
Answer:
[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]
Explanation:
The pressure at a depth of a fluid is
P = ρ g y
where ρ is the density of the fluid, y the depth of the gauge measured from the surface of the fluid.
In this case the pressure for fluid A is
Pa = ρₐ g y
the pressure for fluid B is
P_b = ρ_b g y
depth y not changes as the gauge is stationary
if we look for the relationship between these pressures
[tex]\frac{P_a}{P_b} = \frac{ \rho_a}{\rho_b}[/tex]
[tex]P_b = \frac{\rho_b}{\rho_a} \ P_a[/tex]
therefore we see that the pressure measured for fluid B is different from the pressure of fluid A
if ρₐ < ρ_b B the pressure P_b is greater than the initial reading
ρₐ> ρ_b the pressure in B decreases with respect to the reading in liquid A
A roller coaster has a mass of 1200.0kg. The coaster is going 22.0 m/s at the bottom
of the third loop-the-loop that is 2.5m above the ground. Determine the height of
the first hill that is required, assuming the cart is stationary at the top of the first hill
before it falls.
Answer:
h = 27.17 m
Explanation:
First, we will calculate the total mechanical energy of the system at the bottom point of the third loop:
Mechanical Energy = Kinetic Energy + Potential Energy
[tex]E = \frac{1}{2}mv^2 + mgh[/tex]
where,
E = Total Mechanical Energy = ?
m = mass of the roller coaster = 1200 kg
v = velocity of the roller coaster = 22 m/s
g = acceleration due to gravity = 9.81 m/s²
h = height of roller coaster = 2.5 m
Therefore,
[tex]E = \frac{1}{2}(1200\ kg)(22\ m/s)^2+(1200\ kg)(9.81\ m/s^2)(2.5\ m)\\\\E = 290400 J +29430\ J\\\\E = 319830\ J = 319.83\ KJ[/tex]
Now, the total mechanical energy at the top position of the first hill must also be the same:
[tex]E = \frac{1}{2}mv^2 + mgh[/tex]
where,
v = 0 m/s
h = ?
Therefore,
[tex]319830\ J = \frac{1}{2}(1200\ kg)(0\ m/s)^2+(1200\ kg)(9.81\ m/s^2)(h)\\\\h = \frac{319830\ J}{11772\ N}\\\\[/tex]
h = 27.17 m
Suppose you exert a force of 314 N tangential to a grindstone (a solid disk) with a radius of 0.281 m and a mass of 84.2 kg What is the resulting angular acceleration of the grindstone assuming negligible opposing friction
Answer:
The angular acceleration is 26.6 rad/s^2.
Explanation:
Force, F = 314 N
radius, r = 0.281 m
mass, m = 84.2 kg
The grindstone is a disc.
The torque is given by
torque = force x radius
Torque = 314 x 0.281 = 88.234 Nm
The torque is given by
Torque = Moment of inertia x angular acceleration
[tex]88.234 = 0.5 mr^2 \alpha \\\\88.234 = 0.5\times 84.2\times 0.281\times 0.281\times \alpha \\\\\alpha = 26.6 rad/s^2[/tex]
What are the examples of pulley? Plz tell the answer as fast as possible plz.
Answer:
elevators
Theatre system
construction pulley
lifts
Answer:
elevator,cargo lift system
Is it true that as we gain mass the force of gravity on us decreases
Answer:
No. As we gain mass the force of gravity on us does not decrease
why is it necessary to have end correction in the organ pipe?
Answer:
The vibrating length of the air column is greater than the actual length of the organ pipe
calculate the length of wire.
Answer:
L = 169.5 m
Explanation:
Using Ohm's Law:
V = IR
where,
V = Voltage = 1.5 V
I = Current = 10 mA = 0.01 A
R = Resistance = ?
Therefore,
1.5 V = (0.01 A)R
R = 150 Ω
But the resistance of a wire is given by the following formula:
[tex]R = \frac{\rho L}{A}[/tex]
where,
ρ = resistivity = 1 x 10⁻⁶ Ω.m
L = length of wire = ?
A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²
A = 1.13 x 10⁻⁶ m²
Therefore,
[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]
L = 169.5 m
A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand.
(a) At what angle was the ball thrown if its initial speed was 12.0 m/ s, assuming that the smaller of the two possible angles was used?
(b) What other angle gives the same range, and why would it not be used?
(c) How long did this pass take?
Answer:
a) θ = 14.23º, b) θ₂ = 75.77, c) t = 0.6019 s
Explanation:
This is a missile throwing exercise.
a) the reach of the ball is the distance traveled for the same departure height
R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]
sin 2θ = [tex]\frac{Rg}{v_o^2}[/tex]
sin 2θ = 7.00 9.8 / 12.0²
2θ = sin⁻¹ (0.476389) = 28.45º
θ = 14.23º
the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º
θ ’= 90 -14.23
θ’= 75.77º
b) the two angles that give the same range are
θ₁ = 14.23
θ₂ = 75.77
the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.
C) the time of the pass can be calculated with the expression
x = v₀ₓ t
t = x / v₀ₓ
t = 7 / 11.63
t = 0.6019 s
Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B
Explanation:
Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]
The sum of the two vectors is
[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]
[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]
The difference between the two vectors can be written as
[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]
[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]
Pascal's principle says: a A change in pressure at one point in an incompressible fluid is felt at every other point in the fluid. b The buoyant force equals the weight of the displaced fluid. c Matter must be conserved in a flowing, ideal fluid. d Energy is conserved in a flowing, ideal fluid. e A small input force causes a large output force.
Answer:
A change in pressure at one point in an incompressible fluid is felt at every other point in the fluid.
Explanation:
Pascal's principle states that ''pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.''(Science direct).
The implication of this law is; that a change in pressure at one point in an incompressible fluid is felt at every other point in the fluid. Hence the correct answer chosen above.
The Pascal's principle is applied in hydraulic jacks and automobile brakes.
