TIME REMAINING
45:13
A framed picture hangs from two cords attached to the ceiling.

A picture of a picture frame hanging by two cables at the center of the frame at the same length and angle from the vertical.

Which shows the correct free body diagram of the hanging picture?

A free body diagram with two force vectors, the first pointing downward labeled F Subscript g Baseline, the second pointing upward labeled F Subscript N Baseline.

A free body diagram with three force vectors, the first pointing south labeled F Subscript p Baseline, the second pointing northeast labeled F Subscript T Baseline, and the third pointing northwest labeled F Subscript N.

A free body diagram with three force vectors, the first pointing south labeled F Subscript g Baseline, the second pointing northeast labeled F Subscript T Baseline and the third pointing northwest labeled F Subscript T.

A free body diagram with two force vectors, the first pointing downward labeled F Subscript p Baseline, the second pointing upward labeled F Subscript T Baseline.

Answers

Answer 1

Answer:The answer is C

Explanation:


Related Questions

Which sequence shows the chain of energy transfers that create surface currents on the ocean?

Answers

Answer:

The correct answer is A. The sun is the energy source of the surface currents in the ocean

The energy transfer starts from solar energy , then wind energy and finally wind energy is the cause of surface current .

What is surface current ?

Surface currents are currents that are located in the upper feet of the ocean , they are simply how water moves from one place to another . Pattern of surface current are determined by wind direction .

Surface currents are formed by global wind system that are fueled by energy from the sun . Because of heating effect of sun , the earth's atmosphere gets warmed up . As we know , warm air is lighter then cool air , it rises up and create low pressure near the equator because of this wind causes surface currents the ocean .

hence , The energy transfer starts from solar energy , then wind energy and finally wind energy is the cause of surface current .

learn more about surface current

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The US currently produces about 27 GW of electrical power from solar installations. Natural gas, coal, and oil powered installations produce about 740 GW of electrical power. The average intensity of electromagnetic radiation from the sun on the surface of the earth is 1000 W/m2 . If solar panels are 30% efficient at converting this incident radiation into electrical power, what is the total surface area of solar panels responsible for the 27 GW of power currently produced

Answers

Answer:

The total surface area is "90 km²".

Explanation:

Given:

Power from solar installations,

= 27 GW

Other natural installations,

= 740 GW

Intensity,

[tex]\frac{F}{At}=\frac{P}{A}=1000 \ W/m^2[/tex]

%n,

= 30%

Now,

⇒ %n = [tex]\frac{out.}{Inp.}\times 100[/tex]

then,

⇒ [tex]Inp.=\frac{27}{30}\times 100[/tex]

           [tex]=90 \ GW[/tex]

As we know,

⇒ [tex]I=\frac{P}{A}[/tex]

by substituting the values, we get

[tex]1000=\frac{90\times 10^9}{A}[/tex]

    [tex]A = \frac{90\times 10^9}{10^3}[/tex]

        [tex]=90\times 10^6[/tex]

        [tex]=90 \ km^2[/tex]

why do we go to hospital​

Answers

Answer:

bcz we want to have fun there lolol

Answer:

for emergency, treatment, medicines,etc.....

Difference between scissors and nut cracker​

Answers

but cracked cracks nuts while scripts cut

A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b) How long will it take for this person's temperature to return to 37°c if his mass is 90kg.

Answers

Answer:

(a)  209 Watt

(b) 4482.8 seconds

Explanation:

(a) P = 50×4.18

Where P = rate of heat loss in watt

    P = 209 Watt

Applying,

Q = cm(t₁-t₂)................ Equation 1

Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.

From the question,

Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C

Constant: c = 3470 J/kg.K

Substtut these values into equation 1

Q = 90×3470(40-37)

Q = 936900 J

But,

P = Q/t.............. Equation 2

Where t = time

t = Q/P............ Equation 3

Given: P = 209 Watt, Q = 936900

Substitute into equation 3

t = 936900/209

t = 4482.8 seconds

A boy walks from point C to point D which is 50 m apart. Then, he walks back to point C. what is his displacement of his whole journey ?
A.25 m
B.75 m
C.50 m
D.0 m

Answers

Answer: D. 0 m

Explanation:

Concept:

Here, we need to know the concept of displacement.

Displacement is defined to be the change in position of an object.

The difference between displacement and distance is the total movement of an object without any regard to direction, while displacement is the pure change of position.

If you are still confused, please refer to the attachment below for a graphical explanation.

Solve:

STEP ONE: the boy walks from point C to point D (a distance of 50 m)

C ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ D

                                              50 m

STEP TWO: the boy walks from point D to point C (a distance of 50 m)

D ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ C

                                               50 m

STEP THREE: find the displacement

The boy started with point C

The boy ended with point C

He did not change his position throughout the journey.

Therefore, his displacement is 0 m.

Hope this helps!! :)

Please let me know if you have any questions

Identify each action as a wave erosion war wind erosion

Answers

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

15 . A scientist who studies the whole environment as a working unit .

Botanist
Chemist
Ecologist
Entomologist

Answers

Answer:

Ecologist.

Your answer is Ecologist.

(Ecologist) is a scientist who studies the whole environment as a working unit.

OBJECTI
1. The motion of a liquid inside a U-tube is an
example of what type of motion?
a. Simple Harmonic c. Random
b.Rectilinear
d. Circular

Answers

Answer:

option A

Explanation:

simple harmonic motion

Answer:

random motion I think not sure

Which item will be shipped third?

—-

Answers

Answer:

I know it's groceries

Explanation:

electronics ship before clothing

electronics ship after groceries

urgent items are first so

order:

1.) A/Electronics

2.) Clothing/B

3.) Groceries(since groceries aren't urgent)

thing is it's C or D I'm leaning to D since it says it ships last but i dont know so if I'm wrong sorry.

I NEEED HELP IN PHYSICS PLEASE!

Answers

Answer:

in which topic you need help

A boat is able to move at 7.6 m/s in still water. If the boat is placed on the south shore of a river (water current of 3.4 m/s [SE]), and the captain wants to head straight across to the north shore:

a) In what direction should the captain point the boat?
b) Calculate the time it will take to cross (the river is 212.0 m from the south to the north shore).

Answers

Answer:

I don't get it rewrite please

An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.​

Answers

Answer:

[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].

The unit of both sides of this equation are [tex]\rm s[/tex].

Explanation:

The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].

The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].

On the right-hand side of this equation:

[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].

[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].

Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].

A 2kg ball is rolled along the floor for 0.8 m at a constant speed of 6 m/s. What is the work done by gravity?

A, 0
B, 16 J
C, 72 J
D, 450 J
E, 90 J

Answers

=F×s×cosa=2×g×0,8×cos90°= 0

The work done by gravity on a ball of 2 kg which is moving with a constant speed of 6 meter per second is zero. Thus, the correct option is A.

What is Work?

Work is the energy transfer to or from an object through the application of force along with the displacement. For a constant force aligned with the direction of motion, the work done is equal to the product of the force strength which is applied and the distance traveled by the object.

Work = Force × Displacement

Force = Mass × Acceleration

Acceleration of the ball is zero as it is moving with a constant speed. Therefore, the work done by the gravity is zero.

Therefore, the correct option is A.

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Ibrah open a bottle of perfume infront of the room. After few minutes the smell of perfume reach the whole room. Explain why this happens​

Answers

the particles of the perfume began to spread into the air

calculate the value of 200°C in Kelvin

Answers

Answer:

473.15

Explanation:

1. A 2.7-kg copper block is given an initial speed of 4.0 m/s on a rough horizontal surface. Because of friction, the block finally comes to rest. (a) If the block absorbs 85% of its initial kinetic energy as internal energy, calculate its increase in temperature.

Answers

Answer:

ΔT = 0.017 °C

Explanation:

According to the given condition, the change in internal energy of the block must be equal to 85% of its kinetic energy:

Change in Internal Energy = (0.85)(Kinetic Energy)

[tex]mC\Delta T = (0.85)\frac{1}{2}mv^2\\\\C\Delta T = (0.425)v^2\\\\\Delta T = \frac{0.425v^2}{C}[/tex]

where,

ΔT = increase in temperature = ?

v = speed of block = 4 m/s

C = specific heat capacity of copper = 389 J/kg.°C

Therefore,

[tex]\Delta T = \frac{(0.425)(4\ m/s)^2}{389}\\\\[/tex]

ΔT = 0.017 °C

a beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a focal length of 40 cm

Answers

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

u

1

v

1

=

f

1

+

u

1

v

1

=

20

1

+

12

1

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

u

1

v

1

=

f

1

+

u

1

v

1

=

−16

1

+

12

1

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

What do you understand by moment of inertia and torque?
Word limit 50-60

Please don't copy from any sources. You can rewrite. Plagiarism will be check. Thank you.

Answers

Answer:

Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.

What star is known as the "cold planet"?

Answers

Explanation:

OGLE-2005-BLG-390Lb.

PSR B1620-26 b. Surface Temperature: 72 Kelvin. ...

Neptune. Surface Temperature: 72 Kelvin. ...

Uranus. Surface Temperature: 76 Kelvin. ...

Saturn. Surface Temperature: 134 Kelvin. ...

Jupiter. Image Courtesy: NASA. ...

OGLE-2016-BLG-1195Lb. Surface Temperature: Unknown

After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving

Answers

This question is incomplete, the complete question is;

A bat flies towards a moth at 7.1 m/s while the moth is flying towards the bat at 4.4 m/s. The bat emits a sound wave of 51.7 kHz.

After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving

Answer:

the second moth is moving at 5.062 m/s

Explanation:

Given the data in the question;

Using doppler's effect

[tex]f_{moth[/tex] = f₀( [tex]v_{s[/tex] ± [tex]v_{observer[/tex] / [tex]v_{s[/tex] ± [tex]v_{source[/tex] )

f₁ = f₀( ([tex]v_{s[/tex] + v₂) / ( [tex]v_{s[/tex] - v₁ ) )

frequency reflected from the moth,

Now, moth is the source and the bat is the receiver

f₂ = f₁( ([tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) )

hence, f = f₀[ ( ( [tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) ) ( ( [tex]v_{s[/tex] + u₂ ) / ( [tex]v_{s[/tex] - u₁ ) )

we know that, the velocity of sound [tex]v_{s[/tex] = 343 m/s.

given that v₁ and v₂ { velocity of bat } = 7.1 m/s, f₀ = 51.7 kHz and f = 55.5 kHz.

we substitute

55.5 = 51.7[ ( ( 343 + 7.1 ) / ( 343 - 7.1 ) ) ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 51.7[  ( 350.1 / 335.9 ) ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 51.7[ 1.04227 ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 = 53.885359 ( ( 343 + u ) / ( 343 - u ) ) ]

55.5 / 53.885359 = ( 343 + u ) / ( 343 - u )

1.02996 =  ( 343 + u₂ ) / ( 343 - u )

( 343 + u₂ ) = 1.02996( 343 - u )

343 + u = 353.27628 - 1.02996u

u + 1.02996u = 353.27628 - 343

2.02996u = 10.27628

u = 10.27628 / 2.02996

u = 5.062 m/s

Therefore, the second moth is moving at 5.062 m/s

A standard bathroom scale is placed on an elevator. A 28 kg boy enters the elevator on the first floor and steps on the scale. What will the scale read (in newtons) when the elevator begins to accelerate upward at 0.5 m/s2

Answers

Answer:

Explanation:

Newton's Second Law is pretty much the standard for all motion that involves a force. It applies to gravitational force and torque and friction and weight on an elevator. The main formula for force is

F = ma. We have to adjust that to take into account that when the elevator is moving up, that "surge" of acceleration weighs down a bit on the scale, causing it to read higher than the actual weight until the acceleration evens out and there is no acceleration at all (no acceleration simply means that the velocity is constant; acceleration by definition is a change in velocity, and if there is no change in velocity, there is 0 acceleration). The force equation then becomes

[tex]F_n-w=ma[/tex]  where [tex]F_n[/tex] is normal force. This is what the scale will read, which is what we are looking for in this problem (our unknown). Since we are looking for [tex]F_n[/tex], that is what we will solve this literal equation for:

[tex]F_n=ma+w[/tex] .  m is the mass of the boy, a is the acceleration of the elevator (which is going up so we will call that acceleration positive), and w is weight. We have everything but the unknown and the weight of the boy. We find the weight:

w = mg so

w = 28(9.8) and

w = 274.4 N BUT rounding to the correct number of significance we have that the weight is actually

w = 270 N.

Filling in the elevator equation:

[tex]F_n=28(.50)+270[/tex] and according to the rules of significant digits, we have to multiply the 28(.50) {notice that I did add a 0 there for greater significance; if not that added 0 we are only looking at 1 significant digit which is pretty much useless}, round that to 2 sig fig's, and then add to 170:

[tex]F_n=14+270[/tex] and adding, by the rules, requires that we round to the tens place to get, finally:

[tex]F_n=280N[/tex]  So you see that the surge in acceleration did in fact add a tiny bit to the weight read by the scale; conversely, if he were to have moved down at that same rate, the scale would have read a bit less than his actual weight). Isn't physics like the coolest thing ever!?

Give examples of motion in which the directions of the velocity and acceleration vectors are (a) opposite, (b) the same, and (c) mutually perpendicular​

Answers

Answer:

a) When moving body applies brake then velocity and acceleration would be in opposite direction

b) When body starts to increase velocity then velocity and acceleration would be in same direction

c) When body is circulating then velocity and acceleration would be perpendicular to each other

Explanation:

a) When body applies brake then its velocity starts decreasing, in this case its acceleration would try to stop the moving body. So direction of velocity would be same as direction of motion of body but direction of acceleration would be in opposite direction

b) When body starts to increase velocity, its acceleration would make the body to move faster. So direction of velocity would be the direction of motion of body and acceleration would also be in same direction

c) When body moves in circular path then its acceleration would be towards centre of circle and velocity would try to snap the body out of circle to straight line which in tangent to circle.

An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it the spring is extended 6.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10 cm. While the spring is being extended by the force, the work done by the spring is:
a. -3.6 J
b. -3.3 F
c. -3.4 times 10^-5 J
d. 3.3 J
e. 3.6 J

Answers

Answer:

b) - 3.3 J

Explanation:

Given;

mass, m = 2 kg

initial extension of the spring, x = 6 cm = 0.06 m

The weight of the mass on the spring;

W = mg

where;

g is acceleration due to gravity = 9.81 m/s²

W = 2 x 9.81

W = 19.62 N

The spring constant is calculated as;

W = kx

k = W/x

k = 19.62 / 0.06

k = 327 N/m

The work done by the spring when it is extended to an additional 10 cm;

work done = force x distance

distance = extension, x =  10 cm = 0.1 m

The work done by the spring opposes the applied force by acting in opposite direction to the force.

W = - Fx

W = - (kx) x

W = - kx²

W = - (327) x (0.1)²

W = - 3.27 J

W ≅ - 3.3 J

Therefore, the work done by the spring by opposing the applied force is -3.3 J

A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from the center of the wire. (in Nm2/C, keep 3 significant figures)

Answers

Answer:

[tex]E=35921.96N/C[/tex]

Explanation:

From the question we are told that:

Radius [tex]r=0.321mm[/tex]

Charge Density [tex]\mu=0.100[/tex]

Distance [tex]d= 5.00 cm[/tex]

Generally the equation for electric field is mathematically given by

[tex]E=\frac{mu}{2\pi E_0r}[/tex]

[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]

[tex]E=35921.96N/C[/tex]

A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top of it as shown. The block w2 is attached to avertical wall by a string 6m long. If the coefficient of friction between all surface is 0.25 and the system is in equilibrium find the magnitude of the horizontal force applied to the lower block

Answers

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let T represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, [tex]F_{f2}[/tex] = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = [tex]F_{f2}[/tex] × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

[tex]T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27[/tex]

[tex]Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92[/tex]

The frictional force on the block W₂, [tex]F_{f2}[/tex] ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ [tex]F_{w2}[/tex] = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + [tex]\mathbf{F_{w2}}[/tex]

The frictional force from the ground, [tex]\mathbf{F_{f1}}[/tex] = N×μ + [tex]\mathbf{F_{f2}}[/tex] = P

Where;

P = The horizontal force applied to the block

P = (W₁ + [tex]\mathbf{F_{w2}}[/tex]) × μ + [tex]\mathbf{F_{f2}}[/tex]

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

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Suppose your actual height is 5 feet and 5.2 inches. A tape measure which can be read tothe nearest 1/8 of an inch gives your height as 65 3/8 inches. The laser device at the clinic that givesreadings to the nearest hundredth of an inch says you are 65.31 inches.

Required:
a. Which measuring device is more accurate?
b. Which measuring device is more precise?

Answers

Answer:

a) The laser device

b) The tape

Explanation:

First, there is a need to understand what accuracy and precision mean.

Accuracy is the closeness of a measurement to its true (pre-determined) value.

Precision is the closeness of repeated measurements to each other.

Since 1 feet = 12 inches, then, 5 feet and 5.2 inches would be equivalent to 65.2 inches. This value represents the true value of my height.

The tape measured the height as 65 3/8, which is equivalent to 65.375 inches.

The laser device measured the height as 65.31.

Error = true value - measured value

Absolute error from the tape = 65.2 - 65.375

                                       = -0.175 inches

Absolute error from laser device = 65.2 - 65.31

                                 = -0.11

a) The magnitude of error from the tape is more than that of the laser device. Hence, the laser device is said to be more accurate.

b) Even though there were just single readings from both instruments, the tape can be read to the nearest 1/8 of an inch and as such, can give more precisive measurements than the laser device.

If a car drives 10 mph South, this is an example of a:
A. Displacement
B. Velocity
C. Speed
D. Distance

Answers

Answer:

杰杰伊杜杜杜伊格富尔杰迪耶赫分离福音

Explanation:

莱德利 · 赫耶尔伊 3uritievrirjrirhruebwkwieheoo2hfjcbvi3hd

Answer:

B velocity

Explanation:

A man is pulling a 20 kg box with a rope that makes an angle of 60 with the horizontal.If he applies a force of 150 N and a frictional force of 15 N is present, calculate the acceleration of the box.​

Answers

F (horizontal) = (150 N) cos(60°) - 15 N = (20 kg) a

==>   a = ((150 N) cos(60°) - 15 N)/(20 kg) = 3 m/s²

To calculate the acceleration of the box, we need to consider the net force acting on it. So, the acceleration of the box is 3 m/s².

The net force is the vector sum of the applied force and the force of friction. First, let's find the horizontal and vertical components of the applied force:

Horizontal component of the applied force (F[tex]_{horizontal}[/tex]) = F[tex]_{applied}[/tex] × cos(θ)

F[tex]_{horizontal}[/tex] = 150 N × cos(60°)

F[tex]_{horizontal}[/tex] = 150 N × 0.5

F[tex]_{horizontal}[/tex] = 75 N

Vertical component of the applied force (F[tex]_{vertical}[/tex]) = F[tex]_{applied}[/tex] × sin(θ)

F[tex]_{vertical}[/tex] = 150 N × sin(60°)

F[tex]_{vertical}[/tex] = 150 N × (√3 / 2)

F[tex]_{vertical}[/tex] ≈ 129.9 N

Now, let's calculate the net force in the horizontal direction:

Net Force in the horizontal direction (F[tex]_{net horizontal}[/tex]) = F[tex]_{horizontal}[/tex] - F[tex]_{friction}[/tex]

F[tex]_{net horizontal}[/tex] = 75 N - 15 N

F[tex]_{net horizontal}[/tex] = 60 N

Now, we can calculate the acceleration (a) using Newton's second law of motion, F = ma:

F[tex]_{net horizontal}[/tex] = m × a

60 N = 20 kg × a

Now, solve for acceleration (a):

a = 60 N / 20 kg

a = 3 m/s²

So, the acceleration of the box is 3 m/s².

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How can I solve this?
You have three capacitors of values 40 F, 10 F and 50 F. What would their equivalent capacitance (in F) be if they were connected in parallel with each other? Enter your answer as a number only, to one decimal place.

Answers

Explanation:

The equivalent capacitance of capacitors in parallel can be determined as

[tex]C_{eq} = C_1 + C_2 + C_3[/tex]

[tex]\:\:\:\:\:= 40\:\text{F} + 10\:\text{F} + 50\:\text{F} = 100\:\text{F}[/tex]

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