When a mass of m = 29 kg is hung from the middle of a fixed straight aluminum wire, the wire sags to make an angle of 12∘ with the horizontal as shown in (Figure 1) . The elastic modulus for aluminum is 7.0×1010N/m2. Determine the radius of the wire

When A Mass Of M = 29 Kg Is Hung From The Middle Of A Fixed Straight Aluminum Wire, The Wire Sags To

Answers

Answer 1

The solution to the given question is that the radius of the wire will be 0.378 millimetre approximately

At equilibrium,

Lets assume Tension T in the wire

mass of the block is given as 29 Kg

T sin 12° + T sin 12° = mg  

T sin 12° + T sin 12° = 29 × 9.8

2T sin 12° = 29 × 9.8

T = ( 29 × 9.8 )/2 sin 12°

T = 141.1 / sin 12°

T = 141.1 / 0.2

T = 705.5 Newton

Stress = K × Strain

where K is the elastic modulus of the metal

elastic modulus is given as 7.0 × 10¹⁰ N/m²

T/A = 7.0 × 10¹⁰ × (Δl/l)

A = T × (l/Δl) × (1/ (7.0 × 10¹⁰) )

A = π r² = T / ( (7.0 × 10¹⁰) × ( (1/cos 12°) - 1) )

π r² =  705.5 / ( (7.0 × 10¹⁰) × 0.02234 )

r = √( 705.5 / ( π × (7.0 × 10¹⁰) × 0.02234  ) ) metre

r = √( 705.5 / ( 4910.33 × 10⁶   ) ) metre

r = √( 0.1436 × 10⁻⁶ ) metre

r = ( 0.378 × 10⁻³ ) metre

r = 0.378 millimetre

Thus we have find the radius of the wire by applying stress strain relation which came out to be 0.378 millimetre.

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Related Questions

The figure below shows a man in a boat on a lake. The man's mass is 74 kg, and the boat's is 135 kg. The man and boat are initially at rest when the man throws a package of mass m = 15 kg horizontally to the right with a speed of vi = 5.0 m/s.

What is the velocity of the boat after the package is thrown? Neglect any resistance force from the water. (Give the magnitude in m/s, and select the correct direction from the options given.)

Answers

The velocity of the boat after the package is thrown is 0.36 m/s.

Final velocity of the boat

Apply the principle of conservation of linear momentum;

Pi = Pf

where;

Pi is initial momentumPf is final momentum

v(74 + 135) = 15 x 5

v(209) = 75

v = 75/209

v = 0.36 m/s

Thus, the velocity of the boat after the package is thrown is 0.36 m/s.

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All the forces we see and experience are the result of just a few so-called fundamental forces. What are the fundamental forces in nature? Provide a brief explanation of each and an example of how we ‘see’ the forces in action.

Answers

The fundamental forces in nature are strong force, weak force, electromagnetic force, and the gravitational force.

Fundamental forces

There are four fundamental forces at work in the universe. These fundamental forces include;

The strong force: this force has strong influence or impact on objects. The weak force: this force has little impact on objects.The electromagnetic force: force due to current flowing in a conductor placed in magnetic field.The gravitational force: this is force of attraction between two objects placed in the universe.

Thus, the fundamental forces in nature are strong force, weak force, electromagnetic force, and the gravitational force.

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After brushing, Fluffy's fur has a charge of +8.0 × 10-9 coulombs and her plastic brush has a charge of –1.4 × 10-8 coulombs. If the distance between the fur and brush is roughly 5.0 × 10-1 meters, what is the approximate magnitude of the force between them?

(k = 9.0 × 109 newton·meters2/coulomb2)

Answers

The approximate magnitude of the force between them is [tex]-4.032*10^-^6 N[/tex]

What is columbo's law?

According to Coulomb, the electric force for charges at rest has the following properties:

Like charges repel each other; unlike charges attract.

Thus, two negative charges repel one another, while a positive charge attracts a negative charge.

The equation in Coulomb's law is :

[tex]F =\frac{{kq_1q_2}}{r^2}[/tex]

Where F is the force between charges, k is a constant, q₁ is the charge of the first particle, q₂ is the charge of the second particle and r is the distance between the two particle charges.

According to the question,

The given values are:-

q₁ =  +8.0 × 10⁻⁹

q₂ = –1.4 × 10⁻⁸

Substituting to the equation:

[tex]F =\frac{{kq_1q_2}}{r^2}[/tex]

=[tex]\frac{ { 9*10^{9} * 8* 10^{-9} *-1.4*10^{-8} }}{(5x10^-1)^2}[/tex]

= -4.032 x 10⁻⁶ N

Therefore,

-4.032 x 10⁻⁶ N is the approximate magnitude of the force between them.

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Three resistors with resistances of R1=12 ohms (Ω), R2=8 ohms, and R3=24 ohms are in a series circuit with a 12-volt battery. (A) What is the total resistance of the resistors? (B) How much current can move through the circuit? (C) What is the current through each resistor?

Answers

The total resistance of the resistors is R=44Ω. And the current will be I= 0.27 A.

What is the function of a resistor?

Resistors are electronic components whose main function is to limit the flow of electrical charges by converting electrical energy into thermal energy.

A)The resistances connected in series in an arm of a circuit, total resistance of that arm of is given by algebric sum individual resistances.

R = R1 + R2 + R3 = 12 + 8 + 24 = 44Ω

B) Ohm's law gives that:

V = IR

Where,

Voltage applied to circuit (V)current flowing through the circuit (I) total resistance of the circuit (R)

(I) = V/R = 12 / 44 = 3/11 = 0.27 A

C) In series combination, all the resistances are connected one after other in a linear fashion. So, current flowing through first resistor passes to second, and then third and so on. Hence, the current flowing through all the resistance in series combination is same as the current flowing through that arm of circuit.

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Which notation is commonly used on force diagrams to show forces acting on an object.

Answer: is F^D

Answers

The notation that is commonly used on force diagrams to show forces acting on an object is FD.

What is free body diagram ?

A free body diagram is a diagram that contains all the forces acting on an object.

These forces acting on an object include the following :

Upward force (normal reaction due to weight of the object)Downward force (downward force due to weight of the object)Frictional force (force resisting the motion)

Thus, the notation that is commonly used on force diagrams to show forces acting on an object is FD.

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Calculate the focal length of a spherical mirror whose radius of curvature is 25 cm

Answers

Answer:

The focal length is R/2 = 25 cm / 2 = 12.5 cm

A ray from the center of curvature is focused back on itself (R cm)

As the angle of the ray is changed to being parallel to the axis the ray from the mirror will be brought closer to the mirror      

f(parallel) < focus (back upon itself)

compare  angles of  incidence and reflection

Which statement best explains the movement of electric current from the clouds to the ground during a lightning storm?

Answers

The statement that best explains the movement of electric current is that the ground is positively charged and the clouds are negatively charged.

What is lightning storm?

Lightning storm can be defined as the electrical outlet that occurs due to unsteadiness of the storm clouds and the ground.

Lightning storm occurs when opposite charges build up enough in the air leading to rapid discharge of electricity.

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A transformer consists of 340. primary windings and 980. secondary windings. If the emf of the primary coil is 30.0 V, what is the emf across the secondary coil? 34.7 Ω

Answers

Answer:

86.47 V

Explanation:

It will be multiplied by the ratio of the windings

980/340 * 30 = 86.47 v

Help!!! Asap!!!
The arrows in the diagram show the direction of the magnetic fields of the
atoms that make up an object. Which material could the diagram represent?
A. Paper
B Steel
C Nickel
D Iron

Answers

Answer:

Probs paper cuz the rest are metals

A planet X moves in an elliptical path as shown below. At which point does the planet moves faster. B Assume, Ethiopian institute of space science and technology, allows to launch a space station with you. While you are in orbit, you gets thirsty. The only thing in the space station were a cup of water. you go to the cup of water and pick it up, But unfortunately your hand is twisted and the cup is turned down. Because the cup is turned down, water pour on to the floor. How would you get water?​

Answers

We can get water from the sweat and exhaled breath of the people present in the space station.

How would you get water?​

We can get water from the sweat and exhaled breath. The water we drink is recycled from the sweat and exhaled breath of the people present in the space station which was collected through  condensation on the Space Station's walls.

So we can conclude we can get water from the sweat and exhaled breath of the people present in the space station.

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You cover the following displacements every day going to school: d1=50 m, E and d2=95 m, N. You do this for 12 minutes. a) What is your speed in m/s? b) What is your velocity in m/s?

Answers

The speeds and velcity will be 0.020 m/sec and 0.0149 m/sec.

What is speed?

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while V for the final speed. its si unit is m/sec.

Given data;

d1=50 m E

d2=95 m N
Time period,t = 12 min = 12 × 60 sec = 7200 sec

The displacement is the shortest distance found from the pythogorous theorem as;

Displacement = √[(50 )²+(95)²]

Displacement = 107.35 m

a)The speed is found as;

Speed = distance / time

Speed = 50 +95 / 7200

Speed = 0.020 m/sec

b)The velocity is found as;

Velocity = Displacement / time

Velocity = 107.35 / 7200

Velocity = 0.0149 m/sec

Hence the speed and the velcity will be 0.020 m/sec and 0.0149 m/sec.

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I need help asap!!!

Answers

Answer:

Probably competition with television news-

by 1953 television was becoming common in many houses and people could obtain news from television newscasts

You are raising up a big bucket of water from a 25.9 m deep well. The combined mass of the water and the bucket is 13.9 kg. The bucket is attached to a heavy duty steel chain. The mass of the chain is 19.3 kg.
1. How much work do you perform during the lifting process?
2. If it takes 1.75 minutes for you to raise the bucket of water out of the well, then what was your average power?

Answers

1. The amount of work done during the lifting process is 8426.824 J

2. The average power used in the process is 80.26 Watts

1. How to determine the workdone Combine mass (m) = 13.9 + 19.3 = 33. 2 KgAcceleration due to gravity (g) = 9.8 m/s²Height (h) = 25.9 mWork done (Wd) =?

Wd = mgh

Wd = 33. 2 × 9.8 × 25.9

Wd = 8426.824 J

2. How to determine the powerWork done (Wd) = 8426.824 JTime (t) = 1.75 mins = 1.75 × 60 = 105 sPower (P) =?

P = Wd / t

P = 8426.824 / 105

P = 80.26 Watts

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A prism of angle 60° is of a material of refractive index √√2. Calculate angles of minimum deviation and maximum deviation.​

Answers

Answer:

i don't know if this helps but The refractive index of the material of the prism of an angle 60 degree is 1.62 for sodium light. there you go hope it helps

Explanation:

the main source of this energy is
.

Answers

Answer:motion

Explanation:

kinetic energy of an object that it possess due to its motion.

Kinetic energy or motion; the force that keeps us moving

Particles q1 = -53.0 uc, q2 = +105 uc, and q3 = -88.0 uc are in a line. Particles qı and q2 are separated by 0.50 m and particles q2 and q3 are separated by 0.95 m. What is the net force on particle q3?

Answers

The net force on particle q3 is 112.11 N

What is electrostatic force?

The electrostatic force F between two charged objects placed distance apart is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

F = kq₁q₂/r²

where k = 9 x 10⁹ N.m²/C²

The Particles q1 = -53.0 uc, q2 = +105 uc, and q3 = -88.0 uc are in a line. Particles qı and q2 are separated by 0.50 m and particles q2 and q3 are separated by 0.95 m.

The net force F3 = F13 + F23

Substitute the values, we get the force F on q3 as

F3 = 9 x 10⁹x53.0 x 10⁻⁶x 88.0 x 10⁻⁶ / (0.5+0.95)² + 9 x 10⁹x105.0 x 10⁻⁶x 88.0 x 10⁻⁶ / (0.95)²

F3 = 112.11 N

Thus, the magnitude of the force on charge q3 is 112.11 N.

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A Mercedes-Benz 300SL (m = 1700 kg) is parked on a road that rises 15 degrees above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires?

Answers

Answer: See below

Explanation:

Given:

Mass of the Mercedes-Benz (m) = 1700 kg

Inclination of the road (θ) = 15.0

The free body diagram is shown in figure attached below

a) The normal force is equal to the cos component of the weight of the car.

[tex]\begin{aligned}&f=m g \cos \theta \\&f=1700 \times 9.81 \times \cos 15 \\&f=16108.74 \mathrm{~N}\end{aligned}[/tex]

b) The static force will be equal to the weight's sin component.

[tex]\begin{aligned}&f=m g \sin \theta \\&f=1700 \times 9.81 \times \sin 15 \\&f=4316.32 \mathrm{~N}\end{aligned}[/tex]

A point charge Q1 = +6.0 nC is at the point (0.30 m, 0.00 m); a charge Q2 = -1.0 nC is at (0.00 m, 0.10 m), and a charge Q3 = +5.0 nC is at (0.00 m, 0.00 m). What are the magnitude and direction of the net force on the +5.0-nC charge due to the other two charges? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

Answers

Answer:

My Answer is Answer

Explanation:

We can use Coulomb's equation, which states that the force between two point charges is imparted: due to the other two charges ([tex]\rm Q_1[/tex] and [tex]\rm Q_2[/tex]) to obtain a net force on the +5.0 nC charge ([tex]\rm Q_3[/tex]).

[tex]\rm F = k * |Q_1 * Q_2| / r^2[/tex]

where:

F = force between the charges

k = Coulomb's constant ≈ 9.0 x [tex]\rm 10^9 N.m^2/C^2[/tex]

[tex]\rm Q_1[/tex] and [tex]\rm Q_2[/tex] = magnitudes of the charges

r = distance between the charges

Calculating the forces separately for each pair of charges and then find the net force:

Force between [tex]\rm Q_3[/tex] and [tex]\rm Q_1[/tex]:

[tex]\rm Q_3[/tex] = +5.0 nC = +5.0 x [tex]10^-^9[/tex] C

[tex]\rm Q_1[/tex] = +6.0 nC = +6.0 x [tex]10^-^9[/tex] C

[tex]\rm r_1[/tex] = distance between [tex]\rm Q_3[/tex] and [tex]\rm Q_1[/tex] = √[(0.00 m - 0.30 m[tex])^2[/tex] + (0.00 m - 0.00 m[tex])^2[/tex]] ≈ 0.30 m

[tex]\rm F_1[/tex]= k * |[tex]\rm Q_3[/tex] * [tex]\rm Q_1[/tex]| / [tex]\rm r_1^2[/tex]

[tex]\rm F_1[/tex] = (9.0 x [tex]10^9[/tex] N·[tex]\rm m^2/C^2[/tex]) * |(5.0 x [tex]10^-^9[/tex] C) * (6.0 x [tex]10^-^9[/tex] C)| / (0.30 [tex]\rm m)^2[/tex]

[tex]\rm F_1[/tex]≈ 6.0 N (towards the left)

Force between [tex]\rm Q_3[/tex] and [tex]\rm Q_2[/tex]:

[tex]\rm Q_3[/tex] = +5.0 nC = +5.0 x [tex]10^-^9[/tex] C

[tex]\rm Q_2[/tex] = -1.0 nC = -1.0 x [tex]10^-^9[/tex] C

r2 = distance between Q3 and Q2 = √[(0.00 m - 0.00 m)^2 + (0.00 m - 0.10 m)^2] ≈ 0.10 m

[tex]\rm F_2[/tex] =[tex]\rm k * |Q_3 * Q_2| / r2^2[/tex]

[tex]\rm F_2[/tex] = (9.0 x [tex]\rm 10^9 N.m^2/C^2[/tex]) * |(5.0 x [tex]10^-^9[/tex] C) * (-1.0 x [tex]10^-^9[/tex] C)| / (0.10 [tex]\rm m)^2[/tex]

[tex]\rm F_2[/tex] ≈ 45.0 N (towards the top)

Now we can sum the forces in the x and y directions to determine the net force:

Net force in the x-direction = [tex]\rm F_1[/tex] = 6.0 N (to the left)

Net force in the y-direction = [tex]\rm F_2[/tex] = 45.0 N (upwards)

The Pythagorean theorem can be used to determine the magnitude and direction of the net force.

Magnitude of net force = √(Net force in the x-direction[tex])^2[/tex]+ (Net force in the y-direction[tex])^2[/tex]

Magnitude of net force ≈ √(6.0 N[tex])^2[/tex] + (45.0 N[tex])^2[/tex]≈ √(36 [tex]\rm N^2[/tex] + 2025 [tex]\rm N^2[/tex]) ≈ √(2061 [tex]\rm N^2[/tex]) ≈ 45.4 N

The direction of the net force is the angle θ:

θ = arctan(Net force in the y-direction / Net force in the x-direction)

θ = arctan(45.0 N / 6.0 N) ≈ arctan(7.5) ≈ 82.9°

As a result, the net force exerted by the other two charges on the +5.0 nC charge has a strength of about 45.4 N and is directed about 82.9° above the negative x-axis (in the upper left direction).

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Assume the pipe in the bathroom has a radius 34 cm and the pipe in the chamber has a radius of 45cm.

Answers

The Velocit,y Height and flow rate is is mathematically given as

v2= 1.92m/sh=2.06m F= 700kg/s

What is the flow rate?

Generally, the equation for the Continuity equation is  mathematically given as

A1v1=A2v2

Therefore

[tex]v2=\frac{r1^2}{r2^2}v1[/tex]

v2= 1.92m/s

Generally, the equation for Pressure  is  mathematically given as

the difference is related by

[tex]P=\rho g h[/tex]

Therefore

[tex]h=P / \rho g[/tex]

(1000)(9.81)(h)=2*10100

h=2.06m

Generally, the equation for Flow rate is  mathematically given as

[tex]F=\rho Av[/tex]

Therefore

[tex]F=(1000)\pi (0.45)^2 1.1[/tex]

F= 700kg/s

In conclusion, The resulting solutions are

v2= 1.92m/s

h=2.06m

F= 700kg/s

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3.2 Define conservation of energy.

Answers

Answer:

A principle stating that energy cannot be created or destroyed, but can be altered from one form to another.

Explanation:

(b)) Suggest a reason why pail manufacturers prefer the shape shown to other shapes. 30 cm .... ( b ) ) Suggest a reason why pail manufacturers prefer the shape shown to other shapes . 30 cm ....​

Answers

Manufacturers of pails prefer a cone shape because it has a suitable base as well as more space for its contents.

What is the most common shape of pails?

The most common shape of pails is the shape of the frustum of a cone.

Pails are useful to contain liquids as well as some grainy solids like sand.

Because of their unique shapes, pails are able to rest on their bases and also has enough space to contain enough liquid as their top part spreads outwards.

In conclusion, the choice of shape of pails is to ensure they rest on the base and also have enough space for their contents.

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how is this answer correct

Answers

Answer:

b

Explanation:

c is a velocity time graph, showing how a body moves with constant velocity with change in time

A crate with a mass of M = 62.5 kg is suspended by a rope from the endpoint of a uniform boom. The boom has a mass of m = 116 kg and a length of l = 7.65 m. The midpoint of the boom is supported by another rope which is horizontal and is attached to the wall as shown in the figure.
1. The boom makes an angle of θ = 57.7° with the vertical wall. Calculate the tension in the vertical rope.
2. What is the tension in the horizontal rope?

Answers

The boom makes an angle of θ = 60.2° with the vertical wall and the tension in the horizontal rope is mathematically given as

[tex]T_1=730.85 \mathrm{~N}[/tex]

[tex]T_1'=\frac{1980.51 \mathrm{~N}}{}[/tex]

What is the tension in the vertical rope.?

Generally, the equation for is  mathematically given as

1) Tension in vertical rope,

[tex]T_{1}=74.5 \times 9.81\\T_1=730.85 \mathrm{~N}[/tex]

2) Tension in horizontal rope,

[tex]\sum{Mg} =0\\Mg\frac{l}{2} \sin \theta+T_{1} \frac{l}{2} \sin \theta &=T_{2} l \cos \theta \\[/tex]

[tex]T_{2} &=\frac{M g(l / 2) \sin \theta+T_{1}(l / 2) \sin \theta}{l \cos \theta} \\\\&=\frac{M g \tan \theta}{2}+\frac{T_{1} \tan \theta}{2} \\\\T_{2} &=\frac{142 \times 9.81 \tan 61.8}{2}+\frac{730.85 \times \tan 61.8}{2} \\\\=& 1299+681.51[/tex]

[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]

In conclusion, the tension in the horizontal rope is

[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]

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1. A supersonic airplane is flying horizontally at a speed of 2610 km/h. What is the centripetal acceleration of the airplane, if it turns from North to East on a circular path with a radius of 80.5 km?
2. How much time does the turn take?
3. How much distance does the airplane cover during the turn?

Answers

1)The centripetal acceleration will be 84622.360 km/h².

2)The time the turn take is 0.19 hour.

3)The distance the airplane cover during the turn is 505.54 km.

What is centripetal acceleration?

The acceleration needed to move a body in a curved way is understood as centripetal acceleration.

The direction of centripetal acceleration is always in the path of the center of the course. The total acceleration is the result of tangential and centripetal acceleration.

The given data in the problem;

R is the radius= 80.5 m

v is the linear speed = 2610 km/h

a is the total acceleration.

1)

The centripetal acceleration is found by

[tex]\rm a_c = \frac{v^2}{R} \\\\ a_c = \frac{(2610 \ km/h)^2}{80.5} \\\\ a_c = 84622.360 \ km/h^2[/tex]

2)

The time the turn take is;

[tex]\rm a= \frac{4 \pi^2R}{T^2} \\\\\ 84622.360 = \frac{4 \times (3.14)^2 \times 80.5}{T^2} \\\\\ T = 0.19 \ hr[/tex]

3)

The distance the airplane cover during the turn is;

C= 2πR

C=2 × 3.14 × 80.5 km

C= 505.54 km

Hence centripetal acceleration, time the turn take, and distance the airplane cover during the turn will be 84622.360 km/h²,0.19 sec, and 505.54 km respectively.

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Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 10,600 kg. The thrust of its engines is 25,500 N. (Assume that the gravitational acceleration on the Moon is 1.67 m/s2.)
(a) Calculate its magnitude of acceleration in a vertical takeoff from the Moon.
m/s2

(b) Could it lift off from Earth? If not, why not?
Yes, the thrust of the module's engines is equal to its weight on Earth.
No, the thrust of the module's engines is equal to its weight on Earth.
No, the thrust of the module's engines is less than its weight on Earth.
Yes, the thrust of the module's engines is greater than its weight on Earth.

If it could, calculate the magnitude of its acceleration. (If not, enter NONE.)
m/s2

Answers

(a) The magnitude of acceleration in a vertical takeoff from the Moon will be 0.73 m/s².

(b)Option d is correct.

What is force?

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

Force is defined as the product of mass and acceleration. Its unit is Newton.

Given data;

Force, F =?

Mass of a fully loaded module,m = 10,600 kg

Acceleration due to gravity on moon,g = 1.67 m/s²

Thrust of its engines is,F= 25,500 N.

The net force is found as follows;

[tex]\rm F_{net} = F -W[/tex]

ma = F-W

(10,600) kg × a = 25,500 N - (10,600 kg )(1.67)

a = 0.73 m/s²

b)

Yes, it lifts off from Earth. Because the thrust of the module's engines is greater than its weight on Earth.

F>W

Hence option d is correct.

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Obtain the formula for the focal length of a lens in terms of object distance (u)
and magnification (m)

Answers

Answer:

m=image distance÷object distance

6. (a) Suppose the earth is revolving round the sun in a circular orbit of radius one b astronomical unit (1.5% 10 km). Find the mass of the sun. G = 6.67 x 10" N m² kg 2. (b) If the mass of sun is 2 x 10" kg, distance of earth from the sun is 1.5 x 10¹ m and period of revolution of the former around the latter is 365.3 days, find the value of G.​

Answers

Answer:

tough

ques

Explanation:

Which of the following statements regarding work and retirement is true?

Answers

The statement regarding work and retirement that is true is that most older Americans say they would prefer to enter retirement gradually.

What is retirement?

Retirement is defined as the time in an individuals life that such leaves the Labour market or the workforce.

In some countries such as the United States of America, most elderly citizen prefer taking retirement process slowly.

Therefore, the statement regarding work and retirement is that retirement is a gradual process which one starts to process while actively in service

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An extended body (not shown in the figure) has its center of mass at the origin of the reference frame. In the case below give the direction for the torque τ with respect to the center of mass of the body due to force F acting on the body at a location indicated by the vector r.
Options are:
X
-Y
Z
-Z

Answers

The direction of torque τ this method is mathematically given as

D=X  

Option A is correct.

What is the direction of torque?

Generally, the equation for torque is  mathematically given as

τ = r X F

Since F is the force exerted on the body at r, we need to know the direction of this torque with regard to the body's center of mass.

The dominant hand (the right) is used.Just run the fingers of our right hand along the line r.Our right palm should be placed on the letter F. Then, we'll gradually move r into F. The direction of the thumb's motion will indicate the torque's vector.

In conclusion, the direction of this method is

D=X   Option A.

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A fish rests on the bottom of the bucket of water while the bucket is being weighed on a scale. When the fish begins to swim around, does the scale reading changed? Explain.​

Answers

When the fish begins to swim around, the reading of the scale won't change because mass of the fish and the bucket is constant.

Weight of the fish and bucket

The weight of the fish inside the bucket depends on the mass of the bucket and fish.

W = mg

where;

m is mass of the bucket and fishg is acceleration due to gravity

Thus, when the fish begins to swim around, the reading of the scale won't change because mass of the fish and the bucket is constant.

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