A chef places an open sack of flour on a kitchen scale. The scale reading of
40 N indicates that the scale is exerting an upward force of 40 N on the sack. The magnitude of this force equals the magnitude of the force of Earth’s gravitational attraction on the sack. The chef then exerts an upward force of
10 N on the bag and the scale reading falls to 30 N.Draw a free-body diagram of the latter situation.

The question is incomplete. Here is the complete question.

A current in the long, straight wire, which lies in the plane of rectangular loop, that also carries a current, as shown in the figure.

Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Answer in units of N.

Answer: Net Force = [tex]50.215.10^{-7}[/tex]N

Explanation: Force and Magnetic field are related through the following formula:

F = I.L.B.sinθ

Magnetic field (B) in a straight long wire is given by

[tex]B=\frac{\mu_{0}.I}{2.\pi.r}[/tex]

in which

[tex]\mu_{0}[/tex] is permeability of free space and is [tex]4.\pi.10^{-7}[/tex]T.m/A

I is current in the wire;

r is distance to the wire;

Examining the square loop and using the right hand rule, the top, which we will name it F₂, and the bottom, named F₄, have angle θ = 0, giving sin(0) = 0 and therefore, F₁ = F₃ = 0.

So, for the net force, the relevant forces will be on the sides parallel to the wire.

For the other forces, angle is 90°, sin(90°) = 1, then:

F = I.L.B

Replacing magnetic field:

F = [tex]\frac{\mu_{0}.I_{w}.L.I_{l}}{2.\pi.r}[/tex]

Note: The side closest to the wire is F₁, while the farthest is F₃.

Note2: As the constant unit is in meters, distance and length of side of the square loop are also in meters.

Calculating forces:

F₁ = [tex]\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.082}[/tex]

F₁ = [tex]278.975*10^{-7}[/tex]N

Current in F₃ is flowing thoruhg the negative side of the referential, so:

F₃ = [tex]-\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.1}[/tex]

F₃ = [tex]-228.76*10^{-7}[/tex]N

Net force is total force:

[tex]F_{net} = F_{1}+F_{3}[/tex]

[tex]F_{net}=(278.975-228.76).10^{-7}[/tex]

[tex]F_{net}=50.22.10^{-7}[/tex]

The total force acting on the square loop is [tex]F_{net}=50.22.10^{-7}[/tex]N.

Answer:

it obeys the law of vector addition

Explanation:

In the analytical method, we use the Pythagorean theorem to find the magnitude of the resultant because it obeys the law of vector addition.

The resultant of two vectors A and B is given by;

R' = A' +  B'

[tex]R = \sqrt{A^2 + B^2}[/tex]

Therefore, Pythagorean theorem is used if the two quantities are vectors and vector quantities obey the law of vector addition.

Answer:

D

Explanation:

An open cluster is a group of up to a few thousand stars that were formed from the same giant molecular cloud, and are still loosely gravitationally bound to each other. In contrast, globular clusters are very tightly bound by gravity. ... Open clusters are very important objects in the study of stellar evolution.

Answer:

ns for high-speed rail in the United States date back to the High Speed Ground Transportation Act of 1965. Various state and federal proposals have followed. Despite being one of the world's first countries to get high-speed trains (the Metroliner service in 1969), it failed to spread. Definitions of what constitutes high-speed rail vary, including a range of speeds over 110 mph (180 km/h) and dedicated rail lines. Inter-city railwith top speeds between 90 and 125 mph (140 and 200 km/h) is sometimes referred to in the United States as higher-speed rail.[1]

Amtrak's Acela Express (reaching 150 mph, 240 km/h), Silver Star, Northeast Regional, Keystone Service, Vermonter and certain MARC Penn Line express trains (all five reaching 125 mph, 201 km/h) are the only high-speed services in the country.

As of 2020, the California High-Speed Rail Authority is working on the California High-Speed Rail project and construction is under way on sections traversing the Central Valley. The Central Valley section is planned to open in 2029 and Phase I is planned for completion in 2031.[2]

Contents

1 Definitions in American context

2 History

2.1 Faster inter-city trains: 1920–1941

2.2 Post-war period: 1945–1960

2.3 First attempts: 1960–1992

2.4 Renewed interest: 1993–2008

2.5 Plans for 2008–2013

3 Current state and regional efforts

3.1 The Northeast

3.1.1 Northeast Corridor: Next Generation High-Speed Rail

3.1.1.1 Proposed routes

3.1.2 Northeast Maglev proposal

3.1.3 New Jersey–New York City upgrades

3.1.4 New York

3.1.5 Pennsylvania

3.2 Western States

3.2.1 California

3.2.2 Pacific Northwest

3.2.3 Arizona

3.3 Mid-Atlantic and the South

3.3.1 Florida

3.3.2 Southeast

3.3.3 Texas

3.4 Midwest

3.4.1 Illinois and the Midwest

3.5 The Southwest

4 Federal high-speed rail initiatives

4.1 American Recovery and Reinvestment Act of 2009

4.1.1 Strategic plan

4.2 2009 federal grant funding

4.3 2010 allocation

4.3.1 Cancellation of funds for Wisconsin, Ohio, and Florida

4.4 2011 and 2012 proposals and rejections of funding

5 See also

6 Notes

7 Further reading

8 External links

Explanation:

Answer:

the box is going south at 10n

Explanation:

Answer:

The horizontal displacement is 20 m.

Explanation:

Given that,

Velocity = 10 m/s

Height = 20 m

We need to calculate the time

Using equation of motion

[tex]s=ut+\dfrac{1}{2}gt^2[/tex]

Put the value into the formula

[tex]20=0+\dfrac{1}{2}\times9.8\times t^2[/tex]

[tex]t^2=\dfrac{20\times2}{9.8}[/tex]

[tex]t=\sqrt{\dfrac{20\times2}{9.8}}[/tex]

[tex]t=2.0\ sec[/tex]

We need to calculate the horizontal displacement

Using formula of horizontal displacement

[tex]\Delta x=v_{x}\times t[/tex]

Put the value into the formula

[tex]\Delta x=10\times2.0[/tex]

[tex]\Delta x=20\ m[/tex]

Hence, The horizontal displacement is 20 m.

Answer:

11760 joules

Explanation:

Given

Mass (m) = 75kg

Height (h) = 16m

Required

Determine the increment in potential energy (PE)

This is calculated as thus:

PE = mgh

Where g = 9.8m/s²

Substitute values for m, g and h.

P.E = 75 * 9.8 * 16

P.E = 11760 joules

The potential energy of the man in the fifth floor of the building has increased by 11760J.

Given the data in the question;

Potential energy; [tex]P_E =\ ?[/tex]

Potential energy is the energy possessed by a particle due to its position relative to other particles. It is expressed as:

[tex]P_E = mgh[/tex]

Where m is mass of the particle, h is its height above ground level and g is acceleration due to gravity( [tex]g = 9.8m/s^2[/tex] ).

We substitute our values into the equation

[tex]P_E = 75kg\ *\ 9.8m/s^2\ *\ 16m\\\\P_E = 11760kg.m/s^2\\\\ P_E = 11760J[/tex]

Therefore, the potential energy of the man in the fifth floor of the building has increased by 11760J.

Learn  more: https://brainly.com/question/16850959

Answer:

Explanation:

For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.

The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction

Taking the sum of forces along the y component

Sum Fy = -W+R = ma

Since acceleration is zero

-W+R = m(0)

-W+R = 0

-W = -R

W = R

Hence the Normal reaction force acting on the on the body is equal to normal force

Answer:

Correct choice: D. Object A is the fastest

Explanation:

In a distance vs time graph, the distance is plotted vertically, and the time is plotted horizontally.

The diagram shows three graphs of objects A, B, and C.

The graph of A shows the object traveled 12 meters in 3 seconds, for a speed of 12/3= 4 m/s.

The graph of B shows the object traveled 8 meters in 4 seconds for a speed of 8/4=2 m/s.

Finally, the object C travels 4 meters in 4 seconds, for a speed of 4/4= 1 m/s

Thus, the fastest object is A.

Answer:

communication, if you don't talk you'll never know what's going on.

Explanation:

Answer:

6.7 m/s

Explanation:

Given:

Δx = 5 m

v₀ = 5 m/s

a = 2 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (5 m/s)² + 2 (2 m/s²) (5 m)

v = 6.7 m/s

Answer:

see below

Explanation:

The triangle stands for the change in

We would change the change in x

Answer:

Δ This is the symbol of Delta which means Change

and x is length/distance/position.

Thus, Δx stands for Change in length/distance/position.

-TheUnknownScientist

Answer:

881.84905 LBS

Explanation:

ThErE :p

3922.66 newtons.

This is an exact amount, to get newtons form kg, multiply by 9.8, or in this case, 10.

This gives you 4000 newtons

Answer:

Maybe

Explanation:

I say maybe because it will help them still but not quite

second question: How many seconds after the first snowball

should you throw the second so that they

arrive on target at the same time?

Answer in units of s.

Answer:

Part 1: 28°

Part 2: 1.367

Explanation:

Part 1:

Given: 62°  

Simple

θ = 90°- 62°

θ = 28°

Part 2:

Y-direction

Δy[tex]=v_{yo} t+\frac{1}{2} a_{y} t^{2}[/tex]

[tex]0=[16.2sin(62)]t_{1}+1/2(-9.8)t_{1}^{2} \\[/tex]

[tex]t_{1} =\frac{2[16.2sin(62)]}{9.8}[/tex]

[tex]t_{1}=2.91913s[/tex]

[tex]0=[16.2sin(28)]t_{2}+1/2(-9.8)t_{2}^{2}[/tex]

[tex]t_{2} =\frac{2[16.2sin(28)]}{9.8}[/tex]

[tex]t_{2}=1.55213s[/tex]

Δt[tex]=t_{1}-t_{2}[/tex]

Δt[tex]=2.91913-1.55213[/tex]

Δt= 1.367s

Hope it helps :)

Answer:

Sigmund Freud

Explanation:

Answer:

Wilhelm Wundt

Answer:

  φ = B sin (2π n/a   x)

Explanation:

In quantum mechanics when a particle moves freely it implies that the potential is zero (V = 0), so its wave function is

     φ = A cos kx + B sin kx

we must place the boundary conditions to determine the value of the constants A and B.

In our case we are told that the particle cannot be outside the boundary given by x = ± a / 2

therefore we must make the cosine part zero, for this the constant A = 0, the wave function remains

    φ = B sin kx

the wave vector is

      k = 2π /λ

now let's adjust the period, in the border fi = 0 therefore the sine function must be zero

         φ (a /2) = 0

          0 = A sin (2π/λ  a/2)

therefore the sine argument is

          2π /λ   a/2 = n π

          λ= a / n

we substitute

          φ = B sin (2π n/a   x)

This question is incomplete, the complete question is;

Two blocks, M1M1 and M2,M2, are connected by a massless string that passes over a massless pulley as shown in the figure. M2,M2, which has a mass of 25.0 kg,25.0 kg, rests on a long ramp of angle =25.0∘.θ=25.0∘. Friction can be ignored in this problem.

Determine the mass 1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration).

Answer:

the mass m1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration) is 10.57 kg

Explanation:

Given that;

m₂ = 25 kg

θ = 25°

Now at equilibrium, T = m₁g ------------------lets say equ 1

and also T = m₂gsinθ

therefore

m₁g = m₂gsinθ

m₁ = m₂sinθ

so we substitute

m₁ = 25 × sin(25)

m₁ = 25 × 0.4226

m₁ = 10.565 ≈ 10.57 kg

therefore the mass m1m1 of block M1M1 for which the two blocks are in equilibrium (no acceleration) is 10.57 kg

Answer:

Pressure increases as you move deeper below earth's surface.

Tempurature  increases as you move deeper below earth's surface.

Hope this helps!

Explanation:

Answer:

The magnitude of the net force acting on an object is equal to the mass. and the direction is in 20N

Explanation:

Answer:

N2 + 3H2 ----->  2NH3

Explanation:

Reactants side:

2 Nitrogen

5 Hydrogen

Products Side:

2 Nitrogen

5 Hydrogen

Answer: The parts closer to the equator.

Explanation:

The parts of the earth closer to the equator are similar to the plate, because they receive more sunlights than other parts. Which makes them hotter than any other regions in the earth. Example of such countries are Gabon, Uganda, Kenya, Maldives, the democratic republic of Congo, sao tome and principle  e.t.c

Answer:g

Explanation:jqjqksk

Answer:

[tex]a=0.33\frac{m}{s^2}[/tex]

Explanation:

Hello.

In this case, since the force is defined in terms of the mass and acceleration as follows:

[tex]F=ma[/tex]

Given the force and the mass, we can compute the acceleration as shown below:

[tex]a=\frac{F}{m}=\frac{25N}{75kg}=\frac{25kg\frac{m}{s^2} }{75kg}\\ \\a=0.33\frac{m}{s^2}[/tex]

Best regards.