A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the bucket is gradually increased by the addition of sand. At some point, the bucket will accumulate enough sand to set the block in motion. The coefficients of static and kinetic friction are 0.60 and 0.50 respectively.

Required:
a. Determine the mass of sand in [kg], including the bucket, needed to start the block moving.
b. Find the blocks acceleration, in [m/s^2] up the plane?

Answer:

a = 7.84 m / s²

Explanation:

For this exercise let's use Newton's second law

          F - fr = m a

indicate that the force is F = 609.1 N and the friction force is fr = 107.9 N and is constant

         a = [tex]\frac{F - fr}{m}[/tex]

let's calculate

          a = [tex]\frac{ 609.1 - 107.9 }{63.9}[/tex]

          a = 7.84 m / s²

Answer:

The period the field must be reduced to zero is 9.81 x 10⁻⁵ s

Explanation:

Given;

initial value of the magnetic field, B₁ = 0.276 T

number of turns of the solenoid, N = 517 turns

diameter of the solenoid, d = 10.5 cm = 0.105 m

induced emf, = 12.6 kV = 12,600 V

when the field becomes zero, then the final magnetic field value, B₂ = 0

The induced emf is given by Faraday's law;

[tex]emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s[/tex]

Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s

Answer:

[tex]\boxed {\boxed {\sf 2 \ Newtons}}[/tex]

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

[tex]F= m \times a[/tex]

The mass of the stone is 0.2 kilograms and the acceleration is 10.0 meters per square second.

Substitute the values into the formula.

[tex]F= 0.2 \ kg * 10.0 \ m/s^2[/tex]

Multiply.

[tex]F=2 \ kg*m/s^2[/tex]

Convert the units.

[tex]F= 2 \ N[/tex]

The force is 2 Newtons.

Answer: A nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded and a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field

Explanation:

While interpreting the data in NMR, the positions of signals are studied.

The nucleus/ protons having a higher value of [tex]\delta[/tex] are said to be less shielded. They are said to be upfield.

The nucleus/protons having a lower value of [tex]\delta[/tex] are said to be more shielded. They are said to be downfield.

So, a nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded by the nucleus that absorbs at [tex]4.13\delta[/tex]

Also, the less shielded nucleus/protons will require a weak applied field to come into resonance than the more shielded nucleus/protons

So, a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field to come into resonance than the nucleus that absorbs at [tex]11.45\delta[/tex]

Answer:

Units named after scientists are written in lowercase but their symbols are written as capital

Unit of power is watt, since it is named after a scientist, then symbol will be written as W

Farad, symbol = F

Henry, symbol = H

Units whose names aren't culled from that of scientists are written in lower case and their symbols are also in lower case.

Units such as meter, kilogram should be represented with symbols in small letters as: m and kg respectively.

Explanation:

Kindly check answer

Answer:

a)   μ = 0.0136, b)   F = 22.8 N

Explanation:

This exercise must be solved in parts. Let's start by using conservation of moment.

a) We define a system formed by the downward and the box, therefore the forces during the collision are internal and the momentum is conserved

initial instant. Before the crash

        p₀ = m v₀

final instant. After inelastic shock

        p_f = (m + M) v

the moment is preserved

        p₀ = p_f

        m v₀ = (m + M) v

        v = [tex]\frac{m}{m + M} \ v_o[/tex]

We look for the speed of the block with the bullet inside

        v = [tex]\frac{0.00325}{0.00325 + 2.50 } \ 345[/tex]

        v = 0.448 m / s

Now we use the relationship between work and kinetic energy for the block with the bullet

in this journey the force that acts is the friction

         W = ΔK

          W = ½ (m + M) [tex]v_f^2[/tex]  - ½ (m + M) v₀²

the final speed of the block is zero

the work between the friction force and the displacement is negative, because the friction always opposes the displacement

         W = - fr x

we substitute

           - fr x = 0 - ½ (m + M) vo²

           fr = ½ (m + M) v₀² / x

the friction force is

          fr = μ N

          μ = fr / N

equilibrium condition

          N - W = 0

          N = W

          N = (m + M) g

we substitute

         μ = ½ v₀² / x g

we calculate

          μ = ½ 0.448 ^ 2 / 0.75 9.8

          μ = 0.0136

b) Let's use the relationship between work and the variation of the kinetic energy of the block

          W = ΔK

initial block velocity is zero vo = 0

         F x₁ = ½ M v² - 0

         F = [tex]\frac{1}{2} M \frac{x}{y} \frac{v^2}{x1}[/tex]

         F = ½ 2.50 0.448² / 0.0110

         F = 22.8 N

it's a form of government where people elect their representatives

Answer:

The word democracy itself means rule by the people.

Answer:

b = e > a = c = f > d  

Explanation:

Since the satellites complete one circular orbit in the same amount of time, their speed is the same.

The force needed to maintain the orbit is the centripetal force given by F = mv²/L where m = mass of artificial satellite and L = radius of orbit.

So, for artificial satellite a

a. m=200 kg and L= 5000m

F =  mv²/L

F =  200 kgv²/5000 m

F =  0.04v² N

So, for artificial satellite b

b. m=400 kg and L= 2500m

F =  mv²/L

F =  400 kgv²/2500 m

F =  0.16v² N

So, for artificial satellite c

c. m=100 kg and L= 2500m

F =  mv²/L

F =  100 kgv²/2500 m

F =  0.04v² N

So, for artificial satellite d

d. m=100 kg and L= 10000m

F =  mv²/L

F =  100 kgv²/10000 m

F =  0.01v² N

So, for artificial satellite e

e. m=800 kg and L= 5000m

F =  mv²/L

F =  800 kgv²/5000 m

F =  0.16v² N

So, for artificial satellite f

f. m=300 kg and L= 7500m

F =  mv²/L

F =  300 kgv²/7500 m

F =  0.04v² N

So, the net force are in the order b = e > a = c = f > d  

Answer:

     A = 2,8333  s

Explanation:

El periodo es definido como el tiene que toma de dar una oscilación.

En este caso realiza varias osicilacion por lo cual debemos encontrar el promedio del perdono.

              T = t/n

calculemos

              A = 34,0/ 12,0

              A = 2,8333  s

Answer:

time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.

Explanation:

Height, h =  15 m

Newton's second law

Force = mass x acceleration

The unit of gravitational force is Newton and the value is m x g.

where, m is the mas and g is the acceleration due to gravity.  

Let the time of fall is t.

Use second equation of motion

[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]

Let the final speed is v.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]

Answer:

a. None

b. Both

Explanation:

a. Which rider is traveling faster at the bottom?

Since both riders fall from the same height, h, their potential energy, U at the top equals their kinetic energy, K at the bottom.

U = mgh and K = 1/2mv²

Since U is he same for both water-slide riders, then K will be the same and thus their speed at the bottom will be the same. This is shown below.

K = U

1/2mv² = mgh

v² = 2gh

v =√(2gh) where v = speed of rider at the bottom, g = acceleration due to gravity and h = height of slide.

Since the height is the same, so their speed at the bottom is the same. So, none of the riders travels faster than the other since they have the same speed at the bottom.

b. Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length.

Since the path length of the water slides are the same and friction is neglected, both water-slide rider get to the bottom at the same time since the distance moved is the same and they both start from rest.

So, both riders make it to the bottom at the same time.

The acceleration due to gravity acts vertically downwards, and the component of gravity acceleration is larger when the slope is steeper.

Reasons:

The acceleration of the riders are due to gravity

The component acceleration due to gravity acting on a slope is a = g·sin(θ)

As the steepness of the slope increase, the angle, θ, and sin(θ) increases, therefore, the acceleration increases.

Rider A is on a slide with gentle slope, such that if the slide is flat, rider A will be stationary.

Therefore;

a. The rider that is travelling faster at the bottom is rider B

b. Given that friction is ignored, and the path have the same length to the

bottom, the rider that makes to the bottom first is the rider that is moving

faster, which is rider B

Learn more here:

https://brainly.com/question/13218675

Answer:

The velocity of the fish when it hits the water is:

v = 10.93 m/s and 71.88 ° below the x-direction.

Explanation:

Let's find the velocity of the fish in the y-direction.

[tex]v_{fy}^{2}=v_{iy}^{2}-2gh[/tex]

Here, v(iy) of the fish is zero, and the heigh h = 5.50 m, then the velocity will be:

[tex]v_{fy}^{2}=0-2(9.81)(5.50)[/tex]

[tex]v_{fy}^{2}=-2(9.81)(5.50)[/tex]

[tex]v_{fy}=-10.39 \: m/s[/tex]

Now, we know that the velocity in the x-direction is constant, so we can calculate the velocity of the fish when it hits the water.

[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]

[tex]v=\sqrt{3.40^{2}+(-10.39)^{2}}[/tex]

[tex]v=10.93 \: m/s[/tex]

And the direction will be:

[tex]\theta=tan^{-1}(\frac{10.39}{3.40})[/tex]

[tex]\theta=71.88^{\circ}[/tex]

The angle is 71.88 ° belox the x-direction.

I hope it helps you!

Answer:

the rate of heat transfer after the system achieves steady state is -3.36 kW

Explanation:

Given the data in the question;

mass of water m = 50 kg

N = 300 rpm

Torque T = 0.1 kNm

V = 110 V

I = 2 A

Electric work supplied W₁ = PV = 2 × 110 = 220 W = 0.22 kW

Now, work supplied by paddle wheel W₂ is;

W₂ = 2πNT/60

W₂ = (2π × 0.1 × 300) / 60

W₂ = 188.495559 / 60

W₂ = 3.14 kW

So the total work will be;

W = 0.22 + 3.14

W = 3.36 kW

Hence total work done on the system is 3.36 kW.

At steady state, the properties of the system does not change so the heat transfer will be 3.36 KW.

The heat will be rejected by the system so the sign of heat will be negative.

i.e Q = -3.36 kW

Therefore,  the rate of heat transfer after the system achieves steady state is -3.36 kW

the answer to the question is D

Answer:

According to the data, factors influencing mining investment in South Africa's favour are the availability of labour and skills, the quality of the country's infrastructure, the quality of its geological database, and the State's environmental regulations.

Answer:

The free electrons tend to Negatively Charged body to Positively Charged body

Explanation:

As they have different charges, Electrons are more attracted to the Positive or it's opposite charge.

Answer: The correct answer is a) 3200 m, [tex]-2m/s^2[/tex]

Explanation:

Speed is defined as the ratio of distance travelled to the time taken. The equation follows:

[tex]\text{Speed}=\frac{\text{Distance travalled}}{\text{Time taken}}[/tex]

From the graph:

Speed for the first 40 s, v = 80 m/s

Time taken, t = 40 s

Putting values in above equation, we get:

[tex]\text{Distance travalled}=(80 m/s\times 40s)=3200 m[/tex]

Acceleration is defined as the ratio of change of velocity to the change of time. The equation follows:

[tex]\text{Acceleration}=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{t_2-t_1}[/tex]

From the graph, for the last 40 sec:

Initial velocity, [tex]v_1[/tex] = 80 m/s

Final velocity, [tex]v_2[/tex] = 0 m/s

Initial time, [tex]t_1[/tex] = 40 s

Final time, [tex]t_2[/tex] = 80 s

Putting values in above equation, we get:

[tex]\text{Acceleration}=\frac{(0-80)m/s}{(80-40)s}\\\\\text{Acceleration}=\frac{-80m/s}{40s}=-2m/s^2[/tex]

Hence, the correct answer is a) 3200 m, [tex]-2m/s^2[/tex]

Answer:

45 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 30 m/s

Deceleration (a) = –10 m/s²

Final velocity (v) = 0 m/s

Distance (s) =?

The distance of the dog from the car before the driver hits the brake can be obtained as follow:

v² = u² + 2as

0² = 30² + (2 × –10 × s)

0 = 900 + (–20s)

0 = 900 – 20s

Collect like terms

0 – 900 = –20s

–900 = –20s

Divide both side by –20

s = –900 / –20

s = 45 m

Therefore, the dog was 50 m away when the driver first hit the brake

Answer:

The friction coefficient's minimum value will be "0.173".

Explanation:

The given query seems to be incomplete. Below is the attached file of the complete question.

According to the question,

(a)

The net friction force's magnitude will be:

⇒ [tex]F_{net}=ma[/tex]

           [tex]=5\times 1.7[/tex]

           [tex]=8.5 \ N[/tex]

(b)

For m₃,

⇒ [tex]ma=\mu m_3 g[/tex]

Or,

⇒    [tex]\mu=\frac{a}{g}[/tex]

          [tex]=\frac{1.7}{9.8}[/tex]

          [tex]=0.173[/tex]

Answer:

Copper would experience induced magnestism mostly easily.

Explanation:

Permalloy would experience induced magnetism most easily.

I don't know the options but that is correct too!!

Answer:

130 dB

Explanation:

The equation for decibel level is given by:

[tex]D=10log(\frac{I}{I_n} )\\\\Where\ D\ is\ the \ decibel\ level\ in\ dB, I\ is\ the\ intensity\ in \ W/m^2, \\I_n\ is\ threshold\ intensity\ to\ the\ human\ ear=1*10^{-12}W/m^2\\\\Given\ that\ D=125dB, hence:\\\\125=10log(\frac{I}{1*10^{-12}} )\\\\12.5=log(\frac{I}{1*10^{-12}} )\\\\I=3.2\ W/m^2[/tex]

The intensity for 1 jack hammer is 3.2 W/m², therefore for 3 jack hammers, the intensity = 3 * 3.2 = 9.6 W/m²

[tex]D=10\ log(\frac{I}{I_n} )\\\\D=10*log(\frac{9.6}{1*10^{-12}} )\\\\D=130\ dB[/tex]

Answer:

B.

Explanation:

If both train A and train B are moving in the same direction and the exact same speed then from an observer's viewpoint within either train everything would seem as though it is not moving. Therefore, the clock on train B would appear to be the same width and to run at the same rate. In order for this to be the case the speed of both trains would need to be exactly the same, any difference in speed will cause the clock on the opposite train to appear distorted and run either faster or slower depending on the speed of the train you are on.

Answer:

The trains are moving at the same speed, so the answer is A.

Answer:Solution to 47E Step 1 Angular velocity of the swing=32rpm Weight of the seat =255N Weight of the person =825N Total weight =255+825=1080N Radius =7.5m

Explanation:

Answer:

ans: option A

Explanation:

components along x- axis is -3x and along x- axes is -2y

now use triangle law of vector addition

Answer:

0.94 m³/s

Explanation:

From the question given above, the following data were obtained:

Air flow (in ft³/min) = 2×10³ ft³/min

Air flow (in m³/s) =.?

Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:

35.315 ft³/min = 1 m³/min

Therefore,

2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min

2×10³ ft³/min = 56.63 m³/min

Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:

1 m³/min = 1/60 m³/s

Therefore,

56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min

56.63 m³/min = 0.94 m³/s

Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.