A fan spins at 6.0 rev/s. You turn it off, and it slows at 1.0 rev/s2. What is the angular displacement before it stops

Answer:

2.21 N

Explanation:

The force in this case is the total mass multiplied by the acceleration due to gravity. You are not asked for the solution to be in terms of the torque which is the usual way to solve these problems. That's why you are not given where the fulcrum is.

The fulcrum feels F1 + F2 + 34 * 980

F2 = 141.7 * 980 = 138866

F1 = 50.3 * 980  =  49294

Ruler = 34 * 980=  33320

Total Force = 221480 The units here are dynes

I just saw in the middle of the question that g = 9.80

So the answer becomes 221480 / 1000 = 221.48   because we needed kg

And that answer becomes 221.48/100 2.21 because the force of gravity should be 9.8 not 980

The total force exerted on the fulcrum is

Answer:

Option (A) is correct.

Explanation:

A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is

mass of string, m = 0.00145 kg

Frequency, f = 120 Hz

wavelength = 0.6 m

Speed = frequency x wavelength

speed = 120 x 0.6 = 72 m/s

Let the tension is T.

Use the formula

[tex]v =\sqrt\frac{T L}{m}\\\\72 = \sqrt\frac{T\times 5}{0.00145}\\\\T = 1.5 N[/tex]

Option (A) is correct.

Answer:

periodic

Explanation:

Answer:

a

Explanation:

you take 32,000kg ÷2.0m

Answer:

  r ’= 4 r

Explanation:

Electric potential energy is

          U = [tex]k \frac{q_1q_2}{r_{12}}[/tex]k q1q2 / r12

in this exercise

          q₁ = q₂ = q

          U = k q² / r

for when the charge change

           U ’= k q’² / r’

indicate that

      q ’= 2q

      U ’= U

we substitute

           U = k (2q) ² / r ’

           U = 4 k q² / r ’

we substitute

           [tex]k \ \frac{q^2}{r} = 4 k \ \frac{q^2}{r'}[/tex]k q² / r = 4 k q² / r ’

           r ’= 4 r

Answer: hello your question lacks some vital information below is the complete question

Suppose a power plant uses a Carnot engine to generate electricity, using the atmosphere at 300 K as the low-temperature reservoir. Suppose the power plant produces 1 × 106 J of electricity with the hot reservoir at 500 K during Day One and then produces 1 × 106 J of electricity with the hot reservoir at 600 K during Day Two. The thermal pollution was

answer:

Total thermal pollution = 2.5 * 10^6 J

Explanation:

Low temperature reservoir = 300 K

hot reservoir temperature = 500 K

Electrical energy produced by plant ( W ) = 1 * 10^6 J

lets assume ; Q1 = energy absorbed , Q2 = energy emitted

W = Q1 - Q2  or  Q2 = Q1 - W  ( we will apply this as the formula for determining thermal pollution )

For day 1

T1 = 500k , T2 = 300k

applying Carnot engine formula

W / Q1 = 1 - T2/T1

∴ Q1 = 10^6 / ( 1 - (300/500)) = 2.5 * 10^6 J

thermal pollution ; Q2 = Q1 - W = ( 2.5 * 10^6 - 1 * 10^6 ) = 1.5 * 10^6 J

for Day 2

T1 = 600k,  T2 = 300k

Q1 = 10^6 / ( 1 - (300/600)) = 2 * 10^6 J

Thermal pollution; Q2 = Q1 - W  = 1 * 10^6 J

Therefore the Total thermal pollution =  1 * 10^6  + 1.5 * 10^6  = 2.5 * 10^6 J

Answer:

5N

Explanation:

We have a simple problem of momentum here.

ΔMomentum= mΔv= FΔt

Solve for F

mΔv/Δt=F

Plug in givens

1*(2-1.5)/0.1=F

F=5N

The amount of force that the wall imparts on the ball is 5.0N

According to Newton's second law, the formula for calculating the force applied is expressed as:

[tex]F=ma[/tex]

m is the mass of the object

a is the acceleration of the object

Since acceleration is the change in velocity of an object, hence [tex]a=\frac{\triangle v}{t}[/tex]

The applied force formula becomes [tex]F=\frac{m\triangle v}{t}[/tex]

Given the following parameters

m = 1.0kg

[tex]\triangle v=2.0-1.5\\\triangle v=0.5m/s[/tex]

t = 0.1sec

Substitute the given parameter into the formula

[tex]F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N[/tex]

Hence the amount of force that the wall imparts on the ball is 5.0N

Learn more here: https://brainly.com/question/17811936

Answer is 0.0024

Explanation

divide the length value by 1000.

30.1 N

Explanation:

Given:

[tex]W_1 = 16\:\text{N}[/tex]

[tex]W_2 = 8\:\text{N}[/tex]

Let's write the components of the net forces at the intersections. Note that the system is equilibrium so all the net forces are zero.

Forces involving W1:

[tex]x:\:\:\:-T_1 + T_3\cos \alpha = 0\:\: \\ \text{or}\:\:T_2 = T_3\cos \alpha\:\:\:\:\:(1)[/tex]

[tex]y:\:\:\:T_3\sin \alpha - W_1 = 0\:\:\: \\ \text{or}\:\:\:T_3\sin \alpha = W_1\:\:\:\:\:\:(2)[/tex]

Forces involving W2:

[tex]x:\:\:\:T_1\sin 53 - T_3\sin \alpha = W_2\:\:\:\:\:\:\:(3)[/tex]

[tex]y:\:\:\:T_4 - T_1\cos 53 - T_3\cos \alpha = 0\:\:\:\;(4)[/tex]

Substitute (2) into (3) and we get

[tex]T_1\sin 53 - W_1 = W_2[/tex]

Solving for [tex]T_1[/tex],

[tex]T_1 = \dfrac{W_1 + W_2}{\sin 53} = 30.1\:\text{N}[/tex]

Answer:

A = 26.875 rad/s²

Explanation:

Given the following data;

Initial angular speed, Uw = 150 rads/s.

Final angular speed, Vw = 580 rads/s.

Time = 16 seconds.

To calculate the angular acceleration;

From kinematics equation;

At = Vw - Uw

Where;

Substituting into the formula, we have;

A*16 = 580 - 150

16A = 430

A = 430/16

A = 26.875 rad/s²

Answer:

Look at explanation

Explanation:

a) Kinetic Friction= μmg

μmg=0.53*88.9*9.8=461.75N

b)  -461.75N=ma

a= -5.19m/s^2

v=v0+at

5.19*1.7=v0

v0=8.81m/s^2

(a) The magnitude of the frictional force will be 461.75N

(b)The initial velocity will be 8.81 m/s.

A force that acts among sliding parts is referred to as kinetic friction. A body moving on the surface is subjected to a force that opposes its progressive motion

The size of the force will be determined by the kinetic friction coefficient between the two materials.

The given data in the problem is;

μ is the coefficient of kinetic friction= 0.53.

m is the mass = 88.9 kg

g is the acceleration due to gravity= 9.81 m/s²

v is the speed =?

The formula for friction force is;

[tex]\rm F= \mu R \\\\ R=mg \\\\ F= \mu mg \\\\\ F=0.53 \times 88.9 \times 9.81 \\\\ F= 461.75 \ N[/tex]

Mechanical force is found as;

F=ma

-461.75=(88.9)a

(-ve shows the -ve work done)

a=-5.19 m/s

From the Newton's first equation of motion;

v=u+at

0=u+at

u=-at

u=(- (-5.19)(1.7)

u=8.81 m/s²

To learn more about the kinetic friction refer to;

https://brainly.com/question/13754413

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Answer:

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Explanation:

The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:

[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)

Where:

[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.

[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.

[tex]\Delta P[/tex] - Pressure change, in pascals.

If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:

[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]

[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]

[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Answer:

The arrow will bury itself farther by 3S₁

Explanation:

lets assume; the Arrow shot by me has a speed twice the speed of the arrow fired by the younger shooter

Given that ; acceleration is constant , Frictional force is constant

                    A₂ =   A₁

Vf²₂ - Vi²₂ / 2s₂  = Vf₁² - Vi₁² / 2s₁ ---- ( 1 )

final velocities = 0

Initial velocities : Vi₂ = 2(Vi₁ )

Back to equation 1

0 - (2Vi₁ )² / 2s₂ =  0 - Vi₁² / 2s₁

hence :

s₂ = 4s₁

hence the Arrow shot by me will burry itself farther by :

s₂ - s₁ = 3s₁

Note :  S1 = distance travelled by the arrow shot by the younger shooter

Answer:

Charge on B is 12 uC.

Explanation:

Initial charge on A = 32 uC

Initial charge on B and C = 0

Now A touches to B, so the charge on A and B both is

q = (32 + 0) / 2 = 16 uC

Now A touches to C, so the charge on A and C both is

q' = (16 + 0) / 2 = 8 uC

Now again A touches to B so the charge on B is

q''= (8 + 16) / 2 = 12 uC  

Answer:

[tex]W=2KJ[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=50kg[/tex]

Initial Velocity [tex]v_1=8m/s[/tex]

Final Velocity [tex]v_2=12m/s[/tex]

Generally the equation for Work-done is mathematically given by

W=\triangle K.E

Therefore

 [tex]W=0.5M(v_2^2-v_1^2)[/tex]

 [tex]W=0.5*50(12^2-8^2)[/tex]

 [tex]W=2KJ[/tex]

Answer:

20 seconds

Explanation:

We are given 2 givens in the first statement

v0=0 and a=5

And we are trying to find time needed to cover 1km or 1000m.

So we use

x-x0=v0t+1/2at²

Plug in givens

1000=0+2.5t²

solve for t

t²=400

t=20s

Answer:

B) 1500m/s

Explanation:

Ans is 1500m/s

Answer:

Explanation:

From the given information:

mass = 64 kg

speed = 3.2 m/s

coefficient of friction [tex]\mu =[/tex] 0.70

The mechanical energy touted relates to the loss of energy in the system as a result of friction and this can be computed as:

[tex]W = \Delta K.E[/tex]

[tex]\implies \dfrac{1}{2}m(v^2 -u^2)[/tex]

[tex]= \dfrac{1}{2}(64.0 \kg) (0 - (3.2 \ m/s^2))[/tex]

Thus, the mechanical energy touted = 327.68 J

According to the formula used in calculating the frictional force

[tex]F_r = \mu mg[/tex]

= 0.70 × 64  kg× 9.8 m/s²

= 439.04 N

The distance covered now can be determined as follows:

d = W/F

d = 327.68 J/  439.04 N

d = 0.746 m

Answer:

joule

Explanation:

A, B, and C are hilarious. D is correct.

Charges can be positive or negative, so a pair of charges can be alike or opposite. But so far, we've never seen a negative mass.

Answer:

a=10m/s^2

Explanation:

acceleration= final velocity+ initial velocity/time taken

v-u/t=a

100-0/5=a

100/5=a

a=20m/s^2

case2

100-0/10=a

100/10=a

a=10m/s^2

Don't forget to write the units.

Hope this helps

please mark me as brainliest.

It does not change the chemical composition of water.

Answer:

828 kg/m³ or 0.828 g/cm³

Explanation:

Applying,

D = m/V............. Equation 1

Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.

From the question,

Given: m = 77 g , V = 93 cm³

Substitute these values into equation 1

D = 77/93

D = 0.828 g/cm³

Converting to kg/m³

D = 828 kg/m³

Answer:

The correct answer is "1.2 J".

Explanation:

Seems that the given question is incomplete. Find the attachment of the complete query.

According to the question,

Now,

⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]

⇒      [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]

⇒      [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]

By putting the values, we get

⇒      [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]

⇒      [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]

⇒      [tex]=1.2 \ J[/tex]

Answer:

a) ΔT₁ = -4.68 N,   ΔT₂ = 4.68 N, b) ΔT₂ = 4.68 N, ΔT₁ = 4.68 N

Explanation:

In this exercise we will use Newton's second law.

         ∑F = m a

Let's start with the set of three cars

         F_total = M a

         F_total = M 0.12

where the total mass is the sum of the mass of each charge

          M = m₁ + m₂ + m₃

This is the force with which the three cars are pulled.

Now let's write this law for each vehicle

car 1

         F_total - T₁ = m₁ a

         T₁ = F_total - m₁ a

car 2

         T₁ - T₂ = m₂ a

         T₂ = T₁ - m₂ a

car 3

         T₂ = m₃ a

note that tensions are forces of action and reaction

a) They tell us that 39 kg is removed from car 2 and placed on car 1

         m₂’= m₂ - 39

         m₁'= m₁ + 39

         m₃ ’= m₃

they ask how much each tension varies, let's rewrite Newton's equations

The total force does not change since the mass of the set is the same F_total ’= F_total

car 1

           F_total ’- T₁ ’= m₁’ a

           T₁ ’= F_total - m₁’ a

           T₁ ’= (F_total - m₁ a) - 39 a

           T₁ '= T₁ - 39 0.12

           ΔT₁ = -4.68 N

car 2

           T₁’- T₂ ’= m₂’ a

           T₂ ’= T₁’- m₂’ a

           T₂ '= (T₁'- m₂ a) + 39 a

           T₂ '= T₂ + 39 0.12

           ΔT₂ = 4.68 N

b) in this case the masses remain

            m₁ '= m₁

           m₂ ’= m₂ - 39

           m₃ ’= m₃ + 39

we write Newton's equations

car 3

          T₂ '= m₃' a

          T₂ ’= (m₃ + 39) a

          T₂ '= m₃ a + 39 a

          T₂ '= T₂ + 39 0.12

          ΔT₂ = 4.68 N

car 1

            F_total - T₁ ’= m₁’ a

            T₁ ’= F_total - m₁ a

car 2

            T₁' -T₂ '= m₂' a

            T₁ ’= T₂’- m₂’ a

            T₁ '= (T₂'- m₂ a) + 39 a

            T₁ '= T₁ + 39 0.12

            ΔT₁ = 4.68 N

The tension in each of the coupling bars A, B, and C of the luggage carrier changes as,

Tension is the pulling force carried by the flexible mediums like ropes, cables and string.

Tension in a body due to the weight of the hanging body is the net force acting on the body.

At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 m/s², and friction is negligible.

The acceleration is the same, Tension due to the horizontal component of the forces for car 1, 2 and 3 can be given as,

[tex]\sum F_{1h}=T_A-T_B=m_1a\\\sum F_{2h}=T_B-T_C=m_2a\\\sum F_{3h}=T_C=m_3a[/tex]

On solving the above 3 equation, we get the values of tension in each bar as,

[tex]T_A=(m_1+m_2+m_3)a\\T_B=(m_3+m_2)a\\T_C=m_3a[/tex]

The tension is A and C does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,

[tex]\Delta T_B=39\times0.12\\\Delta T_B=4.68\rm \;N[/tex]

The tension is A and B does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,

[tex]\Delta T_C=39\times0.12\\\Delta T_C=4.68\rm \;N[/tex]

Hence, the tension in each of the coupling bars A, B, and C of the luggage carrier changes as,

Learn more about the tension here;

https://brainly.com/question/25743940

the simple answer is from 61kmph to 120kmph

Explanation:

no explanation is needed

Answer:

(a) Iron > plastic > ice

(b) Same on all

(c) Iron downwards, plastic net force zero, ice upwards.

(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.

Explanation:

Three spheres have same volume , plastic, ice and iron.

(a) The weight is given by

Weight = mass x gravity = volume x density x gravity

As the density of iron is maximum and the density of ice is least so the order of the weight is

Weight of iron > weight of plastic > weight of ice

(b) Buoyant force is given by

Buoyant force = Volume immersed x density of fluid x g

As they have same volume, density of fluid is same so the buoyant force is same on all the spheres.

(c) Net force is

F = weight - buoyant force  

So, the net force on the iron sphere is downwards

On plastic sphere is zero as the density of plastic sphere is same as water. On ice sphere it is upwards.

(d) Iron sphere sinks, plastic sphere is in equilibrium and ice sphere will floats.  

Hi there!

[tex]\large\boxed{\approx 35.29 C}[/tex]

Use the following formula:

E = F / C, where:

E = electric field (N/C)

F = force (N)

C = Charge (C)

Thus:

6.8 × 10⁻⁵ = 2.4 × 10⁻³ / C

Isolate for C:

C = 2.4 × 10⁻³  / 6.8 × 10⁻⁵

Solve:

≈ 35.29 C