A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holding it up at 43 cm from the end of the plank with force F2. If the plank has a mass of 13.7 kg and its center of gravity is at the middle of the plank, what is the force F1

Answer: b force

Explanation:

yes because the world comin g up with more technique

Answer:

E = 16581.6 J

Explanation:

Given that,

Current, I = 4.9 A

Time for which the current is set up, I = 4.7 min = 282 s

The voltage of the battery, V = 12 V

We need to find how much chemical energy of the battery reduced. Let It is E. We know that,

E = P t

Where

P is power of battery, P = VI

So,

[tex]E=VIt[/tex]

Put all the values,

[tex]E=12\times 4.9\times 282\\E=16581.6\ J[/tex]

So, 16581.6 J of chemical energy of the battery is reduced.

Answer:

protons+neutrons= mass number, so if the mass number is 13 and protons are 6 its 13-6=7 neutrons

Explanation:

mass number: the sum of the number of protons and

neutrons in an atom

this is key as it explains that protons+neutrons= mass number, so if the mass number is 13 and protons are 6 its 13-6=7 neutrons

Answer:

22.87 m/s

Explanation:

Since the value of the height at which the divers jump off the cliff is not given,

Assuming that the height of the cliff above the ocean at which the divers jump = 26.7 m

Then;

The speed of the divers when they hit the ocean water can be determined by using the formula:

[tex]v^2 = 2gh \\ \\ v= \sqrt{2gh} \\ \\ v = \sqrt{2 \times 9.8 \times (26.7)} \\ \\ v \simeq 22.87 \ m /s[/tex]

Answer:

I dont. understand the question, maybe insert the picture?

Force is mass into acceleration

and pressure is force applied per unit area.

Answer:

The correct answer is "[tex]0.5\times 10^{-6} \ m[/tex]".

Explanation:

Given:

[tex]\frac{\lambda D}{d} =2.8\times 10^{-3}[/tex]

[tex]d = 0.5\times 10^{-3}[/tex]

[tex]D = 2.80[/tex]

Now,

The wavelength will be:

⇒ [tex]\lambda = 2.8\times 10^{-3}\times \frac{d}{D}[/tex]

By putting the values, we get

⇒    [tex]=\frac{2.8\times 10^{-3}\times 0.5\times 10^{-3}}{2.8}[/tex]

⇒    [tex]=\frac{1.4\times 10^{-6}}{2.8}[/tex]

⇒    [tex]=0.5\times 10^{-6} \ m[/tex]  

Answer:

F = 312 N

Explanation:

Given that,

The mass of a crate, m = 40 kg

Acceleration of the crate, a = 2 m/s²

As the carte is falling downward, the net force exerted by the rope on the carte is given by :

F = m(g-a)

Put all the values,

F = 40(9.8-2)

F = 312 N

Hence, the required force exerted by the rope on the crate is equal to 312 N.

By Newton's second law, the net force on the object is

∑ F = m a

∑ F = (2.00 kg) (8 i + 6 j ) m/s^2 = (16.0 i + 12.0 j ) N

Let f be the unknown force. Then

∑ F = (30.0 i + 16 j ) N + (-12.0 i + 8.0 j ) N + f

=>   f = (-2.0 i - 12.0 j ) N

Given :

Sally travels by car from one city to another. She drives for 30 min at 80 km/hr, 12.0 min at 105 km/hr, and 45 min at 40.0 km/hr.

To Find :

The total distance traveled.

Solution :

Total distance traveled is given as sum :

Distance = sum of  product of speed and time

D = (v₁ × t₁) + (v₂ × t₂) + (v₃ × t₃)

Putting all given value in this we get :

[tex]D = 80\times \dfrac{30}{60} + 105 \times \dfrac{12}{60} + 40\times \dfrac{45}{60}\\\\D = 91\ km[/tex]

Hence, this is the required solution.

Answer:

5.619×10⁶ N

Explanation:

Applying,

F = kqq'/r²................... Equation 1

Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges

From the questiion,

Given: q = 2.5 C, q' = 2.5 C, r = 100 m

Constant: 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.5×2.5×8.99×10⁹)/100²

F = 56.19×10⁵

F = 5.619×10⁶ N

Answer:

The tension in the horizontal string is 25.2 N.

Explanation:

mass, m = 1.2 kg

Angle, A = 65 degree

The tension in the string is T.

So, T cos A = m g

T x cos 65 =  1.2 x 9.8

T x 0.4226 = 11.76

T = 27.83 N

Tension in the horizontal string, T' = T sin A = 27.83 x sin 65 = 25.22 N

Answer:

Explanation:

See the figure attached . T₁ and T₂ are tension in the inclined and horizontal string .

The vertical component of T₁ will balance weight and horizontal component will balance the tension T₂.

T₁ sin 65 = mg and T₁ cos 65 = T₂

Dividing ,

Tan 65 = mg / T₂

T₂ = mg / tan 65

= 1.2 x 9.8 / 2.1445

= 5.5 N

Solution :

The conditions for the maximum in the Young's experiment is :

d sin θ = m λ,     where m = 0, 1, 2, 3, .....

The angle between the central maximum and the 1st order maximum can be determined by setting the m = 1. So,

d sin θ =  λ

[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{d}\right)$[/tex]

Given : d = 100 λ

[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{100 \lambda}\right)$[/tex]

[tex]$\theta = \sin^{-1}\left(\frac{1}{100}\right)$[/tex]

 [tex]$=0.573^\circ$[/tex]

  = 0.01 rad

Answer:

The resulting force on the first object is 800 N.

Explanation:

Given;

force exerted on one of the objects, F₁ = 400 N

let the first charge = q₁

let the second charge = q₂

The force of repulsion between the objects is calculated using Coulomb's law;

[tex]F =\frac{kq_1q_2}{r^2} \\\\\frac{k}{r^2} = \frac{F}{q_1q_2} = \frac{F_1}{q_1\times2q_2} \\\\F_1 = \frac{F(q_1\times2q_2)}{q_1q_2} \\\\F_1 = 2F\\\\F_1 = 2(400 \ N)\\\\F_1 = 800 \ N[/tex]

Therefore, the resulting force on the first object is 800 N.

Answer: See explanation

Explanation:

The evolutionary stages for the formation of planets from earliest to latest will be:

1. Dust keeps matter inside the disk cool enough for planet formation to start

2. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter.

3. Small clumps of matter stick together via the process of accretion to form planetesimals a few hundred kilometers in diameter.

4. Planetesimals begin to accrete, forming protoplanets.

5. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star

Answer:

Option A (69.56 newtons) is the appropriate solution.

Explanation:

According to the question,

On the X-axis,

⇒ [tex]T_1Cos30^{\circ}-T_2Cos60^{\circ}=0[/tex]

or,

    [tex]T_1Cos 30^{\circ}=T_2Cos60^{\circ}[/tex]

On substituting the values, we get

      [tex]T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}[/tex]

      [tex]T_1\times \sqrt{3} =T_2[/tex]....(equation 1)

On the Y-axis,

⇒ [tex]T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N[/tex]

                        [tex]\frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N[/tex]

                    [tex]T_1+\sqrt{3}T_2=139.2\times 2[/tex]

From equation 1, we get

           [tex]T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N[/tex]

                        [tex]T_1+3T_1=278.4 \ N[/tex]

                                [tex]4T_1=278.4 \ N[/tex]

                                  [tex]T_1=\frac{278.4}{4}[/tex]

                                       [tex]=69.6 \ N[/tex]  

Answer:

69.58

Explanation:

Answer:

the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric

Explanation:

The pearls are suspended in a solution, when connecting the power source, it is subjected to an electric shock, the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric

            F = q E

Answer:

a = -0.4m/s²

Explanation:

v_f = v_I + (a)(t)

a(t) = v_f-v_I

a = (v_f-v_I)/t

a = (0m/s-12m/s)/30s

a = -0.4m/s²

Complete Question

The accepted speed of sound at atmospheric pressure and 0 *C is 331.5 m/s. The speed of sound increases 0.607 m/s for every *C. Calculate the speed of sound at the temperature of your room(70F) and compare your measured value to the accepted value.

Answer:

[tex]V_{Tc}=344.314m/s[/tex]

Explanation:

From the question we are told that:

Speed of sound at Temperature [tex]0 \textdegree[/tex] [tex]V_0=331.5m/s[/tex]

Rate of Speed increase [tex]\triangle V_{infty}=0.607[/tex]

Generally the equation for Temperature in Celsius is mathematically given by

 [tex]Tc=\frac{100}{180}(T_f-32)[/tex]

 [tex]Tc=0.56*38[/tex]

 [tex]Tc=21.11 textdegree C[/tex]

Therefore speed at Tc

 [tex]V_{Tc}=V_0+(Tc)( V_{infty})[/tex]

 [tex]V_{Tc}=331.5+(21.11)(0.607)[/tex]

 [tex]V_{Tc}=344.314m/s[/tex]

Answer:

spirit animal?

Explanation:

Answer:

time perion=1/frequency

so here time period =1/262

let wavelength be x

speed of sound=wave length/time period

342=x/1/262

342=262x

342/262=x

1.30=x

therefore wavelength=1.30 metre

Explanation:

Answer:

Explanation:

From the given information:

Let's assume that the missing function is:

s(t) = t³ - 6t², t ≥ 0

From part (b), we are to find the given  required terms when time t = 2

So; from the function s(t) =  t³ - 6t², t ≥ 0

[tex]velocity \ v(t) \ = \dfrac{d}{dt}s(t)[/tex]

[tex]velocity \ v(t) \ = \dfrac{d}{dt}(t^3 - 6t^2)[/tex]

[tex]velocity \ v(t) \ = 3t^2 - 12t[/tex]

[tex]acceleration a(t) = \dfrac{d}{dt}*v(t)[/tex]

[tex]acceleration a(t) = \dfrac{d}{dt}(3t^2 - 12 t)[/tex]

[tex]acceleration\ a(t) = 6t - 12[/tex]

At time t = 2

The position; S(2) = (2)² - 6(2)²

S(2) = 8 - 6(4)

S(2) = 8 - 24

S(2) = - 16 ft

v(2) = 3(2)² - 12 (2)

v(2) = 3(4) - 24

v(2) = 12 - 24

v(2) = - 12 ft/s

speed = |v(2)|

|v(2)|  = |(-12)|

|v(2)| = 12 ft/s

acceleration = 6t - 12

acceleration = 6(2) - 12

acceleration =  12 - 12

acceleration =  0 ft/s²

Answer:

The coefficient of kinetic friction on the floor is 0.138

Explanation:

Given;

mass of the crate, m = 450 kg

force applied by the first worker, F₁ = 380 N

force applied by the second worker in the same direction as the first worker, F₁ = 230 N

frictional force opposing the motion of the box = -[tex]F_k[/tex]

Apply Newton's second law of motion;

∑F = ma

[tex]F_1 + F_2 - F_k = ma[/tex]

If the crate slides with constant speed, acceleration is zero (0).

[tex]F_1 + F_2 - F_k = ma = 0\\\\F_1 + F_2 - F_k = 0\\\\F_k = F_1 + F_2\\\\\mu _kmg= F_1 + F_2\\\\\mu _k = \frac{F_1 + F_2}{mg} \\\\\mu _k = \frac{380 + 230}{450 \times 9.8} \\\\\mu _k = 0.138[/tex]

Therefore, the coefficient of kinetic friction on the floor is 0.138

Answer:

81.47°

Explanation:

The forward force on the tire is the horizontal component of its weight along the incline mgcosФ. The frictional force along the incline is μmgsinΦ where mgsinΦ is the vertical component of the tire's weight along the incline.

Since the tire rolls without slipping, then

mgcosФ = μmgsinΦ where μ = coefficient of kinetic friction = 0.15 (since the tire is in motion) and Φ is the angle of the incline.

So, mgcosФ = μmgsinΦ

mgcosФ/mgsinΦ = μ

μ = cosФ/sinΦ

μ = 1/tanФ

tanФ = 1/μ

Φ = tan⁻¹(1/μ)

substituting into the equation, we have

Φ = tan⁻¹(1/0.15)

Φ = tan⁻¹(6.667)

Φ = 81.47°

Answer:

a) Ink X is likely to be pure because it only contain 1 spot.

b) The chromatography tell us about ink Y that it is a mixture as it contain more than 1 spot.

c) The three different spots are separated out from ink Y at different heights beacaus different substance have different solubility.

The different spots from Y are found at various heights because they represent different compounds.

The term chromatography has to do with a method of separating the component of a substance. The term chromatography originally means color writing.

We can see that the pure ink is the ink marked X. We can see from the chromatogram that Y is a mixture of colors. The different spots from Y are found at various heights because they represent different compounds.

Learn more about chromatography:https://brainly.com/question/26491567?

#SPJ6

Answer:

 F_net = 9.87 10⁻⁴ N

Explanation:

Let's use that force is a vector magnitude

         ∑ F = F₁₃ + F₂₃

De bold arfe vectros. The force is the electric force, we use that charges of the same sign repel and when the charges are of a different sign they attract

the charges q1 and q2 are negative and the charge q3 is positive with the positions y1 = 38 m, y2 = -7m, y3 = 16 m

       ∑ F = F₁₃ - F₂₃

       F_net = [tex]k \frac{q_1q_3}{r_{13}^2 } - k \frac{q_2q_3}{r_{23}^2 }[/tex]

in this case q₁ = q₂ = q

       F_net = k q q₃  (  )

      let's look for the distance

      r₂₃ = y₂ - y₃

      r₂₃ = -7 -16

      r₂₃ = - 23 m

      r₁₃ = 38 - 16

      r₁₃ = 22 m

let's calculate

      F_net = 9 10⁹ 24 26 10⁻¹² ( )

      F_net = 5.616 ( 1.758 10⁻⁴ )

      F_net = 9.87 10⁻⁴ N