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Liula [17]
3 years ago
10

The plank is 5.1 meters long and has a mass of 9.2 kg, ack has a mass of 93.6 kg, and Diane has a mass of 64.2 kg. Assuming Dian

e sits at one end of the plank, how far from his end of the plank should Jack sit for the two of them to be perfectly balanced?

Physics
1 answer:
kodGreya [7K]3 years ago
8 0

Answer:

The answer is 0.8 m

Explanation:

According the attached diagram, if we take the momentum in the point 0:

\frac{m_{D}*5.1 }{2} =m_{J} *X

Where

mD = mass of Diane = 64.2 kg

mJ = mass of Jack = 93.6 kg

Replacing:

X=\frac{\frac{64.2*5.1}{2} }{93.6} =1.75m

\frac{L}{2} -X=Y=\frac{5.1}{2} -1.75\\Y=0.8m

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A capacitor with a very large capacitance is in series with a capacitor that has a very small capacitance. what can we say about
Misha Larkins [42]
<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.

The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)

The capacitance of a series combination is

             1 / (1/A + 1/B + 1/C + 1/D + .....) .

If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is   

             (product of the 2 individuals) / (sum of the individuals)  .

In this problem, we have a humongous one and a tiny one.
Let's call them  1000  and  1 .
Then the series combination is

           (1000 x 1) / (1000 + 1)

        =       (1000) / (1001)

        =         0.999 000 999 . . . 

which is smaller than the smaller individual.

It'll always be that way.   </span>
5 0
3 years ago
The plane of a rectangular coil, 3.7 cm by 3.7 cm, is perpendicular to the direction of a uniform magnetic field B. If the coil
gizmo_the_mogwai [7]

Answer:3.77 T/s

Explanation:

Given

Area of cross-section=3.7\times 3.7 cm^2=13.69 cm^2

N=no of turns=61

Resistance=R=8 \Omega

current (i)=0.04 A

emf induced=i\times R=0.04\times 8=0.32 V

emf induced e=NA\frac{\mathrm{d} B}{\mathrm{d} t}

0.32=62\times 13.69\times 10^{-4}\times \frac{\mathrm{d} B}{\mathrm{d} t}

\frac{\mathrm{d} B}{\mathrm{d} t}=\frac{0.32\times 10^4}{848.78}=3.77 T/s

8 0
3 years ago
A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro
grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

         Φ._{E} = ∫ E. dA = q_{int} / ε₀

For this case we create a Gaussian surface that is a sphere.  We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

        ∫ E dA = q_{int} / ε₀

The area of ​​a sphere is

     A = 4π r²

   

    E 4π r² =q_{int} / ε₀

    E = (1 /4πε₀ )  q / r²

Having the solution of the problem let's analyze the points:

A   ) r = 3R / 4  = 0.75 R.

  In this case there is no charge inside the Gaussian surface therefore the electric field is zero

        E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

    E = (1 /4πε₀ )  Q / (1.25R)²

    E = (1 /4πε₀ )  Q / R2 1 / 1.56²

    E₀ = (1 /4π ε₀ )  Q / R²

   E_{B} =  Eo /1.56 ²

  E_{B}  = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

    E_{B} =(1 /4π ε₀ ) Q    1/(2R)²

    E_{B} = (1 /4π ε₀ ) q/R²   1/4

    E_{B} = Eo  1/4

    E_{B} = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0
3 years ago
What is the centripetal force that would be required to keep a 4.0 kg mass moving in a horizontal circle with a radius of 0.80 m
KIM [24]

Answer:

D. 1.8 × 102 newtons radially inward

Explanation:

The magnitude of the centripetal force is given by:

F=m\frac{v^2}{r}

where

m is the mass of the object

v is the tangential speed

r is the radius of the circular trajector

In this problem, we have m = 4.0 kg, v = 6.0 m/s and r = 0.80 m, therefore substituting into the equation we get

F=(4.0 kg)\frac{(6.0 m/s)^2}{0.80 m}=180 N

The centripetal force is the force that keeps the object in a circular trajectory, so it is a force that is always directed inward (towards the centre of the circular path) and radially. Therefore, the correct answer is

D. 1.8 × 102 newtons radially inward

4 0
4 years ago
Read 2 more answers
A machine accelerates a 5 kg missile from rest to a speed of 5 km/s. The net force accelerating the missile 500,000 N. How long
Gnoma [55]
Final velocity = 5km/sec = 5000 m/sec
Initial velocity = 0, mass = 5kg and force = 500,000 N
F = m(Vf-Vi)/time
Time = m(Vf-Vi)/force = 5*5000/500000 = 0.05 seconds
So, it takes 0.05 sec to arrive at the speed of 5km/s.
3 0
3 years ago
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