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Taya2010 [7]
3 years ago
14

What type of fire can be put out safely with water?

Physics
1 answer:
nignag [31]3 years ago
3 0
Flame of fire could get put out with water
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A bar 4.0m long weights 400 N. Its center of gravity is 1.5m from on end. A weight of 300N is attached at the light end. What ar
Debora [2.8K]

Answer:

The resulting, needed force for equilibrium is a reaction from a support, located at 2.57 meters from the heavy end. It is vertical, possitive (upwards) and 700 N.

Explanation:

This is a horizontal bar.

For transitional equilibrium, we just need a force opposed to its weight, thus vertical and possitive (ascendent). Its magnitude is the sum of the two weights, 400+300 = 700 N, since weight, as gravity is vertical and negative.

Now, the tricky part is the point of application, which involves rotational equilibrium. But this is quite simple if we write down an equation for dynamic momentum with respect to the heavy end (not the light end where the additional weight is placed). The condition is that the sum of momenta with respect to this (any) point of the solid bar is zero:

0=\Sigma_{i}M=400\cdot1.5+300\cdot4-d\cdot700

Where momenta from weights are possitive and the opposed force creates an oppossed momentum, then a negative term. Solving our unknown d:

d=\frac{400\cdot1.5+300\cdot4}{700} =2.57 m

So, the resulting force is a reaction from a support, located at 2.57 meters from the heavy end (the one opposed to the added weight end).

8 0
4 years ago
At a distance of 0.220 cm from the axis of a very long charged conducting cylinder with radius 0.100cm, the electric field is 49
Scorpion4ik [409]

Answer:

At the distance of 0.220cm from the axis.

r = 0.220cm = 0.0022m, E = 490N/C, e0 = 8.854 x 10^-12F/m

Linear charge density = 2*π*e0*r*E = 2 x 3.142 x 8.854x10^-12 x 0.0022 x 490 = 5.998 x 10^-11C/m

Thus, To Calculate the Electric field at the distance r = 0.616cm from the cylinder axis, we substitute the calculated linear change density in the equation

E = (linear charge density)/2*π*e0*r

Here, r = 0.616cm = 0.00616m

E = [(5.998 x 10^-11)/(2 x 3.142 x 8.854 x 10-12 x 0.00616)]

E = 175N/C

Explanation:

The Electric field of a charged conducting cylinder obey the Gauss Law.

Therefore, the Electric field is given as:

E = (linear charge density)/4πe0r,

Where e0 is the permittivity of free space with constant value of 8.854 x 10^-12F/m, r is the radial distance from the axis.

3 0
3 years ago
how will you connect three resistors of 2 ohm centrioles and 5 ohms respectively so as obtain of the result and the resistance o
victus00 [196]

Two resistor of 2Ω in series parallel to resistor 5Ω in series to a 2Ω resistor. This configuration gives to us an equivalent resistor of 2.55Ω.

To solve this problem we have to use the rules of conection of resistor in series and parallel.

A resistor R1 in serie with other resistor R2 gives us an equivalent resistor Req= R1 + R2.

A resistor R1 in parallel with other resistor R2 gives us an equivalent resistor Req = R1.R2/R1+R2.

The circuit that show an arregement of resistor which we obtain a equivalent resistor of 2.5Ω from three resistor of 2Ω and 5Ω respectively is attached in the image:

3 0
4 years ago
A stuntman drives a car with a mass of 1500 kg on a drawbridge. The car accelerates with a constant force of 10,000 N. While he
Viktor [21]

Answer:

1.77 m/s^2

Explanation:

There are two forces acting on the car along the direction parallel to the incline:

- The driving force of 10,000 N, which pushes forward

- The component of the weigth of the car parallel to the incline, which pulls backward

The component of the weight of the car parallel to the incline is:

W_p = mg sin \theta=(1500 kg)(9.8 m/s^2)( sin 30^{\circ})=7350 N

So now we can apply Newton's second law to find the acceleration of the car:

F-W_p = ma\\a=\frac{F-W_p}{m}=\frac{10000 N-7350 N}{1500 kg}=1.77 m/s^2

7 0
3 years ago
Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
sergeinik [125]

Planet Y has rotated by 135.5° through during this time.

To find the answer, we need to know about the relation between angle and radius of orbit.

<h3>What's the expression of angle in terms of radius?</h3>
  • Angle= arc/radius
  • As arc = orbital velocity × time,

            angle= (orbital velocity × time)/radius

  • Orbital velocity= √(GM/radius), G= gravitational constant and M = mass of sun
  • So, angle = (√(GM)× time)/radius^3/2
<h3>What's is the angle rotated by planet Y after 5 years, if ratio of the radius of orbit of planet X and Y is 4:3 and planet X is rotated by 88°?</h3>
  • Let Ф₁= angle rotated by planet Y, Ф₂= angle rotated by planet X
  • As time = 5 years ( a constant)
  • Ф₁/Ф₂= (radius of planet X / radius of planet Y)^(3/2)
  • Ф₁= (radius of planet X / radius of planet Y)^(3/2) × Ф₂

   = (4/3)^(3/2) × 88°

   = 135.5°

Thus, we can conclude that Planet Y has rotated by 135.5° through during this time.

Learn more about the orbital velocity here:

brainly.com/question/22247460

#SPJ1

8 0
2 years ago
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