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3 years ago

Please explain the method of making electricity from non renewable sources like coal?

1 answer:

7 0

The coal is urned to heat up water. this produces steam. the steam turns a turbine that turns a generator which provided energy yhat can be transferred into electrisity

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a 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they

Otrada [13]

According to the given statement Final velocity when they stick together is 8.735i^ + 11.25j^

<h3>What is collision and momentum?</h3>

The unit of momentum is kg ms -1. Momentum is a vector parameter that is influenced by the object's direction. During collisions involving objects, momentum is a relevant concept. The final velocity before a collision between two objects equals the total motion after the impact (in the absence of external forces).

<h3>Briefing:</h3>

From conservation of momentum

Initial momentum = final momentum

m u +M U =(m+M) V

2000×25 i^ +1500×30 j^ =(2000+1500) V

V = 8.735i^ + 11.25j^

Final velocity when they stick together is 8.735i^ + 11.25j^

To know more about Collide visit:

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The complete question is -

A 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they collide at the intersection. Find the direction and magnitude of velocity of the wreckage after the collision, assuming the vehicles stick together after the collision.

6 0

1 year ago

A hockey puck with a mass of 0.175 kg slides over the ice. The puck initially slides with a speed of 5.25 m/s, but it comes to a

Neko [114]

Answer:

1.70 J

Explanation:

The heat dissipated is the difference in the kinetic energies.

This is given by

$E = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

$v_i$ and $v_f$ are the initial and final velocities.

With <em>m</em> = 0.175 kg,

$E = \frac{1}{2}\times0.175(2.85^2 - 5.25^2) = -1.701\text{ J}$

The negative sign appears because energy is lost.

7 0

3 years ago

At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v

Anna007 [38]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. For this purpose we will define the speed as the distance traveled in a given period of time. Here the distance is equivalent to the orbit traveled around the earth, that is, a circle. Approaching the height of the aircraft with the radius of the earth, we will have the following data,

$R= 6370*10^3 m$

$v = 219m/s$

$a = 17m/s^2$

The circumference of the earth would be

$\phi = 2\pi R$

Velocity is defined as,

$v = \frac{x}{t}$

$t = \frac{x}{v}$

Here $x = \phi$ , then

$t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{219}$

$t = 1.82*10^5s$

Therefore will take $1.82*10^5$ s or 506 hours, 19 minutes, 17 seconds

3 0

2 years ago

use the instantaneous center of zero velocity to determine the angular velocity !ab of link ab and the velocity vb of collar b f

allsm [11]

Instantaneous center:

It is the center about a body moves in planer motion. The velocity of Instantaneous center is zero and Instantaneous center can be lie out side or inside the body. About this center every particle of a body rotates.

From the diagram

Where these two lines will cut then it will the I-Center.Point A and B is moving perpendicular to the point I.

If we take three link link1,link2 and link3 then I center of these three link will be in one straight line It means that they will be co-linear.

Therefore, when the mass is at its equilibrium position (which corresponds to x=0), the velocity of the mass will be maximum.

To know more about velocity, refer: brainly.com/question/12413963

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7 0

1 year ago

A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate

exis [7]

Answer:

The velocity of water at the bottom, $v_{b} = 28.63 m/s$

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, $P_{gauge} = 2.90 atm$

Solution:

Now,

Atmospheric pressue, $P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa$

At the top, the absolute pressure, $P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa$

Now, the pressure at the bottom will be equal to the atmopheric pressure, $P_{b} = 1 atm = 1.01\times 10^{5} Pa$

The velocity at the top, $v_{top} = 0 m/s$ , l;et the bottom velocity, be $v_{b}$ .

Now, by Bernoulli's eqn:

$P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}$

where

$h_{t} - h_{b} = 12.8 m$

Density of sea water, $\rho = 1030 kg/m^{3}$

$\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} = v_{b}$

$\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} = v_{b}$

$v_{b} = 28.63 m/s$

5 0

3 years ago