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gogolik [260]
3 years ago
13

A 5,400 W motor is used to do work. If the motor is used for 640 s, about how much work could it do?

Physics
1 answer:
Gemiola [76]3 years ago
4 0

Answer:

3,500,000 J​

Explanation:

WORK = POWER * TIME

WORK= 5400 * 640

=6456000 J = 3,500,000 J​

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A zone reconnaissance involves a directed effort to obtain detailed information on all routes, obstacles, terrain, enemy forces,
son4ous [18]

Answer:

It is True

Explanation:.

A  commander assigns a zone reconnaissance mission when he seeks additional information on a zone before committing other forces in the zone. It is appropriate when the enemy situation is vague,  existing knowledge of the terrain is limited, or combat operations have altered the terrain. A zone  reconnaissance could include several route or area reconnaissance missions assigned to subordinate units.

7 0
2 years ago
A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f
VashaNatasha [74]

Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

(L^{2} - y^{2} ) . w = L^{2} . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

\frac{L^{2} - y^{2}  }{L^{2} } = \frac{5}{9}

Taking L^{2} common and solving for \frac{y}{L}, we will get

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

8 0
3 years ago
Write a question about how changing temperature affects gas
Lera25 [3.4K]

Answer:

"How does the volume of a gas kept at constant pressure change as its temperature is increased?"

Explanation:

One possible question can be:

"How does the volume of a gas kept at constant pressure change as its temperature is increased?"

The answer to this question is contained in Charle's law, which states that for a gas at constant pressure, the volume of the gas is proportional to its absolute temperature:

V\propto T

Or also written as

\frac{V}{T}=const.

By looking at this equation, we can find immediately the answer to our question: as the (absolute) temperature of the gas increases, the volume increases as well, by the same proportion.

3 0
3 years ago
Blind spots of large vehicles are called
AlladinOne [14]
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6 0
3 years ago
Read 2 more answers
At 7.0°c, the volume of a gas is 49 ml. at the same pressure, its volume is 74 ml at what temperature?
Rashid [163]
So i converted everything first; 
<span>7.0 C ---> 280 K </span>
<span>49 mL---> 0.049 L </span>
<span>74mL---> 0.074 L </span>
<span>THEN I tried setting it up by the combined law formula which is P1V1/T1=P2V2/T2 </span>
8 0
3 years ago
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