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ddd [48]
3 years ago
9

A force of 1,500 N is moved through 0.10 m in order to compress a steel cylinder. The cylinder was compressed 0.010 m by a force

of 13,000 N. The efficiency of the compression machine is what % ?
115
0.87
87
8.7
Physics
1 answer:
ValentinkaMS [17]3 years ago
4 0
We first calculate the work effectively done by the force when it compressed the steel cylinder. This is solved by multiplying the distance that the cylinder was compressed by and its causing force.
0.010 m * 13,000 N = 130 N-m

Then, we solve for the work done by the 1,500 N force:
0.10 m * 1,500 N = 150 N-m

We then take the ratio of the efficient:experimental
130 N-m / 150 N-m = 0.87 = 87%

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13. An object, travelling along a straight path, covers 35 m distance in 4 seconds. In next 6
WARRIOR [948]

Answer:

(4) 8.5 m/s

Explanation:

You add both the meters together and both the seconds together and then divide them both.

5 0
3 years ago
Read 2 more answers
Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

3 0
3 years ago
A 75kg man climbs the stairs to the fifth floor of a building. A total hieght of 16m. His potential energy has increased by
yKpoI14uk [10]

Answer:

11760 joules

Explanation:

Given

Mass (m) = 75kg

Height (h) = 16m

Required

Determine the increment in potential energy (PE)

This is calculated as thus:

PE = mgh

Where g = 9.8m/s²

Substitute values for m, g and h.

P.E = 75 * 9.8 * 16

P.E = 11760 joules

8 0
3 years ago
Suppose that instead of being inclined to Earth's orbit around the Sun, the Moonâs orbit was in the same plane as Earthâs orbit
ycow [4]

Answer:

c) 12

Explanation:

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If the Moon's orbit was in the same plane as that of the Earth's orbit. Every new Moon, there would be a Solar Eclipse. The Lunar cycle is of 29.5 Days which means there will be one new Moon every month. So there will be 12 Solar Eclipses every year.

Currently, the orbit of the Moon is tilted at an angle of 5° thus we don't see that many Solar eclipses. Maximum of 5 solar eclipses can occur in an year.

6 0
3 years ago
Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour
Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = \frac{t}{\sqrt{1-\frac{v^2}{c^2} } }

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 10

t₁= 16.7  years

D ) Given t = 10 years , t₁ = ? v = .4c

\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 10

t₁= 10.9  years

E ) Given t = 20 years , t₁ = ? v = .8c

\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 20

t₁= 33.4   years

7 0
2 years ago
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