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Serga [27]
3 years ago
15

a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w

ill be the effort distance ?​
Physics
1 answer:
nataly862011 [7]3 years ago
7 0

Answer:

40 cm

Explanation:

We are given that

Load=800 N

Effort=200 N

Load  distance=10 cm

We have to find the effort distance.

We know that

load\times load\;distance=Effort\times effort\;distance

Using the formula

800\times 10=200\times effort\;distance

Effort distance=\frac{800\times 10}{200}

Effort distance=\frac{8000}{200}

Effort distance=40 cm

Hence,  the effort distance will be 40 cm.

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Bas_tet [7]

Answer:

Exothermic reaction

Explanation:

  • A reaction that releases heat is known as an exothermic reaction.
  • Reactions can be classified as either exothermic or endothermic depending on whether they release or absorb heat from the surrounding.
  • A reaction that absorbs or gains heat from the surrounding is known as endothermic reaction.
  • On the other hand, if a reaction releases heat to the surrounding is known as exothermic reaction.
  • An exothermic reactions results to an increase in temperature of a solution while an endothermic reaction results to a decrease in the temperature of a solution.
4 0
3 years ago
Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres
inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )

Here,  v_{1} is the velocity of water through wide ends of cylindrical pipe and v_{2} is the velocity of water through narrow ends of cylindrical pipe.

Given, v_{1} =1.4 m/s

Now from equation continuity,

v_{1} A_{1} = v_{2} A_{2}.

Here, A_{1} and A_{2} are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}.

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}

v_{2} = 4 \times 1.4 = 5.6 m/s

Substituting these values  with the density of water is 1000 \ kg/m^3 in pressure difference formula we get.

\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa

3 0
3 years ago
Read 2 more answers
An ammeter with a resistance of 5.0 ohm is connected in series with a 3.0V cell and a lamp rated at 300 mA, 3V. Calculate the cu
Alex

Answer:

I = 0.2 A

Explanation:

Lamp is rated at 300 mA

I_lamp = 0.3 A

Voltage is; V = 3V

Thus; Resistance is given by;

R = V/I

R = 3/0.3

R = 10 ohms

Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;

R_eq = 10 + 5

R_eq = 15 ohms

Ammeter current will be;

I = V/R_eq

I = 3/15

I = 0.2 A

3 0
3 years ago
Helpp me pleassee....
lilavasa [31]

Answer:

The fundamental wavelength of the vibrating string is 1.7 m.

Explanation:

We have,

Velocity of wave on a guitar string is 344 m/s

Length of the guitar string is 85 cm or 0.85 m

It is required to find the fundamental wavelength of the vibrating string. The fundamental frequency on the string is given by :

f=\dfrac{v}{2l}\\\\f=\dfrac{344}{2\times 0.85}\\\\f=202.35\ Hz

Now fundamental wavelength is :

\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{344}{202.35}\\\\\lambda=1.7\ m

So, the fundamental wavelength of the vibrating string is 1.7 m.

4 0
3 years ago
What is the kinetic energy of a baseball with a mass of 0.133 kg and moving at a speed of 28.7m/s
blsea [12.9K]

Answer:

54.7 J

Explanation:

kinetic energy formula

3 0
3 years ago
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