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3 years ago

14

When a machine is used to perform a task, work output is always more than work input. please select the best answer from the cho

ices provided t f?

1 answer:

Maurinko [17]3 years ago

5 0

Properly input current above 40 voltage or 100 for example elecrical fan machine is used to perform a task, work output is always more than work

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A 0.350kg bead slides on a curved fritionless wire,

LuckyWell [14K]

Answer:

h2 = 0.092m

Explanation:

From a balance of energy from point A to point B, we get speed before the collision:

$m1*g*h-\frac{m1*V_B^2}{2}=0$ Solving for Vb:

$V_B=\sqrt{2gh}=6.56658m/s$

Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:

$V_{B'}=V_B*\frac{m1-m2}{m1+m2} = \sqrt{2gh}* \frac{m1-m2}{m1+m2}=-1.34316m/s$

Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:

$m1*g*h2-\frac{m1*V_{B'}^2}{2}=0$ Solving for h2:

h2 = 0.092m

6 0

2 years ago

"Influence" between bodies is reduced as the bodies are

azamat

Both particles and big objects will be influenced by each other less if they are moved further apart.

4 0

3 years ago

Read 2 more answers

A.Substitute in the units for each one and combine like terms.

storchak [24]

IDK ghjfnhgfjmrmhjgfhgfmmfh

8 0

2 years ago

Two cylindrical resistors are made of the same material and have the same resistance. The resistors, R1 and R2, have different r

Paladinen [302]

Answer:

Option d is correct.

Explanation:

We know , resistance of a body is directly proportional to its length and inversely proportional to its area.

$R=\dfrac{\rho\ L}{A}=\dfrac{\rho\ L}{\pi r^2}$ ( Here, $\rho$ is constant dependent on object material )

Writing $R_1 \ and\ R_2$ also :

$R_1=\dfrac{\rho\ L_1}{\pi r_1^2}\ , \ R_2=\dfrac{\rho\ L_2}{\pi r_2^2}$ ( since they are of same material therefore, $\rho$ is same.)

Now , if $r_2=2r_1 \ and \ 4L_1=L_2$ .

Then $R_1=R_2$

Therefore, option d. is correct.

Hence, this is the required solution.

7 0

3 years ago

A magnesium oxide component must not fail when a tensile stress of 10.5 MPa is applied. Determine the maximum allowable surface

Aloiza [94]

Answer:

Maximum permitted surface crack length is 1.29 mm

Explanation:

As per the question:

Tensile stress, $\sigma = 10.5\ MPa = 10.5\times 10^{6}\ Pa$

Surface energy of magnesium oxide, SE = $1.0\ J/m^{2}$

Modulus of elasticity of the material, E = 225 GPa = $225\times 10^{9}\ Pa$

Now,

To calculate the maximum allowable surface crack length:

$L = \frac{2E\times SE}{\sigma^{2}\pi }$

$L = \frac{2\times 225\times 10^{9}\times 1.0}{10.5\times 10^{6}\times \pi } = 1.29\times 10^{- 3}\ m = 1.29\ mm$

7 0

3 years ago