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3 years ago

10 PTS! HELP!

1 answer:

6 0

Shortsightedness is corrected using a concave (curved inwards) lens which is placed in front of a myopic eye, moving the image back to the retina and making it clearer. Longsightedness is corrected using a convex Hyperopia is known to you probably as farsightedness(outward facing) lens.

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A box of mass 26 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0

Kazeer [188]

Answer:

$\Delta K = 52J$

Explanation:

The change in kinetic energy will be simply the difference between the final and initial kinetic energies: $\Delta K=K_f-K_i$

We know that the formula for the kinetic energy for an object is:

$K=\frac{mv^2}{2}$

where m is the mass of the object and v its velocity.

For our case then we have:

$\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}$

Which for our values is:

$\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J$

3 0

3 years ago

4. Calculate the total resistance for two 180 ohm resistors connected in parallel

solmaris [256]

Answer:

90 ohms

Explanation:

1/r = 1/180 + 1/180

1/r= 2/180

take the reciprocal of 2/180 which is 180/2 and its 90 ohms

3 0

3 years ago

How fast must an object move before its length appears to be contracted to one-fourth its proper length? (Give your answer in te

Tresset [83]

Answer:

0.97c

Explanation:

From the relativistic equation for length contraction, we have

$l$ = $l_{0}\sqrt{1 - \beta }$

where

$l$ is the final length of the object

$l_{0}$ is the original length of the object before contraction

β = $v^{2} /c^2$

where v is the speed of the object

c is the speed of light in free space = 3 x 10^8 m/s

The equation can be re-written as

$l$ / $l_{0}$ = $\sqrt{1 - \beta }$

For the length to contract to one-fourth of the proper length, then

$l$ / $l_{0}$ = 1/4

substituting into the equation, we'll have

1/4 = $\sqrt{1 - \beta }$

substituting for β, we'll have

1/4 = $\sqrt{1 - v^2/c^2 }$

squaring both side of the equation, we'll have

1/16 = 1 - $v^2/c^2$

$v^2/c^2$ = 1 - 1/16

$v^2/c^2$ = 15/16

square root both sides of the equation, we have

v/c = 0.968

v = 0.97c

3 0

3 years ago

Following could help prevent flood waters from running off too quickly causing further soil erosion

dezoksy [38]

Growing grass plants and trees is the easiest and best way to prevent flood waters from running off too quickly causing further soil erosion.

8 0

3 years ago

You lift a 25-kg child 0.80 m, slowly carry him 10 m to the playroom, and finally set him back down 0.80 m onto the playroom flo

maksim [4K]

To solve this problem we will apply the work theorem which is expressed as the force applied to displace a body. Considering that body strength is equivalent to weight, we will make the following considerations

$\text{Mass of the child} = m = 25kg$

$\text{Acceleration due to gravity} = g = 9.81m/s^2$

$\text{Height lifted} = h = 0.80m (Upward)$

Work done to upward the object

$W = mgh$

$W = (25)(9.81)(0.8)$

$W = 196.2J$

Horizontal Force applied while carrying 10m,

$F = 0N$

$W = 0J$

Height descended in setting the child down

$h' = -0.8m (Downwoard)$

$W = mgh'$

$W = (25)(9.81)(-0.80)$

$W = -196.2J$

For full time, assuming that the total value of work is always expressed in terms of its symbol, it would be zero, since at first it performs the same work that is later complemented in a negative way.

6 0

4 years ago