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3 years ago

In which of the following instrument is the image that is formed erect

Physics

1 answer:

ipn [44]3 years ago

6 0

Answer:

C. microscope

Explanation:

A simple microscope is magnifying glass, an ordinary double convex lens having a short focal length that produces virtual and erect image.

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A 6 ft tall person walks away from a 10 ft lamppost at a constant rate of 5 ft/s. What is the rate (in ft/s) that the tip of the

nalin [4]

Answer:

12.5 ft/s

Explanation:

Height of person = 6 ft

height of lamp post = 10 ft

According to the question,

dx / dt = 5 ft/s

Let the rate of tip of the shadow moves away is dy/dt.

According to the diagram

10 / y = 6 / (y - x)

10 y - 10 x = 6 y

y = 2.5 x

Differentiate both sides with respect to t.

dy / dt = 2.5 dx / dt

dy / dt = 2.5 (5) = 12.5 ft /s

8 0

4 years ago

The water that powers the generators enters and leaves the system at a low speed (thus we can neglect its change of kinetic ener

madam [21]

Answer:

Explanation:

Let density of water be ρ .

During flow , volume of water flowing per second is constant

loss of P. E per unit volume = ρ gh , 83.5 % is lost

Gain of K E per unit volume = 1/2 ρ v²

83.5 % of mgh = ρ 1/2 ρ v²

1/2 ρ v² = .835 x 9.8

v² = 2 x .835 x 9.8

= 16.366

v = 4.04 m /s

4 0

4 years ago

Thylakoids contain chlorophyl that absorb solar energy

steposvetlana [31]

Answer:

True

True statement:

Because pigment molecules absorb solar energy and thylakoids are pigment molecules

3 0

3 years ago

A sarcomere stores calcium within the muscle cell.

professor190 [17]

Sarcomere<span>The Sarcomere is the smallest functional unit of a myofibril (and also of the entire muscle) and contain the contractual elements between each pair of Z-discs.</span>

7 0

3 years ago

) A skier starts down a frictionless 32° slope. After a vertical drop of 25 m, the slope temporarily levels out and then slopes

EastWind [94]

Answer:

The skier’s speed on the two level stretches are 22.13 m/s and 27.29 m/s.

Explanation:

Given that,

Slope down = 32°

Height = 25 m

Before leveling,

Slope down = 20°

Height = 38 m

We need to calculate the skier’s speed on the two level stretches

Using formula of energy

$P.E=K.E$

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

For first stretch,

Height = 25 m

Put the value into the formula

$v=\sqrt{2\times9.8\times25}$

$v=22.13\ m/s$

For second stretch,

Height = 38 m

Put the value into the formula

$v=\sqrt{2\times9.8\times38}$

$v=27.29\ m/s$

Hence, The skier’s speed on the two level stretches are 22.13 m/s and 27.29 m/s.

5 0

4 years ago