__________ is the process by which populations change over time.

What is the total magnification for each lens setting on a microscope with 15x oculars and 4x, 10x, 45x, and 97x objectives lens

KengaRu [80]

Answer:

M = 60x

M = 150x

M = 675x

M = 1455 x

Explanation:

In a microscope the total magnification is found by multiplying the magnification of the ocular (eye piece) and objective lens.

$m_e$ =Ocular magnification = 15x

$m_o$ =Objective magnification = 4x

Total magnification

$M=m_e\times m_o\\\Rightarrow M=15\times 4\\\Rightarrow M=60x$

M = 60x

$M=m_e\times m_o\\\Rightarrow M=15\times 10\\\Rightarrow M=150x$

M = 150x

$M=m_e\times m_o\\\Rightarrow M=15\times 45\\\Rightarrow M=675x$

M = 675x

$M=m_e\times m_o\\\Rightarrow M=15\times 97\\\Rightarrow M=1455x$

M = 1455 x

7 0

3 years ago

How does a perspective drawing differ from an isometric drawing of the same object? when would you use a perspective view in lie

klemol [59]

The quick way to visualize this is to compare isometric and perspective drawings of a cube. In the isometric drawing, all the edges of the cube would be parallel in sets of four. In the perspective drawing, the edges would taper towards one or more vanishing points. The isometric drawing, being easier to construct, perserving all scales and dimensions, is the preferred method for mechanical drawings, and are practical for use in the shop. The perspective drawing, which are trickier to draw properly, and does not preserve scales and dimensions, is the preferred method for architectural drawings, because they illustrate what the eye actually sees.

7 0

3 years ago

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

Assoli18 [71]

Answer:

The tension in the cable when the craft was being lowered to the seafloor is 4700 N.

Explanation:

Given that,

When the craft was stationary, the tension in the cable was 6500 N.

When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N.

The drag force of 1800 N will act in the upward direction. As it was lowered or raised at a steady rate, so its acceleration is 0. As a result, net force is 0. So,

T + F = W

Here, T is tension

F = 1800 N

W = 6500 N

Tension becomes :

$T=W-F\\\\T=6500-1800\\\\T=4700\ N$

So, the tension in the cable when the craft was being lowered to the seafloor is 4700 N.

7 0

3 years ago

Na figura, um bloco de massa igual a 5Kg é abandonado em repouso em um plano inclinado. Supondo que os coeficientes de atrito es

Law Incorporation [45]

The figure mentioned on the question is in the attachment.

Answer: a) $P_{x}$ = - 38.35N

b) $F_{N}$ = 30.5 N

c) $F_{f}$ = 27.45 N

d) a = - 13.16 m/s²

Explanation: A block on an inclined plane has 3 forces acting on it:

Force due to gravity $F_{g}$ = m.g;
Normal Force due to the plane;
Force of Friction $F_{f}$ = µ.N;

Since the plane is inclined, Normal Force is equal the y-component of the force due to gravity and Force of friction and the x-component of the force due to gravity are opposite forces.

The second attachment ilustrate the forces acting on the block.

Calculating:

A) The magnitude of the x-component of Force due to gravity:

According to the second image:

$P_{x}$ = P.sinθ

$P_{x}$ = 5.9.8.sin(36.8)

$P_{x}$ = - 38.35 N

B) $F_{N}$ = $P_{y}$ = m.g.cosθ

$F_{N}$ = 5.9.8.cos(36.8)

$F_{N}$ = 30.5 N

C) $F_{f}$ = 0.9.30.5

$F_{f}$ = 27.45 N

D) For the acceleration, use Newton's Law: $F_{R}$ = m . a

If there is movement, it is only on x-axis, so the net force is:

$P_{x}$ - $F_{f}$ = m.a

- 38.35 - 27.45 = 5a

a = - 13.16 m/s²

The value of acceleration shows there is no movement on the x-axis due to the friction.

7 0

3 years ago

A boy kicks a football with an initial velocity of 28.0 m/s at an angle of 30.0o above the horizontal. what is the highest eleva

sladkih [1.3K]

As the boy kicks the football with an angle, due to the effect of the gravitational force, the ball would follow a projectile path which is parabolic in nature. From this idea, we can derive equations pertaining to the maximum height that the ball would reach. At the maximum height of the ball, the velocity of the ball would be equal to zero. From the equations for projectile motion, we would obtain the equation as follows:

Maximum height = v0^2 sin^2 (theta) / 2g
Maximum height = (28.0 m / s )^2 sin^2 (30.0) / 2(9.8 m / s^2)
Maximum height = 10 m

The maximum height that the ball would reach would be 10 m.

7 0

3 years ago