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andrew-mc [135]

4 years ago

6

A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

tion of 400x with an objective lens that has a focal length of 0.60 cm. The distance between the eyepiece and objective lenses is 16 cm. Find the focal length of the eyepiece lens assuming a near point of 25 cm (the closest an object can be and still be seen in focus).

1 answer:

grin007 [14]4 years ago

3 0

Answer:

$f_e = 1.51 cm$

Explanation:

given.

magnification(m) = 400 x

focal length (f_0)= 0.6 cm

distance between eyepiece and lens (L)= 16 cm

Near point (N) = 25 cm

focal length of the eyepiece (f_e)= ?

using equation

$m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}$

$400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}$

$9.6 = \dfrac{16-f_e}{f_e}$

$9.6f_e = 16-f_e$

$10.6 f_e = 16$

$f_e = 1.51 cm$

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Which of the following objects would have the strongest gravitational pull on a third object positioned at an equal distance fro

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I believe Box B will have a greater gravitational pull because the gravitational pull of an object depends on its mass. The more mass an object has, the greater its gravitational pull will become.

For example, we can take planets. Naturally, they are round because once upon a time there was a larger piece of rock that attracted others. But the size of the rock won't matter, it's the weight that matters. If the rock weighed nothing, the other rocks would just rebound upon contact. But if the rock weighed a lot, then things wouldn't so easily rebound and might actually stick to it.

8 0

3 years ago

A cyclist rides at 6.20 m/s through a intersection. A stationary car begins to

Xelga [282]

Answer:

The width of the intersection is 20 meters

Explanation:

The speed with which the cyclist is riding, v₁ = 6.20 m/s

The rate at which the car starts to accelerate, a = 3.844 m/s²

The initial velocity of the car = The car is stationary at the start = 0 m/s

The time at which the cyclist and the car reach the other side of the intersection = The same time;

Let 't' represent the time at which the cyclist and the car both reach the other side of the intersection, we have;

The distance travelled by the cyclist = The distance traveled by the car

∴ v₁ × t = 1/2 × a × t²

Plugging in the values for 'v₁', and 'a' in the above equation, we get;

6.20 × t = 1/2 × 3.844 × t²

∴ 1.922·t² - 6.20·t = 0

∴ t·(1.922·t - 6.20) = 0

t = 0, or t = 6.20/1.922 = 100/31

The time at which the cyclist and the car both reach the other side of the intersection, t = 100/31 seconds

The with of the intersection, w = v₁ × t

∴ w = 6.20 × 100/31 = 100/5 = 20

The width of the intersection, w = 20 meters.

8 0

3 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

2 years ago

Imagine a 15 kg block moving with a velocity of 20 m/s to the left. Calculate the kinetic Energy of this block.

ivann1987 [24]

Answer:

3000 J

Explanation:

Kinetic energy is:

KE = ½ mv²

If m = 15 kg and v = -20 m/s:

KE = ½ (15 kg) (-20 m/s)²

KE = 3000 J

3 0

3 years ago

2. A person applies a force of 66 N to a fridge as they push it across the length of a standard tennis court. So far today, the

Lubov Fominskaja [6]

Answer:

$P=39.2205\, watt$

$E=374.948 \,cal$

Explanation:

Given that:

force applied, $F=66\,N$

displacement, $s=23.77\,m$ (length of a tennis court)

time taken for pushing, t = 40 s

Since, work is given by:

$W=F.s$

$W=66\times 23.77$

$W=1568.82\,J$

Now, power is given as:

$P=\frac{W}{t}$

$P=\frac{1568.82}{40}$

$P=39.2205 \,watt$

Calories consumed is:

$E= 1568.82\times 0.239$

$E=374.948\, cal$

3 0

4 years ago