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3 years ago

An object with a mass of 5.0 kg is accelerating toward the surface of Earth at a rate of 9.8 m/s^2

1 answer:

4 0

The answer is A. Force is equal to mass x acceleration. The acceleration due to gravity is a constant so multiplying the mass by 2 would double the force.

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A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

3 years ago

Please help me I have been struggling

disa [49]

High/low notes have high/low frequency, not high/low amplitude.
Heartz measures the number of waves per second, not per hour.
Sounds do not travel in space because it's empty
Sounds make our eardrum vibrate. The diaphragm is the muscle we use to breathe.
Soft, pale materials are best for sound proofing because they absorb the waves.
In the next paragraph, ultrasound and infrasound are inverted (ultra means "over, above" and infra means "below").

4 0

3 years ago

A 1.3-kg model airplane flies in a circular path on the end of a 23-m line. The plane makes

storchak [24]

(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in

(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s

(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:

(2π (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s

(c) The plane accelerates toward the center of the path with magnitude

a = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²

(d) By Newton's second law, the tension in the line is

F = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N

4 0

3 years ago

A rifle fires a bullet into a giant slab of butter on a frictionless surface. The bullet penetrates the butter, but while passin

Vera_Pavlovna [14]

Answer:

V= 3.48 m/s

Explanation:

8 0

3 years ago

How much pressure is created when you apply a

Alekssandra [29.7K]

Pressure
= Force/Area

Area = π(d^2)/4
= π(0.4^2)/4
=0.126 m2
Pressure
= 50/0.126
= 396.825 Pa

5 0

2 years ago