a beaker has mass of 125 g. what is the mass of liquid if the beaker plus liquid have a mass of 232 g?

two identical steel balls mounted on wooden posts initially have different amounts of charge, one -14цC and the other +2цC. the

larisa86 [58]

-6 and -6 (each)

....

4 0

3 years ago

Elephants can communicate over distances as far as 6 km by using very low-frequency sound waves. What is the wavelength of a 10

harkovskaia [24]

The characteristics of the sound waves allow to find the result for the wavelength is:

λ = 34.29 m

The sound wave is a longitudinal traveling wave, where the speed of the wave depends on the air temperature.

v = $v_o \sqrt{1 + \frac{T}{273} }$

where v₀ is the speed of sound for T = 0ºc v₀ = 331 m / s.

indicate that the temperature is T = 20ºC, let's find the speed of the traveling wave.

v = 331 $\sqrt{1 + \frac{20}{273} }$

v = 342.9 m / s

The speed of sound is related to wavelength and frequency.

v = λ f

λ = $\frac{v}{f}$

Let's calculate

$\lambda = \frac{342.29}{10}$

λ = 34.29 m

In conclusion using the characteristics of sound waves we can find the result for the wavelength is:

λ = 34.29 m

Learn more here: brainly.com/question/21995826

3 0

2 years ago

A projectile lands at the same height from which it was launched. which initial velocity will result

Serhud [2]

The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin θ.

The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.

If the <u>air resistance is ignored</u>, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.

Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

Learn more here; brainly.com/question/12870645

5 0

3 years ago

Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l

Masja [62]

Answer:

a) I2 = 3 (o-10) / (o- 30) , b) h ’/h= 3 (o-10) / o (o-30)

Explanation:

The builder's equation is

1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

1 / i1 = 1 / f - 1 / o

1 / i1 = 1/15 - 1 / o

i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

o2 = d-i1

We write the builder equation

1 / f2 = 1 / o2 + 1 / i2

1 / i2 = 1 / f2 -1 / o2

1 / i2 = 1 / f2 - 1 / (d-i1)

1 / i2 = 1/20 - 1 / (d-i1) (1)

Let's evaluate the last term

d-i1 = d - 15 o / (o-15)

d-i1 = (d (o-15) - 15 o) / (o-15)

d- i1 = (30 or -30 15 -15 o) / (o-15)

d-i1 = (15 or - 450) / (o- 15)

d-i1 = = (15 or -450) / (o-15)

replace in 1

1 / i2 = 1/20 - (or - 15) / (15 or -450)

1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

1 / i2 = (-5 or -150) / (15 or -150)

1 / i2 = (or -30) / (3 or - 30)

I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

m = h ’/ h = - i / o

h ’= - h i / o

In our case

h ’= h i2 / o

h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0

3 years ago

The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud

motikmotik

<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = $\frac{Gm_1m_2}{r^2}$

Here G is the gravitational constant

m₁ is the mass of one student

m₂ is the mass of second student .

and r is the distance between them

Thus r = $\sqrt{\frac{Gm_1m_2}{F} }$

If we substitute the values in the above equation

r = $\sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }$

= 2.46 m

3 0

3 years ago

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