A box weighing 4.0 N is lifted 3.0 meters. How much work is done on the box?

Một khối lập phương bằng nhôm cạnh 20cm nổi trên thuỷ ngân. Hỏi khối nhôm sẽ chìm xuống bao nhiêu khi nhiệt độ tăng từ 270K đến

Nady [450]

Answer:

English or spanish please

Explanation:

7 0

3 years ago

Anna Litical analyzes the force between a planet and its moon, varying the mass of

Travka [436]

Answer:

Trial 1 is the largest, trial 3 is the smallest

Explanation:

Given:

<em>Trial 1</em>

M₁ = 6·10²² kg

d₁ = 3 500 km = 3.5·10⁶ м

<em>Trial 2</em>

M₂ = 6·10²² kg

d₂ = 7 000 km = 7·10⁶ м

<em>Trial 3</em>

M₃ = 3·10²² kg

d₃ = 7 000 km = 7·10⁶ м

___________

F - ?

Gravitational force:

F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)

F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)

F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)

Trial 1 is the largest, trial 3 is the smallest

5 0

1 year ago

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

3 years ago

The drawing shows 6 point charges arranged in a rectangle. The value of q is 2.83 uC and the distance d is 0.123 m. Find the tot

vova2212 [387]

the total electric potential at location P, which is at the center of the rectangle is 0V.

The charges placed at the corner of the rectangle are same in magnitude but different in charge. hence the total electric potential will be same in magnitude but different in charge and will be cancelled. Similarly, all the total electric potential will be cancelled and resultant will be zero.

<h3>What is total electric potential?</h3>

The amount of labor required to convey a unit of electric charge from a reference point to a given place in an electric field is known as the electric potential (also known as the electric field potential, potential drop, or the electrostatic potential).
More specifically, it is the energy per unit charge for a test charge that is negligibly disruptive to the field under discussion. In order to prevent the test charge from gaining kinetic energy or radiating, the travel across the field is also meant to occur with very little acceleration.
The electric potential at the reference location is, by definition, zero units. Any point may be used as the reference point, but typically it is earth or a point at infinity.

To learn more about total electric potential with the given link

brainly.com/question/14776328

#SPJ4

3 0

2 years ago

An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, tu

Doss [256]

Answer:

$400000\ \text{N/C}$