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3 years ago

List all the forces that act on a hot air ballon when it is in the sky

2 answers:

6 0

Down ... gravity
Up ....... buoyancy

When these are equal, then the balloon either stays at the same altitude,
or moves up or down at a constant rate.

Horizontal ... wind

When wind is not zero, the balloon accelerates in the direction of the wind.

nikdorinn [45]3 years ago

3 0

Gravity, acceleration, kinetic energy, the atmosphere

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Defination wheel and axle

Maurinko [17]

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3 years ago

What shape is the unit cell of ruby?

sleet_krkn [62]

Ruby, a variety of the mineral corundum is in the trigonal crystal system, with hexagonal scalenohedra crystals

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3 years ago

Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa

lyudmila [28]

Answer:

No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field $\vec{E}$ is related to the gradient of the electric potential V with :

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})$

This means that for constant electric potential the electric field must be zero:

$V(\vec{r}) = k$

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k$

$\vec{E} (\vec{r}) = - (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k$

$\vec{E} (\vec{r}) = - (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})$

$\vec{E} (\vec{r}) = - (0,0,0)$

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

$V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4$

give an electric field of zero at point (0,0,0)

8 0

3 years ago

Find the mass of a football player who has 1420 N of force and an acceleration of 4 m/s^2 *

Hatshy [7]

The answer is 833.3 kg

7 0

3 years ago

A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm

Galina-37 [17]

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

$v =\sqrt{\dfrac{T}{\mu}}$

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

$v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}$

v = 1074.49 m/s

The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

λ=2 L

λ=2 x 0.577 = 1.154 m

we now,

v = f λ

$f = \dfrac{v}{\lambda}$

$f = \dfrac{1074.49}{1.154}$

f = 931.1 Hz

The frequency for first harmonic is equal to f = 931.1 Hz

7 0

3 years ago