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2 years ago

Protect your plants against ________ off disease

Physics

2 answers:

erastovalidia [21]2 years ago

7 0

Attackinggggggggggggg

Kazeer [188]2 years ago

5 0

Answer:

attacking

!!!!!! !!!!!

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Asteroid Ida was photographed by the Galileo spacecraft in 1993, and the photograph revealed that the asteroid has a small moon,

Nady [450]

Answer:

The orbital speed of Dactyl is $5.55m/s$

Explanation:

The orbital speed can be determined by the combination of the universal law of gravity and Newton's second law:

$F = G\frac{M \cdot m}{r^{2}}$ (1)

Where G is gravitational constant, M is the mass of the asteroid, m is the mass of the moon and r is the distance between them

In the other hand, Newton's second law can be defined as:

$F = ma$ (2)

Where m is the mass and a is the acceleration

Then, equation 2 can be replaced in equation 1

$m\cdot a = G\frac{M \cdot m}{r^{2}}$ (2)

However, a will be the centripetal acceleration since the moon Dactyl describe a circular motion around the asteroid

$a = \frac{v^{2}}{r}$ (3)

$m\frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$ (4)

Therefore, v can be isolated from equation 4:

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$ (5)

Finally, the orbital speed can be found from equation 5:

Notice, that it is necessary to express r in units of meters.

$r = 95km \cdot \frac{1000m}{1km}$ ⇒ $95000m$

$v = \sqrt{\frac{(6.672x10^{-11}N.m^{2}/kg^{2})(4.4x10^{16}kg)}{95000m}}$

$v = 5.55m/s$

Hence, the orbital speed of Dactyl is $5.55m/s$

3 0

2 years ago

When a person plucks a guitar string, the number of half wavelengths that fit into the length of the string determines the _____

Vedmedyk [2.9K]

Answer:

Frequency

Explanation:

Each half wavelength has a point of largest amplitude (aka a node). Depending on the wavelength each node oscillates at a certain rate of swings per unit of time. The latter is referred to as frequency and measure in Hertz [Hz].

8 0

3 years ago

1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position

antiseptic1488 [7]

Answer:

The value of the distance is $\bf{14.52~cm}$ .

Explanation:

The velocity of a particle(v) executing SHM is

$v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)$

where, $\omega$ is the angular frequency, $A$ is the amplitude of the oscillation and $x$ is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., $x = 0$ .

The maximum velocity( $\bf{v_{m}}$ ) is

$v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Divide equation (1) by equation(2).

$\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$

Given, $v = 0.25 v_{m}$ and $A = 15~cm$ . Substitute these values in equation (3).

$&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm$

6 0

3 years ago

A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a

sleet_krkn [62]

When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is

∑ F = R - mg = 0

where mg = weight of the mass = (7.00 kg) g = 68.6 N.

It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that

k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m

5 0

2 years ago

A two-phase, liquidâ€“vapor mixture of h2o, initially at x 5 30% and a pressure of 100 kpa, is contained in a pistonâ€“ cylinder

Andrew [12]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
Read more on Brainly.com - brainly.com/question/1581851#readmore

7 0

2 years ago

Protect your plants against ________ off disease​

Protect your plants against ________ off disease