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Murljashka [212]
3 years ago
7

The amplitude of a paricular wave is 4.0 m. The crest to trough distance

Physics
1 answer:
kozerog [31]3 years ago
6 0

Answer:

The crest to trough distance = 8 m

Explanation:

Given that,

The amplitude of a particular wave is 4.0 m.

We need to find the crest to trough distance.

We know that,

Amplitude = The distance from the base line to the crest or the the distance from the baseline to the trough.

It means,

Distance from crest to trough = 2(Amplitude)

= 2(4)

= 8 m

Hence, the crest to trough distance is equal to 8 m.

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What animals were delisted in july 2019?
andrezito [222]

The animals which were delisted from the endangered species in July 2019 is Grey Wolf

Answer: Grey wolf

<u>Explanation:</u>

The ultimate goal of the Endangered Species Act was to preserve the species that are in the verge of extinction. The species are conserved and taken care of till they can be left free in the wild for it to survive of it's own.If the species has crossed all the recovery goals then they are delisted.

The grey wolf in July 2019 was delisted from the endangered species list. Now it is calculated that there are 6000 wolves that are found. But yet the grey wolf has not returned back to their livelihood and also to their sutable habitat.

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3 years ago
Crews at the International Space Station are researching the effects of the weightlessness of space on ________.
andrezito [222]

A: Human Body

C is wrong because they don’t have the tools to test it on another planet

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3 years ago
Read 2 more answers
While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 5.65 m/s. The stone subs
xxTIMURxx [149]

Answer:

1. 20.54m/s

2. 1.52s

Explanation:

QUESTION 1:

The speed the stone impact the ground is the final speed/velocity, which can be calculated using the formula:

v² = u² + 2as

Where;

v = final velocity (m/s)

u = initial velocity (m/s)

a = acceleration due to gravity (m/s²)

s = distance (m)

From the provided information, u = 5.65m/s, v = ?, s = 19.9m, a = 9.8m/s²

v² = 5.65² + 2 (9.8 × 19.9)

v² = 31.9225 + 2 (195.02)

v² = 31.9225 + 390.04

v² = 421.9625

v = √421.9625

v = 20.5417

v = 20.54m/s

QUESTION 2:

Using v = u + at

Where v = final velocity (m/s) = 20.54m/s

t = time (s)

u = initial velocity (m/s) = 5.65m/s

a = acceleration due to gravity (m/s²)

v = u + at

20.54 = 5.65 + 9.8t

20.54 - 5.65 = 9.8t

14.89 = 9.8t

t = 14.89/9.8

t = 1.519

t = 1.52s

3 0
3 years ago
What is the difference between kinetic and gravitacional energy?
kondaur [170]

Answer:

In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion

In classical mechanics, the gravitational potential at a location is equal to the work (energy transferred) per unit mass that would be needed to move an object to that location from a fixed reference location. It is analogous to the electric potential with mass playing the role of charge. The reference location, where the potential is zero, is by convention infinitely far away from any mass, resulting in a negative potential at any finite distance.

In mathematics, the gravitational potential is also known as the Newtonian potential and is fundamental in the study of potential theory. It may also be used for solving the electrostatic and magnetostatic fields generated by uniformly charged or polarized ellipsoidal bodies

8 0
3 years ago
Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e
Nataliya [291]

Answer:

q=1.95*10^{-7}C

Explanation:

According to the free-body diagram of the system, we have:

\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)

So, we can solve for T from (1):

T=\frac{mg}{cos(15^\circ)}(3)

Replacing (3) in (2):

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e

The electric force (F_e) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)

Finally, we replace (5) in (4) and solving for q:

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