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3 years ago

What is the Deepest point of the ocean and where is it?

Physics

2 answers:

qaws [65]3 years ago

5 0

Answer:

marinas trench in Pacific Ocean

<h2>mark it as brainliest</h2>

Rashid [163]3 years ago

3 0

Explanation:

Located in the western Pacific Ocean, the Marina Trench

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Which biome's yearly rainfall mainly evaporates? A. taiga B. desert C. tropical rainforest D. temperate grassland

worty [1.4K]

The answer is B. desert. Deserts don't get much rainfall to begin with and most of it evaporates.

7 0

3 years ago

How much 8 grams will weigh after a physical change

Ber [7]

I would look this one up on Google

5 0

3 years ago

What is the velocity of a wave with a wavelength of 1.5m and a frequency of 500 Hz

Schach [20]

Wave speed = (wavelength) x (frequency)

= (1.5 m) x (500 / sec)

= 750 m/s .

4 0

3 years ago

Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

kramer

Answer:

$v_x = -6\ m/s$

Explanation:

initial velocity

magnitude of velocity, v = 12 m/s

angle made of velocity with negative x-axis,θ = 60°

We need to calculate x- component of v

$v_x = v cos \theta$

velocity is in negative x-direction, v = -12 m/s

now,

$v_x = -12\times cos 60^0$

$v_x = -12\times 0.5$

$v_x = -6\ m/s$

Hence, the velocity x-component is equal to -6 m/s.

5 0

3 years ago

The 9 kg block is then released and accelerates to the right, toward the 5 kg block. The surface is rough and the coefficient of

Ahat [919]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The velocity is $=4.51m/s$

Explanation:

The kinetic energy of the 9 kg can be determined by these expression

Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction

Now to obtain the initial energy stored

Let U denote the initial energy stored and

$U = \frac{1}{2} kx^2$

Where x is the length the spring is displaced

k is the force constant of the string

$U = \frac{1}{2} * 627 * (0.6)^2$

$= 112.86 J$

Now referring to the formula above

i.e Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction

$\frac{1}{2} mv^2 = 112.86 - \mu_kmgx$

$v^2 = \frac{2(112.86 -\mu_kmgx)}{m}$

$v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}$

and we are told that coefficient of friction = 0.4 and the mass is 9 kg ,the acceleration due to gravity $= 9.8m/s^2$ this displacement length of spring = 0.6

Therefore $v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }$

$=4.51m/s$

8 0

3 years ago