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3 years ago

Most of the life is spent in a cell development called?

Physics

2 answers:

lisov135 [29]3 years ago

8 0

The Interphase is the most.

tatuchka [14]3 years ago

6 0

It's called cellular differentiation. I think.

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Data:

____ [38]

a. The speed of the tsunami at ocean depth of 8000 m is 1,019.66 km/h.

b. The speed of the tsunami at ocean depth of 3500 m is 660.75 km/h.

c. The speed of the tsunami at ocean depth of 70 m is 89.66 km/h.

d. The speed of the tsunami at ocean depth of 5 m is 30.625 km/h.

<h3>Speed of the tsunami at the given depths</h3>

The speed of the tsunami is calculated from method of extrapolation or interpolation as shown below;

<h3>Speed at 8000 meters</h3>

8000 m -------- ?

7000 m -------- 943 km/h

4000 m -------- 713 km/h

(8000 - 7000)/(7000 - 4000) = (? - 943)/(943 - 713)

0.3333 = (? - 943)/230

230(0.3333) = ? - 943

76.66 = ? - 943

? = 1,019.66 km/h

<h3>Speed at 3500 m</h3>

4000 m ---------- 713 km/h

3500 m ---------- ?

2000 m ----------- 504 km/h

(4000 - 3500)/(4000 - 2000) = (713 - ?) / (713 - 504)

0.25 = (713 - ?) /209

0.25(209) = 713 - ?

52.25 = 713 - ?

? = 660.75 km/h

<h3>Speed at 70 m</h3>

200 m ------ 159 km/h

70 m --------- ?

50 m ------- 79 km/h

(200 - 70)/(200 - 50) = (159 - ?)/(159 - 79)

0.8667 = 159 - ? / 80

80(0.8667) = 159 - ?

69.336 = 159 - ?

? = 89.66 km/h

<h3>Speed at 5 meters</h3>

50 m ------- 79 km/h

10 m --------- 36 km/h

5 m ---------- ?

(50 - 5)/(50 - 10) = (79 - ?)/(79 - 36)

1.125 = 79 - ?/43

43(1.125) = 79 - ?

48.375 = 79 - ?

? = 30.625 km/h

Learn more about speed of tsunami here: brainly.com/question/23136161

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8 0

1 year ago

A 1500kg car is travelling at v=30m/s. The cars kinetic energy is? *

Anuta_ua [19.1K]

Hi there!

$\large\boxed{\text{D. 675000J}}$

Use the following formula to solve:

KE = 1/2mv², where:

KE = kinetic energy

m = mass (kg)

v = velocity (m/s)

Therefore:

KE = 1/2(1500)(30)²

KE = 1/2(1500)(900)

KE = 675000 J

4 0

3 years ago

A solid sphere of radius R carries a fixed, uniformly distributed charge q. Obtain an expression for the magnitude of the electr

NISA [10]

Answer:

The electric field outside the sphere will be $\dfrac{qr}{4\pi\epsilon_{0}R^3}$ .

Explanation:

Given that,

Radius of solid sphere = R

Charge = q

According to figure,

Suppose r is the distance between the point P and center of sphere.

If $\rho$ be the volume charge density,

Then, the charge will be,

$q=\rho\times\dfrac{4}{3}\pi R^3$ .....(I)

Consider a Gaussian surface of radius r.

We need to calculate the electric field outside the sphere

Using formula of electric field

$\oint{\vec{E}\cdot \vec{dA}}=\dfrac{Q}{\epsilon_{0}}$

$E\times4\pi r^2=\dfrac{\rho\dotc \dfrac{4}{3}\pi r^3}{\epsilon_{0}}$

Put the value from equation (I)

$E\times4\pi r^2=\dfrac{qr^3}{\epsilon_{0}R^3}$

$E=\dfrac{qr}{4\pi\epsilon_{0}R^3}$

Hence, The electric field outside the sphere will be $\dfrac{qr}{4\pi\epsilon_{0}R^3}$ .

4 0

3 years ago

represents the space-time, speed-time and acceleration-time graphs for a ball pulled upwards with a speed of 10 m / s from 1 met

alexandr1967 [171]

Gyhugugygubuhuhuhubu

5 0

3 years ago

When object 1 came into contact with object 2, object 2 got warmer. what can be concluded about the objects before they came int

luda_lava [24]

The statement that is true regarding the objects is that object 1 had faster-moving particles, which is option 2. Details about matter can be found below.

<h3>What is matter?</h3>

Matter can be defined as any particle that has weight and occupies space.

Certain particles are present in the different states of matter. According to this question, when object 1 came into contact with object 2, object 2 got warmer.

This suggests that the particles in the object 1 is moving faster due to an increase in temperature.

Learn more about matter at: brainly.com/question/4513444

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7 0

2 years ago