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3 years ago

Definition oceanographer

Physics

1 answer:

Art [367]3 years ago

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Oceanography covers a wide range of topics, including marine life and ecosystems, ocean circulation, plate tectonics and the geology of the seafloor, and the chemical and physical properties of the ocean. Just as there are many specialties within the medical field, there are many disciplines within oceanography.

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If the man pushes with a force of 2000N and friction is 500N, what is the resultant force?

alexandr402 [8]

Answer:

Explanation:

If the force of 2000 N is directed towards the right and the friction is directed towards the left, the 2000 N force is positive and the other is negative. To find the resultant force:

2000 - 500 = 1500 N to the right

3 0

3 years ago

What is the basic unit of each type of elements

Lynna [10]

A Atom is the basic unit of each type of element

8 0

3 years ago

A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is appl

Taya2010 [7]

Answer:

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

Given that,

Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

$F-f=ma$

Put the value into the formula

$340-f=60\times2.0$

$f=340-60\times2.0$

$f=220\ N$

We need to calculate the coefficient of friction

Using formula of friction force

$f= \mu mg$

$\mu=\dfrac{f}{mg}$

$\mu=\dfrac{220}{60\times9.8}$

$\mu =0.37$

We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

$F = ma$

$a=\dfrac{F}{m}$

$a=\dfrac{220}{100}$

$a=2.2\ m/s^2$

Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

3 0

4 years ago

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

katrin2010 [14]

Answer:

2.1x10^6m/s

Explanation:

One electron has a charge of –1.602e-19 C

mass of electron is 9.1e-31 kg

mass of proton is 1.6726e−27 kg

mass ratio is 1.6726e−27 / 9.1e-31 = 1838

The force is constant, F

distance is constant, d

a = F/m

a increases by a factor 1838, as m decreases by that factor

a = a₀1838

v₀² = 2a₀d

v² = 2a₀d1838

v²/v₀² = 2a₀d1838 / 2a₀d = 1838

v² = 1838v₀² = 1838(45000)²

v = 45000√1838 = 2.1e6 m/s

5 0

3 years ago

Two spherically symmetric planets with no atmosphere have the same average density, but planet B has twice the radius of planet

Nina [5.8K]

A period of a satellite is the time taken by the satellite to travel round a

body.

The comparison between the periods $T_B$ , and $T_A$ is $\underline {T_B = \dfrac{\sqrt{2} }{4 } \cdot T_A}$

Reason:

The period, T, of a satellite is given as follows;

$T = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \cdot M} }$

Volume of the planet A = $\dfrac{4}{3} \cdot \pi \cdot r^3$

Mass of planet A, $m_A$ = $\dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho$

Volume of the planet B = $\dfrac{4}{3} \cdot \pi \cdot (2 \cdot r)^3 = \dfrac{32}{3} \cdot \pi \cdot r^3$

Mass of planet B, $m_B$ = $\dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho$

Period of the satellite on planet A, $T_A$ , is given as follows;

$T_A = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{4}{3} \cdot \pi \times \rho} }$

Period of the satellite on planet B, $T_B$ , is given as follows;

$T_B = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{32}{3} \cdot \pi \times \rho} }$

Therefore, get;

$\dfrac{T_A}{T_B} = \dfrac{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 4 \cdot \pi \times \rho} }}{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 32 \cdot \pi \times \rho} }} = \sqrt{\dfrac{32}{4} } = \sqrt{8} = 2 \cdot \sqrt{2}$

Therefore, $T_A$ = (2·√2)· $T_B$

$T_B = \dfrac{T_A}{2 \cdot \sqrt{2} } = \dfrac{\sqrt{2} \cdot T_A}{4 }$

The comparison between $T_A$ and $T_B$ is therefore;

$\underline {T_B = \dfrac{\sqrt{2} }{4 } \cdot T_A}$

Learn more here:

brainly.com/question/1448749

7 0

3 years ago