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3 years ago

What are the factors that affect ocean density? Please help!!

2 answers:

6 0

The density of seawater plays a vital role in causing ocean currents and circulating heat because of the fact that dense water sinks below less dense. long story short, seawater is the problem because its denser than pure water.

Fudgin [204]3 years ago

6 0

The density of ocean water at the sea surface is about 1027 kg/m3. The two main factors that affect density of ocean water are the temperature of the water and the salinity of the water. The density of ocean water continuously increases with decreasing temperature until the water freezes. Hope this helps girl

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What causes wind how does the process reverse itself between day and night?

iris [78.8K]

<span>Wind is nature's way of balancing the temperature between hot and cold. Wind always flows from heat to cool. When night falls, the air cools. And since it gets cooler at night it reverses.</span>

5 0

3 years ago

An unknown charged particle passes without deflection throughcrossed electric and magnetic fields of strengths 187,500 V/m and0.

UNO [17]

Explanation:

The given data is as follows.

Electric field strength (E) = 187,500 V/m

Magnetic field strength (B) = 0.125 T

Diameter (d) = 25.05 cm = 0.2505 m (as 1 m = 100 cm)

Radius (r) = $\frac{d}{2}$

= $\frac{0.2505}{2}$

= 0.12525 m

Formula to calculate the magnetic force ( $F_{M}$ ) is as follows.

$F_{M} = Bqv$ ............ (1)

Electrical force is calculated as follows.

$F_{E} = qE$ ............ (2)

On both electric and magnetic fields the velocity is perpendicular.

$F_{M} - F_{E}$ = 0

or, $F_{M} = F_{E}$

Hence, from equations (1) and (2)

Bqv = qE

or, v = $\frac{E}{B}$ ............. (3)

= $\frac{187500 V/m}{0.125 T}$

= 1,500,000 m/s

As the particle is moving in a semi-circular trajectory and motion of charged particle is given by the electric field as follows.

$F_{c} = \frac{mv^{2}}{r}$ ........... (4)

where, $F_{c}$ = centripetal force

$F_{M} = F_{c}$

Using equation (1) and (4) as follows.

$F_{M} = F_{c}$

Bqv = $\frac{mv^{2}}{r}$

$\frac{q}{m} = \frac{v}{Br}$

= $\frac{15 \times 10^{5}}{0.125 \times 0.12525}$

= $958.08 \times 10^{5} C/kg$

Thus, we can conclude that charge-to-mass ratio of the given particle is $958.08 \times 10^{5} C/kg$ .

8 0

3 years ago

Which best describes a chemical reaction? the rearrangement of atoms by breaking and reforming chemical bonds a reaction that ab

Ede4ka [16]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers

A 7500 kg truck traveling at 5.0 m/s east collides with a 1500 kg car moving

julia-pushkina [17]

Since the collision is inelastic, both vehicle will move with a common velocity of 2.5 m/s in south east direction after collision.

<h3>Conservation of Linear Momentum</h3>

It states that, the sum of the momentum before collision is equal to the sum of the momentum after collision.

Given that a 7500 kg truck traveling at 5.0 m/s east collides with a 1500 kg car moving at 20 m/s in a direction 30° south of west. After collision, the two vehicles remain tangled together.

The given parameters are;

M1 = 7500 kg

U1 = 5 m/s

M2 = 1500

U2 = 20 sin 30 = 10 m/s

V = ?

The formula to find the speed will be

M1U1 - M2U2 = (M1 + M2)V

7500 × 5 - 1500 × 10 = (7500 + 1500)V

37500 - 15000 = 9000V

22500 = 9000V

V = 22500/9000

V = 2.5 m/s

Therefore, the common speed will be 2.5 m/s and the direction of the car will be toward the south east.

Learn more about Collision here: brainly.com/question/7694106

#SPJ1

5 0

1 year ago

A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$