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valkas [14]
3 years ago
8

How do hydrothermal minerals form?

Physics
1 answer:
Katena32 [7]3 years ago
6 0
<span>Hydrothermal mineral deposit, any concentration of metallic minerals formed by the precipitation of solids from hot mineral-laden water. The solutions are thought to arise in most cases from the action of deeply circulating water heated by magma.</span>
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What work??? I don’t see anything
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I don’t know how to answer this question? Can anyone help?
VMariaS [17]

Answer:

F=ma

here F is force, m is mass and a is accelaration,

According to the question,

F=3*F= 3F

m= 1/3 of m= m/3

a= ?

so the equation becomes,

3F= m/3*a

3F*3= ma

9F=ma

F= ma/9

Therefore accelaration reduces by 1/9.

I am not very sure.

7 0
3 years ago
List two elements that have the same reactivity
timama [110]

Answer:

electron and neutron

Explanation:

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3 years ago
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A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends 30 psi, but it's been a
Rama09 [41]

Answer:

21 psi

Explanation:

The weight of the car is:

W = mg

W = 1000 kg * 9.8 m/s²

W = 9800 N

Divided by 4 tires, each tire supports:

F = W/4

F = 9800 N / 4

F = 2450 N

Pressure is force divided by area, so:

P = F / A

P = (2450 N) / (0.13 m × 0.13 m)

P ≈ 145,000 Pa

101,325 Pa is the same as 14.7 psi, so:

P ≈ 145,000 Pa × (14.7 psi / 101,325 Pa)

P ≈ 21 psi

3 0
3 years ago
A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary
Snezhnost [94]

Explanation:

Let N_p\ and\ N_s are the number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p}{N_s}=\dfrac{1}{3}

A resistor R connected to the secondary dissipates a power P_s=100\ W

For a transformer, \dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}

V_s=(\dfrac{N_s}{N_p})V_p

V_s=3V_p...............(1)

The power dissipated through the secondary coil is :

P_s=\dfrac{V_s^2}{R}

100\ W=\dfrac{V_s^2}{R}

V_p^2=\dfrac{100R}{9}.............(2)

Let N_p'\ and\ N_s' are the new number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p'}{N_s'}=\dfrac{1}{24}

New voltage is :

V_s'=(\dfrac{N_s'}{N_p'})V_p'

V_s'=24V_p...............(3)

So, new power dissipated is P_s'

P_s'=\dfrac{V_s'^2}{R}

P_s'=\dfrac{(24V_p)^2}{R}

P_s'=24^2\times \dfrac{(V_p)^2}{R}

P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}

P_s'=6400\ Watts

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0
3 years ago
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