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3 years ago

13

__________ is extremely resistant to oxidation and rusting.

1 answer:

Nookie1986 [14]3 years ago

3 0

ALUMINUM. give me brainliest if i’m right i need one more to rank higher

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What is kirchoff s law???

kolbaska11 [484]

There are two laws named for Kirchhoff. The both concern electrical circuits.
Here they are in my own words:

1). The sum of the voltage drops around any closed loop in a circuit is zero.

2). The sum of the currents at any single point in a circuit is zero.

8 0

2 years ago

Read 2 more answers

What would happen to a moving object if no unbalanced force acts upon it?<br> Help please?

Kisachek [45]

Answer:

B

Explanation:

if no force acts upon it it will just continue moving

hope this helps!! have a wonderfull day!

7 0

3 years ago

If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu

ioda

Given :

Number of operations move through a pocket calculator during a full day's operation , $n=1.34 \times 10^{20}$ .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in one electron is :

$e^-=-1.6\times 10^{-19}\ coulombs$

So , charge on n electron is :

$C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C$

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

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2 years ago

How much tension is in a rope if it pulls a 5-kg bucket filled with water with an upward acceleration of 1m/s^2

liraira [26]

F=ma
Tension - weight = mass x acceleration
T - 5(9.81) = 5 x 1
T = 5 + 5(9.81)
T = 54.05 N
T ≈ 54 N

4 0

3 years ago

A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If

inessss [21]

Answer:

the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

v² = v₀² + 2 a₁ x

indicate that the initial velocity is zero

v² = 2 a₁ x

let's calculate

v = $\sqrt {2 \ 15.0 \ 645}$

v = 143.666 m / s

now for the second interval let's find the distance it takes to stop

v₂² = v² - 2 a₂ x₂

in this part the final velocity is zero (v₂ = 0)

0 = v² - 2 a₂ x₂

x₂ = v² / 2a₂

let's calculate

x₂ = $\frac{ 143.666^2 }{2 \ 18.2}$

x₂ = 573 m

as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

3 0

2 years ago