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Naddik [55]
4 years ago
11

Which activity is an example of a chemical change?

Physics
2 answers:
saul85 [17]4 years ago
5 0

C lighting a match is a chemical reaction


melomori [17]4 years ago
3 0

The answer is C because when the chemicals on the tip of a match are heated up from friction it ignites.

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How would you determine how much error there is between a vector addition and the real results
chubhunter [2.5K]
Desired operation: A + B = C; {A,B,C) are vector quantities. 

<span>Issue: {A,B} contain error (measurement or otherwise) </span>

<span>Objective: estimate the error in the vector sum. </span>

<span>Let A = u + du; where u is the nominal value of A and du is the error in A </span>
<span>Let B = v + dv; where v is the nominal value of B and dv is the error in B </span>
<span>Let C = w + dw; where w is the nominal value of C and dw is the error in C [the objective] </span>

<span>C = A + B </span>

<span>w + dw = (u + du) + (v + dv) </span>

<span>w + dw = (u + v) + (du + dv) </span>

<span>w = u+v; dw = du + dv </span>

<span>The error associated with w is the vector sum of the errors associated with the measured quantities (u,v)</span>
6 0
3 years ago
Unstable atmospheric conditions lead to the formation of lightning and thunder in
Fantom [35]
Hi the answer is cumulonimbus clouds. Hope this helps.
6 0
3 years ago
3 A jet plane accelerated along a straight runway from rest to a
aleksklad [387]

Explanation:

v=140 m/s

t=50s

u=0

change of velocity = v-u

140-0=140 m/s

for acceleration:

from Newton's first law of motion:

v=u+at

140=0+a50

50a=140

a=140/50

a=2.8 m/s²

8 0
3 years ago
If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric
poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0
3 years ago
A potential difference of 3.00 nV is set up across a 2.00 cm length of copper wire that has a radius of 2.00 mm. How much charge
Anvisha [2.4K]

The number of charge drifts are 3.35 X 10⁻⁷C

<u>Explanation:</u>

Given:

Potential difference, V = 3 nV = 3 X 10⁻⁹m

Length of wire, L = 2 cm = 0.02 m

Radius of the wire, r = 2 mm = 2 X 10⁻³m

Cross section, 3 ms

charge drifts, q = ?

We know,

the charge drifts through the copper wire is given by

q = iΔt

where Δt = 3 X 10⁻³s

and i = \frac{V}{R}

where R is the resistance

R = \frac{pL}{r^{2} \pi }

ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm

So, i = \frac{\pi(r)^{2}V  }{pL}

q = \frac{\pi(r^{2} )Vt }{pL}

Substituting the values,

q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02

q = 3.35 X 10⁻⁷C

Therefore, the number of charge drifts are 3.35 X 10⁻⁷C

3 0
3 years ago
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