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3 years ago

8

Where does the energy in an unburned match come from?

1 answer:

choli [55]3 years ago

5 0

A matchstick has a lot of chemical energy stored in it. When the match is struck, it burns and the chemical energy in it produces heat energy and light energy.

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2) Um gás ideal sofre uma determinada transformação, conforme mostra o gráfico abaixo. Considere

Roman55 [17]

Answer:

yes

Explanation:

5 0

3 years ago

For a wire has a circular cross section with a radius of 1.23mm.

Mila [183]

Answer:

$5.731\times 10^{-5}\ m/s$

Decrease

Explanation:

I = Current = 3.7 A

e = Charge of electron = $1.6\times 10^{-19}\ C$

n = Conduction electron density in copper = $8.49\times 10^{28}\ electrons/m^3$

$v_d$ = Drift velocity of electrons

r = Radius = 1.23 mm

Current is given by

$I=neAv_d\\\Rightarrow v_d=\dfrac{I}{neA}\\\Rightarrow v_d=\dfrac{3.7}{8.49\times 10^{28}\times 1.6\times 10^{-19}\times \pi (1.23\times 10^{-3})^2}\\\Rightarrow v_d=5.731\times 10^{-5}\ m/s$

The drift speed of the electrons is $5.731\times 10^{-5}\ m/s$

$v_d=\dfrac{I}{neA}$

From the equation we can see the following

$v_d\propto \dfrac{1}{n}$

So, if the number of conduction electrons per atom is higher than that of copper the drift velocity will decrease.

5 0

3 years ago

A burglar attempts to drag a 108 kg metal safe across a polished wood floor Assume that the coefficient of static friction is 0.

V125BC [204]

Answer:

2.00 m/s²

Explanation:

Given

The Mass of the metal safe, M = 108kg

Pushing force applied by the burglar, F = 534 N

Co-efficient of kinetic friction, $\mu_k$ = 0.3

Now,

The force against the kinetic friction is given as:

$f = \mu_k N = u_k Mg$

Where,

N = Normal reaction

g= acceleration due to the gravity

Substituting the values in the above equation, we get

$f = 0.3\times108\times9.8$

or

$f = 317.52N$

Now, the net force on to the metal safe is

$F_{Net}= F-f$

Substituting the values in the equation we get

$F_{Net}= 534N-317.52N$

or

$F_{Net}= 216.48$

also,

$F_{Net}= M\times$ acceleration of the safe

Therefore, the acceleration of the metal safe will be

acceleration of the safe= $\frac{F_{Net}}{M}$

or

acceleration of the safe= $\frac{216.48}{108}$

or

acceleration of the safe= $2.00 m/s^2$

Hence, the acceleration of the metal safe will be 2.00 m/s²

3 0

2 years ago

I need help with 1-7 ASAP

elena55 [62]

Refraction. ... Diffraction. ... EM spectrum. ... Intensity. ... Transverse wave. ... Frequency. ... Compression wave.

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3 years ago

What are the strengths and limitations of the doppler and transit methods? What kind of planets are easiest to detect with each

Arturiano [62]

$\huge\mathfrak\red{✔Answer:-}$

Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on

Limitations: yield only planet's mass and orbital properties

3 0

3 years ago