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2 years ago

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Caelyn wanted to find out what shampoo made her hair the shiniest . Everyday she washed her hair with different shampoos and the

n rated how shiny her hair was, on a scale from 1-10. She used Pantene, Herbal Essences, L’Oreal, and just water

1 answer:

Arte-miy333 [17]2 years ago

8 0

Answer:

IV: type of shampoo used

DV: what shampoo made her hair the skinniest

control: water

Constant: rated from scale from 1-10

Explanation:

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A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g

Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

$Q_{cube}=Q_{water}$ (1)

But Q is related with mass (m), specific heat (c) and changes in temperature ( $\varDelta T$ )in the next way:

$Q=cm\varDelta T$ (2)

Using (2) on (1):

$c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}$

$(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

$(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})$

Because we have an equality and 0.385 < 4.186 then $\varDelta T_{cooper}>\varDelta T_{waterer}$ to conserve the equality

4 0

2 years ago

The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper

meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, $v_{d}$ , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

$v_{d}$ = $\frac{I}{qnA}$ ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = $\frac{I}{A}$ = $\frac{E}{p}$

Equation (i) can then be written as follows;

$v_{d}$ = $\frac{J}{qn}$ = $\frac{E}{qnp}$

$v_{d}$ = $\frac{E}{qnp}$ ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

$v_{d}$ = $\frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}$

$v_{d}$ = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

4 0

2 years ago

What would the speed of each particle be if it had the same wavelength as a photon of yellow light (????=575.0 nm)? Proton (mass

PilotLPTM [1.2K]

Answer:

Proton: v=0.689 m/s

Neutron: v=0.688 m/s

Electron: v=1265.078 m/s

Alpha particle: v=0.173 m/s

Explanation:

De Broglie equation allows you to calculate the “wavelength” of an electron or any other particle or object of mass m that moves with velocity v:

λ= $\frac{h}{mv}$

h is the Planck constant: 6.626×10⁻³⁴ $\frac{kg.m^2}{s}$

We know that the wavelength of the particle is 575 nm (575×10⁻⁹m), so we find the velocity v for each particle:

λ= $\frac{h}{mv}$

v=h÷(mλ)

<u>Proton:</u>

m=1.673×10⁻²⁴ g · $\frac{1kg}{1000g}$ =1.673×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴ $\frac{kg.m^2}{s}$ ÷(1.673×10⁻²⁷ kg×575×10⁻⁹m)

v=0.689 m/s

<u>Neutron:</u>

m=1.675×10⁻²⁴ g · $\frac{1kg}{1000g}$ =1.675×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴ $\frac{kg.m^2}{s}$ ÷(1.675×10⁻²⁷ kg×575×10⁻⁹m)

v=0.688 m/s

<u>Electron:</u>

m= 9.109×10⁻²⁸ g · $\frac{1kg}{1000g}$ =9.109×10⁻³¹ kg

v=h÷(mλ)

v=6.626×10⁻³⁴ $\frac{kg.m^2}{s}$ ÷(9.109×10⁻³¹ kg×575×10⁻⁹m)

v=1265.078 m/s

<u>Alpha particle:</u>

m=6.645×10⁻²⁴ g · $\frac{1kg}{1000g}$ =6.645×10⁻²⁷ kg

v=h÷(mλ)

v=6.626×10⁻³⁴ $\frac{kg.m^2}{s}$ ÷(6.645×10⁻²⁷ kg×575×10⁻⁹m)

v=0.173 m/s

3 0

2 years ago

Read 2 more answers

Describe the energy transformations that take place when a skier starts skiing down a hill but after a time is brought to rest b

Andrews [41]

<span>The skier will transform their gravitational energy into mostly kinetic energy (with a minor amount transformed into heat from the friction of the skis across the snow and air friction). Once the skier hits the snowdrift, their kinetic energy is transferred into the snow which moves when they strike it due to the kinetic energy that is now in the snow. Along with again a minor amount of heat energy transferred as they move through the snowdrift.</span>

6 0

2 years ago

A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational

a_sh-v [17]

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: $F = 49.06N$

Pebble: $F = 29.44N$

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: $a =9.8m/s^{2}$

Pebble: $a =9.8m/s^{2}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{m1m2}{r^{2}}$ (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em> $m = 5.0 Kg$ <em>:</em>

m1 will be equal to the mass of the Earth $m1 = 5.972×10^{24} Kg$ and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth $r = 6371000m$ .

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}$

$F = 49.06N$

Newton's second law can be used to know the acceleration.

$F = ma$

$a =\frac{F}{m}$ (2)

$a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}$

$a =9.8m/s^{2}$

<em>Case for the pebble </em> $m = 3.0 Kg$ <em>:</em>

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}$

$F = 29.44N$

$a =\frac{F}{m}$

$a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}$

$a =9.8m/s^{2}$

3 0

2 years ago

Read 2 more answers