PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS AND I NEED ALL CORRECT ANSWERS ONLY!!!

• What value is used to identify each element in the periodic table?

bearhunter [10]

Answer:

protons

Explanation:

5 0

2 years ago

An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piec

lutik1710 [3]

Answer:

The temperature of the metal is $T_m = 376.8 ^o C$

Explanation:

From the question we are told that

The mass of the metal is $M = 60 \ kg$

The specific heat of the metal is $c_p = 0.1027 kcal/(kg \cdot ^oC)$

The mass of the oil is $M_o = 810 \ kg$

The temperature of the oil is $T_o = 35^oC$

The specific heat of oil is $c_o = 0.7167 kcal/(kg \cdot ^oC )$

The equilibrium temperature is $T_e = 39 ^oC$

According to the law of energy conservation

Heat lost by metal = heat gained by the oil

So

The quantity of heat lost by the metal is mathematically represented as

$Q = - Mc_p \Delta T$

=> $Q = -Mc_p (T_m - T_c)$

Where $T_ m$ the temperature of metal before immersion

The negative sign show heat lost

The quantity of gained t by the metal is mathematically represented as

$Q = M_o c_o \Delta T$

=> $Q = M_o c_o (T_c - T_o)$

So

$Mc_p (T_m - T_c) = M_o c_o (T_c - T_o)$

substituting values

$- 60 * 0.1027 (T_m - 39) = 810 * 0.7167 * (39 - 35)$

=> $T_m = 376.8 ^o C$

6 0

3 years ago

What is power, and what is its relationship to voltage and amperage?

kvasek [131]

The relationship between amperage, voltage, and power is that power equals the amperage quantity times the amount of voltage.

Power: is the amount of energy transferred or converted per unit time.

7 0

3 years ago

A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg

saveliy_v [14]

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

Following formulas will be used.

$x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2$

$F =ma$

$V(t) = V_{o} +at$

$x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2$

(a) Calculating velocity right before going up the ramp.

Wooden block is going on a straightaway and has net for on it.

$F_{n} =F-F_{s} = F-uF_{n} = 100N-0.4*9.8m/s^2*5kg$ = $81N$

and this force produces acceleration of

$a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .$

With this acceleration, wooden block would reach at the foot of ramp in.

$x(t) = 12m = 16.2m/s^2*t^2$

$t = 1.217s$

and final velocity will be

$v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.$

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

$V= V_{h}cos(20) = 18.5288m/s$

and deceleration along the ramp is

$a = \frac{F_{s} }{m}$

Where $F_{s}$ force of friction along the inclined plane.

$F_s = uF_n = u*m*a$

$a = 9.8m/s^2*cos(20) = 9.2089m/s^2$

is a component of g along normal of the inclined plane.

$F_{s} = 0.25*5kg*9.2089m/s^2$

$= 11.5112N$

$a = \frac{11.5112N}{5kg} = 2.3022m/s^2$

And with this deceleration time needed to get wooded block to stop is.

$v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0$

$t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s$

and in that time wooden block would travel

$x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m$

This is how up wooden box will go before coming to stop.

3 0

3 years ago

When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a

sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = $\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}$

= $4.59\times 10^{-19} \ J$

or,

= $\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}$

= $2.87 \ eV$

(b)

As we know,

⇒ $Vq=\frac{hc}{\lambda}-\Phi_0$

By substituting the values, we get

⇒ $1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0$

⇒ $\Phi_0=2.3\times 10^{-19} \ J$

or,

⇒ $=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}$

⇒ $=1.4375 \ eV$

5 0

3 years ago