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3 years ago

When a mule stops suddenly, the packages

1 answer:

5 0

The answer here would be the first law of motion. An object in motion will stay in motion unless acted upon by another force.

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A motorcycle is moving at 30 m/s when the rider applies the brakes, giving the motorcycle a constant deceleration. During the 3.

shutvik [7]

Answer:

The motorcycle travelled 69.73 m during these 3.1 s.

Explanation:

In order to calculate the distance that the motorcycle travelled we first need to obtain the acceleration rate that was used to brake the vehicle. We do that by using the following formula:

a = (V_final - V_initial)/(t) = (15 - 30)/(3.1) = -4.84 m/s^2

The distance is given by the following formula:

S = (V_final^2 - V_initial^2)/(2*a)

S = (15^2 - 30^2)/[2*(-4.84)] = (225 - 900)/(-9.68) = -675/(-9.68) = 69.73 m

The motorcycle travelled 69.73 m during these 3.1 s.

5 0

3 years ago

A photon ionizes a hydrogen atom from the ground state. The liberated electron 11. recombines with a proton into the first excit

anygoal [31]

Answer:

a) 23.2 e V

b) energy of the original photon is 36.8 eV

Explanation:

given,

energy at ground level = -13.6 e V

energy at first exited state = - 3.4 e V

A photon of energy ionized from ground state and electron of energy K is released.

h ν₁ - 13.6 = K

K combine with photon in first exited state giving out photon of energy

$h\nu_2 =\dfrac{hc}{\lambda}=\dfrac{12400}{466}$

= 26.6 e V

h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°

K + ( 3.4 ) = 26.6 e V

a) energy of free electron

K = 26.6 - 3.4 = 23.2 e V

b) energy of the original photon

h ν₁ - 13.6 = K

h ν₁ = 23.2 + 13.6

= 36.8 e V

energy of the original photon is 36.8 eV

3 0

3 years ago

What are the physical and chemical properties of sodium?

Hitman42 [59]

Answer:

Physical Properties of Sodium

Atomic number 11

Melting point 97.82°C (208.1°F)

Boiling point 881.4°C (1618°F)

Volume increase on melting 2.70%

Latent heat of fusion 27.0 cal/g

Lenntech Water treatment & purification

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Home Periodic table Elements Sodium

Sodium - Na

Chemical properties of sodium - Health effects of sodium - Environmental effects of sodium

Atomic number

Atomic mass

22.98977 g.mol -1

Electronegativity according to Pauling

0.9

Density

0.97 g.cm -3 at 20 °C

Melting point

97.5 °C

Boiling point

883 °C

Vanderwaals radius

0.196 nm

Ionic radius

0.095 (+1) nm

Isotopes

Electronic shell

[Ne] 3s1

Energy of first ionisation

495.7 kJ.mol -1

8 0

2 years ago

A truck is driving over a scale at a weight station. When the front wheels drive over the scale, the scale reads 5800 N. When th

aev [14]

Answer:

$x_2=1.60m$

Explanation:

From the Question We are told that

Initial Force $F_1=5800N$

Final Force $F_2=6500N$

Distance between the front and rear wheels \triangle x=3.20 m

Since

$\triangle x=3.20 m$

Therefore

$x_1+x_2=3.20$

$x_1=3.20-x_2$

Generally the equation for The center of mass is at x_2 is mathematically

given by

$x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}$

$x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}$

$2*F_1*x_2 =3.20F_1$

$x_2=1.60m$

6 0

2 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

3 years ago