Mitch holds a pumpkin at waist level. How can he add potential energy to the pumpkin?

when light falls obliquely, there is a change in direction as it moves from one medium to another but when it falls perpendicula

exis [7]

Answer:

wait in comments....................

3 0

1 year ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

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

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

3 years ago

A glass tube 1.50 meters long and open at one end is weighted to keep it vertical and is then lowered to the bottom of a lake. W

Lilit [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The total pressure is $P_T = 10.79*10^{5} N/m^2$

The temperature at the bottom is $T_b = 284.2 \ K$

Explanation:

From the question we are told that

The length of the glass tube is $L = 1.50 \ m$

The length of water rise at the bottom of the lake $d = 1.33 \ m$

The depth of the lake is $h = 100 \ m$

The air temperature is $T_a = 27 ^oC = 27 +273 = 300 \ K$

The atmospheric pressure is $P_a = 1.01 *10^{5} N/m$

The density of water is $\rho = 998 \ kg/m^3$

The total pressure at the bottom of the lake is mathematically represented as

$P_T = P_a + \rho g h$

substituting values

$P_T = 1.01*10^{5} + 998 * 9.8 * 100$

$P_T = 10.79*10^{5} N/m^2$

According to ideal gas law

At the surface the glass tube not covered by water at surface

$P_a V_a = nRT_a$

Where is the volume of

$P_a *A * L = nRT_a$

At the bottom of the lake

$P_T V_b = nRT_b$

Where $V_b$ is the volume of the glass tube not covered by water at bottom

and $T_b$ i the temperature at the bottom

So the ratio between the temperature at the surface to the temperature at the bottom is mathematically represented as

$\frac{T_b}{T_a} = \frac{d * P_T}{P_a * h}$

substituting values

$\frac{T_b}{27} = \frac{0.133 * 10.79 *10^5}{1.01 *10^{5} * 1.5}$

=> $T_b = 284.2 \ K$

3 0

3 years ago

What is the correct order of the steps in the scientific method

stira [4]

<h2>what is the correct order of the steps in the scientific method</h2>

Make a hypothesis, test the hypothesis, analyze the results, ask a question, draw conclusions, communicate results.

hope it helps

6 0

2 years ago

How much work does a 65 kg person climbing a 2000 m high cliff do?

qaws [65]

The answer for the following question is explained below.

Therefore the work done is 130 kilo Joules.

Explanation:

Work:

A force causing the movement or displacement of an object.

Given:

mass of the person (m) = 65 kg

height of the cliff (h) = 2000 m

To calculate:

work done (W)

We know;

According to the formula:

W = m × g × h

Where;

m represents mass of the person

g represents the acceleration due to gravity

where the value of g is;

g = 10 m/ s²

h represents the height of the cliff

From the above formula;

W = 65 × 10 × 2000

W = 130,000 J

W = 130 Kilo Joules

Therefore the work done is 130 kilo Joules.

3 0

2 years ago