Answer:

**Answer:**

306.8264448 m

47.0016 m/s

**Explanation:**

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled by car

Distance traveled by truck

In order to overtake both distances should be equal

**The distance the car has to travel is 306.8264448 m**

**The speed of the car when it overtakes the truck is 47.0016 m/s**

1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position will the speed of the mass be 25% of its maximum speed?

State the following forms of electromagnetic radiation in increasing order of wavelength.Radiowaves, gamma rays, x-rays, infrared radiation, visible light

Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Rutherford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus. What is the strength of the nucleus' electric field at the orbital radius of the electrons? What is the kinetic energy of an electron in a circular orbit around the gold nucleus?

If half the kinetic energy of the initially moving object (m1m1) is transferred to the other object (m2m2), what is the ratio of their masses? Express your answers using three significant figures separated by a comma.

At what partial pressure are argon atoms expected to have a free travel of approximately 5 µm, if the gas is at a temperature of 400 K? The cross section of collision, σ, or Argon is 0.28 nm2Ar molar mass is 39.9 g/mole

State the following forms of electromagnetic radiation in increasing order of wavelength.Radiowaves, gamma rays, x-rays, infrared radiation, visible light

Before the development of quantum theory, Ernest Rutherford's experiments with gold atoms led him to propose the so-called Rutherford Model of atomic structure. The basic idea is that the nucleus of the atom is a very dense concentration of positive charge, and that negatively charged electrons orbit the nucleus in much the same manner as planets orbit a star. His experiments appeared to show that the average radius of an electron orbit around the gold nucleus must be about 10−1010−10 m. Stable gold has 79 protons and 118 neutrons in its nucleus. What is the strength of the nucleus' electric field at the orbital radius of the electrons? What is the kinetic energy of an electron in a circular orbit around the gold nucleus?

If half the kinetic energy of the initially moving object (m1m1) is transferred to the other object (m2m2), what is the ratio of their masses? Express your answers using three significant figures separated by a comma.

At what partial pressure are argon atoms expected to have a free travel of approximately 5 µm, if the gas is at a temperature of 400 K? The cross section of collision, σ, or Argon is 0.28 nm2Ar molar mass is 39.9 g/mole

b. With what speed does an electron exit the electron gun if its entry speed is close to zero? [Note: ignore relativity]

c. If the capacitance of the plates is 1 nF, how much charge is stored on each plate? How many extra electrons does the cathode have?

d. If you wanted to push an electron from the anode to the cathode, how much work would you have to do?

Answer:

A. 2.083 MV/m from anode to cathode.

B. 93648278.15 m/s

C. 2.5x10^-5 C and there are about 1.56x10^14 electrons

D. 4x10^-15 Joules

Explanation:

Voltage V across plate is 25 kV = 25x10^3 V

Distance apart x = 1.2 cm = 1.2x10^-2 m

A. Electric field strength is the potential difference per unit distance

E = V/x = 25x10^3/1.2x10^-2 = 2083333.3 V/m

= 2.083 MV/m

B. Energy of electron is electron charge times the voltage across

i.e eV

Charge on electron = 1.6x10^-19 C

Energy of electron = 1.6x10^-19 x 25x10^3 = 4x10^-15 Joules

Mass of electron m is 9.12x10^-31 kg

Kinetic energy of electron = 0.5mv^2

Where v is the speed

4x10^-15 = 0.5 x 9.12x10^-31 x v^2

v^2 = 8.77x10^15

v = 93648278.15 m/s

C. From Q = CV

Q = charge

C = capacitance = 1 nF 1x10^-9 F

V = voltage = 25x10^3 V

Q = 1x10^-9 x 25x10^3 = 2.5x10^-5 C

Total number of electrons = Q/e

= 2.5x10^-5/1.6x10^-19 = 1.56x10^14 electrons

D. To push electron from cathode to anode, I'll have to do a work of about

4x10^-15 Joules

b. Wavelength of the wave?

c. Write down a mathematical expression for the wave, substituting numbers for the variables

**Answer:**

(a) The speed of the wave, v is 4.2 m/s

(b) Wavelength of the wave, λ is 0.35 m

(c) mathematical expression of the wave, Y = 0.036sin(5.71πx - 24πt)

**Explanation:**

Given;

tension on the string, T = 15 N

Linear density, μ = 0.85 kg/m

amplitude of the wave, A = 3.6 cm = 0.036 m

frequency of the wave, f = 12 Hz

(a) The speed of the wave, v is calculated as;

(b) Wavelength of the wave, λ

v = fλ

λ = v / f

λ = 4.2 / 12

**λ = 0.35 m**

(c) mathematical expression of the wave;

**Answer:**

a) h=3.16 m, b) v_{cm }^ = 6.43 m / s

**Explanation:**

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

Em₀ = U = mg h

final point. Lowest point

= K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

K = ½ m + ½ w²

angular and linear speed are related

v = w r

w = v / r

K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

Em₀ = Em_{f}

mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)

h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

I_{cm} = ⅔ m r²

we substitute

h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

h = ½ v_{cm }^{2}/g 5/3

h = 5/6 v_{cm }^{2} / g

let's calculate

h = 5/6 6.1 2 / 9.8

h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

I_{cm} = ½ m r²

we substitute

v_{cm } = √ [2gh / (1 + ½)]

v_{cm } = √(4/3 gh)

let's calculate

v_{cm } = √ (4/3 9.8 3.16)

v_{cm }^ = 6.43 m / s

**Answer:**

When you jump off a train, you jump off a certain height and your downwards (vertical) velocity is zero. But your forward (horizontal) velocity is not. You will hit the ground on split second with your horizontal velocity practically the same as the train.

**Explanation:**

you be in serious injury.

If its initial angular speed was 3890rpm , what is the magnitude of its angular deceleration? (|?| in revs/s^2 )

Part B

How many revolutions did the centrifuge complete after being turned off?

**Answer:**

**Explanation:**

Given:

time of constant deceleration,

A.

initial angular speed,

__Using equation of motion:__

B.

Using eq. of motion for no. of revolutions, we have:

(b) Find the ball's speed at impact.

(c) Find the horizontal range of the ball.

**Answer:**

B

**Explanation:**