Calculate the maximum kinetic energy of electrons ejected from this surface by electromagnetic radiation of wavelength 236 nm.

ENERGY SAVERS RACE, BRAIN BURNER. This question is about solar cars at the Chuck Norris Institute of Technology, CNIT, in Ocala,

Hunter-Best [27]

Answer:

a) d = 6.0 m

Explanation:

Since car is accelerating at uniform rate then here we can say that the distance moved by the car with uniform acceleration is given as

$d = \frac{(v_f + v_i)}{2} \times t$

here we know that

$v_f = 10 m/s$

$v_i = 0$

$t = 1.2 s$

now we will have

$d = \frac{(10 + 0)}{2}\times 1.2$

$d = 5 \times 1.2$

$d = 6.0 m$

4 0

3 years ago

A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w

amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

10 x 3 + 25 x 0 = v(10 + 25)

30 = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

$K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\$

the final kinetic energy of the package;

$K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J$

the fraction of the initial energy lost;

$= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71$

7 0

3 years ago

A,b, e are complete. Help on the others would be so appreciated!!

bixtya [17]

Answer:

serie Ceq=0.678 10⁻⁶ F and the charge Q = 9.49 10⁻⁶ C

Explanation:

Let's calculate all capacity values

a) The equivalent capacitance of series capacitors

1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5

1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2

1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2

1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225

1 / Ceq = 1,147

Ceq = 0.678 10⁻⁶ F

b) Let's calculate the total system load

Dv = Q / Ceq

Q = DV Ceq

Q = 14 0.678 10⁻⁶

Q = 9.49 10⁻⁶ C

In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C

c) The potential difference

ΔV = Q / C5

ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶

ΔV = 1,725 V

d) The energy stores is

U = ½ C V²

U = ½ 0.678 10-6 14²

U = 66.4 10⁻⁶ J

e) Parallel system

Ceq = C1 + C2 + C3 + C4 + C5

Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶

Ceq = 22.7 10⁻⁶ F

f) In the parallel system the voltage is maintained

Q5 = C5 V

Q5 = 5.5 10⁻⁶ 14

Q5 = 77 10⁻⁶ C

g) The voltage is constant V5 = 14 V

h) Energy stores

U = ½ C V²

U = ½ 22.7 10-6 14²

U = 2.2 10⁻³ J

8 0

3 years ago

The Cassegrain design provides more compact (shorter) telescopes. Why? (Examine figures 2.4.2 and 2.4.3). The shorter design is

Ipatiy [6.2K]

Answer:

Because the light reflects multiple times until it gets to the Cassegrain focus.

Explanation:

The Cassegrain design can be seen in a reflecting telescope. In this type of design the light is collected by a concave mirror, and then intercepted by a secondary convex mirror, and sends it down to a central opening in the primary mirror (concave mirror), in which a detector is placed (Cassegrain focus)

Since, the light is reflected many times due to Cassegrain design, that leads to shorter telescopes.

5 0

3 years ago

Please help

iVinArrow [24]

Answer is 4,400,000 kg • m/s

4 0

3 years ago