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Black_prince [1.1K]
4 years ago
10

When the tube is filled with mercury vapor, as in this case, a sharp drop in the collected current is observed when the accelera

ting potential reaches 4.9 VV. The current drops suddenly to a negligible value and slowly increases again when the accelerating potential is above 4.9 VV. What is the energy EEE absorbed by the atomic electrons in the mercury atom when the accelerating potential is 4.9 VV
Physics
1 answer:
adell [148]4 years ago
3 0

Answer:

4.9 eV .

Explanation:

It is the case of discharge through mercury tube light . In it , mercury  atoms are exited due to which electrons are sent to higher energy  level . Here current drops to zero because electrons are excited to higher level . Energy are absorbed in quantised manner . Energy absorbed by electrons will be 4.9 eV . That means , difference in energy  between two energy level is  4.9 eV .

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Triss [41]

Answer:

C. 3.375 m/s^2

Explanation:

The acceleration of an object can be found using the equation:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time it takes for the velocity to change from u to v

In this problem:

u = 30 m/s is the initial velocity of Angelica

v = 84 m/s is the final velocity

t is the time

Substituting into the equation, we find the acceleration:

a=\frac{84-30}{16}=3.375 m/s^2

4 0
4 years ago
In order to move a object must have energy.<br><br> True or false and why
andriy [413]

Answer:

True

Explanation:

4 0
2 years ago
3
Misha Larkins [42]

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

KE = \frac{1}{2}mv^2

KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)

Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:

U_s = \frac{1}{2}kx^2

k = Spring Constant (1.14 × 10⁷ N/m)

x = compressed distance of bumper (? m)

Since energy is conserved:

E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2

We can simplify and solve for 'x'.

mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}

Plug in the givens and solve.

x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}

3 0
2 years ago
To find the specific heat capacity of a solid of mass 600 g whose temperature was 40oC, it was placed in a calorimeter that cont
RSB [31]

Explanation:

The specific heat capacity is the heat or energy required to change one unit mass of a substance of a constant volume by 1 °C. The formula is Cv = Q / (ΔT ⨉ m) .

4 0
3 years ago
Read 2 more answers
A force of 140 140 newtons is required to hold a spring that has been stretched from its natural length of 40 cm to a length of
icang [17]

Answer:

The work done in stretching the spring is 0.875 J.

Explanation:

Given that,

Force = 140 N

Natural length = 60-40 = 20 cm

Stretch length of the spring = 65-60 = 5 cm

We need to calculate the spring constant

Using formula of Hooke's law

F= kx

140=k\times20\times10^{-2}

k=\dfrac{140}{20\times10^{-2}}

k=700

We need to calculate the work done

W=\int_{a}^{b}{kx}dx

=\int_{0}^{0.05}{700x}dx

On integration

W=700\times(\dfrac{x^2}{2})_{0}^{0.05}

W=700\times(\dfrac{(0.05)^2}{2}-0)

W=0.875\ J

Hence, The work done in stretching the spring is 0.875 J.

3 0
4 years ago
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