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2 years ago

8

It takes 146./kjmol to break an oxygen-oxygen single bond. calculate the maximum wavelength of light for which an oxygen-oxygen

single bond could be broken by absorbing a single photon.

1 answer:

erastova [34]2 years ago

8 0

To answer this question, we will use the following rule:
E = hc / lambda where:
E is the energy = 146 kJ/mol = 146000 J/mol
h is Planck's constant = 6.62*10^-34 kg m^2 / sec
c is the speed of light = 3*10^8 m/sec
lambda is the wavelength that we need to calculate

Substitute with the given values in the above equation to get lambda as follows:
146000 = (6.62 * 10^-34 * 3 * 10^8) / lambda
lambda = 1.36*10^-30 meters

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You’ve been given the challenge of balancing a uniform, rigid meter-stick with mass M = 95 g on a pivot. Stacked on the 0-cm end

Mariulka [41]

Answer: d = 4750n/3.1+95n

Explanation:

Using the principle of moment to solve the question.

Sum of clockwise moments = sum of anti clockwise moments

Since there are n identical coins with mass 3.1g placed at point 0cm, 1 coin will have mass of 3.1/n grams

Taking moment about the pivot,

Mass 3.1/n grams will move anti-clockwisely while the mass 95g will move in the clockwise direction.

Since its a meter rule (100cm) the distance from the center mass(95g) to the pivot will be 50-d (check attachment for diagram).

To get 'd'

We have 3.1/n × d = 95 × (50-d)

3.1d/n = 4750-95d

3.1d = 4750n-95dn

3.1d+95dn=4750n

d(3.1+95n) = 4750n

d = 4750n/3.1+95n

6 0

2 years ago

A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

KengaRu [80]

a) 219.8 rad/s

b) $20.0 rad/s^2$

c) $2.9 m/s^2$

d) $7005 m/s^2$

e) Towards the axis of rotation

f) $0 m/s^2$

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

$\alpha = \frac{\Delta \omega}{t}$

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

$a_t=(20.0)(0.145)=2.9 m/s^2$

d)

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

$a_c=(219.8)^2(0.145)=7005 m/s^2$

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

$a_t=(0)(0.145)=0 m/s^2$

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

$v=(219.8)(0.145)=31.9 m/s$

7 0

2 years ago

A student attaches a block to a vertical spring so that the block-spring system will oscillate if the block-spring system is rel

vodka [1.7K]

Answer:

Time period of the motion will remain the same while the amplitude of the motion will change

Explanation:

As we know that time period of oscillation of spring block system is given as

$T= 2\pi\sqrt{\frac{M}{k}}$

now we know that

M = mass of the object

k = spring constant

So here we know that the time period is independent of the gravity

while the maximum displacement of the spring from its mean position will depends on the gravity as

$mg = kx$

$x = \frac{mg}{k}$

so we can say that

Time period of the motion will remain the same while the amplitude of the motion will change

4 0

2 years ago

Which of the following best describes a capacitor?

galben [10]

Answer:

B

Explanation:

The capacitor is a component which has the ability to store energy in the form of an electrical charge making a potential difference on those two metal plates

A capacitor consists of two or more parallel conductive (metal) plates. They are electrically seperated by an insulating material (ex: air, mica,ceramic etc.) which is called as Dielectric Layer

Due to this insulating layer, DC current can not flow through the capacitor.But it allows a voltage to be present across the plates in the form of an electrical charge.

4 0

2 years ago

Read 2 more answers

a box weighing 155 N is pushed horizontally down the hall at constant velocity. the applied force is 83 n what is the coefficien

meriva

Answer:

μ = 0.535

Explanation:

On a level floor, normal force = weight.

N = W

Friction force = normal force × coefficient of friction.

F = Nμ

Substitute:

F = Wμ

83 = 155μ

μ = 0.535

Round as needed.

8 0

2 years ago