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Lisa [10]
3 years ago
10

as a mercury atom absorbs a photon of energy as electron in the atom changes from energy level B to energy level E. calculate th

e frequency of the absorb photon.
Physics
2 answers:
Setler79 [48]3 years ago
8 0

Answer:

2.00x 10 14th Hz

Explanation:

Alchen [17]3 years ago
5 0

Answer:

2.99 x 10^14 Hz

Explanation:

E photon= hf (you have to solve for f)

f= E photon/h

f= 1.98 x 10^-19 J / 6.63 x 10^-34 J x s

f=2.99 x 10^14 Hz

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Which statement is true for both types of transistors?
lara31 [8.8K]

For both NPN and PNP this is true:

The base is between the collector and the emitter.

8 0
3 years ago
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Write the answer:<br>physics ... i need help ​
Mariana [72]

Answer:

6 gallons

Explanation:

At 30 mph, the fuel mileage is 25 mpg.

After 5 hours, the distance traveled is:

30 mi/hr × 5 hr = 150 mi

The amount of gas used is:

150 mi × (1 gal / 25 mi) = 6 gal

7 0
3 years ago
An electric wall clock has a second hand 15 cm long. at the tip of his hand, what is the magnitude of the velocity?
Mila [183]

The velocity of the tip of the second hand is 0.0158 m/s

Explanation:

First of all, we need to calculate the angular velocity of the second hand.

We know that the second hand completes one full circle in

T = 60 seconds

Therefore, its angular velocity is:

\omega = \frac{2\pi}{T}=\frac{2\pi}{(60)}=0.105 rad/s

Now we can calculate the velocity of a point on the tip of the hand by using the formula

v=\omega r

where

\omega=0.105 rad/s is the angular velocity

r = 15 cm = 0.15 m is the radius of the circle (the distance of the point from the centre of rotation)

Substituting,

v=(0.105)(0.15)=0.0158 m/s

Learn more about angular motion here:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

7 0
3 years ago
uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station
Alex787 [66]

Answer:

The magnitude of the tangential velocity is v= 0.868 m/s

The magnitude of the resultant acceleration at that point is  a = 4.057 m/s^2

Explanation:

From the question we are told that

      The mass of the uniform disk is m_d = 40.0kg

       The radius of the uniform disk is R_d = 0.200m

       The force applied on the disk is F_d = 30.0N

Generally the angular speed i mathematically represented as

             w = \sqrt{2 \alpha  \theta}

Where \theta is the angular displacement given from the question as

           \theta  = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}

                 =1.257\  rad

   \alpha is the angular acceleration which is mathematically represented as

                    \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

    The moment of inertial is mathematically represented as

                     I = \frac{1}{2} m_dR^2_d

Substituting values

                    I = 0.5 * 40 * 0.200^2

                        = 0.8kg \cdot m^2

Considering the equation for angular acceleration

               \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

Substituting values

               \alph\alpha = \frac{(30.0)(0.200)}{0.8}

                   = 7.5 rad/s^2

Considering the equation for angular velocity

    w = \sqrt{2 \alpha  \theta}

Substituting values

     w =\sqrt{2 * (7.5) * 1.257}

         = 4.34 \ rad/s

The tangential velocity of a given point on the rim is mathematically represented as

                 v = R_d w

Substituting values

                    = (0.200)(4.34)

                     v= 0.868 m/s

The radial acceleration at hat point  is mathematically represented as

            \alpha_r = \frac{v^2}{R}

                  = \frac{0.868^2}{0.200^2}

                 = 3.7699 \ m/s^2

The tangential acceleration at that point is mathematically represented as

               \alpha _t = R \alpha

Substituting values

           \alpha _t = (0.200) (7.5)

                 = 1.5 m/s^2

The magnitude of resultant acceleration at that point is

                 a = \sqrt{\alpha_r ^2+ \alpha_t^2 }

Substituting values

                a = \sqrt{(3.7699)^2 + (1.5)^2}

                   a = 4.057 m/s^2

         

7 0
3 years ago
a body starts from rest with a uniform acceleration of 2m s-2 find the distance covered by the body in 2s
Evgen [1.6K]
Finding acceleration= final velocity-initial velocity/ time taken (or A= V-U/T)

Final speed= 2m
Initial speed= 0m
Time taken= 2 seconds

2-0/2 so it’ll be 1m/s

2-0=0
2/2=

8 0
3 years ago
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