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andreyandreev [35.5K]
3 years ago
13

A 3.0-kg block moves up a 40o incline with constant speed under the action of a 26-N force acting up and parallel to the incline

. What magnitude force must act up and parallel to the incline for the block to move down the incline at constant velocity?
Physics
1 answer:
CaHeK987 [17]3 years ago
3 0

Answer:

it is very hard question for me sorry i cant solve it

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An electron accelerates through a 12.5 V potential difference, starting from rest, and then collides with a hydrogen atom, excit
Yanka [14]

Answer:

Initial state    Final state

     3           ⇒        2

     3           ⇒        1

     2          ⇒         1

Explanation:

For this exercise we must use Bohr's atomic model

         E = - 13.606 / n²

where is the value of 13.606 eV is the energy of the ground state and n is the integer.

The energy acquired by the electron in units of electron volt (eV)

          E = e V

          E = 12.5 eV

all this energy is used to transfer an electron from the ground state to an excited state

        ΔE = 13.6060 (1 / n₀² - 1 / n²)

the ground state has n₀ = 1

       ΔE = 13.606 (1 -  1/n²)

        1 /n² = 1 - ΔE/13,606

         1 / n² = 1 - 12.5 / 13.606

         1 / n² = 0.08129

          n = √(1 / 0.08129)

          n = 3.5

 since n is an integer, maximun is

         n = 3

because it cannot give more energy than the electron has

From this level there can be transition to reach the base state.

 

Initial state    Final state

     3           ⇒        2

     3           ⇒        1

     2          ⇒         1

8 0
3 years ago
In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vi
irinina [24]

To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where,

m_{1,2}= Mass of each object

v_{1,2} = Initial velocity of each object

v_f= Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

m_2v_2 = (m_1+m_2)v_f

(0.42)(21) = (90+0.42)v_f

v_f = 0.0975m/s

Therefore the velocity right after catching the ball is 0.0975m/s

8 0
3 years ago
A 5c charge experiences a force of 40n when put at a certain location in space. What is the electric field at that location?
stepladder [879]
The 'strength' of the electric field is the force on 1C of charge at that point.

At this 'certain location', the field is 40/5 = 8 newtons per coulomb = <u>8 volts</u>
3 0
3 years ago
If you walk 30 meters forwards, and then turn around and walk 25 meters backwards, what is the distance that you walked? What di
xeze [42]

Given :

Walk in forward direction is 30 m .

Walk in backward direction is 25 m .

To Find :

The distance and displacement .

Solution :

We know , distance is total distance covered and displacement is distance between final and initial position .

So , distance travelled is :

D = 30 + 25 m = 55 m .

Now , we first move 30 m in forward direction and then 25 m in backward direction .

So , displacement is :

D = 30 - 25 m = 5 m .

Therefore , distance and displacement covered is 55 m and 5 m respectively .

Hence , this is the required solution .

5 0
3 years ago
What is the relationship between increase speed of a car and energy in a<br> collision?
sweet [91]
The relationship between a car and energy is that the car uses gas to produce speed within energy needs to be powered
6 0
3 years ago
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