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allsm [11]
3 years ago
11

A skateboarder rolls horizontally off the top of a staircase and lands at the bottom. The staircase has a horizontal length of 1

2.0 with a vertical drop of 2.00. We can ignore air resistance.
Physics
1 answer:
TiliK225 [7]3 years ago
3 0

Answer: 18.8 m/s

Explanation:

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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
A jogger runs by a river with a velocity of 7 m s relative to the ground. A leaf floating on the river moves with a velocity of
Hunter-Best [27]

Answer:

The velocity of the leaf relative to the jogger is 5 m/s.                    

Explanation:

Given that,

Velocity of jogger wrt to the ground, V_j=7\ m/s

velocity of leaf wrt the ground, v_i=2\ m/s

We need to find the velocity of the leaf relative to the jogger. Let it is equal to V. So, it is given by :

V=v_j-v_i\\\\V=7-2\\\\V=5\ m/s

So, the velocity of the leaf relative to the jogger is 5 m/s. Hence, this is the required solution.

3 0
3 years ago
The reaction is at dynamic equilibrium.
DiKsa [7]

Answer:

Nitrogen and hydrogen combine at the same rate that ammonia breaks down.

Explanation:

7 0
3 years ago
If 50 km thick crust having an average density of 3.0 g/cm3 has a surface elevation of 2.5 km above sea level, what would you pr
RUDIKE [14]

Answer:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

Explanation:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

5 0
3 years ago
You want to slide a 0.39 kg book across a table. If the coefficient of kinetic friction is .21, what force is required to move t
uranmaximum [27]
Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together.  This is done because the friction force is going to have to be compensated for.  We will need that much more force than we otherwise would to achieve the desired acceleration:

F_{NoFric}=ma=0.39kg \times0.18 \frac{m}{s^2}  =0.0702N

The friction force will be given by the normal force times the coefficient of friction.  Here the normal force is just its weight, mg

F_{Fric}=0.39kg \times 9.8 \frac{m}{s^2} \times 0.21=0.803N

Now the total force required is:

0.0702N+0.803N=0.873N

5 0
3 years ago
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