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likoan [24]
3 years ago
9

(pleases show work)

Physics
1 answer:
Feliz [49]3 years ago
4 0

a. 46 m/s east

The jet here is moving with a uniform accelerated motion, so we can use the following suvat equation to find its velocity:

v=u+at

where

v is the velocity calculated at time t

u is the initial velocity

a is the acceleration

The jet in the problem has, taking east as positive direction:

u = +16 m/s  is the initial velocity

a=+3 m/s^2  is the acceleration

Substituting t = 10 s, we find the final velocity of the jet:

v=16 + (3)(10)=46 m/s

And since the result is positive, the direction is east.

b. 310 m

The displacement of the jet can be found using another suvat equation

s=ut+\frac{1}{2}at^2

where

s is the displacement

u is the initial velocity

a is the acceleration

t is the time

For the jet in this problem,

u = +16 m/s  is the initial velocity

a=+3 m/s^2  is the acceleration

t = 10 s is the time

Substituting into the equation,

s=(16)(10)+\frac{1}{2}(3)(10)^2=310 m

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3 years ago
Block A can slide relative to block B which, in turn, can slide on a perfectly smooth horizontal plane. If the initial velocity
Anna11 [10]

Answer:

the final velocity of the two blocks is v = \frac{mv_o}{m+M}

the distance that A slides relative to B is S = \frac{v_o^2M}{2 \mu g (M+m)}  

Explanation:

From the diagram below;

acceleration of A relative to B is : a = - ( \mu g  + \frac{ \mu mg}{M})

where

v = u + at

0 = v_o + ( - \mu g - \frac{\mu m g }{M})t

Making t the subject of the formula; we have:

t = \frac{v_o M}{(\mu g )(M+m)}

v^2 = u^2 +2 as\\\\0^2 = v_o^2 - 2 (\mu g ) (\frac{M+m}{M})S\\\\

S = \frac{v_o^2M}{2 \mu g (M+m)}  which implies the distance that A slides relative to B.

The final velocities of the two blocks can be determined as follows:

v = u + at

v = v_o - \mu g \frac{v_oM}{\mu g (M+m)}\\\\v = \frac{\mu g mv_o}{m+M}\\\\

v = \frac{mv_o}{m+M}

Thus, the final velocity of the two blocks is v = \frac{mv_o}{m+M}

4 0
2 years ago
A 10 g bullet moving horizontally with a speed of 2000 m/s strikes and passes through a 4.0 kg block moving with a speed of 4.2
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Answer:

The answer is "512 J".

Explanation:

bullet mass m_1 = 10 g= 10^{-2} \ kg\\\\

initial speed u_1 = 2\ \frac{Km}{s}= 2000\ \frac{m}{s}\\\\

block mass m_2 = 4\ Kg

initial speed v_2 =-4.2 \frac{m}{s}

final speed v_2= 0

Let v_1 will be the bullet speed after collision:

throughout the consevation the linear moemuntum

\to M_1V_1+m_2v_2=M_1U_1+m_2u_2\\\\\to (10^{-2} kg) V_1 +0 = (10^{-2} kg)(2000 \frac{m}{s}) + (4 \ kg)(-4.2 \frac{m}{s}) \\\\\ \to 10^{-2} v_1 = 20 -16.8\\\\

                = 320 \frac{m}{s}

The kinetic energy of the bullet in its emerges from the block

k=\frac{1}{2} m_1 v_1^2

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3 0
3 years ago
An electron accelerates through a 12.5 V potential difference, starting from rest, and then collides with a hydrogen atom, excit
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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

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