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dlinn [17]
2 years ago
8

A projectile of mass M, initially at rest, is acted upon by a net force [including gravity] that increases quadratically with ti

me as the projectile accelerates in a vertical gun barrel of length L, Fy = bt2 (1) where b is a constant. After leaving the gun barrel, the time-dependent force vanishes and the projectile rises under gravitational acceleration to a maximum height h above the end of the barrel.
Physics
1 answer:
Elden [556K]2 years ago
7 0

Answer:

maximum height is y = b²/18g √ (12L/b)³

Explanation:

Let's analyze the situation first we have a projectile subjected to an acceleration depends on time, so we must use the definition of acceleration to find the speed when it is at distance L, then we will use the projectile launch equations

Acceleration dependent on t

     a = dv / dt

     dv = adt

     ∫dv =∫ (b t²) dt

     v = b t³ / 3

The initial speed is zero for zero time

 

we use the definition of speed

     v = dy / dt

     dy = v dt

     ∫dy = ∫b t³ / 3 dt

     y = b/3   t⁴ / 4

     y = b/12 t⁴

we evaluate from the initial point where the height is zero for the zero time

Let's calculate the time to travel the length (y = L) of the canyon

     t = (12 y / b) ¼

     t = (12 L / b) ¼

Taking the time, we can calculate the projectile's output speed

     v = b/3  ( (12 L / b)^{3/4}

 

This is the speed of the body, which is the initial speed for the projectile launch movement. Let's calculate the highest point where the zero speed

      Vy² = v₀² - 2 g y

       0 = Vo² - 2 g y

      2 g y = v₀²

      y =  v₀²/ 2g

      y = 1/2g    [b/3 (12L / b^{3/4}) ] 2

      y = 1 / 2g [b²/9  (12L/b)^{3/2}]

      y = b²/18g √ (12L/b)³

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Temka [501]

Total time in between the dropping of the stone and hearing of the echo = 8.9 s

Time taken by the sound to reach the person = 0.9 s

Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds

Initial speed (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s^2

Time taken (t) = 8 seconds

Let the depth of the well be h.

Using the second equation of motion:

h = ut + \frac{1}{2}\times a \times t^2

h = 0 \times 8 + \frac{1}{2} \times 9.8 \times 8^2

h = 313.6 m

Hence, the depth of the well is 313.6 m

4 0
2 years ago
An electron is at the origin. (a) Calculate the electric potential VA at point A, x 5 0.250 cm. (b) Calculate the electric poten
saw5 [17]

Answer:

a)  V_a = -5.7536 10⁺⁷ V , b) Vb = -1.92 10⁻⁷ V  c) the sign of the potential change

Explanation:

The electrical potential for a point charge

     V = k q / r

Where k is the Coulomb constant that you are worth 8.99 10⁹ N m² / C²

a) potential At point x = 0.250 cm = 0.250 10-2m

    V_a =  -8.99 10⁹ 1.6 10⁻¹⁹ /0.250 10⁻²

    V_a = -5.7536 10⁺⁷ V

b) point x = 0.750 cm = 0.750 10-2

    Vb = 8.99 10⁹ (-1.6 10⁻¹⁹) /0.750 10⁻²

    Vb = -1.92 10⁻⁷ V

potemcial difference

    ΔV = Vb- Va

    V_ba = (-5.7536 + 1.92) 10⁻⁷

    V_ba = -3.83 10⁻⁷ V

c) To know what would happen to a particle, let's use the relationship between the potential and the electric field

     ΔV = E d

The force on the particle is

     F = q₀ E

     F = q₀ ΔV / d

We see that the force on the particle depends on the sign of the burden of proof. Now the burden of proof is negative to pass between the two points you have to reverse the sign of the potential, bone that the value should be reversed

          V_ba = 0.83 10⁻⁷ V

5 0
2 years ago
Four different observers are standing in a straight line on a street and hear a siren from a police car. Each person recorded th
Amanda [17]

Answer:

Wycleff is on block 1, Lilly is on block 4, Emilia is on block 12, and Quincy is on block 17.

Explanation:

Wycleff was at block 1 and heard a low pitch sound the whole time, so the police car must have been moving away from him.

Lilly observed was in block 4 change in pitch first.  So the car must have passed her first.

Emilia was at block 12 observed a Doppler effect after Lilly.  So the car passed her after passing Lilly

Quincy was at block 17 so she heard a high pitch sound the whole time, so the police car must have been moving toward him.

7 0
2 years ago
Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the s
Flura [38]

Answer:

the correct answer is A, the object goes 4 times as far

Explanation:

This is a projectile launching approach. Where the parameter we are controlling is the initial speed and they ask us how far it goes from the initial one. Let's calculate the range with a speed (vo)

        R1 = v₀² sin 2θ / g

Now let's double vo, the new speed is

         v = 2 v₀

We calculate the scope

         R2 = (2v₀)² sin 2θ / g

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Therefore the correct answer is A, the object goes 4 times further

5 0
2 years ago
A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° below the
Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, (\theta) = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ F_x=Fcos(\theta)          and             ⇒ F_y=Fsin (\theta)

⇒ F_x=15cos(15)                           ⇒ F_y=15sin (15)

⇒ Applying concept of  forces.

⇒ \sum F_x=F_n_e_t =F-f

⇒ F_n_e_t =F-f

⇒ ma =F-f       <em>  ...Newtons second law Fnet = ma</em>

⇒ a =\frac{F-f}{m}              

⇒ Plugging the values.

⇒ a =\frac{15cos(15)-0}{40}     <em>...f is the friction which is zero here.</em>

⇒ a =\frac{14.48}{40}

⇒ a=0.362\ ms^-^2

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0
2 years ago
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