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algol [13]
3 years ago
10

A projectile rolls off a cliff with a velocity of 40 m/s. The cliff is 60 meters high.

Physics
1 answer:
masya89 [10]3 years ago
6 0

Answer:

1) t = 3.45 s, 2)  x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

      y = y₀ + v_{oy} t - ½ g t²

When leaving the cliff the speed is horizontal  v_{oy}= 0 and at the bottom of the cliff y = 0

      0 = y₀ - ½ g t2

      t = √ 2y₀ / g

      t = √ (2 60 / 9.8)

      t = 3.45 s

2) The horizontal distance traveled

     x = v₀ₓ t

     x = 40 3.45

     x = 138 m

3) The vertical velocity at the point of impact

     v_{y} = I go - g t

     v_{y} = 0 - 9.8 3.45

     v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

   v = √ (vₓ² + v_{y}²)

   v = √ (40² + 33.8²)

   v = 52.37 m / s

5) angle of impact

    tan θ = v_{y} / vx

    θ = tan⁻¹ v_{y} / vx

    θ = tan⁻¹ (-33.81 / 40)

    θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis

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Answer:

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While bucket is falling downwards we have force equation of the bucket given as

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so we have

T = \frac{1}{2}MR^2(\frac{a}{R^2})

T = \frac{1}{2}Ma

now we have

mg = (\frac{M}{2} + m)a

a = \frac{mg}{(\frac{M}{2} + m)}

a = \frac{15 \times 9.81}{(6 + 15)}

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now we have

T = \frac{12 \times 7}{2}

T = 42 N

Part b)

speed of the bucket can be found using kinematics

so we have

v_f^2 - v_i^2 = 2 a d

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Part c)

now in order to find the time of fall we can use another equation

v_f - v_i = at

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Part d)

as we know that cylinder is at rest and not moving downwards

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First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:

W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2

where :

K_f = \frac{1}{2}mv^2 is the final kinetic energy of the car, with

m = 2000 kg is the mass of the car

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K_i = \frac{1}{2}mu^2 is the initial kinetic energy of the car, with

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Soolving:

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Solving:

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A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic
xxTIMURxx [149]

Answer:

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the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2

 =2.5m

on the axis of x at x = 2.5

r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}

r = 2.5m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T

(Along z direction)

B)r = 5.00m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2

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on the axis of x at x = 5.0

r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}

r = 5.00m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T

(Along x direction)

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3 years ago
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