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Kipish [7]
3 years ago
8

Birdman is flying horizontally at a

Physics
1 answer:
den301095 [7]3 years ago
5 0

Answer:

68 m

Explanation:

Given that the horizontal velocity of the birdman = 17 m/s and

the height, h= 78 m.

The gravitational force is acting in the downward direction, so it will not change the horizontal speed.

The horizontal speed will remains be constant and will be equal to the initial horizontal speed of the turd.

Initially, the turd was also flying horizontally with the birdman, so the initial velocity of the turd is the same as the horizontal velocity of the birdman, i.e In the horizontal direction, u_0=17 m/s.

In the vertical direction, u = 0,

The distance to be traveled, in the direction of application of force, is equals to the height of the turd, i.e

s= 78 m

Let t be the time taken to cover a distance of 78 m.

Now, applying the equation of motion in the vertical direction,

s=ut+\frac 12 at^2

where u is the initial velocity and a is the acceleration due to gravity in the direction of displacement,s.

Here, a=g=9.81 m/s^2, so

78=0\times t +\frac 12 (9.81)t^2

\Rightarrow t^2=(78\times2)/9.81

\Rightarrow t = 4 seconds.

Hence, the time taken to reach the ground is 4 seconds.

As the horizontal speed, u_0=17 m/s, is constant throughout the journey, so

the horizontal distance covered by turd

= u\times t

= 17 \times 4 = 68 m.

So, the distance of landing from the start of the field is 68 m as the birdman releases a turd directly  above the start of the field.

Hence, the robot must hold the bucket at a distance of 68 m from the start of the field.

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Sidana [21]

consider the motion of the tennis ball in downward direction

Y = vertical displacement = 400 m

a = acceleration = acceleration due to gravity = 9.8 m/s²

v₀ = initial velocity of the ball at the top of building = 10 m/s

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using the kinematics equation

v² = v²₀ + 2 a Y

inserting the values

v² = 10² + 2 (9.8) (400)

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3 years ago
Problem 3: Thermal expansionThe steel rod has the length 2 m and cross-section area 200 cm2at the room temperature 20◦C. Weapply
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Answer:

a) 2.00024 m

b) 0.036%

c) 436.67°C

Explanation:

Given

Initial length = L₀ = 2 m

Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²

We can obtain initial volume = V₀ = A₀L₀ = 0.02 × 2 = 0.04 m³

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Coefficient of linear expansivity = α = (2 × 10⁻⁶) (°C)⁻¹

a) New length of the rod after heating to 80°C

Linear expansion is given as

ΔL = L₀ × α ×ΔT

ΔL = 2 × 2 × 10⁻⁶ × (80 - 20) = 0.00024 m = 0.24 mm

New length = old length + expansion = 2 + 0.00024 = 2.00024 m

b) The percentage of the volume change of the rod.

Volume expansion is given by

ΔV = V₀ × (3α) × ΔT

Volume expansivity ≈ 3 × (linear expansivity)

ΔV = 0.04 × (3×2×10⁻⁶) × (80 - 20) = 0.0000144 m³

Percentage change in volume = 100% × (ΔV/V₀) = 100% × (0.0000144/0.04) = 0.036%

c) The maximal temperature we can allow if the volume should not increase by more than half percent.

For a half percent increase in volume, the corresponding change in volume needs to be first calculated.

Percentage change in volume = 100% × (ΔV/V₀)

0.5 = 100% × (ΔV/0.04)

(ΔV/0.04) = 0.005

ΔV = 0.0002 m³

Then we now investigate the corresponding temperature that causes this.

ΔV = V₀ × (3α) × ΔT

0.0002 = 0.04 × (3×2×10⁻⁶) × ΔT

ΔT = (0.0002)/(0.04 × 3 × 2 × 10⁻⁶) = 416.67°C

Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C

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3 years ago
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Answer:

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Answer:

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