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anastassius [24]
3 years ago
14

A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreci

able air resistance. When it has reached a height of 575 m , its engines suddenly fail so that the only force acting on it is now gravity. Part A What is the maximum height this rocket will reach above the launch pad
Physics
1 answer:
Simora [160]3 years ago
4 0

Answer:

Explanation:

We shall first calculate the velocity at height h = 575 m .

acceleration a = 2.2 m /s²

v² = u² + 2 a s

u is initial velocity , v is final velocity , s is height achieved

v² = 0 + 2 x 2.2 x 575

v = 50.3 m /s

After 575 m , rocket moves under free fall so g will act on it downwards

If it travels further by height H

from the relation

v² = u² - 2 g H

v = 0 , u = 50.3 m /s

H = ?

0 = 50.3² - 2 x 9.8 H

H = 129.08 m

Total height attained by rocket

= 575 + 129.08

= 704.08 m .

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A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete
Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is 6.25\times 10^{- 3} ft/s^{2}

Solution:

As per the question:

Constant velocity of the horse in the horizontal, v_{x} = 1 ft/s

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

20^{2} + 10^{2} = l^{2}

l^{2}= 500

Now,

x^{2} + y^{2} = 500                            (1)

Differentiating equation (1) w.r.t 't':

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

x\frac{dx}{dt} = - y\frac{dy}{dt}

where

\frac{dx}{dt} = Rate of change of displacement along the horizontal

\frac{dy}{dt} = Rate of change of displacement along the vertical

v_{x} = velocity along the x-axis.

v_{y} = velocity along the y-axis

xv_{x} = -yv_{y}

v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s

|v_{y}| = 0.5\ ft/s

Acceleration of the hay bale is given by the kinematic equation:

v_{y}^{2} = u_{y} + 2ay

(-0.5)^{2} =0 + 2ay

0.25 = 2ay

\frac{0.25}{2y} = a

a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}

7 0
3 years ago
A bird takes off from the tree and after 5 s is flying at a speed of 10 m/s. Calculate
Volgvan

Answer:

10miles/second

Explanation:

Change in velocity of the bird is expressed as the difference between the final velocity and initial velocity of the body.

Change in velocity = Final velocity - initial velocity

Since the bird takes off from the tree, the initial velocity of the bird = 0miles/sec

Final velocity = 10miles/secs

Change in velocity = 10-0

Change in velocity = 10miles/second

6 0
2 years ago
What is an example of kinetic energy
Olenka [21]

An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity.

6 0
3 years ago
During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 * t2 to so
Ulleksa [173]

8/9 score.....................

3 0
3 years ago
Read 2 more answers
What is not an essential part of isolated soldier guidance
mel-nik [20]

Answer:

Option B, visual sightings

Explanation:

Options for the question are

A) accountability mechanism

B) visual sightings

C)  intelligence

D) surveillance

E) reconnaissance  operations, or communications

Solutions

The Fundamentals of Army Personnel Recovery (PR) outlines certain circumstances under which a person has to undergo survival situation thereby taking the necessary steps to avoid capture and return safely to their respective unit.

An isolated soldier is expected to know where they are, upcoming route and rally points. They are supposed to know the near and far recognition signals, recovery site protocols, challenge and password etc. A proper preparation is to be done for this including planning, medicines, kits, etc.  

Visual sightings is not an essential part of isolated PR

Hence, option B is correct

4 0
3 years ago
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