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Degger [83]
2 years ago
12

A 10000 kg rocket blasts off vertically from the launch pad with a constant upward of 2.25 m/s2 and feels no appreciable air res

istance. When it has reached a height of 525 m, its engine suddenly fails so that only gravity acts on it after that. What is the maximum height that the rocket reaches?
Physics
1 answer:
Sloan [31]2 years ago
4 0

Answer:

The maximum height that the rocket reaches is 645.5 m.

Explanation:

Given that,

Mass = 10000 kg

Acceleration = 2.25 m/s²

Distance = 525 m

We need to calculate the velocity

Using equation of motion

v^2=u^2+2as

Put the value in the equation

v^2=0+2\time2.25\times525

v=\sqrt{2\times2.25\times525}

v=48.60\ m/s

We need to calculate the maximum height with initial velocity

Using equation of motion

v^2=u^2-2gh

h=\dfrac{v^2-u^2}{-2g}

Put the value in the equation

h=\dfrac{0-(48.60)^2}{-2\times9.8}

h=120.50\ m

The total height  reached by the rocket is

h'=s+h

h'=525+120.50

h'=645.5\ m

Hence, The maximum height that the rocket reaches is 645.5 m.

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wariber [46]

Answer:

The magnitude of the tension in the cable, T is 1,064.315 N

Explanation:

Here we have

Length of beam = 4.0 m

Weight = 200 N

Center of mass of uniform beam = mid-span = 2.0 m

Point of attachment of cable = Beam end = 4.0 m

Angle of cable = 53° with the horizontal

Tension in cable = T

Point at which person stands = 1.50 m from wall

Weight of person = 350 N

Therefore,

Taking moment about the wall, we have

∑Clockwise moments = ∑Anticlockwise moments

T×sin(53) = 350×1.5 + 200×2

T = 850/sin(53)  = ‭1,064.315 N.

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2 years ago
What is one standard kilogramun si system<br><br><br><br><br>​
Phoenix [80]

Answer:

The kilogram (kg) is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015 ×10−34 when expressed in the unit J s, which is equal to kg m2 s−1, where the meter and the second are defined in terms of c and ∆νCs.

3 0
2 years ago
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Use the terms "force", "weight", "mass", and "inertia" to explain why it is easier to tackle a 220 lb football player than a 288
Tomtit [17]
<span><u>Answer </u>
The mass of 220 lb football has less than 288 lb football. So, it will be easier to move it since it will require less force. The heavy football will have a bigger momentum. Since 288 lb has more weight than 220 lb, it will have bigger inertia making it difficult for the players to stop it.
This makes it easier to tackle 220 lb football than 288 lb football. 
</span>
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3 years ago
Chapter 36, Problem 007 Light of wavelength 586 nm is incident on a narrow slit. The angle between the first diffraction minimum
svp [43]

Answer:

d =4.77\times 10^{-5}\ m

Explanation:

given

wavelength of light  λ = 479 nm

                                   = 479 x 10⁻⁹ m

the angle

θ = 1.15 / 2 = 0.575°                

using                                              

condition for diffraction minimum ,

         d sinθ = m λ                        

for first minimum m = 1

       d sinθ = λ                      

therefore ,

slit width                                  

d =\dfrac{\lambda}{sin\theta}          

d =\dfrac{479\times 10^{-9}}{sin 0.575^0}

d =\dfrac{479\times 10^{-9}}{0.01}              

d =4.77\times 10^{-5}\ m                                                    

hence, the width of the slit is equal to d =4.77\times 10^{-5}\ m

3 0
2 years ago
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