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Ray Of Light [21]
3 years ago
6

0.10-kilogram model rocket’s engine is designed to deliver an impulse of 6.0 newton-seconds. If the rocket engine burns for 0.75

second, what average force does it produce?
Physics
1 answer:
UkoKoshka [18]3 years ago
6 0

Answer:

8.0 N

Explanation:

Force: This can be defined as the mass of a body and its acceleration. The S.I unit of Force is Newton (N).

Mathematically, Fore is expressed as

F = ma ........................... equation 1

Where F = force, m = mass, a = acceleration.

and

I = mΔv

Δv = I/m ............................ Equation 2

Where I = impulse, m = mass, Δv = change in velocity

Given: I = 6.0 Newton-seconds, m = 0.1 kilogram.

Substituting into equation 2

Δv = 6.0/0.1

Δv = 60 m/s.

But

a = Δv/t

where t = time = 0.75 seconds.

a = 60/0.75

a = 80 m/s²

Substitute the values of a and m into equation 1.

F = 0.1(80)

F = 8.0 N.

Thus the average force produced = 8.0 N

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Answer:

S=2.693m

Explanation:

Let point O be the origin of the room and the fly is crawling at point P as shown in the attached photo:

#Distance of the fly from the corner of the room is defined as:

OP=S=\sqrt{{(x_1-x_0)^2+(y_1+y_0)^2}

#Substitute the coordinate values in the above equation:

S=\sqrt{{(2.6m-0m)^2+(0.7m-0m)^2}}\\=\sqrt{7.25}\\=2.693m

Hence, the distance of the fly from the corner of the room is 2.693m

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4 years ago
How do amps correlate to brightness?
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4 years ago
________________ is the tendency of an object to resist a change in its motion.
Sliva [168]
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3 years ago
A missile is moving 1350 m/s at 25.0° angle. It needs to hit in a 55.0° direction in 10.20 s. What is the direction of its final
77julia77 [94]

Answer:

final velocity = 3504 m/s  

Explanation:

<em>Given data:</em>

velocity of missile = Vi = 1350m/s

angle at which missile is moving = 25degree

distance between missile and targets = 23500m

angle between target and missile=55degree

time=10.2s

<em>To find:</em>

Final velocity: ?

<em>Formula:</em>

x = Vx*t + ½*ax*t²  

Let x be the horizontal component of distance

x = ertical component of distance

t-time

ax = horizontal component of acceleration

ay = Vertical component of acceleration

Vx = horizontal component of velocity

Vy = Vertical component of velocity

<em>Solution:</em>

x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²  

ax = 19.2 m/s²  

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s  

<em>similarly vertically:</em>

y = Vy*t + ½*ay*t² 

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²  

ay = 258 m/s²  

V'y = Vy + ay*t

     = 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s  

V = √(V'x² + V'y²)

   = 3504 m/s  

8 0
3 years ago
In an earlier chapter you calculated the stiffness of the interatomic "spring" (chemical bond) between atoms in a block of lead
fiasKO [112]

Answer:

8.01e-22

Explanation:

7 0
3 years ago
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