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inna [77]
3 years ago
14

A rocket's acceleration is 6.0 m/s2. Assuming it starts at 0 m/s, how long will it take for the rocket to reach a velocity of 42

m/s? 250 s 0.14 s 7.0 s 290 s
Physics
1 answer:
elena-s [515]3 years ago
7 0
You said "<span>A rocket's acceleration is 6.0 m/s2.".

That just means that its speed increases by 6 m/s every second.
Whenever you look at it, its speed is 6 m/s faster than it was
one second earlier.

If it starts out with zero speed, then its speed is 6 m/s after 1 second,
12 m/s after 2 seconds, 18 m/s after 3 seconds . . . etc.

How long does it take to reach 42 m/s ?

Well, how many times does it have to go 6 m/s FASTER 
in order to build up to 42 m/s ?

That's just (42/6) = 7 times.

Writing it correctly, with the units and everything, it looks like this:


(42 m/s) / (6 m/s</span>²)

= (42/6)  (m/s) / (m/s²)

= (42/6)  (m/s · s²/m)

=  7 seconds
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   for angular velocity, \omega_{f}^{2}  = \omega _{i}^{2}+2.a.S

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4 0
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An object completes one and half revolution of a circle of radius R calculate the displacement and distance
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8 0
3 years ago
A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c
GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
  • Let's draw the free body diagram of the system using the given data.
  • From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
  • For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

                           N_x=86.62N

  • We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

                           N_y=F_V=mg-Tsin59\\

  • To find Ny, we need to find the tension T.
  • For this, we can equate the net horizontal force.

                           F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N

  • Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

                    N_y= (40*9.8)-(169.8*sin59)=246.4N

  • Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

                 N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

Learn more about the tension here:

brainly.com/question/28106871

#SPJ1

4 0
2 years ago
Read 2 more answers
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