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AysviL [449]
3 years ago
5

The electromagnetic spectrum includes radio waves inferared waves and x rays

Physics
1 answer:
olya-2409 [2.1K]3 years ago
8 0

Answer:

True

Explanation:

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Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke
umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

        Thrust = v_{e}  \frac{dM}{dt}

        v_{f}-v₀ = v_{e} ln ( \frac{M_{o} }{M_{f}} )

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

       M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

     

The final mass is the mass of the engines + the mass of the rocket

      M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

       thrust = 5.26 N

       t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

          thrust = v_{e} \frac{M_{f} - M_{o}  }{t_{f} - t_{o}  }

          v_{e} = thrust  \frac{\Delta t}{\Delta M}

          v_{e} = 5.26 \frac{1.90}{0.080 -0.0927}

          v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

            v_{f}-0 = v_{e} ln ( \frac{M_{o} }{M_{f} } )

we calculate

            v_{f} = 786.93 ln (0.0927 / 0.080)

            v_{f} = 115.95 m / s

5 0
3 years ago
A hand pushes two blocks, A and B, along a frictionless table for a distance d. When the hand starts to push, the blocks are mov
Kitty [74]
  <span>net work = change in kinetic energy 

for Block B, we just have the force from block A acting on it 
F(ab)d= .5(1)vf² - .5(1)(2²) 
F(ab)d= .5vf² - 2 

Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction 

10-F(ba)d = .5(4)vf² - .5(4)(2²) 
10-F(ba)d = 2vf² - 8 
F(ba)d = 18 - 2vf² 

now we have two equations: 
F(ba)d = 18 - 2vf² 
F(ab)d= .5vf² - 2 

since the magnitude of F(ba) and F(ab) is the same, substitute and find vf (I already took into account the direction when solving for F(ab) 

10-.5vf² + 2 = 2vf² - 8 
12 - .5vf² = 2vf² - 8 
20 = 2.5vf² 
vf² = 8 

they both will have the same velocity 
KE of block A= .5(4)(2.828²) = 16 J 
KE of block B=.5(1)(2.828²) = 4 J</span>
5 0
3 years ago
If you double the period of a pendulum. what happens to its length?​
victus00 [196]

Answer:

Time period is directly proportional to the length so if the length is doubled the time period is doubled

hope it is helpful....

5 0
3 years ago
Read 2 more answers
How much kinetic energy will an electron gain if it accelerates through a potential difference of 23,000 volts in a cathode ray
Sauron [17]

Answer:

3.68x10-15 J

Explanation:

I hope this is the right answer sorry if im wrong

8 0
2 years ago
Vertically polarized light with an intensity of 36.8 lux passes through a polarizer whose transmission axis is an angle of 51.0°
Annette [7]

Answer:

0.01774 lux

Explanation:

I_0 = Polarized light intensity = 36.8 lux

\theta = Angle

Through first filter

I_1=I_0cos^2\theta\\\Rightarrow I_1=36.8cos^2{51}\\\Rightarrow I_1=14.57\ lux

Through second filter

I_2=I_1cos^2\theta\\\Rightarrow I_2=14.57cos^2{88}\\\Rightarrow I_2=0.01774\ lux

intensity of transmitted light is 0.01774 lux

5 0
3 years ago
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