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wariber [46]
2 years ago
11

A rocket ship has several engines and thrusters. While the Solid Rocket Booster (SRB) and main engines only work together during

the first 2 minutes of flight, the main engines operate for a total of 8.5 minutes after the launch. Once the SRBs are released, the main engines alone accelerate the rocket from about 1341 m/s to 7600 m/s.
What is the acceleration of the SRB and main engine during the first 2.0 minutes of flight?

A. 52 m/s2
B. 13 m/s2
C. 9.8 m/s2
D. 11 m/s2
Physics
1 answer:
Katarina [22]2 years ago
3 0

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:

  • From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
  • From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.

We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:

  • The speed increases from 0 m/s to 1341 m/s.
  • The time elpased is 2.0 min.
  • 1 min = 60 s.

The acceleration of the ship during the first 2.0 minutes is:

a = \frac{\Delta v }{t} ) \frac{(1341m/s-0m/s)}{2.0min} \times \frac{1min}{60s}  = 11 m/s^{2}

A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).

Learn more: brainly.com/question/16274121

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3 0
4 years ago
Consider a metal single crystal oriented such that the normal to the slip plane and slip direction are at angles of 60° and 35°,
Vedmedyk [2.9K]

Answer:

19.5324 MPa

Explanation:

Information provided

Angle between the normal to the slip plane with tensile axis, \alpha=60^{o}


Angle by slip direction with tensile axis, \beta=35^{o}

Critical resolved shear stress, \tau_{c}=8 MPa

Applied stress \sigma=12 MPa

Shear stress at slip plane

\tau=\sigma cos\alpha cos\beta

\tau=12cos60^{o}cos35^{o}=4.915 MPa

\tau hence crystal won’t yield

Applied stress, \sigma for crystal to yield is given by

\sigma=\frac {\tau_{c}}{cos\alpha cos\beta}

\sigma=\frac {8}{cos60cos35}=19.53239342 MPa

\sigma=19.5324 MPa


7 0
3 years ago
Do you think a population besides the moon jellies’ consumer population (sea turtles) and resource population (zooplankton) coul
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Answer:

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8 0
4 years ago
Three liquids that will not mix are poured into a cylindrical container. The volumes and densities of the liquids are 0.50 L, 2.
aleksklad [387]

Answer:

13.524 N

Explanation:

Volume and densities are given as:

ρ1 = 2.6 g/cm³ => 2600 kg/m³ ; V1 = 0.50 L => 0.5 x 10^-3 m³

ρ2 = 1.0 g/cm³ => 1000 kg/m³ ; V2= 0.25 L => 0.25 x 10^-3 m³

ρ3 = 0.7 g/cm³ => 700 kg/m³ ; V3 = 0.4 L => 0.4 x 10^-3 m³

Next is to calculate force exerted on the bottom of the container due to these liquids:

F= ρ1V1g + ρ2 V2 g+ ρ 3 V3g

where ,

ρ= density

V= volume

g= 9.8m/s²

F= g( 2600 x 0.5 x 10^-3 + 1000 x 0.25 x 10^-3 + 700 x 0.4 x 10^-3)

F= 9.8 (1.38)

F=  13.524 N

Therefore,  the force on the bottom of the container due to these liquids is 13.524 N

6 0
4 years ago
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