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klio [65]
3 years ago
5

A jet lands with a speed of 100 m/s and can accelerate uniformly at a rate of -5.0 m/s^2 as it comes to rest. What is the minimu

m length runway to land this plane? A. 1000 m
B. 20 m
C. 675 m
D. 500 m
Physics
1 answer:
schepotkina [342]3 years ago
3 0

Answer:

A ) 1000 m.

Explanation:

Here initial velocity u = 100 m /s

Final velocity v = 0

Acceleration a = -5 ms⁻²

Distance travelled = S

v² = u² + 2aS

0 = (100)² -2 x 5 S

S = 10000/ 10

=1000 m.

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What is the speed of an object that travels 20 meters in 4 seconds?
Iteru [2.4K]

Given:-

  • Distance covered by the particle (s) = 20 m
  • Time taken (t) = 4 seconds

To Find: Speed of it.

We know,

v = s/t

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  • v = Speed,
  • s = Distance &
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v = (20 m)/(4 s)

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4 years ago
Far out in space, a 100,000-kg rocket and a 200,000-kg rocket are docked at opposite ends of a motionless 90m-long tunnel. The t
postnew [5]

Answer:

a) X_{cm} = 60m

b) I = 5.4*10^8Kg*m^2

c) T= 4.5*10^6 N*m

d) a = 8.3*10^{-3}rad/s^2

e) W= 0.25 rad/s

Explanation:

a) we know that:

X_{cm} = \frac{m_1x_1+m_2x_2}{m1+m2}

where X_cm is the ubicaton of the center of mass, m_1 the mass of the first rocket, x_1 its distances with the rocket 1, m_2 the mass of the second rocket and x_2 its distance with the rocket 1. So, replacing values, we get:

X_{cm} = \frac{(100,000kg)(0)+(200,000kg)(90m)}{100,000kg+200,000kg}

X_{cm} = 60m

So, the center of mass is at 60m from the rocket 1.

b) we know that:

I = M_1R_1^2 +M_2R_2^2

where I is the moment of inertia, M_1 is the mass of the rocket 1, R_1 its distance from the center of mass, M_2 the mass of the second rocket and R_2 the distance between the rocket 2 and the center of mass. So, replacing values, we get:

I = (100,000kg)(60m)^2 +(200,000kg)(30m)^2

I = 5.4*10^8Kg*m^2

c) We know that:

T = Fr

where T is the net torque, F is the force and r is the distance between the rocket and the radius. Then:

FR_1+FR_2 = T

Replacing values, we get:

50,000N(60m)+50,000N(30m) = T

T= 4.5*10^6 N*m

d) We know that:

T = Ia

where T is the net torque, I the moment of inertia and a is the angular aceleration. So, replacing values, we get:

4.5*10^6Nm = 5.4*10^8Kg*m^2a

solving for a:

a = 8.3*10^{-3}rad/s^2

e) Finally, using:

W = at

where W is the angular velocity, a is the angular aceleration and t is the time.

Then, replacing values. we get:

W = (8.3*10^{-3}rad/s^2)(30s)

W = 0.25 rad/s

3 0
3 years ago
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