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MrRissso [65]
2 years ago
5

A glider moving with a speed of 200 kilometers/hour experiences a cross wind of 30 kilometers/hour. What is the resultant speed

of the glider relative to the ground?
Physics
1 answer:
liraira [26]2 years ago
5 0
You didn't say so, but we must assume that the "200 km/hr" is
the glider's air-speed, that is, speed relative to the air. 

If the air itself is moving at 30 km/hr relative to the ground and
across the glider's direction, then the glider's speed relative to
the ground is

                 √(200² + 30²)

             =  √(40,000 + 900)

             =  √(40,900)  =  202.24... km/hr   (rounded) 
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Inertia= inactivity
The factor involved in increasing inertia would be "mass". The more mass something has, the more inertia. 
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A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x=0 marking the city limits. His
adoni [48]

Answer:

1) v_2=23\ m.s^{-1}              &     x_2=43\ m east of sign post

2) x'=55\ m east of sign post

3) x_n=205\ m east of the signpost.

4) v_z=35\ m.s^{-1}

Explanation:

Given:

  • position of motorcyclist on entering the city at the signpost, x_0=0\ m
  • time of observation after being at x=5m east of the signpost, t_m=0\ s
  • constant acceleration of the on entering the city, a=4\ m.s^{-2}
  • distance of the motorcyclist moments later after entering, s_m=5\ m
  • velocity of the motorcyclist moments later after entering, u_m=15\ m.s^{-1}

<u>Now the initial velocity on at the sign board:</u>

u_m^2=u^2+2.a.x_m

where:

u= initial velocity of entering the city at the signpost

Putting respective values:

15^2=u^2+2\times 4\times 5

u=13.6015\ m.s^{-1}

1)

Position at time t_2=2\ s sec.:

Using equation of motion,

x_2=u_m.t_2+\frac{1}{2} a.(t_2)^2+5 because it has already covered 5m before that point

x_2=15\times 2+0.5\times 4\times 2^2+5

x_2=43\ m east of sign post

Velocity at time t_2=2\ s sec.:

v_2=u_m+a.t_2

v_2=15+4\times 2

v_2=23\ m.s^{-1}

2)

Position when the velocity is v'=25\ m.s^{-1}:

using equation of motion,

v'^2=u_m^2+2.a.x'+5

25^2=15^2+2\times 4\times x'+5

x'=55\ m east of sign post

3)

Given that:

acceleration be, a_n=2\ m.s^{-2}

time, t_n=5\ s

Position after the new acceleration and the new given time:

using equation of motion,

x_n=u_m.t_n+\frac{1}{2} a_n.(t_n)^2+5

x_n=15\times 5+0.5\times 2\times 5^2+5

x_n=205\ m east of the signpost.

4)

now time of observation, t_z=5\ s

v_z=u_m+a.t_z

v_z=15+4\times 5

v_z=35\ m.s^{-1}

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Answer:

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Explanation:

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