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malfutka [58]
3 years ago
10

Now switch to “Position vs. Time #2” on the Choose a graph menu. Once again, experiment until you are able to reproduce the grap

h.
What did the caterpillar have to do during the horizontal segments of the graph?
Physics
2 answers:
seraphim [82]3 years ago
6 0

Answer:

The caterpillar had to move across the horizontal segments. I hope this helps :)

Explanation:

Alex17521 [72]3 years ago
3 0
The caterpillar had to had the horizontal segments of the graphs because
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In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. the sharp upward
ladessa [460]

<u>Answer:</u>

 The skier’s launch speed = 17.08 m/s

<u>Explanation:</u>

The motion of skiing is projectile motion.

 Projectile motion has two types of motion Horizontal and Vertical motion.  

Vertical motion:  

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.  

Considering upward vertical motion of projectile.  

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.  

0 = u sin θ - gt  

t = u sin θ/g  

Total time for vertical motion is two times time taken for upward vertical motion of projectile.  

So total travel time of projectile = 2u sin θ/g  

Horizontal motion:  

We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.  

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g  

So range of projectile, R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}  

Vertical motion (Maximum height reached, H) :

We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.  

Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

We have θ = 63° and H = 11.8 meter.

Substituting

   11.8=\frac{u^2sin^263}{2*9.81}\\ \\ u^2=\frac{11.8*2*9.81}{sin^263}=\frac{231.516}{0.794} =291.58\\ \\ u=17.08m/s

 The skier’s launch speed = 17.08 m/s

3 0
4 years ago
Read 2 more answers
Which force diagram accurately represents a satellite in orbit around Earth?
Anit [1.1K]

Answer:

First choice

Explanation:

A satellite in orbit around Earth experiences only one force: the gravitational attraction exerted by the Earth on it. This force is labelled with F_g. In space, there are no other forces acting on the satellite.

The force of gravity acts as centripetal force, "pulling" the satellite towards the centre of its circular orbit. The inertia of the satellite (which has an initial velocity) tends to keep it moving straight, so the combination of these two effects (inertia and force of gravity) results into the circular motion of the satellite.

5 0
3 years ago
Read 2 more answers
A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 5.
Kobotan [32]

Explanation:

The given data is as follows.

           F = 5.75 \times 10^{-16} N

          q = 1.6 \times 10^{-19} C

          v = 385 m/s

       sin (63.9^{o}) = 0.876

Now, we will calculate the magnitude of magnetic field as follows.

              B = \frac{F}{qv sin (\theta)}

                  = \frac{5.75 \times 10^{-16} N}{1.6 \times 10^{-19} C \times 385 m/s \times 0.876}

                  = 0.01065 \times 10^{3} T

                  = 10.65 T

Thus, we can conclude that magnitude of the magnetic field is 10.65 T.

4 0
3 years ago
How many minutes will it take to drive to los angeles from san francisco if an average speed of 72mi/hr is maintained?
Sholpan [36]
The distance between Los Angeles and San Francisco is approximately 381.9mi. 
Divide this distance by the speed.
381.9/72 = 5.3
You are left with the time in hours. To convert to minutes, multiply by 60. 
5.3*60 = 318.25

Therefore, it will take 318.25 minutes.
6 0
3 years ago
An 85 kg object is moving at a constant speed of 15 m/s in a circular path, which has a radius of 20 meters. What centripetal fo
UNO [17]
As we know that centripetal force =mv^2/r
given data is 
m = mass
v = speed
r = radius
putting values we get

= 85 x 15^2 / 20

= 956.25 N
option d is correct
7 0
3 years ago
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