I need help please it’s for a lab

Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and

mario62 [17]

Horizontal distance covered by a projectile is X = Vix *T

where Vix is the initial horizontal component of velocity and T is time taken by the projectile

Vix = ViCos theta

In question they said that initial velocity and angle is same on earth and moon

so Vix would remains same

now let's see about time taken T

time taken to reach the highest point

Vfy = Viy +gt

at highest point vertical velocity become zero so Vfy =0

0 = Vi Sin theta + gt

t = Vi Sintheta /g

Total time taken to land will be twice of that

On earth

Te= 2t

Te = 2Sinθ/g

on moon g is one-sixth of g(earth)

Tm = 2Sinθ/(g/6)

Tm = 6(2Sinθ/g)

Tm = 6Te

so total time taken by the projectile on moon will be six times the time taken on earth

From first equation X = Vix*T

we can see that X will also be 6 times on moon than earth

so projectile will cover 6 times distance on moon than on earth

4 0

2 years ago

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What is an example of a stable system

oksian1 [2.3K]

Some examples of stable system are:

1) functions of sine

2) DC

3) signum

4) unit step

5) cosine.

Happy Studying! ^^

6 0

3 years ago

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A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37 above horizontal]. It gets blocked just after re

stepan [7]

Refer to the diagram shown below.

Neglect air resistance.
The horizontal component of the launch velocity is
(20 m/s)*cos(37°) = 15.973 m/s
The vertical component of the launch velocity is
(20 m/s)*sin(37°) = 12.036 m/s

The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by
(12.036 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (3 m)
12.036t - 4.9t² - 3 = 0
2.4543t - t² - 0.6122 = 0
t² - 2.4563t + 0.6122 = 0
Solve with the quadratic formula.
t = (1/2)[2.4563 +/- √(6.0334 - 2.4488)]
t = 2.1748 or 0.2815 s
The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.

Part a.
The vertical velocity when t = 0.2815 s is
Vy = 12.036 - 9.8*0.2815
= 9.2773 m/s
The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is
√(9.2773² + 15.973² ) = 18.47 m/s
Answer:
The velocity at a height of 3.0 m is 18.5 m/s (nearest tenth)

Part b.
The horizontal distance traveled is
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)

6 0

3 years ago

HELP I NEED THIS ANSWERED AS FAST AS POSSIBLE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

WITCHER [35]

Displacement is d

Vf² = Vi² + 2 g d

(-20²) = (+10²) + 2 (-9.8) d

-19.6 d = 300

d = -15.3 m

negative means lower

time is t

d = Vi t + 1/2 g t²

-15.3 = 10 t + (-4.9) t²

4.9 t² - 10 t -15.3 = 0

t = 3.06 s

Hope this helps -John

8 0

3 years ago

The acceleration due to gravity on the moon is 1.6 m/s2, about a sixth that of Earth’s. Which accurately describes the weight of

tatyana61 [14]

The answer is C.

An object on the moon is six times lighter than on Earth.

6 0

2 years ago

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