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baherus [9]
3 years ago
8

PLEASE HELP What is it called when the moon passes through the penumbra of Earth’s shadow?(1 point)

Physics
1 answer:
bogdanovich [222]3 years ago
8 0
Partial. a penumbral lunar eclipse is when the moon passes through the earths penumbra
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In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.
HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is 41.9^{\circ}

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

s=u_y t + \frac{1}{2}at^2 (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

u_y = u sin \theta is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

\theta=40.0^{\circ} is the angle of projection

So

u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s

a=g=-9.8 m/s^2 is the acceleration due to gravity (downward)

Substituting the numbers, we get

-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

d=u_x t

where

u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

d=(9.19)(1.778)=16.34 m

B)

In this second case,

\theta=42.5^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s

So the equation for the vertical motion becomes

4.9t^2-8.1t-1.80=0

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s

So, the range of the shot is

d=u_x t = (8.85)(1.851)=16.38 m

C)

In this third case,

\theta=45^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s

So the equation for the vertical motion becomes

4.9t^2-8.5t-1.80=0

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s

So, the range of the shot is

d=u_x t = (8.49)(1.925)=16.34 m

D)

In this 4th case,

\theta=47.5^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s

So the equation for the vertical motion becomes

4.9t^2-8.8t-1.80=0

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s

So, the range of the shot is

d=u_x t = (8.11)(1.981)=16.07 m

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is \theta=42.5^{\circ}.

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

s-u sin \theta t + \frac{1}{2}gt^2=0

The solutions of this quadratic equation are

t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}

This is the time of flight: so, the horizontal range is

d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta

It can be found that the maximum of this function is obtained when the angle is

\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})

Therefore in this problem, the angle which leads to the maximum range is

\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0
3 years ago
The orientation of which of the following does not influence the phases of the moon? a. Earth c. Sun b. the moon d. Stars Please
beks73 [17]
I'm pretty sure it's D. The stars don't influence the moon's phases.
6 0
3 years ago
A force of 3 newtons moves a 10 kilogram mass horizontally a distance of 3 meters. The mass does not slow down or speed up as it
Mashutka [201]

Answer:

9 joules of heat energy was produced

Explanation: there is no acceleration therefore its not a kinetic energy

Energy= force × distance

= 3×3

=9

3 0
3 years ago
How much force is needed to accelerate a 1000Kg car at a rate of 3m/s2?
ikadub [295]

According to Newton (2nd law),   Force = (mass) x (acceleration)

Substitute what we know :           Force = (1,000 kg) x (3 m/s²)

Do the arithmetic:                        Force = 3,000 kg-m/s²  =  3,000 newtons

6 0
2 years ago
Read 2 more answers
Please I need help with these 2 questions. Thank you.
Lemur [1.5K]
First one is D and Second one is B
7 0
3 years ago
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