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2 years ago

If a football is thrown from rest with an acceleration of 8.5 m/s2, and had an final

Physics

1 answer:

AnnyKZ [126]2 years ago

4 0

Answer:

t = 2.94 seconds

Explanation:

Given that,

The initial velocity of the football, u = 0

The final velocity of the football, u = 25 m/s

Acceleration, a = 8.5 m/s²

We need to find the time for which the football is accelerating. Let the time is t. We know that,

$a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{25-0}{8.5}\\\\t=2.94\ s$

So, the football is accelerating for 2.94 seconds.

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A straight trail with a uniform inclination of 15 degrees leads from a lodge at an elevation of 600 feet to a mountain lake at a

9966 [12]

Answer:

The length of the trail = 22796 ft

Explanation:

From the ΔABC

AC = length of the trail = x

AB = 6100 - 600 = 5500 ft

Angle of inclination $\theta$ = 15°

$\sin \theta = \frac{AB}{AC}$

$\sin 15 = \frac{5900}{x}$

$x = \frac{5900}{0.2588}$

x = 22796 ft

Since x = AC = Length of the trail.

Therefore the length of the trail = 22796 ft

7 0

2 years ago

Carbon-14 has a half-life of 5,700 years. How long will it take for 6.25% of the Carbon-14 to be remaining?

IRISSAK [1]

Answer:

22,800 years

Explanation:

Half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.0625 = (½)^(t / 5700)

log 0.0625 = (t / 5700) log 0.5

4 = t / 5700

t = 22,800

It takes 22,800 years.

4 0

3 years ago

Explain, using your own words, how noise cancelling headphones work, (physics)

Afina-wow [57]

Answer:

They use noise control, creating a wave that negates outside or ambient sound and replaces it with the desired sound that listeners request.

Explanation:

I hope this helped

4 0

3 years ago

I NEED HELP!!! PLS HELP ME MARK U BRAINLIEST!!

FinnZ [79.3K]

Answer:

The answer is X

Explanation:

Cause the highest points will most likely have the most potential energy

8 0

2 years ago

The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu

BartSMP [9]

Answer:

hello your question is incomplete attached below is the missing part

answer : short period oscillations frequency = 0.063 rad / sec

phugoid oscillations natural frequency ( $w_{np}$ ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= $( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns} ) = 0$

where :

$w_{np} = Natural frequency of plugiod oscillation$

$\alpha _{p} = damping ratio of plugiod oscilations$

comparing the general form with the given equation

$w^{2} _{np}$ = 18.2329

$w^{2} _{ns} = 0.003969$

hence the short period oscillation frequency ( $w_{ns}$ ) = 0.063 rad/sec

phugoid oscillations natural frequency ( $w_{np}$ ) = 4.27 rad/sec

8 0

2 years ago