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2 years ago

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Jorge visits Panama city, Florida during the month of may. he feels a shore breeze blowing from the ocean on the beach. what cau

ses this shore breeze?
a.) the ocean is colder than the land.
b.) the land is colder than the ocean
c.) the ocean is denser than the land
d.) the land is denser than the ocean

1 answer:

Wittaler [7]2 years ago

3 0

The answer is a.) the ocean is colder than the land.
Thus, Jorge visits Panama city, Florida during the month of may. he feels a shore breeze blowing from the ocean on the beach because Jorge visits Panama city, Florida during the month of may. he feels a shore breeze blowing from the ocean on the beach.

Hope it helps!

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A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from i

murzikaleks [220]

Answer:

f" = 40779.61 Hz

Explanation:

From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;

from the Doppler effect equation, we can calculate the initial observed frequency as:

f' = f(1 - (v_o/v))

We are given;

f = 46.2 kHz = 46200 Hz

v_o = 21.8 m/s

v is speed of sound = 343 m/s

Thus;

f' = 46200(1 - (21/343))

f' = 43371.4285 Hz

In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;

Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;

f" = f'/(1 + (v_o/v))

f" = 43371.4285/(1 + (21.8/343))

f" = 40779.61 Hz

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2 years ago

A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio

ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is $v_0$

Horizontal acceleration is $a_x=a$

At maximum height velocity is zero therefore

$v_f=v_i-gt$

$0=v_0\sin u-gt$

$t=\frac{v_0\sin u}{g}$

Total time of flight $T=2t=\frac{2v_0\sin u}{g}$

During this time horizontal range is

$R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}$

$R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}$

For maximum range $\frac{\mathrm{d} R}{\mathrm{d} u}=0$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0$

$\tan 2u=\frac{g}{a}$

$u=\frac{1}{2}tan ^{-1}\frac{g}{a}$

(b)If a =10% g

$a=0.1g$

thus $u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}$

$u=42.14^{\circ}$

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2 years ago

(NEED HELP PLEASE) A physics student goes to the roof of the school, 24.15 m above the ground, and drops a pumpkin straight down

slavikrds [6]

Answer:

t = 2.2 s

Explanation:

Given that,

Height of the roof, h = 24.15 m

The initial velocity of the pumpkin, u = 0

We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :

$h=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s$

So, it will take 2.22 s for the pumpkin to hit the ground.

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2 years ago

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Trava [24]

The distance quantity/ measurement must be squared.

3 0

2 years ago

Read 2 more answers

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

5 0

2 years ago