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3 years ago

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g You heard the sound of a distant explosion (3.50 A/10) seconds after you saw it happen. If the temperature of the air is (15.0

B) oC, how far were you from the site of the explosion

1 answer:

Soloha48 [4]3 years ago

8 0

Answer:

The answer is "1557 meters".

Explanation:

speed of sound in ( $\frac{m}{s}$ ) $= 331.5 + 0.60 \ T^{\circ}\ C\\\\$

$\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters$

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Convert 5.7 miles to km

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5.7 kilometers is equal to 3.5418157957528034 miles

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3 years ago

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How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 n/m so that it wil

mrs_skeptik [129]

The frequency of a simple harmonic oscillator such as a spring-mass system is given by
$f= \frac{1}{2 \pi} \sqrt{ \frac{k}{m} }$
where
k is the spring constant
m is the mass attached to the spring.

Re-arranging the formula, we get:
$m= \frac{k}{4 \pi^2 f^2}$
and since we know the constant of the spring:
$k=39.5 N/m$
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
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3 years ago

You're carrying a 3.2-m-long, 24kg pole to a construction site when you decide to stop for a rest. You place one end of the pole

Aleksandr [31]

Answer:

Tension of 132N

Explanation:

We need to apply Summatory of Force to find the tension in the hand.

We define te tensión in the hand as $F_2$ and the Tension in fence post as $F_1$ , then

$\sum F = 24(9.8)$

$F_1 + F_2= 24(9.8)$

We apply summatory of moments then

$F_2*1.25 = F_1*1.6$

Where the Force 2 is 1.25m from the center of summatory,

We can note that,

$1.6 m - 0.35m=1.25m$

We have two equation and two incognites, then replacing (1) in (2)

$1.6(235.2 -F_2) = 1.25F_2$

$376.32 = F2(1.6+1.25)$

$F_2= \frac{376.32}{2.85}$

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3 years ago

What is the term for substances that have several unpaired electrons and are strongly magnetic

Murljashka [212]

Answer:

Scientific definitions for ferromagnetic

The property of being strongly attracted to either pole of a magnet. Ferromagnetic materials, such as iron, contain unpaired electrons, each with a small magnetic field of its own, that align readily with each other in response to an external magnetic field.

Explanation:

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3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

$v_{avg} = \frac{6000}{20*60}$

$v_{avg} = 5m/s$

now since effect of wind is in opposite direction then we can say

$V_{net} = v_{bird} - v_{wind}$

$5 = 9 - v_{wind}$

$v_{wind}= 4 m/s$

Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

$time = 7.7 minutes$

Part c)

Total time of round trip when wind is present

$T = t_1 + t_2$

$T = 20 + 7.7 = 27.7 min$

now when there is no wind total time is given by

$T = \frac{6000}{9} + \frac{6000}{9}$

$T = 22.22 min$

So due to wind time will be more

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3 years ago