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scoundrel [369]

3 years ago

15

In which medium would sound travel the fastest? a)across a room b)in a swimming pool c)through outer space d)through a railroad

track

1 answer:

lora16 [44]3 years ago

7 0

d)through a railroad track

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Escribo dentro del paréntesis la letra (F) si el enunciado es falso o (V) si el enunciado es verdadero. Fundamento si mi elecció

KatRina [158]

Answer:

El primero es (V) y el segundo es(f)

4 0

3 years ago

A block starting from rest slides down the length of an 18 plank with an acceleration of 4.0 meters per second. How long does th

Snowcat [4.5K]

Answer:

$\boxed{\text{\sf \Large 3.0 s}}$

Explanation:

Use distance formula

$\displaystyle d=ut+\frac{1}{2} at^2$

$u= \text{\sf initial velocity}\\d= \text{\sf distance}\\a= \text{\sf acceleration}\\t= \text{\sf time taken}$

$\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2$

$t=3$

4 0

3 years ago

If a ball is thrown horizontally with a speed of 65 mph, how far will it fall while traveling 90 ft of horizontal distance?

kow [346]

Answer:

install socrati it give you all answers

Explanation:

3 0

3 years ago

As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil

galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

$-F_3sin(b) + F_1 = 0$

For the address and we have:

$-F_3cos(b) + F_2 = 0$

The forces $F_1$ and $F_2$ are known

$F_1 = 5.7\ N\\\\F_2 = 1.9\ N$

We have 2 unknowns ( $F_3$ and b) and we have 2 equations.

Now we clear $F_3$ from the second equation and introduce it into the first equation.

$F_3 = \frac{F_2}{cos (b)}$

Then

$-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°$

Then we find the value of $F_3$

$F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N$

Finally the answer is 6.3 N at 162° counterclockwise from

F1->

7 0

3 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

3 years ago