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mafiozo [28]
2 years ago
6

The velocity of the transverse waves produced by an earthquake is 5.08 km/s, while that of the longitudinal waves is 8.3312 km/s

. A seismograph records the arrival of the transverse waves 55.9 s after that of the longitudinal waves. How far away was the earthquake?
Physics
1 answer:
ASHA 777 [7]2 years ago
7 0

Answer:

727.67 km

Explanation:

Sine they have Same distance D

distance = speed * time

D = 5.08t

D = 8.3312(t+55.9)

so

5.08t = 8.3312(t+55.9) t in

3.2512t = 465.71

t = 143.2s

Subtitute t

D=5.08 t

= 5.08 × 143.2

= 727.67km

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A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what hei
tekilochka [14]
Assume no air resistance, and g = 9.8 m/s².

Let
x =  angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity

The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x)  s

With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
 55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0

Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°

The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s

If t = 5.8096 s,
  u*t = 9.467*5.8096 = 55 m (Correct)
or
 u*t = 28.469*15.8096 = 165.4 m (Incorrect)

Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s

The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m

Answer: h = 110.4 m

7 0
3 years ago
You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the eart
hichkok12 [17]

Answer:

v = 3.5 \times 10^5 m/s

Explanation:

At some distance from the Earth the force of attraction due to moon is balanced by the force due to Moon

so we will have

\frac{GM_em}{r^2} = \frac{GM_m}{(d-r)^2}

now we have

\frac{d - r}{r} = \sqrt{\frac{M_m}{M_e}}

\frac{3.844\times 10^8 - r}{r} = \sqrt{\frac{7.36 \times 10^{22}}{5.9742\times 10^{24}}}

so we will have

r = 3.46 \times 10^8 m

Now by energy conservation

-\frac{GM_e}{R_e} - \frac{GM_m}{d - (R_e + R_m)} + \frac{1}{2}v^2 = -\frac{GM_e}{r} - \frac{GM_m}{d - r}

-6.26 \times 10^{8} - 13046 + \frac{1}{2}v^2 = -1.15 \times 10^6 - 1.28 \times 10^5

v = 3.5 \times 10^5 m/s

7 0
3 years ago
Un automóvil acelera de 0 a 140 km/h en 9.8 segundos, determina su aceleración
guajiro [1.7K]

Answer:

The acceleration is 14.28 km/h^2

Explanation:

Step one:

Given data

initial speed u= 0 km/h

final speed v= 140km/h

time t= 9.8 seconds

Required

The acceleration of the car

Step two:

From a= v-u/t

substitute

a= 140-0/9.8

a=140/9.8

a=14.28 km/h^2

6 0
3 years ago
Chemical changes occur as energy is transferred when molecules- - or - to form one or more new substances
timurjin [86]

Answer:

Correct options are C. As food is digested, molecule bonds are broken, allowing new molecules to form that can be used by our bodies.

D. Energy stored in organic fuels are used for heat and electricity/ When batteries are connected to a circuit, a reaction occurs inside the cell producing electrical energy.

Explanation:

The breakdown of food leads to the release of chemical energy which was first stored in food. This energy is used to drive necessary functions of the body. A series of chemical reactions are involved in the breakdown of food and changing them into useful components for the cell.

Chemical energy is stored in circuits. When the battery is connected to a circuit, the chemical energy gets transferred into electrical energy.

4 0
3 years ago
Read 2 more answers
g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the
MrRa [10]

Answer:

Approximately -1.58\; \rm rad \cdot s^{-2}.

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

v^2 - u^2 = 2\, a\, x,

where

  • v is the final velocity of the moving object,
  • u is the initial velocity of the moving object,
  • a is the (linear) acceleration of the moving object, and
  • x is the (linear) displacement of the object while its velocity changed from u to v.

The angular analogue of that equation will be:

(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta, where

  • \omega(\text{final}) and \omega(\text{initial}) are the initial and final angular velocity of the rotating object,
  • \alpha is the angular acceleration of the moving object, and
  • \theta is the angular displacement of the object while its angular velocity changed from \omega(\text{initial}) to \omega(\text{final}).

For this object:

  • \omega(\text{final}) = 0\; \rm rad\cdot s^{-1}, whereas
  • \omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}.

The question is asking for an angular acceleration with the unit \rm rad \cdot s^{-1}. However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}.

Rearrange the equation (\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta and solve for \alpha:

\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}.

7 0
3 years ago
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