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3 years ago

At which region of a state's geology would an earthquake be more likely?

Physics

1 answer:

arlik [135]3 years ago

3 0

Transform boundary USJSJSSOAKAMAMXMXHDJEUSUS

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Football is kicked at 23.4 m/s at a 38.5 angle. How far away does it land?

natulia [17]

Answer: go0gle will know the answer

Explanation: hey I’m not giving fault answers so yea

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3 years ago

Read 2 more answers

A 12.0 kg block rests on an inclined plane. The plane makes an angle of 31.0° with the horizontal, and the coefficient of fricti

allochka39001 [22]

Answer:

The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.

Explanation:

Given that,

Mass = 12.0 kg

Angle = 31.0°

Friction coefficient = 0.158

Mass of second block = 38.0 kg

Using formula of frictional force

$f_{\mu} = \mu N$ ....(I)

Where, N = normal force

$N = mg\cos\theta$

Put the value of N into the formula

$N =12\times9.8\times\cos 31^{\circ}$

$N=100.80\ N$

Put the value of N in equation (I)

$f_{mu}=0.158\times100.80$

$f_{mu}=15.9264\ N$

Now, Weight of second block

$W = mg$

$W=38.0\times9.8$

$W=372.4\ N$

The horizontal force is

$F = mg\sintheta$

$F=12\times9.8\times\sin 31^{\circ}$

$F=60.5684\ N$ ....(II)

(I). We need to calculate the acceleration

$a=m_{2}g-\dfrac{f_{\mu}+mg\sin\theta}{m_{1}+m_{2}}$

$a=\dfrac{372.4-(15.9264+60.5684)}{12+38}$

$a=5.92\ m/s^2$

(II). We need to calculate the tension in the rope

$m_{2}g-T=m_{2}a$

$-T=38\times5.92-38\times9.8$

$T=147.44\ N$

Hence, The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.

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4 years ago

How are different types of radiation arranged along the electromagnetic spectrum

Leona [35]

In order of increasing frequency: Radio, Microwaves, Infrared, Visible light, Ultra-violet, X-rays, Gamma rays. To remember this try:
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3 years ago

A force of attraction would exist between

siniylev [52]

Answer:

A. positively charged object and a negatively charged object.

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4 years ago

Achargingbullelephantwithamassof5240kgcomesdi- rectly toward you with a speed of 4.55 m/s. You toss a 0.150-kg rubber ball at th

Ierofanga [76]

Answer:

Given:

Mass of elephant = 5240 kg

The initial speed of the elephant = 4.55 m/s

Mass of the rubber ball, m, = 0.15 kg

Inital speed of the rubber ball, v = 7.81 m/s

On substitution in $V_{2f}$ = $[\frac{2m_{1} }{m_{1} +m_{2} } ]V_{1f}$ + $[\frac{m_{2}-m_{1}}{m_{1}+m_{2} } ] v_{2f}$

$V_{2f}$ = $[\frac{2*5240_{} }{5240_{} +0.15_{} } ](-4.55_{})$ + $[\frac{0.15_{}-5240_{}}{5240_{}+0.15_{} } ] (7.81_{})$

a) The negatıve sign shows that the ball bounces back in the direction opposite to the incident

b) it is clear that the velocity of the ball increases and therefore it is kinetic energy . The ball gains kinetic energy from the elephant.

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3 years ago