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3 years ago

Where is the epicenter (it's about earthquakes)

Physics

2 answers:

jasenka [17]3 years ago

5 0

Answer:

epic centre is the location on the surface of the earth directly above where the earthquake start s

sweet [91]3 years ago

4 0

Answer:

The epicenter is the point on the earth's surface vertically above the hypocenter (or focus), point in the crust where a seismic rupture begins.

<h2><u>PLEASE MARK ME BRAINLIEST, PLEASE</u></h2>

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If 6000 J of heat is added to 200 gm of water at 25° C. What will be its final<br>temperature?

Debora [2.8K]

Answer:

T₂ = 305.17 K

Explanation:

Given that,

Heat, Q = 6000 J

Mass, m = 200 gram

Initial temperature, T₁ = 25° C

We need to find its final temperature. Let it is T₂.

We know that,

$Q=mc\Delta T$

Where

c is the specific heat of water, c = 4.18 J/g°C

So,

$6000=200\times 4.18\times (T_2-298)\\\\\dfrac{6000}{200\times 4.18}=(T_2-298)\\\\7.17=(T_2-298)\\\\7.17+298=T_2\\\\T_2=305.17\ K$

So, the final temperature is equal to 305.17 K.

3 0

3 years ago

Kyle lays a mirror flat on the floor and aims a laser at the mirror. The laser beam reflects off the mirror and strikes an adjac

fgiga [73]

Answer:

θr = 55.2 °

Explanation:

The law of reflection states that the angle of incidence and the angles of reflection is the same.

Let's use trigonometry to find the angle between the reflected ray, where the horizontal distance x = 55.9 cm and the vertical distance is y = 38.9 cm

tan θ = y / x

tan θ = 38.9 / 55.9

θ = tan⁻¹ (0.6959)

θ = 34.8°

This angle is measured with respect to the x-axis (horizontal), but in general the angles in optics are measured from the y-axis so that the angle is

θ = 90 - 34.8

θr = 55.2 °

6 0

3 years ago

In a ballistics test, a 24 g bullet traveling horizontally at 1300 m/s goes through a 25-cm-thick 360 kg stationary target and e

MaRussiya [10]

Answer:

A)0.00022s b)40363.6N c) 0.025m/s

Explanation:

Mass = 24g = 0.024kg, distance though the target = thickness of the target = 25cm = 0.25m

Initial speed of the bullet = 1300m/s, final speed = 930m/s

Using equation of motion

Distance = 1/2(vf+vi)*t (time in seconds)

t = 0.25*2/(1300+930) = 0.00022s

B) force exerted on the body

F = ma = m* (vf-vi)/t = 0.024*(930-1300)/0.00022

F = -40363N, it is negative because the body decelerated during this motion

C) using law of conservation of momentum,

M1*U1+ M2*U2(M2and U1 are the mass and initial speed of the body) = M1V1+ M2V2

The target was at rest so initial speed U2 = 0

0.024*1300 + 360*0 = 0.024*930 + 360*V2

31.2 = 22.32+360*V2

31.2-22.33 = 360*V2

V2 = 8.88/360 = 0.025m/s

8 0

3 years ago

How do I solve this???

Naya [18.7K]

The projectile maintains its horizontal component of speed because there's nothing exerting any horizontal force on it. <em>(b) </em>

Gravity has no effect on horizontal motion.

4 0

3 years ago

Read 2 more answers

Calculate the electrostatic force that a small sphere A, possessing a net charge of 2.0 x 10^-6 Coulombs exerts on another small

Anni [7]

Answer:

F = 5.33*10^-4N

Explanation:

to find the electrostatic force you use the Coulomb's law, given by the formula:

$F=k\frac{q_Aq_B}{r^2}$

k: Coulomb's constant = 8.89*10^9 Nm^2/C^2

q_a: charge of A = 2.0*10^{-6}C

q_B: charge of B = -3.0*10^{-6}C

r: distance between the spheres = 10.0m

By replacing all these values you obtain:

$F=(8.89*10^9Nm^2/C^2)\frac{(2.0*10^{-6}C)(-3.0*10^{-6}C)}{(10.0m)^2}=5.33*10^{-4}N$

hence, the forcebetween the spheres is about 5.33*10^-4N

3 0

3 years ago