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3 years ago

The seafloor spreading process at ridges produces what kind of faults?

Physics

1 answer:

irga5000 [103]3 years ago

4 0

Active transform faults are between two tectonic structures or faults.

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Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

aev [14]

Answer:

Explanation:

Given

charge of first body $q_1=2.5\ mu C$

charge of second body $q_2=-3.5\ mu C$

Particle 1 is at origin and particle 2 is at $x=0.6\ m$

third Particle which charge +q must be placed left of $2.5\mu C$ because it will repel the q charge while $-3.5\mu C$ will attract it

suppose it is placed at a distance of x m

$F_{1q}=\frac{kq(2.5)}{x^2}$

$F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}$

$F_{1q}+F_{2q}=0$

$\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0$

$\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}$

$\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}$

$0.6+x=1.1832x$

$x=3.27\ m$

5 0

3 years ago

If a body having mass 40kg started moving initially with rest and it takes a velocity of 20m/sec in time 4 seconds. Find the val

baherus [9]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}$

$\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \ body \ (v) = 20 \ m/s}$

$\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \ body \ (u) = 0}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \ body \ ( F)}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

☯ Using 1st equation of motion

$\\$

$\dashrightarrow\:\: \sf{v = u + at}$

$\\$

$\dashrightarrow\:\: \sf{20 = 0 + a(4)}$

$\\$

$\dashrightarrow\:\: \sf{20 = 4a}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}$

$\\$

$\dashrightarrow\:\: \sf{a = 5}$

$\\$

☯ Now, Finding the force exerted

$\\$

$\dashrightarrow\:\: \sf{F = ma}$

$\\$

$\dashrightarrow\:\: \sf{F = 40 \times 5}$

$\\$

$\dashrightarrow\:\: \sf{F = 200 \ N}$

$\\$

☯ Hence, $\\$

$\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}$

8 0

3 years ago

A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),

labwork [276]

Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

θ = Angle between force and displacement

FOR FORCE (P):

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

W = P d(1)

W = Pd

Therefore, work done by force (P) is Positive.

FOR NORMAL FORCE (Fn) AND WEIGHT (W):

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.

FOR KINETIC FRICTIONAL FORCE (fk):

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

Therefore, work done by Kinetic Frictional Force (fk) is Negative.

8 0

3 years ago

Calculate the acceleration using the formula acceleration = (final velocity - initial velocity) / time.

PtichkaEL [24]

Answer:

10 km/hr/s

Explanation:

The acceleration of an object is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time

For the car in this problem:

u = 0

$v=60 km/h$

t = 6 s

Substituting in the equation,

$a=\frac{60 km/h-0}{6s}=10 km/h/s$

6 0

3 years ago

Can we use a clinical thermometer to measure the temperature of a candle flame

miv72 [106K]

In theory, yes. The 2 problems are the materials used for clinical thermometers, & the temperature capacity of the clinical thermometer. If anything, change the material & extend the measurement threshold. At that point, it wouldn´t be used for clinical garbage anymore.

3 0

3 years ago