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3 years ago

HELP ASAP PLEASE!!!

Physics

2 answers:

Firlakuza [10]3 years ago

7 0

The answer is A. Earthquakes usually occur in fault lines

blagie [28]3 years ago

4 0

Answer: A

Explanation:Earthquakes occur on faults - strike-slip earthquakes occur on strike-slip faults, normal earthquakes occur on normal faults, and thrust earthquakes occur on thrust or reverse faults. When an earthquake occurs on one of these faults, the rock on one side of the fault slips with respect to the other.

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On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze

aleksklad [387]

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4 years ago

Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400

hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

$v^2=u^2+2as$

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

$v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s$

The launching speed of the beetle is 6.4 m/s.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

$v=u+at$

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

$v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms$

The time taken by the beetle to launch itself upwards is 1.62 ms.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

$v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m$

The beetle can jump to a height of 2.1 m

7 0

3 years ago

A toy plane is flying in a horizontal circle by being attached to a 0.75 meter string. The plane has a mass of 101.7 grams and m

nadezda [96]

Answer:

F = 2,894 N

Explanation:

For this exercise let's use Newton's second law

F = m a

The acceleration is centripetal

a = v² / r

Angular and linear variables are related.

v = w r

Let's replace

F = m w² r

The radius r and the length of the rope is related

cos is = r / L

r = L cos tea

Let's replace

F = m w² L cos θ

Let's reduce the magnitudes to the SI system

m = 101.7 g (1 kg / 1000g) = 0.1017 kg

θ = 5 rev (2π rad / rev) = 31,416 rad

w = θ / t

w = 31.416 / 5.1

w = 6.16 rad / s

F = 0.1017 6.16² 0.75 cos θ

F = 2,894 cos θ

The maximum value of F is for θ equal to zero

F = 2,894 N

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3 years ago

Which statement best describes the movement of electrical current when a capacitor is used in a circuit?

Masteriza [31]

Answer: B.

Capacitors prevent current from moving through a circuit

Explanation:

If a direct voltage is applied on the capacitor, no conduction current flows through the capacitor if its insulating medium is perfect insulator. This is due to the fact that there are no free charge carriers in such medium. Basically the real insulator contains very few charge carriers and therefore a very small leakage current passes in the capacitor depending on the conductivity of the insulator.

If an alternating voltage is applied on the capacitor, a displacement current passes through the capacitor irrespective of the insulating medium. This current is termed also the capacitive current. It flows because of changing electric displacement with time.

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A rocket is launched at an angle of = 49° above the horizontal with an initial speed vi = 54 m/s, as shown below. It moves for 2

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