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erik [133]
4 years ago
12

A rock is thrown downward from an unknown height above the ground with an initial speed of 10m/s. It strikes the ground 3s later

. Determine the initial
Physics
1 answer:
bazaltina [42]4 years ago
6 0

At the beginning of the 3 sec, its speed is 10 m/s.

Over the course of 3 sec, gravity adds (9.8 x 3) = 29.4 m/s
to its speed, so its speed after 3 sec is (10 + 29.4) = 39.4 m/s.

Its average speed during the 3 sec is

                               (1/2) (10 + 39.4) 

                           =  (1/2) (49.4)  =  24.7 m/s .

In 3 sec, at an average speed of 24.7 m/s,
the rock travels

                     (3) x (24.7) = 74.1 meters.

If there's no air resistance in its path, then it was thrown
from 74.1 meters above the ground.
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Answer:

\theta=8.922714^{o} South-West direction

Explanation:

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therefore,

V_{pg}=V_{pa}+V_{ag}

From the triangle attached, the angle characteristics to the direction of relative velocity of the plane can be found

Sin \theta=\frac {V_{ag}}{V_{pa}}

\theta=sin^{-}\frac {38.0m/s}{245 m/s}

\theta=sin^{-1}0.155102

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4 0
3 years ago
Tlo
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Answer:

9.949 m

Explanation:

From the question,

L' = L+LαΔT................. Equation 1

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L' = 10+10(0.00017)(50-80)

L' = 10-0.051

L' = 9.949 m

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When gravitational forces and air resistance equalize on an object that is falling toward earth and the object stops acceleratin
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A freely suspended magnet does not point exactly at the geographical N-S direction​
elena-s [515]
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Write the differential equation that governs the motion of the damped mass-spring system, and find the solution that satisfies t
melisa1 [442]

This question is incomplete, the complete question  is;

Write the differential equation that governs the motion of the damped mass-spring system, and find the solution that satisfies the initial conditions specified. Units are mks; γ is the damping coefficient, with units of kg/sec

m = 0.2, γ = 1.6 and k = 4

Initial displacement is 1 and initial velocity is -2

x" + _____ x' ____x = 0

x(t) =

Answer:

the solution that satisfies the initial conditions specified is;

x(t) = c_1e^{-4t}cos(2t) + c_2e^{-4t}sin(2t)

Explanation:

Given the data in the question ;

m = 0.2, γ = 1.6, k = 4

x(0) = 1, x'(0) = -2

Now, the differential equation that governs the motions of spring mass system is;

mx" + γx' + kx = 0

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hence, characteristics equation will be;

m² + 8m + 20 = 0

we find m using; x = [ -b±√(b² - 4ac) ] / 2a

m = [ -8 ± √((8)² - 4(1 × 20 )) ] / 2(1)

m = [ -8 ± √( 64 - 80 ) ] / 2

m = [ -8 ± √-16 ) ] / 2

m = ( -8 ± 4i ) / 2

m = -4 ± 2i

Hence, the general solution of the differential equation is;

x(t) = c_1e^{-4t}cos(2t) + c_2e^{-4t}sin(2t)

From the initial conditions;

c₁ = 1, c₂ = 1

the solution that satisfies the initial conditions specified is;

x(t) = c_1e^{-4t}cos(2t) + c_2e^{-4t}sin(2t)

6 0
3 years ago
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