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Ksenya-84 [330]
3 years ago
5

If the frequency of a system undergoing simple harmonic motion doubles, by what factor does the maximum value of acceleration ch

ange?
a. 4
b. 2/pi
c. 2
d. (2)^1/2
Physics
1 answer:
faltersainse [42]3 years ago
6 0

Answer:

the answers the correct one is a  4

Explanation:

The centripetal acceleration is by

           a = v² / R

angular and linear velocities are related

           v = w R

let's substitute

           a = w² R

for initial condition

           a₀ = w₀² R

suppose the initial angular velocity is wo, suppose the angular velocity doubles

           a = (2w₀)² R

           a = 4 (w₀² R)

           a = 4 a₀

when reviewing the answers the correct one is a

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A rocket rises vertically, from rest, with an acceleration of 3.6m/s^2 until it runs out of fuel at an altitude of 1500m. After
Montano1993 [528]

A. 103.9 m/s

The motion of the rocket until it runs out of fuel is an accelerated motion with constant acceleration a = 3.6 m/s^2, so we can use the following equation

v^2 -u^2 = 2ad

where

v is the velocity of the rocket when it runs out of fuel

u = 0 is the initial velocity of the rocket

a = 3.6 m/s^2 is the acceleration

d = 1500 m is the distance covered during this first part

Solving for v, we find

v=\sqrt{u^2+2ad}=\sqrt{(0^2+2(3.6 m/s^2)(1500 m)}=103.9 m/s

B. 28.9 s

We can calculate the time taken for the rocket to reach this altitude with the formula

d=\frac{1}{2}at^2

where

d = 1500 m

a = 3.6 m/s^2

Solving for t, we find

t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(1500 m)}{3.6 m/s^2}}=28.9 s

C. 2050.8 m

We can calculate the maximum altitude reached by the rocket by using the law of conservation of energy. In fact, from the point it runs out of fuel (1500 m above the ground), the rocket experiences the acceleration due to gravity only, so all its kinetic energy at that point is then converted into gravitational potential energy at the point of maximum altitude:

K_i = U_f\\\frac{1}{2}mu^2 = mgh

where h is distance covered by the rocket after it runs out of fuel, and v=103.9 m/s is the velocity of the rocket when it starts to decelerate due to gravity. Solving for h,

h=\frac{v^2}{2g}=\frac{(103.9 m/s)^2}{2(9.8 m/s^2)}=550.8 m

So the maximum altitude reached by the rocket is

h' = d+h=1500 m +550.8 m=2050.8 m

D. 39.5 s

The time needed for the second part of the trip (after the rocket has run out of fuel) can be calculated by

h=\frac{1}{2}gt^2

where

h = 550.8 m is the distance covered in the second part of the trip

g = 9.8 m/s^2

Solving for t,

t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(550.8 m)}{9.8 m/s^2}}=10.6 s

So the total time of the trip is

t'=28.9 s+10.6 s=39.5 s

E. 200.5 m/s

When the rocket starts moving downward, it is affected by gravity only. So the gravitational potential energy at the point of maximum altitude is all converted into kinetic energy at the instant the rocket hits the ground:

\frac{1}{2}mv^2 = mgh

where

v is the final velocity of the rocket

h = 2050.8 m is the initial altitude of the rocket

Solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(2050.8 m)}=200.5 m/s

F. 60 s

We need to calculate the time the rocket takes to fall down to the ground from the point of maximum altitude, and that is given by

h'=\frac{1}{2}gt^2

where

h' = 2050.8 m

g = 9.8 m/s^2

Solving for t,

t=\sqrt{\frac{2h'}{g}}=\sqrt{\frac{2(2050.8 m)}{9.8 m/s^2}}=20.5 s

So the total time of the trip is

t''=39.5 s+20.5 s=60 s

6 0
4 years ago
A block of mass M slides down a frictionless plane inclined at an angle (theta) with the horizontal. The normal reaction force e
WINSTONCH [101]

Answer: N = Mgcos(theta)

Therefore, the Normal reaction force is equal to Mgcos(theta)

Explanation:

See attached for a sketch.

From the attachment.

.

N = normal reaction force on block

W = weight of the block

theta = angle of the inclined plane to the horizontal

From the sketch, we can see that

N is equal in magnitude but opposite direction to Wy

N = Wy

And

Wy = Wcos(theta)

Wx = Wsin(theta)

Then,

N = Wy = Wcos(theta)

But W = mass × acceleration due to gravity = mg

N = Mgcos(theta)

Therefore, the Normal reaction force is equal to Mgcos(theta)

7 0
4 years ago
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How do chemical, physical, geological, and biological factors influence an aquatic ecosystem?
Elis [28]

Answer:

Aquatic organisms and the physical and chemical components of their environment are inseparably inter-related and interact with other. Flow and water chemistry are the primary factors governing life in riverine habitats, and both are closely related to seasonal variations.

Explanation:

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3 years ago
Turbo chargers helps to increase the fuel economy? O True O False
larisa [96]

Answer:

false

Explanation:

5 0
3 years ago
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A 3.04 kg particle is located on the x-axis at xm = −8 m, and a 5.61 kg particle is on the x-axis at xM = 3.56 m. Find the coord
Murljashka [212]

Answer:

center of mass = −0.50 m

Explanation:

given data

mass m1 = 3.04 kg

distance xm = -8 m

mass m2 = 5.61 kg

distance xM = 3.56 m

solution

we get here center of mass for n mass of system that is express as

center of mass = \frac{m_1x_1+m_2x_2......m_nx_n}{m_1+m_2...m_n}     ......................1

but we have only 2 particle system so we will get

center of mass = \frac{m1 \times xm+m2 \times xM}{m1+m2}      .................2

put here value and we will get

center of mass = \frac{3.04 \times (-8 )+5.61 \times 3.56}{3.04 + 5.61}

solve it we will get

center of mass = −0.50 m

8 0
4 years ago
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