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An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 14.0

Reptile [31]

Answer:

28 m/s^2

Explanation:

distance, s = 14 m

time, t = 2 - 1 = 1 s

initial velocity, u = 0 m/s

Let a be the acceleration.

Use third equation of motion

$s = ut + \frac{1}{2}at^{2}$

$14 = 0 + \frac{1}{2}a\times 1^{2}$

a = 28 m/s^2

Thus, the acceleration is 28 m/s^2.

7 0

3 years ago

A simple pendulum having a length of 1.02 m and a mass of 6.66 kg undergoes simple harmonic motion when given an initial speed o

attashe74 [19]

Answer:

Time period of pendulum is 2.02 s.

Explanation:

A simple pendulum is a device which consists of mass m hanging from the string of length L attached to the some point.When displaced and released its swings back and forth with periodic motion.

The time period of pendulum is defined as time taken by the pendulum to complete one full oscillation . it is denoted by T.

By Huygens law of period of pendulum,

T = 2π $\sqrt{\frac{L}{g} }$ eqn 1

where L is the length of pendulum,

g is acceleration due to gravity

Period of pendulum is independent of the mass of pendulum,

Substituting values in eqn 1

T = 2π $\sqrt{\frac{1.02}{9.8\\} }$

T = 2.02 s

Time period of pendulum is 2.02 s.

4 0

3 years ago

While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find th

Dovator [93]

Answer:

Explanation:

From the given information:

radius = 15 m

Time T = 23 s

a) Speed (v) = $\dfrac{2 \pi r}{T}$

$v = \dfrac{2\times \pi \times 15}{23}$

v = 4.10 m/s

b) The magnitude of the acceleration is:

$a = \dfrac{v^2}{r} \\ \\ a = \dfrac{(4.10)^2}{15}$

a = 1.12 m/s²

c) True weight = mg

Apparent weight = normal force

From the top;

the normal force = upward direction,

weight is downward as well as the acceleration.

true weight - normal force = ma

apparent weight =mg - ma

$\dfrac{apparent \ weight}{true \ weight} = \dfrac{(mg - ma)}{(mg)}$

$=1- \dfrac{1.12}{9.8}$

= 0.886 m/s²

d)

From the bottom;

acceleration is upward, so:

apparent weight - true weight = ma

apparent weight = true weight + ma

$\dfrac{apparent \ weight }{true \ weight} =\dfrac{ mg + ma}{mg}$

$\dfrac{apparent \ weight }{true \ weight} =1+ \dfrac{a}{g} \\ \\ = 1 + \dfrac{1.12}{9.8}$

$= 1 + \dfrac{1.12}{9.8} \\ \\$

= 1.114 m/s²

6 0

3 years ago

The smallest particle of an element that retains the characteristics of the element is a(n) ________

Novay_Z [31]

//////////“atom”//////////

5 0

2 years ago

A5 cm object is 18.0 cm from a convex lens, which has a focal length of 10.0 cm.

Masja [62]

Explanation:

We have,

Height of object is 5 cm

Object distance from a convex lens is 18 cm

Focal length of convex lens is 10 cm

i.e. h = 5 cm

u = -18 cm

f = +10 cm

Let v is distance of the image from the lens. Using lens formula :

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{10}+\dfrac{1}{(-18)}\\\\v=22.5\ cm$

The magnification of lens is :

$m=\dfrac{v}{u}=\dfrac{h'}{h}$ , h' is height of the image

$\dfrac{v}{u}=\dfrac{h'}{h}\\\\h'=\dfrac{vh}{u}\\\\h'=\dfrac{22.5\times 4}{(-18)}\\\\h'=-5\ cm$

h' = -5.00 cm (in three significant figures)

6 0

3 years ago