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vfiekz [6]
3 years ago
7

A 1.1-kg object is suspended from a vertical spring whose spring constant is 160 N/m. (a) Find the amount by which the spring is

stretched from its unstrained length. (b) The object is then pulled straight down by an additional distance of 0.26 m and released from rest. Find the speed with which the object passes through its original position on the way up.
Physics
1 answer:
Eva8 [605]3 years ago
8 0

Answer:

3.14 m/s

Explanation:

We are given that

Mass of object=1.1 Kg

Spring constant=160 N/m

a.We have to find the amount by which the spring is stretched from its unstrained length(it means x)

We know that F=kx ,g=9.8m/s^2

x=\frac{F}{k}=\frac{1.1\cdot 9.8}{160}=0.0674 m

x=0.0674\times 100=6.74 cm (1m=100 cm)

b.When an object is pulled straight down by an additional distance 0.26 m and released from rest.

x=0.26 m

We have to find the speed with which the object passes through its original position on the way up.

K.E=\frac{1}{2}kx^2 (Law of conservation of energy)

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

Substitute the values then we get

1.1\cdot v^2=160\cdot (0.26)^2

v=\sqrt{\frac{160\cdot(0.26)^2}{1.1}}

v=3.14 m/s

Hence, the speed of an object with which the object passes through its original position on the way up=3.14 m/s

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The graph shows a heating curve for water. Between which points on the graph would condensation occur?​
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Read 2 more answers
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1,
mash [69]

Answer:

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 − t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ − 44 t₁²

18350 = 3960 t₁ − 22 t₁²

9175 = 1980 t₁ − 11 t₁²

11 t₁² − 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.

3 0
3 years ago
in a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency 105.0 mhz in opposite direction
Romashka [77]

As a result, the hollow is 10.90 meters long and the distance between the nodal planes is 1.36 meters.

<h3>Explain electromagnetic waves.</h3>

The oscillations between an electric field and a magnetic field produce waves known as electromagnetic waves, or EM waves.

By definition, we understand that the frequency equals,

f = c/λ

where,

λ = wavelength

c= Speed of light

λ = 2L / n

While the wavelength is equal to,

Where,

L = Length

n = Number of antinodes/nodes

PART A) We know that the first component's wavelength is 110 MHz, so

λ = c/ f

λ = 3*10^8 / 11*10^6

λ = 1.36m

Therefore the distance between the nodal planes is 1.36m

PART B) For this part we need to find the Length through the number of nodes (8) and the wavelength, that is,

λ` = 2l /n

L = 8*2.72/ 2

L = 10.90m

Therefore the length of the cavity is 10.90m.

To know more about electromagnetic waves visit:-

brainly.com/question/3101711

#SPJ4

7 0
1 year ago
A 0.50 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict
Genrish500 [490]

Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

The object is released from rest when the spring is compressed 0.19 m.

(A) The force acting on the object is given by :

F = kx

F=157\times 0.19\\\\F=29.83\ N

(B) The force is simply given by :

F = ma

a is acceleration at that instant

a=\dfrac{F}{m}\\\\a=\dfrac{29.83}{0.5}\\\\a=59.66\ m/s^2

6 0
3 years ago
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