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Dahasolnce [82]
3 years ago
10

A 0.415-kg mass suspended from a spring undergoes simpleharmonic oscillations with a period of 1.4 s. How much mass, inkilograms

, must be added to the object to change the period to2.2 s?
Physics
1 answer:
koban [17]3 years ago
6 0

Answer:

m=0.893kg

Explanation:

time period of oscillations is given by= 2π√(m/k)

m: mass of the object

k: spring constant

when T=1.5 and m=0.415

1.5= 2π√(0.415/k)

k= 7.27 N/m

when T= 2.2s

2.2= 2π√(m/7.27)

m=0.893kg

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Juliana walks 45 meters East, 45 meters south and 45 meters north. What is her resultant displacement?
katrin2010 [14]

Her resultant displacement is (45 Meters East.)

She originally walks 45 meters east, then she goes south 45 meters, then north 45 meters.  The south and north placements  just make her go back to where her previous placements were before them. Making her 45 meters east.

6 0
3 years ago
How do spectroscopes help with studying of distance stars?
Svetradugi [14.3K]

Answer:

You take the light from a star, planet or galaxy and pass it through a spectroscope, which is a bit like a prism letting you split the light into its component colours. "It lets you see the chemicals being absorbed or emitted by the light source. From this you can work out all sorts of things," says Watson

8 0
3 years ago
Two identical twins hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about the
777dan777 [17]

Answer:

Part a)

L = 2683.2 kg m^2/s

Part b)

v' = 8.60 m/s

Part c)

W = 4326.7 J

Explanation:

Part a)

As we know that there is no external torque on the system of two twins

so here we will use

L = mv r + mvr

L = 2(78 \times 4.30 \times 4)

L = 2683.2 kg m^2/s

Part b)

Since angular momentum is conserved here as there is no external torque

so we will have

2(m v r) = 2( m v' \frac{r}{2})

v' = 2v

v' = 8.60 m/s

Part c)

Work done by both of them = change in kinetic energy

so we have

W = 2(\frac{1}{2}mv'^2 - \frac{1}{2}mv^2)

W = m(v'^2 - v^2)

W = 78(8.60^2 - 4.3^2)

W = 4326.7 J

5 0
3 years ago
A sound from a source has an intensity of 270 db when it is 1 m from the source. what is the intensity of the sound when it is 3
Lina20 [59]
Sound intensity is inversely proportional to the square of the distance between the source and the receiver.
That is 
I = k/r^2
 where
 k = constant
 r = radius

When r=1,  the intensity is I₁ = k/1 = k
When r=3, the intensity I₂ = k/3² = k/9
Therefore
 I₂ = I₁ /9

In decibels,
I = 10 log₁₀(I/I₀)
where I₀ = reference intensity

When r=1,
10 log₁₀ (I₁/I₀) = 270

When r =3,
10 log₁₀ (I₂/I₀) = 10 log₁₀ [(I₂/I₁)*(I₁/I₀)]
                     = 10 log₁₀ [(1/9)*(I₁/I₀)]
                     = 10 log₁₀(1/9) + 270
                     = 260.5

Answer: 260.5 dB (nearest tenth)
4 0
3 years ago
a rocket has a mass 250(10^3) slugs on earth. Specify its mass in si units and its weight in si unites. if the rocket is on the
Katena32 [7]

Answer:

W_{earth} = 35.74 * 10^6 N : Rocket weight on earth

W_{moon} = 5.91 * 10^6 N : Rocket weight on moon

Explanation:

Conceptual analysis

Weight is the force with which a body is attracted due to the action of gravity and is calculated using the following formula:

W = m × g Formula (1)

W: weight

m: mass

g: acceleration due to gravity

The mass of a body on the moon is equal to the mass of a body on the earth

The acceleration due to gravity on a body is different on the moon and on the earth

Equivalences

1 slug = 14.59 kg

Known data

m_{earth} = m_{moon} = 250 * 10^3 slug = 250* 10^3slug * \frac{14.59kg}{1slug} = 3647.5* 10^3 kg

g_{moon}= 1.62 \frac{m}{s^2}

g_{earth}= 9.8 \frac{m}{s^2}

Problem development

To calculate the weight of the rocket on the moon and on earth we replace the data in formula (1):

W_{earth} = 3647.5* 10^3 kg * 9.8 \frac{m}{s^2} = 35.74 * 10^6 N : Rocket weight on earth

W_{moon} = 3647.5* 10^3 kg * 1.62 \frac{m}{s^2} = 5.91 * 10^6 N : Rocket weight on moon

7 0
3 years ago
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