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sveta [45]
2 years ago
13

Suppose a 2662 kg giant seal is placed on a scale and produces a 20.0 cm compres- sion. If the seal and spring system are set in

to simple harmonic motion, what is the period of the oscillations
Physics
1 answer:
Yuki888 [10]2 years ago
5 0

Answer:

  0.8976 seconds

Explanation:

The period of oscillation for the simple harmonic motion can be found using the formula ...

  T = 2π√(d/g)

where d is the displacement of the spring due to the attached weight, and g is the acceleration due to gravity.

__

For d = 0.20 meters, the period is ...

  T = 2π√(0.20/9.8) ≈ 0.8976 . . . . seconds

_____

<em>Additional comment</em>

The formula for the oscillator period is usually seen as ...

  T = 2π√(m/k)

where m is the mass in the system and k is the spring constant. The value of the spring constant is calculated from ...

  k = mg/d

Using that in the formula, we find it simplifies to ...

  T=2\pi\sqrt{\dfrac{m}{k}}=2\pi\sqrt{\dfrac{m}{\left(\dfrac{mg}{d}\right)}}=2\pi\sqrt{\dfrac{d}{g}}

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An arrow of mass 20 g is shot horizontally into a bale of hay, striking the hay with a velocity of 60 m/s. It penetrates a depth
goblinko [34]

From the question, The kinetic energy of the fired arrow is equal to the work done by the bale of hale in stopping the arrow.

We make use of the following formula

mv²/2 = F'd................... Equation 1

Where

  • m = mass of the arrow
  • v = velocity of the arrow
  • F' = average stopping force acting on the arrow
  • d = distance of penetration

Make F' the subject of the equation

F' = mv²/2d.................. Equation 2

From the question,

Given:

  • m = 20 g = 0.02 kg
  • v = 60 m/s
  • d = 40 cm = 0.4 m

Substitute these values into equation 2

  • F' = 0.02(60²)/(0.4×2)
  • F' = 72/0.8
  • F' = 90 N

Hence, The average stopping force acting on the arrow is 90 N

Learn more about average stooping force here: brainly.com/question/13370981

5 0
2 years ago
In the absence of a gravitational force, the weight of a body is?
azamat

Answer:

0

Explanation:

Weight = mass x gravity

if gravity = 0 then Weight =0 as well

5 0
2 years ago
Consider four point charges arranged in a square with sides of length L. Three of the point charges have charge q and one of the
nydimaria [60]

Answer:F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

Explanation:

Given

Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge

Force due to the charge placed at diagonally opposite end on -q charge

F_1=\frac{kq(-q)}{(L\sqrt{2})^2}

where  L\sqrt{2}=Distance between the two charges

F_1=-\frac{kq^2}{2L^2}

negative sign indicates that it is an attraction force

Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

F_2=\frac{kq(-q)}{(L)^2}

The magnitude of force by both the  charge is same but at an angle of 90^{\circ}

thus combination of two forces at 2 and 3 will be

F'=\sqrt{2}\frac{kq^2}{2L^2}

Now it will add with force due to 1 charge

Thus net force will be

F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]

6 0
3 years ago
A(n) 83 kg fisherman jumps from a dock into a 139 kg rowboat at rest on the West side of the dock. If the velocity of the fisher
Anna71 [15]

here we can say that there is no external force on fisherman and dock

so here we will use momentum conservation theory

As per momentum conservation

initial momentum of fisherman + boat = final momentum of fisherman + boat

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now we will have

83(3.1) + 139(0) = 83 v + 139 v

257.3 = 222v

v = 1.16 m/s

so the speed of boat and fisherman will be 1.16 m/s

3 0
4 years ago
What is specific heat? List the formula
sladkih [1.3K]

An example of a high specific heat is water’s specific heat, which requires 4.184 joules of heat to increase the temperature of 1 gram of water 1 degree Celsius. Scientifically, water’s specific heat is written as: 1 calorie/gm °C = 4.186 J/gm °C.

5 0
3 years ago
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