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dalvyx [7]
3 years ago
15

The rings of Saturn occupy the region inside Saturn's Roche limit.

Physics
1 answer:
Oliga [24]3 years ago
7 0
I suppose this is a true or false question and that sentence is true.
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An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the
poizon [28]

Answer:

The value is A  =  0.014 \  m

Explanation:

From the question we are told that

    The mass of the object is  m  =  2.0 \  kg

    The unstressed length of the string is  l  =  0.08 \  m

    The length of the spring when it is  at equilibrium is  l_e = 5.9 \  cm  =  0.059 \  m

      The initial speed (maximum speed)of the spring when given a downward blow v  =  0.30 \  m/s

Generally the maximum speed  of the spring  is mathematically represented as

           u =  A *  w

Here A is maximum height above the floor (i.e the maximum amplitude)

            and w is the angular frequency which is mathematically represented as

       w = \sqrt{\frac{k}{m} }

So

        u =  A *   \sqrt{\frac{k}{m} }

=>      A  =  u *   \sqrt{\frac{m}{k} }

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

           b  =  l -l_e

=>         b  = 0.08 - 0.05 9

=>         b  = 0.021 \  m

Generally at equilibrium position the net force acting on the spring is  

            k *  b  -  mg  =  0

=>         k *  0.021   -   2 * 9.8  =  0

=>        k =  933 \  N/m

So

            A  =  0.30  *   \sqrt{\frac{2}{933} }

=>          A  =  0.014 \  m

8 0
3 years ago
Physics, calculating net force. Please work it out for me
Margarita [4]

Answer:

F = 2985.125 N

Explanation:

Given that,

The radius of curvature of the roller coaster, r = 8 m

Speed of Micheal, v = 17 m/s

Mass of body, m = 65 kg  We need to find the net force acting on Micheal. Net force act the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg\\\\F=m(\dfrac{v^2}{r}+g)\\\\=65(\dfrac{17^2}{8}+9.8)\\\\=2985.125\ N

So, the net force is 2985.125 N.

3 0
3 years ago
A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod.
mestny [16]

Answer:

a = 17.68 m/s²

Explanation:

given,

length of the string, L = 0.8 m

angle made with vertical, θ = 61°

time to complete 1 rev, t = 1.25 s

radial acceleration = ?

first we have to calculate the radius of the circle

 R = L sin θ

 R = 0.8 x sin 61°

 R = 0.7 m

now, calculating at the angular velocity

\omega =\dfrac{2\pi}{T}

\omega =\dfrac{2\pi}{1.25}

  ω = 5.026 rad/s

now, radial acceleration

 a = r ω²

 a = 0.7 x 5.026²

a = 17.68 m/s²

hence, the radial acceleration of the ball is equal to 17.68 rad/s²

7 0
3 years ago
Edward with it gathering physical evidence at the scene of a crime. He finds a fingerprint pressed into the wax of the candle. W
muminat

Answer:trace

Explanation:

3 0
3 years ago
Read 2 more answers
A 2.00-gram air inflated balloon is given an excess negative charge, q1 = -3.25 × 10-8 C, by rubbing it with a blanket. It is fo
Llana [10]

Answer:

The rod`s charge must be positive,  because the gravity force is pointing downwards and the electrostatic force must be pointing upwards (in order to balance the gravity force)

The charge is q_2 = 1.667 times 10^(-7) C

Explanation:

F_e = F_g

where F_g = m g and F_e= (1/4 pi e_0)*(q_1*q_2)/d^2,

please see the file attached for more details.

Download pdf
3 0
3 years ago
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