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stealth61 [152]
3 years ago
10

The case in Latin used for indirect objects is ______. nominative accusative dative

Physics
2 answers:
Aloiza [94]3 years ago
5 0
<h3><u>Answer</u>;</h3>

Dative case

The case in Latin used for indirect objects is <u>dative case.</u>

<h3><u>Explanation</u>;</h3>
  • A noun or pronoun is in the Dative Case when it is used as an indirect object.

Example. Susan gave me an orange.

  • This sentence contains two objects, a direct object and an indirect object.
  • Orange is the direct object. It receives the action of the verb.
  • Me is the indirect object.
  • Me is a pronoun in the dative case. It does not receive the action of the verb directly, but it does receive it indirectly.
sashaice [31]3 years ago
5 0

Answer:

dative

Explanation:

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A box sits at the edge of a spinning disc. The radius of the disc is 0.5 m, and it is initially spinning at 5 revolutions per se
Furkat [3]

Answer:

a) α = 0.375 rad/s²

b) at = 0.1875 m/s²

c) ac =79 m/s²  

d) θ = 52 rad

Explanation:

The uniformly accelerated circular movemeis a circular path movement in which the angular acceleration is constant.

Tangential acceleration is calculated as follows:

at = α*R     Formula (1)

Centripetal acceleration is calculated as follows:

ac =ω² *R   Formula (2)

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t  Formula (3)

θ=  ω₀*t + (1/2)*α*t² Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed  ( rad/s)

R : radius of the circular path (m)

at:  tangential acceleration, (m/s²)

ac: centripetal acceleration, (m/s²)

Data:

R= 0.5 m  : radius of the disk

t₀=0 , ω₀ = 5 rev/s  

1 revolution = 2π rad

ω₀ = 5*(2π)rad/s  =10π rad/s  = 31.42 rad/s

ωf = 2*(2π)rad/s  =4π rad/s  = 12.57 rad/s

t = 8 s

(a) angular acceleration of the box

We replace data in the formula (3)

ωf= ω₀ + α*t

2 = 5 + α*(8)

2 -5 = α*(8)

-3 = (8)α

α=3 /8

α = 0.375 rad/s²

(b) Tangential acceleration of the box

We replace data in the formula (1)z

at =(α)*R

at = (0.375)*(0.5)

at = 0.1875 m/s²

c) Centripetal acceleration of the box at  t = 8 s

We replace data in the formula (2)

ac =ω² *R

ac =(12.57)² *(0.5)

ac = 79 m/s²  

d) Radians that the box has rotated over after t = 8 s

We replace data in the formula (4)

θ = ω₀*t + (1/2)*α*t²

θ = (5)*(8)+ (1/2)*( 0.375)*(8)²

θ = 52 rad

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