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olga nikolaevna [1]
3 years ago
7

A ball is thrown vertically upwards at an initial speed of

Physics
1 answer:
sveta [45]3 years ago
6 0

Answer:

114m

Explanation:

s=ut + 1/2 at^2

s=23 m/s ×3s + 1/2×3×3s^2×10m/s^2

= (69+45)m

= 114m

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They move in a waves motion

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3. Provide two examples of static electric charge.
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Answer: 1.  walking across a carpet and touching a metal door handle            2. pulling your hat off and having your hair stand on end.

Explanation

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Answer:

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What is kepler's law??​
Elan Coil [88]
<h2>QUESTION:- </h2>

➜what is kepler's law??

\huge\red{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W}\orange{ER}}}}

Kepler gave the three laws or theorems of motion of the orbitals bodies

{\huge {\bold{ \red{ \star}}}}{ \blue{ \bold{FIRST \: \: \: LAW}}}

This law state that the celestial bodies revolves around the stars in elliptical orbit and star as a single focus.

Example :- Earth revolves around the Sun as assuming it as single focus

This also shows that earth revolves around the sun in elliptical orbit.

{\huge {\bold{ \blue{ \star}}}}{ \green{ \bold{SECOND \: \: \: LAW}}}

Area covered by the planet is equal in equal duration of time irrespective of the position of the planet.

It also states that Angular momentum is constant

As Angular momentum is constant it means areal velocity is also constant.

\frac{ \triangle \: A}{ \triangle \: T} = \frac{L}{2m}△T△A=2mL

where:-

A is the area.

T is the time.

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M is the mass of the body.

{\huge {\bold{ \green{ \star}}}}{ \purple{ \bold{THIRD \: \: \: LAW}}}

square of the time of the revolution is directly proportional to the cube of the distance between the planet and star in Astronomical unit.

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8 0
3 years ago
Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the
yaroslaw [1]

Answer:

 K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

          p₀ = 0

final moment

          p_f = m_a v_a + m_b v_b

          p₀ = p_f

          0 = m_a v_a + m_b v_b

tells us that

          m_a = 8 m_b

         

           0 = 8 m_b v_a + m_b v_b

           v_b = - 8 v_a                    (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

           Em₀ = K₀ = 73 J

final point. After separating the body

          Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

           K₀ = K_f

           73 = ½ m_a (v_a² + v_b² / 8)

           

we substitute equation 1

           73 = ½ m_a (v_a² + 8² v_a² / 8)

           73 = ½ m_a (9 v_a²)

           73/9 = ½ m_a (v_a²) = K_a

            K_a = 8,111 J

3 0
3 years ago
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