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erma4kov [3.2K]
3 years ago
8

A player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. How long was

the ball in the air?
Physics
1 answer:
RSB [31]3 years ago
6 0
2.6 seconds (If acceleration due to gravity is taken 10 m/s^2) or 2.65  (If acceleration due to gravity is taken 9.8 m/s^2)
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Dennis_Churaev [7]
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7 0
3 years ago
A block of mass 27.00 kg sits on a horizontal surface with, coefficient of kinetic
zhannawk [14.2K]

Answer:

The force is  F  = 172 \ N

Explanation:

From the question we are told that

    The  mass of the block is  m_b  = 27.0 \ kg

     The  coefficient of  static friction is  \mu_s  =  0.65

     The coefficient of kinetic friction is  \mu_k  =  0.50

The  normal force acting on the block is  

      N  =  m *  g

substituting values

     N  =   27 *  9.8

     N  =   294.6  \  N

Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as

        F_f  =  \mu_s  *  N

substituting values

        F_f  =   0.65 *  264.6

        F_f  =   172 \ N

Now for this  block to move the force require is  equal to F_f i.e

       F= F_f

=>    F  = 172 \ N

       

   

5 0
3 years ago
A __________is a large cool star located in the top right of the HR diagram.
crimeas [40]

Answer:

Vermeer star is located at the top of large Venus

5 0
2 years ago
7.) What is the relationship between kinetic and potential energy?
Arte-miy333 [17]
We know that potential energy is the energy that is stored within an object while kinetic energy is the energy that is in motion. The connection between the two is that potential energy transforms into kinetic energy
4 0
2 years ago
A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin
gogolik [260]

Answer:

a) 23 m/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

       p_{o} = p_{f}  (1)

  • The initial momentum p₀, can be written as follows:

       p_{o} =  m_{1}  * v_{1o} + m_{2}* v_{2o} =   6.0 kg * 5.0 m/s + 2.0 kg * v_{2o}  (2)

  • The final momentum pf, can be written as follows:

        p_{f} = (m_{1} + m_{2} )* v_{f}  = 8.0 kg* (-2.0 m/s)  (3)

  • Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

       v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg}  = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s  (4)

  • This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.
6 0
3 years ago
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