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Maslowich
2 years ago
8

A basketball is tossed upwards with a speed of 5.0 m/s. ​We can ignore air resistance.

Physics
1 answer:
FinnZ [79.3K]2 years ago
4 0

We have that  the maximum height reached by the basketball from its release point is

S=1.3m

From the question we are told

  • A basketball is tossed upwards with a speed of 5.0 m/s. ​We can ignore air resistance.  
  • What is the maximum height reached by the basketball from its release point?

Generally the Newtons equation for Motion  is mathematically given as

V^2=u^2+2as\\\\Therefore\\\\0=25-19.6*S\\\\S=\frac{25}{19.6}\\\\S=1.276

S=1.3m

Therefore

The maximum height reached by the basketball from its release point is

S=1.3m

For more information on this visit

brainly.com/question/23366835

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The apparent change is the Doppler shift
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A 21.0 kg shopping cart is moving with a velocity of 4.0 m / s. It strikes a 7.0 kg box that is initially at rest. They stick to
jonny [76]

Explanation:

Given that,

Mass of the shopping cart, m_s=21\ kg

Initial speed of the shopping cart, u_s=4\ m/s

Mass of the box, m_b=7\ kg

Initial speed of the box, u_b=0 (at rest)

They stick together and continue moving at a new velocity. It is a case of inelastic collision.

(a) The momentum of the shopping cart before the collision is given by :

p_s=m_s\times u_s\\\\p_s=21\times 4\\\\p_s=84\ kg-m/s

(b) The momentum of the box before the collision is given by :

p_b=m_b\times u_b\\\\p_b=7\times 0\\\\p_b=0

(c) The velocity of the combined shopping cart/box wreckage after the collision is given by using the conservation of momentum as :

m_su_s+m_bu_b=(m_s+m_b)V\\\\V=\dfrac{m_su_s+m_bu_b}{(m_s+m_b)}\\\\V=\dfrac{21\times 4+0}{(21+7)}\\\\V=3\ m/s

Hence, this is the required solution.

8 0
3 years ago
A 10 Kg dog is running with a speed of 5.0 m/s what is the minimum work required to stop the dog
ra1l [238]

Answer:

25J

Explanation:

Given parameters:

Mass of the dog  = 10kg

Speed of the dog  = 5m/s

Unknown:

The minimum energy required to stop the dog  = ?

Solution:

The dog is moving with a kinetic energy and to stop the dog, an equal amount of kinetic energy generated must be applied to the dog.

 To find the kinetic energy;

        K.E  = \frac{1}{2} m v²

m is the mass

v is the velocity

Now insert the parameters and solve;

      K.E  =  \frac{1}{2}  x 10 x 5  = 25J

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14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards
quester [9]

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

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Which planet has the GREATEST attraction to the sun?
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I think its Mercury because it's the closest to the sun.
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