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Charra [1.4K]
2 years ago
14

A robot standing on a cliff shoots a ball upwards with an initial speed of 30 m/s. What is the height of the cliff given that th

e ball reaches the bottom of the cliff 8 s after the shoot? (Take g = 10 m/s^2 and the height of the robot is negligible.)
A 25 m
B 45 m
C 80 m
D 145 m​

Physics
1 answer:
valina [46]2 years ago
5 0

Answer:

C 80 m

Explanation:

Given:

v₀ = 30 m/s

a = -10 m/s²

t = 8 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²

Δy = -80 m

The ball lands 80 m below where it started.  So the height of the cliff is 80 m.

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Two identical small charged spheres hang in equilibrium with equal masses (0.02kg). The length of the strings is equal (0.18m) a
Yuliya22 [10]

Answer:

The value is q = 3.4 *10^{-6} \ C

Explanation:

From the question we are told that

    The mass of each sphere is m_1 = m_2  = m  =  0.020 \ kg

     The length of the string is  l = 0.18 \  m

     The angle of with the vertical is \theta  =  7^o

      The acceleration due to gravity is g = 9.8 \ m/s^2

Generally the force acting between the forces is mathematically represented as

       F  =  T cos \theta =  \frac{k*  q^2}{ r^2}

=>     T cos \theta =  \frac{k*  q^2}{ r^2}

Generally from Pythagoras theorem the radius of the circular curve created by the force is

         r = 2 L sin (\theta )

=>      r = 2* 0.180 sin (7)

=>      r = 0.043 \ m  

=>     q = tan \theta * \frac{m * g * r^2 }{k}

=>      q = tan(7)* \frac{ 0.02 * 9.8 * 0.043^2 }{9*10^{9}}

=>      q = 3.4 *10^{-6} \ C

7 0
2 years ago
A train traveling at 6.4 m/s accelerates at 0.10 m/s 2 over a distance of 100 m. How large is the train’s final velocity?
klasskru [66]

The final velocity of the train at the end of the given distance is 7.81 m/s.

The given parameters;

  • initial velocity of the train, u = 6.4 m/s
  • acceleration of the train, a = 0.1 m/s²
  • distance traveled, s = 100 m

The final velocity of the train at the end of the given distance is calculated using the following kinematic equation;

v² = u² + 2as

v² = (6.4)² + (2 x 0.1 x 100)

v² = 60.96

v = √60.96

v = 7.81 m/s

Thus, the final velocity of the train at the end of the given distance is 7.81 m/s.

Learn more here:brainly.com/question/21180604

4 0
2 years ago
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