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Monica [59]
3 years ago
11

A ball is dropped from 215m. How long will it take to reach the ground? Use 9.8

Physics
1 answer:
valkas [14]3 years ago
7 0

Answer:

The time is  t = 6.62 \  s

Explanation:

From the question we are told that

   The height of the platform where the ball was dropped from is s  =  215 m

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

Generally from kinematic equations

     s =  ut + \frac{1}{2} gt^2

Here u is the initial velocity of the ball and the value is  u = 0 m/s given that the ball was at rest before it was dropped

So

     21 5 =  0 * t + \frac{1}{2} *  9.8t^2

=>  21 5 =  4.9t^2

=>   t^2  = \frac{215}{4.9}

=>  t = 6.62 \  s

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Assume that the Styrofoam slab and the fur are both initially neutral, and that the slab charged negatively after it is rubbed w
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Here it is given that slab get negatively charged so correct options are

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4 0
3 years ago
Name the physical quantity measured by the velocity-time graph?
VARVARA [1.3K]
The so-called "velocity-time" graph is actually a "speed-time" graph.  At any point
on it, the 'x'-coordinate is a time, and the 'y'-coordinate is the speed at that time.

'Velocity' is a speed AND a direction.  Without a direction, you do not have a velocity,
and these graphs never show the direction of the motion.  It seems to me that it would be
pretty tough to draw a graph that shows the direction of motion at every instant of time,
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At best, it would need a second line on it, whose 'y'-coordinate referred to a second
axis, calibrated in angle and representing the 'bearing' or 'heading' of the motion at
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7 0
2 years ago
According to the law of conservation of energy, which statement must be true?
Nitella [24]

Answer:

B. A system cannot take in additional matter.

Explanation: The total amount of energy in the universe remains constant, it can merely change from one form to another.

Hope it helps you:)))

have a good day

5 0
1 year ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

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        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

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we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

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liraira [26]

Answer:

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