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Serga [27]
3 years ago
9

With what minimum speed must you toss a 200 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release

the ball 1.3 m above the ground? solve this problem using energy.
Physics
1 answer:
Brut [27]3 years ago
8 0
When they say use energy, you want to use

Total energy = potential energy + kinetic energy = mgh + 1/2mv²

I assume you mean 200 g ball,

so, we know the total distance traveled is going to be 13 - 1.3 = 11.7 m

If the ball just makes it to the top ( 13 m ) , then the ball will stop moving and the kinetic energy will be 0,

therefore, the potential energy at the top will be the total energy of the system = mgh

from this, we say that mgh = 1/2mv² solve for v

<span> v = sqrt (2gh) = 15.2 m/s </span>
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The mcb of rupa's room is tripped and keeps on tripping again and again . if it is a domestic circuit, what could be the reason
Tamiku [17]

The MCB of a rupas room is tripped and keeps on tripping again and again, and if it is a domestic circuit, what could be the reason for this phenomenon?

The reason could be a short circuit which is resulting in higher level of currents to pass through the MCB which is resulting in trip every time.

OR

The MCB is faulty and might need a replacement.

To Diagnose the problem further more.

Turn off all the switches in rupas room and then try turning on the MCB. If it trips again then MCB is faulty (Subjective to the fact there everything was normal before this issue and no signs of short circuit or spark in wiring were observed)

If MCB does not trip in point 1 then Turn ON all the switches one by one. This shall give you the cause of problem.

4 0
3 years ago
What are three things you already know about the game of baseball?
Tatiana [17]

Answer:

Positions

Batting

Pitching

Explanation:

4 0
3 years ago
How does the density of fluid affect the magnitude of buoyancy acting on an object immersed in it
Oksi-84 [34.3K]

Answer:

Explanation:

more the density , more will be the buoyant force acting on it , less the density less will be the buoyant force acting on it. This is why people float in dead sea and sink in other seas

5 0
3 years ago
A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the
Ierofanga [76]

Answer:

The discharge rate is Q = 0.0192 \  m^3 /s

Explanation:

From the question we are told that

   The  diameter is  d =  60 \ mm   =  0.06 \ m

    The  head is  h  =  6 \ m

     The  coefficient of contraction is  Cc  =  0.68

     The  coefficient of  velocity is  Cv  =  0.92

The radius is mathematically evaluated as

         r =  \frac{d}{2}

substituting values

        r =  \frac{ 0.06 }{2}

        r =  0.03 \ m

The  area is mathematically represented as

      A =  \pi r^2

substituting values

      A =  3.142 *  (0.03)^2

      A = 0.00283 \ m^2

 The  discharge rate is mathematically represented as

        Q =  Cv *Cc  *  A  *  \sqrt{ 2 * g *  h}

substituting values

       Q = 0.68 *  0.92*   0.00283  *  \sqrt{ 2 * 9.8 *  6}

       Q = 0.0192 \  m^3 /s

6 0
3 years ago
A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus
babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

I = m\cdot (\vec{v}_{2} - \vec{v_{1}}) (1)

Where:

I - Impulse, in kilogram-meters per second.

m - Mass, in kilograms.

\vec{v_{1}} - Initial velocity of the hockey park, in meters per second.

\vec{v_{2}} - Final velocity of the hockey park, in meters per second.

If we know that m = 0.2\,kg, \vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right] and \vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right], then the impulse applied by the stick to the park is approximately:

I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]

I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

8 0
3 years ago
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