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statuscvo [17]
3 years ago
9

Baseballs pitched by a machine have a horizontal velocity of 30 meters/second. The machine accelerates the baseball from 0 meter

s/second to 30 meters/second in 0.5 seconds. If a baseball has a mass of 0.15 kilograms, the force the machine exerts is _______ newtons. Use F = ma, where a= v-u/t.
Physics
1 answer:
miv72 [106K]3 years ago
4 0

using the formula

F = ma

Where F is the force applied by the machine

A is the acceleration which is also equal to v/t where v is the velocity and t is time

M is the mass

 

F = mv/t

F = (0.15kg) (30 – 0 m/s)/ 0.5 s

<span>F = 9 N</span>

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What do these compounds have in common H2O cl2 NH3
nasty-shy [4]

Answer:

All the compounds are covalent compounds . This means that they are formed by the sharing of pair of electrons.

7 0
3 years ago
An ice skater is spinning at 6.00 rev/s with his moment of inertia being 0.400 kg/m2. Calculate his new moment of inertia if he
labwork [276]

Answer:

New moment of inertia will be I=1.92kgm^2

Explanation:

It is given initially angular velocity \omega =6rev/sec=6\times 2\pi =37.68rad/sec

Moment of inertia I=0.4kgm^2

Angular momentum is equal to L=I\omega =37.68\times 0.4=15.072kgm^2/sec

Now angular velocity is decreases to \omega =1.25rev/sec=1.25\times 2\times 3.14=7.85rad/sec

As we know that angular momentum is conserved

So 15.072=I\times 7.85

I=1.92kgm^2

So new moment of inertia will be I=1.92kgm^2

4 0
3 years ago
Calculate the amount of heat liberated (in kj) from 411 g of mercury when it cools from 88.0°c to 12.0°c.
Katen [24]
You should note that the melting point of mercury is -38.83°C, while the boiling point is at 356.7°C. Then, that means that there is no latent heat involved here. We only compute for the sensible heat.

ΔH = mCpΔT
The Cp of mercury is 0.14 J/g·°C
Thus,
ΔH = (411 g)(0.14 J/g·°C)(88 - 12°C)
<em>ΔH = 4,373.04 J</em>
5 0
3 years ago
A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu
dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

      T_1 =  0.341 * 9.8

      T_1 =  3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

5 0
3 years ago
A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik
shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle=40^{\circ}

x =12\pm 0.27

y=1.05

equation of trajectory of ball is given by

y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }

for x=12.27

1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }

u=10.69

for x=11.73

1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }

u=11.436 m/s

Thus range of speed is (10.69, 11.436)

3 0
3 years ago
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