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Alexxandr [17]
3 years ago
8

An outfielder throws a baseball with an initial speed of 81.8mi/h. Just before an infielder catches the ball at the same level,

the ball’s speed is 110 ft/s. In foot-pounds, by how much is the mechanical energy of the ball– Earth system reduced because of air drag? (The weight of a baseball is 9.0 oz.)

Physics
1 answer:
agasfer [191]3 years ago
3 0

Answer:

-20.158ft-lb

Explanation:

Check the attached files for the explanation.

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the formula to calculate the force is, F = MA

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Please help me it's URGENT (#10 btw and also how do I solve for time​
Afina-wow [57]

Answer:

t = 12s

Explanation:

Given:

v-initial = 0 m/s

x = 360 m

a = 5.0 m/s^2

Solve:

x = (v-initial)t + 1/2(a*t^2)

360 = 0t + 1/2 (5.0t^2)

360 = 2.5 t^2

144 = t^2

t = sqrt(144) = 12

Therefore, it takes 12 seconds.

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The magnetic lines of force always travel from north to south true or false
kumpel [21]

Answer:

True

Explanation:

Magnetic field lines outside of a permanent magnet always run from the north magnetic pole to the south magnetic pole. Therefore, the magnetic field lines of the earth run from the southern geographic hemisphere towards the northern geographic hemisphere.

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2 years ago
An 8-passenger Learjet has a force of gravity of 6.6x10^4 [down] acting on it as it travels at a constant velocity of 6.4x10^2 k
IgorC [24]
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3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
3 years ago
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