Physics: Aprojectile is fired straight up from ground level. after t seconds it s height is above the ground is h feet where h=-16t^2+144t. for what time period is the projectile at least 288 ft off the ground?

Aprojectile is fired straight up from ground level.

Benthega

01.01.2022

3 sec or 6 sec.

Explanation:

Given that, from the ground level a projectile is fired.

The equation which is given after t sec is,

$h=-16t^{2} +144t$

The height of the projectile is given after t seconds is, h=288 ft

Now put this value in height equation will get.

$288=-16t^{2} +144t\\t^{2}-9t+18=0\\t^{2}-(6+3)t+18=0\\t^{2}-6t-3t+18=0\\t(t-6)-3(t-6)=0$

Therefore the value of t will be 3 sec or 6 sec when projectile is at least 288 ft off from the ground.

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kendratorres200

13.01.2021

The average binding energy per nucleon of $^{24} _{12} Mg$ is 8.00 MeV/nucleon

Explanation:

Here we have the molar mass of $^{24} _{12} Mg$ is 23.985042 amu

The mass of a proton m_p = 1.007276 amu

The mass of a neutron m_n = 1.008665 amu

From the notation, 1 Mg = has 12 protons and 12 neutrons

Therefore, we find the binding energy as follows;

$E_{bind}$ = 12 × (1.007276) amu + 12 × (1.008665) amu - 23.985042 amu

$E_{bind}$ = 24.191292 amu - 23.985042 amu = 0.20625 amu

1 amu = 931 MeV

Therefore, 0.20625 amu = 0.20625 amu × 931 MeV = 192.01875 MeV

The average binding energy per nucleon in $^{24} _{12} Mg$ which has 24 nucleons, is thus;

$E_{bind (ave)}$ = 192.01875 MeV ÷ 24 = 8.00078 MeV/nucleon ≈ 8.00 MeV/nucleon.

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