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3 years ago

How can people reach their full potential?

Physics

1 answer:

PolarNik [594]3 years ago

6 0

If they have self motivation or others motivation, they will show their full potential.

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Can sumbody help me wit dis

stepladder [879]

C, exothermic reaction. These types of reaction releases heat so that you can heat up your ready-to-eat meals.

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2 years ago

What is the acceleration of the object?

mars1129 [50]

Resultant force is 400-600=-200N.
-200=60a solve for a.

4 0

3 years ago

What is the relation between centre of gravity and stability

maw [93]

Explanation:

tilting it will raise the height of its center of gravity.

8 0

2 years ago

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with

Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = $\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2$

= $\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2$

= 31 J

Final KE = $\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J$

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0

3 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

3 years ago