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2 years ago

A 325 N force applied to an object for 5.5 s. What’s the impulse?

1 answer:

4 0

According to Newton second law of motion, the resultant force is directly proportional to the rate of change in momentum while maintaining other factors constant. Therefore, F = (mv-mu)/t where F is the resultant force , m is the mass of the object, v is the final velocity and u is the initial velocity.
Hence, Ft = mv-mu, but impulse is given by force multiplied by time, thus, impulse is equivalent to the change in momentum.
Impulse = Ft
= 325 × 2.2 sec
= 715 Ns

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(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

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where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

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where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

$\rho_a = 1.29 kg/m^3$ is the air density

$V=329 m^3$ is the volume of the balloon

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The mass of the helium gas inside the balloon is

$m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg$

where $\rho_h$ is the helium density; so we the total mass of the balloon+helium gas inside is

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(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

$W'=(m'+m)g=B$

where m' is the additional mass. Re-arranging the equation for m', we find

$m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg$

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