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Ostrovityanka [42]
2 years ago
6

A 4.0 kg mass has a velocity of 10 m/s to the EAST. The 4.0 kg mass is subjected to a constant net force of 16 N to the WEST for

3.0 sec. What is the velocity of the 4.0 kg mass at the end of the 3.0 sec interval?​
Physics
1 answer:
julia-pushkina [17]2 years ago
5 0

Answer:

v = 2.0 m/s West

Explanation:

Let East be the positive direction

The initial momentum of the mass is p = mv = 4.0(10) = 40 kg•m/s East

This is acted on by an impulse of p = Ft = 16(3.0) = 48 N•s west

An impulse will result in a change of momentum

so final momentum of the mass is

40 - 48 = -8 kg•m/s or 8 kg•m/s West

-8 = 4.0v

v = -2.0 m/s or 2.0 m/s West

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Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the
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Answer:

distance of 2nd team from 1st team will be:  58.2

Direction of 2nd team from 1st team will be:  14.90 deg North of east

Explanation:

ASSUME Vector is R and  makes angle A with +x-axis,

therefore component of vector R is

R_x = Rcos A

R_y = Rsin A

From above relation

Assuming base camp as the origin, location of 1st team is

R_1 = 37 km away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km

R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km

location of 2nd team is at

R_2 = 32 km, at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km

R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km

Now position of 2nd team with respect to 1st team will be given by:

R_3 = R_2 - R_1

R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j

Using above values:

R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j

R_3 = 51.49 i + 13.71 j

distance of 2nd team from 1st team will be:

\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)

\left | R_3 \right | = 53.28 km = 58.2 km

Direction of 2nd team from 1st team will be:

Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]

Direction = 14.90 deg North of east

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