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Vlad [161]
3 years ago
14

A horse trots 17 miles north and then turns and trots 24 miles at an angle of 40° west of north. Find the resultant displacement

and direction mathematically. Hint: This is a non-perpendicular vector problem.
Physics
1 answer:
DanielleElmas [232]3 years ago
8 0

Answer:

 Resultant displacement = 38.58 miles at an angle of 23.57° north of west.

Explanation:

Let north represent positive y direction and east represent positive x direction. Unit vector along x direction is represented by i and vector along unit vector along y direction is represented by j

Initial displacement = 17 miles north = 17 j miles

Final displacement = 24 miles at an angle of 40° west of north = 24 miles at an angle of 130° to positive x -axis

                               = (24 cos 130 i + 24 sin 130 j )

                               = (-15.43 i + 18.36 j ) miles

Total displacement = 17 j + (-15.43 i + 18.36 j) = -15.43 i + 35.36 j

Magnitude of displacement =\sqrt{(-15.43)^2+35.36^2} =38.58 miles

Angle to positive x -axis = tan⁻¹(35.36/-15.43) = 113.57°

Resultant displacement = 38.58 miles at an angle of 23.57° west of north.

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A picture of a baseball being thrown tworad a batter at home plate.

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3 years ago
A man in a gym is holding an 8.0-kg weight at arm's length, a distance of 0.55 m from his shoulder joint. What is the torque abo
kipiarov [429]

Answer:

  \tau =37.34\ N m

Explanation:

given,

mass of the weight = 8 Kg

distance = 0.55 m

angle below horizontal = 30°

torque about shoulder

  \tau = \vec{r} \times \vec{F}

  \tau = r \times F cos \theta

  \tau = 0.55 \times 8 \times 9.8 \times cos 30^0

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torque about his shoulder join is equal to   \tau =37.34\ N m

5 0
4 years ago
A football player kicks a ball horizontally off a hill with an initial velocity of 42.0 m/s. It travels a horizontal distance of
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3 years ago
In Example 2.7, we investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on sol
igor_vitrenko [27]

Answer:

a) 20 seconds

b) No.

Explanation:

t = Time taken for jet to stop

u = Initial velocity = 100 m/s (given in the question)

v = Final velocity = 0 (because the jet will stop at the end)

s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)

a = Acceleration = -5.00 m/s² (slowing down, so it is negative)

a) Equation of motion

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-100}{-5}\\\Rightarrow t=20\ s

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

s=ut+\frac{1}{2}at^2\\\Rightarrow s=100\times 20+\frac{1}{2}\times -5\times 20^2\\\Rightarrow s=1000\ m

The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.

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A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se
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Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

v = \frac{Bqr}{m}

v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}

v = 661098.9 = 661099 m/s

(b)

\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}

e / m = 1.76 x 10^14 C / kg

(c) Let K be the kinetic energy

K = 0.5 x mv²

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K = 1.99 x 10^-19 J

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So, the potential difference is

V = 1.24 V

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V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

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Let v is the speed

K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

r=\frac{mv}{Bq}

r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}

r = 0.0113 m = 1.13 cm

7 0
3 years ago
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