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Klio2033 [76]
3 years ago
15

A displacement vector is 34.0 m in length and is directed 60.0° east of north. What are the components of this vector? Northward

Eastward choice component component 1 29.4 m 17.0 m 2 18.2 m 28.1 m 3 22.4 m 11.5 m 4 17.0 m 29.4 m 5 25.2 m 18.2 m A) Choice 1 B) Choice 2 C) Choice 3 D) Choice 4 E) Choice 5 79) If I walk 8.0 meters in a straight line and then walk 5.0 meters in another straight line, the total displacement cannot have a magnitude of A) 8 m. B) 2 m. C) 13 m. D) 5 m. E) 3 m.
Physics
1 answer:
PolarNik [594]3 years ago
8 0

Answer:

A)  29.4 m 17.0 m; B) 2 m

Explanation:

If a vector is 34.0 m in length and is directed 60.0° east of north (which means 30.0° over the horizontal), then its coordinates will be:

Horizontal: (34.0 m)cos(30.0°)=29.4 m

Vertical: (34.0 m)sin(30.0°)=17 m

If one person walks 8.0 meters in a straight line and then walks 5.0 meters in another straight line, then the minimum displacement would be to go back over his tracks, displacing himself 8m-5m=3m, while the maximum displacement would be going straight ahead, displacing himself 8m+5m=13m. Any answer outside this interval is impossible (2m).

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garik1379 [7]

Answer:

Explanation:

Work: This can be defined as the product of force and distance. The unit of work is Joules (J). it can be expressed mathematically as

W = F×d

or

W = \int\limits^b_a {Fx} \, dx.................................. Equation 1

Where b = upper limit, a = lower limit, Fx = expression of force.

<em>Given: a = 0 , b = 1.3 m, Fx = 4 + 15.7x - 1.5x²</em>

Substituting these values into equation 1

<em>W = \int\limits^a_b {(4 + 15.7x - 1.5x^{2} )dx} \,</em>

W = ᵇ[4x + 15.7x²/2-1.5x³/3 +C]ₐ

Work = upper limit - lower limit

Work = ᵃ[4x + 15.7x²/2 - 1.5x³/3 +C] - [4x + 15.7x²/2 + 1.5x³/3 +C]ᵇ............... Equation 2

Substituting the values of a and b into equation 2

Work = [4(1.3) + 15.7(1.3)²/2-1.5(1.3)³/3 + C] - [0 +C]

Work = [5.2 + 26.53 -3.29 + C] - C

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Work done by the force = 28.44 J.

8 0
3 years ago
A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity
Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

\omega_i = 2\pi f_1

\omega_1 = 2\pi(\frac{610}{60})

\omega_1 = 63.88 rad/s

similarly final angular speed is given as

\omega_f = 2\pi f_2

\omega_2 = 2\pi(\frac{837}{60})

\omega_2 = 87.65 rad/s

angular acceleration of the turbine is given as

\alpha = 5.9 rad/s^2

now we have to find the angular displacement is given as

\theta = \omega t + \frac{1}{2}\alpha t^2

\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)

\theta = 234.62 radian

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3 years ago
A mother and daughter are on a seesaw in the park. How far from the center must the 160.9lb mother sit in order to balance the 6
Soloha48 [4]

Answer:

The mother has to sit 2.17 ft from the center on the other side of the seesaw.

Explanation:

We are trying to find the sum of torques given by the weights of mother and daughter to be zero.

If the torque of the daughter on one side of the pivoting point is given by:

5.5 ft x 63.5 lb x g = 349.25 g ft lb

we need that the absolute value of the torque exerted by the mom (160.9 lb) to be the same in magnitude (and of course opposite direction). So we assume that "d" is the distance at which the mother locates to make this torque equal in magnitude to her daughter's torque:

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Explanation:

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