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2 years ago

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A train goes up a hill with a 15º incline. If the train has constant speed of 22 m/s, what are the vertical and horizontal compo

nents of the train's velocity? (Remember to use the law of sine and cosines for this problem)

1 answer:

Kobotan [32]2 years ago

8 0

Any two-dimensional vector in cartesian (x,y) coordinates can be broken down into individual horizontal and vertical components using trigonometry. If a train goes up a hill with 15 degree incline at a speed of 22 m/s, the horizontal component is 22cos(15)=21.3 m/s and the vertical component is 22sin(15)=5.5 m/s.

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Charra [1.4K]

To solve the problem it is necessary to apply the Torque equations and their respective definitions.

The Torque is defined as,

$\tau = I \alpha$

Where,

I=Inertial Moment

$\alpha =$ Angular acceleration

Also Torque with linear equation is defined as,

$\tau = F*d$

Where,

F = Force

d= distance

Our dates are given as,

R = 30 cm = 0.3m

m = 1.5 kg

F = 20 N

r = 4.0 cm = 0.04 m

t = 4.0s

Therefore matching two equation we have that,

$d*F = I\alpha$

For a wheel the moment inertia is defined as,

I= mR2, replacing we have

$d*F= \frac{mR^2a}{R}$

$d*F= mRa$

$a = \frac{rF}{ mR}$

$a = \frac{0.04*20}{1.5*0.3}$

$a=1.77 m/s^2$

Then the velocity of the wheel is

$V = a *t \\V=1.77*4 \\V=7.11 m/s$

Therefore the correct answer is D.

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2 years ago

A resultant vector is 8.00 units long and makes an angle of 43.0 degrees measured �� – �� with respect to the positi

Komok [63]

Answer:

223 degree

Explanation:

We are given that

Magnitude of resultant vector= 8 units

Resultant vector makes an angle with positive -x in counter clockwise direction

$\theta=43^{\circ}$

We have to find the magnitude and angle of the equilibrium vector.

We know that equilibrium vector is equal in magnitude and in opposite direction to the given vector.

Therefore, magnitude of equilibrium vector=8 units

x-component of a vector= $v_x=vcos\theta$

Where v=Magnitude of vector

Using the formula

x-component of resultant vector= $v_x=8cos43=5.85$

y-component of resultant vector= $v_y=vsin\theta=8sin43=5.46$

x-component of equilibrium vector= $v_x=-5.85$

y-component of equilibrium vector= $-v_y=-5.46$

Because equilibrium vector lies in III quadrant

$\theta=tan^{-1}(\frac{v_x}{v_y})=tan^{-1}(\frac{-5.46}{-5.85})=43^{\circ}$

The angle $\theta'$ lies in III quadrant

In III quadrant ,angle = $\theta'+180^{\circ}$

Angle of equilibrium vector measured from positive x in counter clock wise direction=180+43=223 degree

4 0

1 year ago

Determine the scalar components Ra and Rb of the force R along the nonrectangular axes a and b. Also determine the orthogonal pr

Liula [17]

Answer: Hello your question is incomplete attached below is the complete question

Ra = 1132 N

Rb = 522.6 N

Pa = 679.7 N

Explanation:

To determine the scalar components Ra

$\frac{Ra}{Sin 120^o} = \frac{750}{sin 35^o}$

therefore : Ra = $\frac{sin120^o * 750}{sin 35^o}$ = 1132 N

To determine the scalar component Rb

$\frac{Rb}{sin 25^o} = \frac{750}{sin 35^o}$

therefore : Rb = $\frac{sin 25^o * 750}{sin 35^o}$ = 522.6 N

To determine the orthogonal projection Pa of R onto

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Using Ohm's Law, we can derived from this the value of resistance. If I=V/R, therefore, R = V/I
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R = V/I = 2.5/ (750 x 10^-3)
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Calculating the power, we have P = IV

P = (750 x 10^-3)(2.5)
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The power consumption is the power consumed multiply by the number of hours. In here, we have;
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