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Katyanochek1 [597]
3 years ago
7

An elevator car weighs 5500 N. If the car accelerates upwards at a rate of 4.0 m/s2, what is the tension in the support cable li

fting the car? Use g = 10 m/s².
Physics
1 answer:
nadya68 [22]3 years ago
4 0

Answer:

Explanation:

The equation for this, since we are talking about weight on an elevator, is Newton's 2nd Law adjusted to fit our needs:

F_n=ma+w where the Normal Force needed to lift that elevator car is the tension. So the equation then becomes

T = ma + w where T is the tension in the cable to lift the elevator, m is the mass of the elevator (which we have to solve for), a is the acceleration of the elevator (positive since it's going up), and w is the weight of the elevator (which we have as 5500 N). Solving first for mass:

w = mg and

5500 =- m(10) so

m = 550 kg. Now we have what we need to solve for the tension:

T = 550(4.0) + 5500 and

T = 2200 + 5500 so

T = 7700 N

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Answer:

   f = 6.37 Hz,       T = 0.157 s

Explanation:

The expression you have is

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this is the equation of a traveling wave, the general form of the expression is

      y = A sin (kx - wt)

where A is the amplitude of the motion, k the wave vector and w the angular velocity

Angle velocity and frequency are related

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from the equation w = 40 rad / s

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3 years ago
A driver slows down her car from 32.7 km/hr at a constant rate of 0.63 m/s2 just by taking her foot of the accelerator (the gas
prohojiy [21]

Answer: She moves 5.616 meters in that second.

Explanation:

If we define t = 0s as the moment when she starts decelerating we can write the function of acceleration as:

a(t) = -(0.63 m/s^2)

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The velocity equation can be found if we integrate over time:

v(t) = -(0.63m/s^2)*t + v0

Where v0 is the constant of integration, that represents the initial velocity, in this case is:

v0 = 32.7 km/h

Now, because the acceleration is in m/s^2, we should write this velocity in m/s.

in one km we have 1000 meters, and in one hour we have 3600 seconds, then we have that:

32.7 km/h = 32.7 *(1000/3600) m/s = 9.08 m/s

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v(t) = -(0.63m/s^2)*t  + 9.08 m/s

And for the position equation, we should integrate again to get:

p(t) = -(1/2)*(0.63m/s^2)*t^2 + (9.08m/s)*t + p0

Where p0 is the initial position.

For this problem, we want to find the distance that she moved between t = 5s and t = 6s, and that can be calculated as:

D = p(6s) - p(5s)

D = -(1/2)*(0.63m/s^2)*(6s)^2 + (9.08m/s)*6s + p0 +(1/2)*(0.63m/s^2)*(5s)^2 - (9.08m/s)*(5s) - p0

D = -(1/2)*(0.63m/s^2)*((6s)^2 - (5s)^2) + (9.08m/s)*(6s - 5s)

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In this problem, the velocity of the particle is perpendicular to the magnetic field, so

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