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3 years ago

10

(will give brainliest to whoever is correct and shows reasoning) What is the acceleration of an object that has a velocity of 60

m/s and is moving in a circle of radius 50m?

2 answers:

hoa [83]3 years ago

5 0

Answer:

5.0/s

Explanation:

expeople1 [14]3 years ago

5 0

Answer:

b and a it is this that abewsr

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Someone please help me

professor190 [17]

The answer is A. conduction

7 0

3 years ago

A pendulum of unknown mass is attached to the ceiling and nearly touches the floor as it completes 6 full cycles in 45 s. At its

lys-0071 [83]

Answer:

a) h = 14 m

b) h = 88 cm

c) f = 0.054 Hz

d) f = 0.13 Hz

Explanation:

a) T = 2π√(L/g)

L = T²g/4π²

L = (45/6)²(9.8) / 4π² = 13.963...

b) ½mv² = mgh

h = v²/2g

h = 4.15²/ (2(9.8)) = 0.87869

c) f = 1/T = 1 / (2π√(14 / 1.62)) = 0.0542

d) f = 6/45 = 0.13333...

6 0

3 years ago

A particle moves according to a law of motions=f(t),t≥0 Wheretismeasured in seconds andsin feet.a) Find the velocity at timet.b)

Ray Of Light [21]

Answer:

a) $v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}$

b) 3s

c) t < 3s

d) 1.8ft

e) particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s

Explanation:

a)Suppose the equation for motion is:

$f(t) = \frac{9t}{t^2+9}$

Then the velocity is the derivative of the motion function

$v(t) = \frac{df(t)}{dt} = (9t(t^2+9)^{-1})^'$

From here we can apply product rule

$v(t) = (9t)^{'}(t^2+9)^{-1} + (9t)((t^2+9)^{-1})^'$

$v(t) = \frac{9}{t^2+9} - \frac{9t(2t)}{(t^2+9)^2}$

$v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}$

b) The particle is at rest when v(t) = 0:

$\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} = 0$

$\frac{9}{t^2+9} = \frac{18t^2}{(t^2+9)^2}$

Multiply 2 sides by $(t^2+9)^2$ we have:

$t^2 + 9 = 2t^2$

$t^2 = 9$

$t = 3s$

(c) The particle is moving in positive direction when v(t) > 0:

$\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} > 0$

$\frac{9}{t^2+9} > \frac{18t^2}{(t^2+9)^2}$

Multiply 2 sides by $(t^2+9)^2$ we have:

$t^2 + 9 > 2t^2$

$t^2 < 9$

$t < 3s$

(d) As particle is moving in positive direction when t < 3s and negative direction when t > 3s, we can calculate the distance it's moving up to 3s and then after 3s

$f(3) = \frac{9*3}{3^2+9} = \frac{27}{18} = 1.5 ft$

At 3s, particle is changing direction to negative, so its position at 6s is

$f(6) = \frac{9*6}{6^2+9} = \frac{54}{45} = 1.2 ft$

Therefore from 3s to 6s it would have moved a distance of 1.5 - 1.2 = 0.3 ft

Then the total distance it has moved in the first 6 s is 1.5 + 0.3 = 1.8 ft

e) Acceleration is the derivative of velocity function:

$a(t) = \frac{dv(t)}{dt} = (\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2})^'$

$a(t) = -\frac{9(2(t^2+9))(2t)}{(t^2+9)^2} - \frac{36t}{(t^2+9)^2} + \frac{18t^2(2(t^2+9))(2t)}{(t^2+9)^3}$

$a(t) = -\frac{36t}{t^2+9} - \frac{36t}{(t^2+9)^2} + \frac{72t^3}{(t^2+9)^2}$

$a(t) = -\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2}$

Particle is speeding up when a(t) > 0:

$-\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2} > 0$

$\frac{72t^3 - 36t}{(t^2+9)^2} > \frac{54t}{t^2+9}$

as $t \& (t^2 + 9)^2 \geq 0$ we can multiply/divide both sides by it:

$8t^2 - 4 > 6(t^2+9)$

$8t^2 > 6t^2 + 58$

$t^2 > 29$

$t > \sqrt(29) \approx 5.385 s$

so particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s

8 0

3 years ago

A —————— error is the perceived shift in a objective position as it is viewed from

Citrus2011 [14]

Answer:

Hi student

Your name ???

3 0

3 years ago

If the truck has a mass of 2,500 kilograms , what is its momentum ? (V=45m/s) Express your amswer

vlabodo [156]

Answer:

The answer is 2,500 * 45 = 112500kgm/s

6 0

2 years ago