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3 years ago

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An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. how high does it rise (v = 0 cm

/s at top of rise). remember, acceleration is -9.80 m/s2.

2 answers:

maw [93]3 years ago

7 0

Answer: The maximum height is 20.4 meters

Explanation: We have that the initial velocity is 20m/s and the initial position is 0m.

The acceleration of the object is:

a(t) = -9.8m/s^2

for the velocity, we integrte over time.

v(t) = -9.8m/s^2*t + 20m/s

and the position is:

p(t) = -4.9m/s^2*t^2 + 20m/s*t + 0

And we want to do the maximum height (in this case the maximum value of p(t))

First, we need to find the time in wich te velocity is equal to zero, which means that the object stops to going up and starts moving down.

v(t) = 0 = -9.8m/s^2*t + 20m/s

t = (20/9.8)s = 2.04s

Now, we put this time in the position equation:

p(2.04s) = -4.9m/s^2*(2.04s)^2 + 20m/s*2.04s = 20.4m

Aliun [14]3 years ago

5 0

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A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 m/s2 for t1 = 20 s. At that point the

arsen [322]

Total displacement of the car: 405 m

Explanation:

The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

$s_1=ut_1+\frac{1}{2}a_1 t_1^2$

where:

u = 0 is the initial velocity (the car starts from rest)

$t_1 = 20 s$ is the time elapsed in the 1st part

$a_1=1.62 m/s^2$ is the acceleration of the car in the 1st part

$s_1$ is the displacement of the car in the 1st part

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We can also find the velocity of the car after these 20 seconds using the equation:

$v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s$

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

$s_2 = (\frac{v_1 + v_2}{2})t_2$

where:

$v_1=32.4 m/s$ is the initial velocity at the beginning of the 2nd phase

$v_2=0$ is the final velocity (the car comes to a stop)

$t_2=5 s$ is the time elapsed in the 2nd phase

Substituting,

$s_2=\frac{32.4+0}{2}(5)=81 m$

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3 years ago

What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your an

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Complete Question

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the wheel starts turning, you are at the very top.

What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘. Express your answer in degrees.

Answer:

$\phi=123.75$

Explanation:

From the question we are told that:

Height $h=27m$

Period $T=32sec$

Time $t=75sec$

Generally the equation for angular velocity is mathematically given by

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$\omega=\frac{2 \pi}{32}$

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Therefore

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Therefore

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natulia [17]

Answer:

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Here velocity is constant so acceleration is zero.

It means the net force acting on the vehicle is zero.

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marta [7]

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