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3 years ago

5

A mysterious rocket-propelled object of mass 49.5 kg is initially at rest in the middle of the horizontal, frictionless surface

of an ice-covered lake. Then a force directed east and with magnitude F(t)?

1 answer:

Romashka-Z-Leto [24]3 years ago

8 0

The distance covered by the rocket in the first 5 seconds is 7.13 m

Explanation:

In order to solve the problem, we have to find the acceleration of the rocket. We can do it by using Newton's second law, which states that the net force on the rocket is equal to the product between its mass (m) and its acceleration (a):

$F(t)=ma(t)$

Where here we have:

$F(t)=16.8t$ is the time-dependent force

m = 49.5 kg is the mass

Solving for a(t),

$a(t)=\frac{F(t)}{m}=\frac{16.8t}{49.5}=0.34t[m/s^2]$

This is the time-dependent acceleration of the rocket.

By integrating this expression, we find the velocity of the rocket at time t:

$v(t)=\int a(t)dt=\int (0.34t)dt=\frac{0.34t^2}{2}=0.17t^2$

where we didn't add the constant term since the velocity at t = 0 is zero (the rocket starts from rest).

By integrating this expression, we then find an expression for the position of the rocket at time t:

$x(t)=\int v(t) dt = \int (0.17t^2)dt=\frac{0.17t^3}{3}=0.057t^3$

Therefore, the distance covered by the rocket in the first t = 5.00 s is:

$x(5)=0.057(5.0)^3=7.13 m$

Learn more aboiut force and acceleration:

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3 years ago

A swimming pool has the shape of a right circular cylinder with radius 21 feet and height 10 feet. Suppose that the pool is full

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Answer:

The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

Explanation:

Given that,

Radius = 21 feet

Height = 10 feet

Weighing = 62.5 pounds/cubic

Work = 4329507.37572

Height = 2 feet

Let's look at a horizontal slice of water at a height of h from bottom of pool

We need to calculate the area of slice

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times21^2$

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Thickness of slice $t=\Delta h\ ft$

The volume is,

$V=(441\pi\times\Delta h)\ ft^3$

We need to calculate the force

Using formula of force

$F=W\times V$

Where, W = water weight

V = volume

Put the value into the formula

$F=62.5\times(441\pi\times\Delta h)$

$F=27562.5\pi\times\Delta h\ lbs$

We need to calculate the work done

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs$

We do this by integrating from h = 0 to h = 10

We need to find the total work,

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh$

$W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}$

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$W=4329507.37572$

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh$

$W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}$

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$W=1929375\pi$

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kipiarov [429]

Answer:

Explanation:

Given that,

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Let find the transverse component of the final speed using

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Using the conservation of angular momentum,

(HA)1 = (HA)2

Mb • Vb • r1 = Mb • Vθ • r2

Mb cancels out

Vb • r1 = Vθ • r2

2 × 1 = 3.97 × r2

r2 = 2/3.97

r2 = 0.504m

The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m

The required time,

Using equation of motion

V = ∆r/t

Then,

t = ∆r/Vr

t = (r1—r2) / Vr

t = (1—0.504) / 0.5

t = 0.496/0.5

t = 0.992 second

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3 years ago