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3 years ago

14

A bolt falls off an airplane high above the ground. How far does the bolt have to fall before its speed reaches 100m/s (about 20

0 miles per hour)? You should ignore air friction in your calculation.

1 answer:

omeli [17]3 years ago

3 0

We can use the equation vf (the final velocity) =vi (the initial velocity) +at (aceleration times time)

We know the final velocity 100m/s, the initial velocity 0, and the acceleration (gravity) 9.8m/s^2. So, 100=0+9.8t. t=100/9.8

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A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

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3 years ago

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$

The box is initially at rest, so $v_i=0$ . Solving for $v_f$ :

$v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}$

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2 years ago

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s 1045 rad/s ). If a particular disk is

Scilla [17]

Answer:

The magnitude of the average angular acceleration of the disk is $4139.74\ rad/s^2$ .

Explanation:

Given that,

Angular velocity, $\omega_i=968.7\ rad/s$

The disk comes to rest, $\omega_i=0$

Time, t = 0.234 s

We need to find the magnitude of the average angular acceleration of the disk. It is given by change in angular velocity per unit time. So,

$\alpha =\dfrac{\omega f}{t}\\\\\alpha =\dfrac{968.7\ rad/s}{0.234\ s}\\\\\alpha =4139.74\ rad/s^2$

So, the magnitude of the average angular acceleration of the disk is $4139.74\ rad/s^2$ .

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2 years ago