I need help on 7 &8

Calculate the work done to push a 200-N object 5-meters

AVprozaik [17]

1000 joules
jskskmxk dnd

6 0

3 years ago

Brainliest if right

mixas84 [53]

They traveling at -0.37/ms^

3 0

3 years ago

How we can best share the observation of the light through one hole the paper

zhenek [66]

Drawing is an easy way to connect to your inner child

5 0

3 years ago

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f

Hitman42 [59]

(a) The angular acceleration of the wheel is given by
$\alpha = \frac{\omega_f - \omega_i }{t}$
where $\omega_i$ and $\omega_f$ are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
$\alpha = \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2$

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
$\theta (t) = \omega_i t + \frac{1}{2} \alpha t^2$
where $\omega_i = 0$ is the initial angular speed. So, the angle covered after a time t=3.07 s is
$\theta= \frac{1}{2} \alpha t^2 = \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad$

6 0

3 years ago

A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple s

posledela

Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

$P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m$

The spring constant of the net is 20130.76 N

From Hooke's Law

$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m$

The net would strech 0.03167 m

If h = 35 m

From energy conservation

$65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0$

Solving the above equation we get

$x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45$

The compression of the net is 1.52 m

4 0

3 years ago