Hooke's Law Practice

You are in your car driving on a highway at 23 m/s when you glance in the passenger-side mirror (a convex mirror with radius of

Irina-Kira [14]

Answer:

v = 26. 88 m/s +23 m/s

Explanation:

u = 23 m/s, r = 150 cm, u₁ = 2.0 m/s, s =2.0 m

$\frac{1}{s} +\frac{1}{s'} = \frac{2}{R}$

$\frac{1}{2.0 m} +\frac{1}{s'} = \frac{2}{1.50 m}$

Solve s'

$\frac{1}{s'} = \frac{2}{1.50 m} - \frac{1}{2.0 m}$

$\frac{1}{s'} = 1.833 m$

$s' = - 0.545 m$

To determine the speed of the trick to the highway

$\frac{ds}{dt}= \frac{s^2* \frac{ds}{dt}}{s' ^2} =\frac{2.0 ^2m * 2.0 m/s}{0.545^2m}$

$\frac{ds}{dt} = 26.88 m/s$

Now to determine the velocity highway is going to be

v = ds/dt + u

v = 26. 88 m/s +23 m/s

8 0

3 years ago

Which statement best describes the relationship between mass and gravitational attraction (pull)?

Gnesinka [82]

Answer:

The first option, "The more mass an object has, the greater the gravitational pull. "

Explanation:

Newton's Law of Gravity states that $F_G=\frac{Gm_1m_2}{r^{2}}$ , where G is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and <em>r</em> is the distance between the objects' centers. Because the objects' masses are in the fraction's numerator, they are directly proportional to $F_G$ , and increasing the mass of one or both objects increases the gravitational pull.

Please have a great day! I hope this helps you understand the question!

7 0

3 years ago

An 89.0 kg fullback moving east with a speed of 5.6 m/s is tackled by an 85.0 kg opponent running west at 2.84 m/s, and the coll

Korolek [52]

Answer:

a. $v_f=1.477m/s$

b. Δ $K=1558.3J$

c. $E_k=1034.7 J$

Explanation:

a).

Momentum conserved

$p_{ix}=p_{fx}$

$m_1*v_1+m_2*v_2=(m_1+m_2)*v_f$

$v_f=\frac{m_1*v_1+m_2*v_2}{m_1+m_2}$

$v_f=\frac{89.0kg*5.6m/s+85.0kg*-2.84m/s}{(89.0+85.0)kg}$

$v_f=1.477m/s$

b).

Δ $K=K_i-K_f$

$\frac{1}{2}*m_1*v_1^2+\frac{1}{2}*m_2*v_2^2=\frac{1}{2}*(m_1+m_2)*v_f^2$

$\frac{1}{2}*89.0kg*(5.6m/s)^2+\frac{1}{2}*85.0kg*(2.84m/s)^2=\frac{1}{2}*(89.0+85.0)kg*(1.447m/s)^2$

Δ $K=1558.3J$

c).

$E_k=\frac{1}{2}*89kg*(5.8m/s)^2-\frac{1}{2}*(85+89)kg*(1.44m/s)^2$

$E_k=1034.7 J$

d).

All of which has been lost as mechanical energy, and is now thermal energy warmer football players, noise a loud crunch for example.

8 0

3 years ago

How do professional bicycle riders reduce friction. Name two ways.<br> Thx

wlad13 [49]

A well maintained bicycle is key. Oil the chain so that it's less likely to get hung up. It will have a smoother flow. Another could be in their clothing. Notice they usually wear good fitting clothing? I believe the more from fitting pants help clothing friction. Another way could be keeping the tires well inflated so that the tires aren't dragging. Rolling along smoothly on properly inflated tires seems like a must. Get a few more answers along with mine, so that you have a variety to choose from.

4 0

3 years ago

Why is a shoe considered individual evidence

Advocard [28]

Footprints and tireprints

As shoes and tires are used, individual characteristics such as nicks, cuts, and wear patterns develop. These characteristics may show up in prints and impressions and can be compared with a suspect's shoes or tires.

7 0

3 years ago