The wind blows a leaf at 37.9 m/s for 118 s left. How are does the leaf go in that period of time.

A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the

const2013 [10]

Answer:

The equation for the object's displacement is $u(t)=0.583cos11.35t$

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

$k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in$

The angular speed is:

$w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s$

The damping force is:

$F_{D} =cu$

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

$c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in$

The critical damping is equal:

$c_{c} =\frac{2k}{w} =\frac{2*5.33}{11.35} =0.94lbs/in$

Like cc>c the system is undamped

The equilibrium expression is:

$u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t$

3 0

3 years ago

If there is a difference in air pressure between two locations, what happens?

tino4ka555 [31]

Answer:

High in the atmosphere, air pressure decreases. ... A low pressure system has lower pressure at its center than the areas around it. Winds blow towards the low pressure, and the air rises in the atmosphere where they meet. As the air rises, the water vapor within it condenses, forming clouds and often precipitation.

Explanation:

4 0

3 years ago

Read 2 more answers

At a given location from a source charge, the electric field______________.a. strength is dependent on the source charge and the

noname [10]

Answer:

b. The electric field points away from the source charge, if the source charge is positive

d. The electric field points toward the source charge, if the source charge is negative.

Explanation:

A positive source charge would create an electric field that would exert a repulsive effect upon a positive test charge. Thus, the electric field vector would always be directed away from positively charged objects. On the other hand, a positive test charge would be attracted to a negative source charge. Therefore, electric field vectors are always directed towards negatively charged object.

Also electric field strength depends only on test charge

The correct options include b and d

The electric field points away from the source charge, if the source charge is positive.

Also, the electric field points toward the source charge, if the source charge is negative.

3 0

3 years ago

You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought g

bezimeni [28]

Answer:

It's only 1.11 m/s2 weaker at 400 km above surface of Earth

Explanation:

Let Earth radius be 6371 km, or 6371000 m. At 400km above the Earth surface would be 6371 + 400 = 6771 km, or 6771000 m

We can use Newton's gravitational law to calculate difference in gravitational acceleration between point A (Earth surface) and point B (400km above Earth surface):

$g = G\frac{M}{r^2}$

where G is the gravitational constant, M is the mass of Earth and r is the distance form the center of Earth to the object

$\frac{g_B}{g_A} = \frac{GM/r^2_B}{GM/r^2_A}$

$\frac{g_B}{g_A} = \left(\frac{r_A}{r_B}\right)^2$

$\frac{g_B}{g_A} = \left(\frac{6371000}{6771000}\right)^2$

$\frac{g_B}{g_A} = 0.94^2 = 0.885$

$g_B = 0.885 g_A$

So the gravitational acceleration at 400km above surface is only 0.885 the gravitational energy at the surface, or 0.885*9.81 = 8.7 m/s2, a difference of (9.81 - 8.7) = 1.11 m/s2.

6 0

3 years ago

The accuracy of a series of measurements is best understood by looking at the

bonufazy [111]

The accuracy of a series of measurements is best understood by looking at the known value.

The values obtained in a series of measurement can be precise or accurate.

The precision of the measurement is obtained by looking at the trend of the values obtained.

If the values are close to each other, the measurement is precise but if the values are far from each other, the measurement is not precise.

The accuracy of a measurement is determine by comparing the determine value against a known value.

If the measured value is close to a <em>known value</em>, then the measurement is accurate, but if the measured value is from from a known value then the measurement is not accurate.

Thus, the accuracy of a series of measurements is best understood by looking at the known value.

Learn more here: brainly.com/question/13688896

4 0

2 years ago