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2 years ago

Earth's magnetic compass and its effect on birds?

1 answer:

5 0

Changes in the dip of the earth's magnetic field correspond to changes in latitude. Displacing migrating birds to areas of smaller dip angles (= lower magnetic latitudes) results in predeterminable shifts in the birds migratory direction .

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Suppose that a spring with a larger spring constant was used in this same apparatus. If a given mass were rotated at the same ra

malfutka [58]

Answer:

The new period of rotation using the new spring would be less than the period of rotation using the original spring

Explanation:

Generally the period of rotation of the mass is mathematically represented as

$T = 2 \pi \sqrt{\frac{I}{k} }$

Here I is the moment of inertia of the mass about the rotation axis and k is the spring constant

Now looking at the equation we can tell that T is inversely proportional to the square root of the spring constant which means that for a larger spring constant the time period would be lesser

6 0

2 years ago

The whooping crane (Grus americana) is the tallest bird native to North America. It almost went extinct in the 1900s; in 1938, t

grin007 [14]

Answer:

The value is $a = 2.7183$

Explanation:

From the question we are told that

The relationship between the number of whooping cranes and the number of decades is $ln N = 2.85 + 0.039t$

The exponential relationship is $N = na^{kt}$

Now from the given equation we have that

$N = e^{2.85 + 0.039t}$

$N = 17.29 e^{0.039t}$

So comparing this equation obtained an the given exponential relationship we have that

$n = 17.29$

$k = 0.039$

$a = e = 2.7183$

$a = 2.7183$

7 0

2 years ago

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

tiny-mole [99]

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

${v}^{2} = {z}^{2} + 2ax$

$v = z + at$

First let's find the final velocity the plane will have at the end of the runway using the first equation:

${v}^{2} = {0}^{2} + 2(5)(1800)$

$v = 60 \sqrt{5}$

Now we can plug this into the second equation to find t:

$60 \sqrt{5} = 0 + 5t$

$t = 12 \sqrt{5}$

Then using 3 significant figures we round to 26.8 seconds

3 0

2 years ago

You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend

Bogdan [553]

Answer:

a. $\tau=200J$ b. $\alpha=0.44 \frac{rad}{s^2}$ c. $\alpha=0.33\frac{rad}{s^2}$ d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by $\tau=rF\sin \theta$ , being r the radius, F the force aplied and $\theta$ the angle between the vector force and the vector radius. Since $\theta=90^{\circ}, \, \sin\theta=1$ and so $\tau=rF=2m100N=200Nm=200J$ .

b. Since the relation $\tau=I\alpha$ hols, being I the moment of inertia, the angular acceleration can be calculated by $\alpha=\frac{\tau}{I}$ . Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is $I=\frac{1}{2}Mr^2$ , being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by $I_p=m_pr_p^{2}$ , being $m_p$ the mass of the person and $r_p^{2}$ the distance from the person to the center. Given all of this, we have

$\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}$ .

c. Similar equation to b, but changing $r_p=2m$ , so

$alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}$ .

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0

3 years ago

When a postitive charge is placed in a electric field the field will

Dafna11 [192]

the electric field direction about a positive source charge is always directed away from the positive source. And the electric field direction about a negative source charge is always directed toward the negative source.

5 0

3 years ago