All categories

3 years ago

The master gland of the endocrine system is the: (2 points)

Physics

2 answers:

galben [10]3 years ago

8 0

Answer: Pituitary Gland

Explanation: The pituitary gland produces hormones that control other glands and body functions, therefore, it is sometimes called the master endocrine gland.

Ahat [919]3 years ago

4 0

The answer is pituitary hopes this helps

You might be interested in

Suppose two carts, one twice as massive as the other, fly apart when the

disa [49]

Answer:

the question this morning

Explanation:

90+70 look than

8 0

3 years ago

The Old Faithful geyser found in Yellowstone Park is an example of _____.

katen-ka-za [31]

Answer:

cone geyser

Explanation:

The Old Faithful geyser is the oldest discovered geyser in the Yellowstone national park. The eruptions of the geyser are particularly predictable. It is a cone type geyser.

Cone geysers generally have a spout through which the water ejects out. When super heated water in the tube then the water starts to boil and form bubbles of steam, after this process the eruption takes place.

5 0

3 years ago

A copper wire has a square cross section 2.1 mm on a side. The wire is 4.3 m long and carries a current of 3.6 A. The density of

aleksley [76]

Explanation:

Below is an attachment containing the solution.

4 0

3 years ago

A charge of -8.5 µC is traveling at a speed of 9.0 106 m/s in a region of space where there is a magnetic field. The angle betwe

Sindrei [870]

Answer:

The magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ .

Explanation:

Given that,

Charge, $q=-8.5\ \mu C=-8.5\times 10^{-6}\ C$

Speed of the charged particle, $v=9\times 10^6\ m/s$

The angle between the velocity of the charge and the field is 56°.

The magnitude of force, $F=5.9\times 10^{-3}\ N$

We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

$F=qvB\ \sin\theta$

B is the magnetic field.

$B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T$

So, the magnitude of the magnetic field is $9.3\times 10^{-5}\ T$ . Hence, this is the required solution.

4 0

3 years ago

7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ba

Dima020 [189]

$\\ \tt \longmapsto \: h = - 4.8 {t}^{2} + 8.2t + 1.8 \\ \\ \tt \longmapsto \: h = - 4.8(1.7) {}^{2} + 8.2 \times 1.7 + 1.8 \\ \\ \tt \longmapsto \: - 4.8 \times 2.89 + 1.39 + 1.8 \\ \\ \tt \longmapsto \: 13.8 + 1.39 + 1.8 \\ \\ \tt \longmapsto \: 17.06$

6 0

3 years ago