All categories

3 years ago

Type your answer in the box.

1 answer:

4 0

An organ is a group of two or more tissue. They function to maintain homeostasis in the body (keep the person alive) all body systems work together to perform functions (switching between throat and windpipe etc)

You might be interested in

Two objects with a mass of 10,000-kg each and a distance of 1 meter between them.

Fofino [41]

Answer:

d) What is the force if we doubled both the masses AND we doubled the distance

4 0

3 years ago

An electric toy with a resistance of 2.50 Ω is operated by a 3.00-V battery. (a) What current does the toy draw? (b) Assuming th

Leno4ka [110]

Answer:

a) The current is i = 1.2 A

b) The charge is Q = 17280 C

c) The energy is E = 43200 J

Explanation:

a) The current is given by the ohm's law wich is:

i = V/R = 3/2.5 = 1.2 A

b) Since the charge is steady we can use the following equation to find the charge amount in that time:

i = Q/t

Q = t*i

Where t is in seconds, so we have 4h * 3600 = 14400 s

Q = 1.2*14400 = 17280 C

c) The energy is the power delivered to the toy multiplied by the time:

P = 1.2*2.5 = 3 W

E = P*t = 3*14400 = 43200 J

7 0

3 years ago

Please help meeeee!!!!!!

Vinil7 [7]

Answer:

Net force

Explanation:

Bruh, easy question

3 0

2 years ago

A box is being pulled to the right. The free body diagram is shown.What is the magnitude of the kinetic frictional force?

kirill115 [55]

Answer:

A - 0 N

plz mark 5 stars, thanks, and brainliest

4 0

3 years ago

A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway

Andru [333]

Answer:

63.44 rad/s

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet $v_{1}$ = 250 m/s

final velocity of bullet $v_{2}$ = 140 m/s

loss of kinetic energy of the bullet = $\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})$

==> $\frac{1}{2}*0.0033*(250^{2} - 140^{2} )$ = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = $\frac{1}{2} mv^{2}$

70.785 = $\frac{1}{2}*0.25*v^{2}$

566.28 = $v^{2}$

$v= \sqrt{566.28}$ = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 = 63.44 rad/s

5 0

3 years ago