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2 years ago

(GIVING BRAINLIEST) do what the following image says.

Physics

2 answers:

Delicious77 [7]2 years ago

8 0

Answer:

that pink part that circles around is the pancreas

Nostrana [21]2 years ago

8 0

Answer:

Explanation:

red colour organ is liver

organ attached to oesophagues or brown type coloured organ is stomach

light yellow colur organ is pancreas.

organ which is like boarder at the bottom is large intestine.

inside large intestine is small intestine

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A 3 kg bowling ball is thrown onto a mattress. If it takes 0.3 seconds to stop the ball using a force of 24 N, what was the init

enyata [817]

Answer:

-24 m/s

Explanation:

mass of the bowling ball = 3 kg

time (t) = 0.3 seconds

Force = 24 N

initial velocity u = ???

We know that;

Force = mass × acceleration (a)

So;

24 = 3 × a

a = 24/3

a = 8 m/s²

Also;

From equation of motion; acceleration is given by the relation;

$a =\frac{v-u}{t}$

if v = 0

then ;

$8 = \frac{0-u}{0.3}$

24 = 0- u

u = -24 m/s

Thus; the initial velocity of the bowling ball when it first touched the mattress = -24 m/s

7 0

3 years ago

A 61.0-kg person jumps from rest off a 10.0-m-high tower straight down into the water. Neglect air resistance. She comes to rest

9966 [12]

Answer:

Explanation:

In this case, law of conservation of energy will be implemented. It states that "the energy of the system remains conserved until or unless some external force act on it. Energy of the system may went through the conversion process like kinetic energy into potential and potential into kinetic energy.But their total always remain the same in conserved systems."

Given data:

Height of tower = 10.0 m

Depth of the pool = 3.00 cm

Mass of person = 61.0 kg

Solution:

Initial energy = Final energy

$U_{i} = (K.E) + U_{f}$

As the person was at height initially so it has the potential energy only.

$mg(h_{1} +h_{2}) = K.E + mgh_{2}$

$K.E = mgh_{1}$

$K.E = (61.0)(9.8)(10)\\K.E = 5978 J$

Lets find out the magnitude of the force that the water is exerting on the diver.

W =ΔK.E

$F.h_{2} = 5978\\$

$F = \frac{5978}{3}$

F = 1992.67 N

7 0

2 years ago

A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.

Viefleur [7K]

Answer:

Explanation:

a )

change in the gravitational potential energy of the bear-Earth system during the slide = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J - F x 11 = 756.9 J

F x 11 = 4851 J - 756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N

4 0

2 years ago

A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?

dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

$I = MR^2$