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Alex777 [14]
4 years ago
15

In mammals the weight of the heart is approximately 0.5%. Write a linear model that gives the heart weight in terms of whole bod

y weight?
Physics
1 answer:
sesenic [268]4 years ago
8 0

Answer: h=0.05w

Let the whole body weight of a whale be w

Let the weight of the heart be h

It is given that weight of the heart is 5% weight of the whole body weight.

\Rightarrow h=5 \% \hspace{1mm} of \hspace{1mm} w\\ \Rightarrow h=\frac{5}{100}\times w\\ \Rightarrow h=0.05w

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Write one example situation of Newton's Third Law involving mass that are the same.
puteri [66]

Answer:

Examples of Newton's third law of motion are ubiquitous in everyday life. For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air. Engineers apply Newton's third law when designing rockets and other projectile devices.

6 0
3 years ago
Physicist ____ was found to have completely made up some of the data on which his astounding discoveries were made.
GaryK [48]

Answer:

Jan Hendrik Schön I believe

6 0
3 years ago
Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea
Setler [38]

Answer:

a)  M = 4,997 10²⁰ kg ,  b)   T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

          v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

         a = (v₀ - v) / t

         a = (15 - (-15)) /9.00 = 30/9

         a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

          F = m a

         G m M / r² = m a

         M = a r² / G

Let's calculate

         M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

         M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

         R = R_{m} + h

         R = 1 10⁵ + 2.00 10⁴

         R = 12 10⁴ m

         F = m a

        G m M / R² = m a

Centripetal acceleration is

         a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

        d = 2π R

        v = d / t

        v = 2π R / T

Let's replace

        G m M / R² = m (2π R / T)² / R

        G M = R³ 4π² / T²

        T² = 4π² R³ / G M

       T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

       T² = 6.82 10¹⁶ / 3.33 10¹⁰

       T = √ (2,048 10⁶)

       T = 1.43 10³ s

3 0
3 years ago
A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric f
zalisa [80]

Explanation:

Charge=5.00 E-5

Force=0.751N

F=qE

0.751=5.00 E-5*E

E=1.502*10⁴

option a

8 0
3 years ago
Read 2 more answers
The first mechanized industry was
Oxana [17]
C. Textiles

It was the first thing mechanized in the Industrial Revolution

3 0
3 years ago
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