Your pencil is 11 cm long . How long is it in millimeters?

In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

Yuliya22 [3333]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as $w_{h}$ .

We aim to determine the heart's weight for a human

Using the provided information

$w_{h}=0.5\times w$

Where, h = heart weight

w = human weight

$w_{h}=\dfrac{0.5}{100}\times 185$

$w_{h}=0.93\ lbs$

The final weight of a human heart is 0.93 lbs.

8 0

4 months ago

A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio

Ostrovityanka [3204]

Answer:

Explanation:

According to the parameters provided,

mass of the clay lump, m₁ = 0.05 kg

initial velocity of the lump, u₁ = 12 m/s

mass of the cart, m₂ = 0.15 kg

initial speed of the cart, u₂ = 0

As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}$

The resultant speed is v = 3 m/s.

Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.

3 0

3 months ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Yuliya22 [3333]

Details that are not provided: the problem figure is included.

We can address the exercise by applying Poiseuille's law. This law indicates that for a fluid flowing in a laminar manner within a confined pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ represents the pressure difference across the two ends

$\mu$ denotes the viscosity of the fluid

L signifies the length of the pipe

$Q=Av$ indicates the volumetric flow rate, where $A=\pi r^2$ is the cross-sectional area of the tube and $v$ refers to the fluid's velocity

r stands for the pipe's radius.

This law can be utilized for the needle, allowing us to compute the pressure difference between point P and the needle's end. In this scenario, we have:

$\mu=0.001 Pa/s$ is the dynamic viscosity of water at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Substituting these values into the formula yields:

$\Delta P = 3200 Pa$

This pressure difference specifies the value between point P and the needle's termination. As the end of the needle is under atmospheric pressure, the gauge pressure at point P, relative to atmospheric pressure, is exactly 3200 Pa.