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dem82
9 days ago
13

(a) A submarine descends to a depth of 70 m below the surface of water. The density of the water is 1050 kg/m3. Atmospheric pres

sure is 1.0 × 10^5 Pa (1.0 × 10⁵ Pa). Calculate
Physics
1 answer:
ValentinkaMS [3.4K]9 days ago
5 0

Answer:

sttouyietETwe2e664yrwtwwteuwtrwruwuuwwuwtwuw7w7w5w7w772253536464647

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A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w
Keith_Richards [3271]
25.82 m/s Explanation: Given: Force applied by the baseball player; F = 100 N Distance the ball travels; d = 0.5 m Mass of the ball; m = 0.15 kg To find the velocity at which the ball is released, we will equate the work done with the kinetic energy involved. It's important to recognize that work done reflects the energy the baseball player has used. Thus, the relationship can be represented as follows: F × d = ½mv² 100 × 0.5 = ½ × 0.15 × v² Solving gives: v² = (2 × 100 × 0.5) / 0.15 v² = 666.67 v = √666.67 v = 25.82 m/s.
4 0
27 days ago
Radiation emitted from human skin reaches its peak at λ = 940 µm. (a) What is the frequency of this radiation? (b) What type of
inna [3103]

Answer:

a) The frequency is 3.191x10^11.

b) The type of radiation is microwave radiation.

c) The energy of one quantum of this radiation amounts to 1.32x10^-3ev.

Explanation:

For further details and calculations, please refer to the image below.

8 0
1 month 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.

8 0
1 month ago
A person who climbs up something (e.g., a hill, a ladder, the stairs) from the ground gains potential energy. a person's weight
kicyunya [3294]

The following values have been provided:

weight w = 240 lb = 1,067.52 N

energy E = 3,000 J

 

The equation for potential energy is:

E = w h

where h indicates the height that the person needs to ascend, therefore:

h = 3000 / 1067.52

h = 2.81 m

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<span>Thus, he must ascend 2.81 meters</span>

3 0
27 days ago
Mosses don't spread by dispersing seeds; they disperse tiny spores. The spores are so small that they will stay aloft and move w
Keith_Richards [3271]

Solution:

Em_{f} / Em₀ = 0.30

Explanation:

In this problem, we apply the connection between mechanical energy, kinetic energy, and gravitational potential energy.

      K = ½ m v²

      U = mgh

We assess the mechanical energy at two positions:

Initial. Lower

    Em₀ = K = ½ m v²

At its highest point

    Em_{f} = U = mg and

Now let's compute

    Em₀ = ½ m 3.6²

    Em₀ = m 6.48

    Em_{f} = m 9.8 × 0.2

    Em_{f} = m 1.96

Thus the energy lost is given by:

    Em_{f} / Em₀ = m 1.96 / m 6.48

   Em_{f} / Em₀ = 0.30

This means that 30% of the sun's energy is transformed into potential energy.

There are various conversion possibilities.

This energy changes into thermal energy affecting the spores and air, since it cannot be regained.

8 0
1 month ago
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