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kobusy [5.1K]
3 years ago
9

A man runs at 7.5 m s-1 due east across the deck of a ship which moves at 10.0 m s-1 due north. Determine the resultant velocity

of the man( his velocity relative to the water)
Physics
1 answer:
german3 years ago
7 0

Answer:

12.5 m/s North/east

Explanation:

I thought of it as a graph. I basically just found the distance between the points (0,0) and (10, 7.5)!

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A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to
Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa

On the lower part of the hatch, there is a pressure equal to

p_{bot}=p_{atm}=1.013\cdot 10^5 Pa

So, the net pressure acting on the hatch is

p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa

which acts from above.

The area of the hatch is given by:

A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N

8 0
3 years ago
You and your little cousin sit on a see-saw. You sit 0.5 m from the fulcrum, and your cousin sits 1.5 m from the fulcrum. You we
svetoff [14.1K]

Answer:

200N

Explanation:

0.5/1.5=x/600N

1/3=x/600

x=200N

7 0
3 years ago
One negative change you may encounter as a student or an employee​
matrenka [14]

Answer:

you may get bullied or teased for being a differrent race, ethnic.

7 0
3 years ago
Read 2 more answers
Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?
eduard

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = \frac{1}{2}mv^{2}

Plug in the numbers = \frac{1}{2}(4.0)(8^{2})

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

7 0
3 years ago
Which situation can strengthen a community's stability?
nadya68 [22]
B. Interdependence among specie
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