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Anna71 [15]
3 years ago
7

Where is the best place to put pfds while you are out on your boat?

Physics
1 answer:
Alexeev081 [22]3 years ago
3 0
The best place to put PFDs while you are out on your boat is at the top deck of your boat. PFDs must be placed at an area where it is easily accessible, especially in times of accidents, without any unnecessary gear or objects covering the PFDs, but it is best to wear a life jacket during the whole duration of the boat trip.
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What is the uncertainty of the position of the bacterium? express your answer with the appropriate units?
lbvjy [14]
For two un-related quantities, the Heisenberg uncertainty equations holds: the prduct of the two uncertainty quantities is greater than \hbar/2
Example of unrelated quantities are position and momentum, energy and time. 
Thus
\Delta x*\Delta p  \ \textgreater \ \hbar/2
Knowing the speed of the bacteria the uncertainty in its position is
\Delta x \ \textgreater \ \hbar/(2 \Delta p) =\hbar/(2mv)
4 0
3 years ago
This diagram shows the process that powers stars. This process is called?
viktelen [127]
The process that powers stars is C)fusion
4 0
3 years ago
Read 2 more answers
I really need help can anyone help
sergiy2304 [10]

Answer:

percentage w a slug preference (start) - 35% (0.35)

percentage w a fish preference (start) - 65% (0.65)

percentage w a slug preference (end) - 25% (0.25)

percentage w a fish preference (end) - 75% (0.75)

hope this helps kind stranger

5 0
2 years ago
3. Una cuerda de guitarra tiene 60 cm de longitud y una masa de 0.05 kg de masa. Si se tensiona mediante una fuerza de 20 N. La
jok3333 [9.3K]

Answer:

f1 = 12.90 Hz

Explanation:

To calculate the first harmonic frequency you use the following formula for n = 1:

f_n=\frac{n}{2L}\sqrt{\frac{T}{M/L}}

f_1=\frac{1}{2L}\sqrt{\frac{T}{M/L}}    ( 1 )

It is necessary that the unist are in meters, then you have:

L: length of the string = 60cm = 0.6m

M: mass of the string = 0.05kg

T: tension on the string = 20 N

you replace the values of L, M and T in the expression (1) for getting f1:

f_1=\frac{1}{2(0.6m)}\sqrt{\frac{20N}{0.05kg/0.6m}}=12.90\ Hz

Hence, the first harmonic has a frequency of 12.90 Hz

4 0
3 years ago
24 A uniform electric field of magnitude 1.1×104 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the elect
Tatiana [17]

Answer:

44,000 Nm^2/C

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

\Phi = EA cos \theta

where:

E is the magnitude of the electric field

A is the area of the surface

\theta is the angle between the direction of the field and of the normal to the surface

In this problem, we have:

E=1.1\cdot 10^4 N/C is the electric field

L = 2.0 m is the side of the sheet, so the area is

A=L^2=(2.0)^2=4.0 m^2

\theta=0^{\circ}, since the electric field is perpendicular to the surface

Therefore, the electric flux is

\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C

4 0
3 years ago
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