How far away is a cliff if an echo is heard 0.486 s after the original sound? Assume that sound traveled at 343 m/s on that day.

My buddy and I have just finished a dive to 15 metres/50 feet for 60 minutes. We want to return to the same site and depth and s

marishachu [46]

Answer:

1) Periodically check the no stop or NDL time on their computers

2) The dive computer planning mode can be used if available

3) Make use of a dive planning app

4) Check data from the RDP table or an eRDPML

Explanation:

The no stop times information from the computer gives the no-decompression limit (NDL) time allowable which is the time duration a diver theoretically is able to stay at a given depth without a need for a decompression stop

The dive computer plan mode or a downloadable dive planning app are presently the easiest methods of dive planning

The PADI RDP are dive planners based on several years of experience which provide reliable safety limits of depth and time.

7 0

3 years ago

Vector A → has magnitude 8.78 m at 37.0 ∘ from the + x axis. Vector B → has magnitude 8.26 m at 135.0 ∘ from the + x axis. Vecto

kodGreya [7K]

Answer:

R = (- 3.72î + 8.29j)

Magnitude of R = 9.09 m

Explanation:

Let î and j represent unit vectors along the x and y axis respectively.

Vector A --> magnitude 8.78 m, direction 37.0° from the +x-axis

Let the x and y components of this vector be Aₓ and Aᵧ

A = (Aₓî + Aᵧj) m

The components given magnitude and direction from the +x-axis are calculated as

Aₓ = A cos θ and Aᵧ = A sin θ

Aₓ = (8.78 cos 37°) = 7.01 m

Aᵧ = (8.78 sin 37°) = 5.28 m

A = (7.01î + 5.28j) m

Vector B has magnitude 8.26 m and direction 135° from the +x-axis

B = (Bₓî + Bᵧj) m

Bₓ = (8.26 cos 135°) = - 5.84 m

Bᵧ = (8.26 sin 135°) = 5.84 m

B = (-5.84î + 5.84j) m

Vector C has magnitude 5.65 m and direction 210° from the +x-axis

C = (Cₓî + Cᵧj) m

Cₓ = (5.65 cos 210°) = - 4.89 m

Cᵧ = (5.65 sin 210°) = - 2.83 m

C = (- 4.89î - 2.83j) m

The resultant force is a vector sum of all the forces. Let the resultant force be R

R = (Rₓî + Rᵧj) m

R = A + B + C = (7.01î + 5.28j) + (-5.84î + 5.84j) + (- 4.89î - 2.83j)

Summing the î and j components seperately,

R = (- 3.72î + 8.29j) m

To get its magnitude,

Magnitude of R = √(Rₓ² + Rᵧ²) = √((-3.72)² + (8.29)²) = 9.09 m

8 0

4 years ago

Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia

Crank

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

3 0

3 years ago

What force is needed to move a 5 kg mass with an acceleration of 5 m/s²?

ElenaW [278]

Answer:

b)25N

Explanation:

F=ma

F=(5kg)(5m/s^2)

F=25N

4 0

2 years ago

Read 2 more answers

The gauge pressure at the bottom of a cylinder of liquid is 0.30atm. The liquid is poured into another cylinder with twice the r

likoan [24]

Answer:

$P_g' = 0.075 atm$

Explanation:

Gauge pressure at the bottom of the cylinder depends on the height of water in the cylinder

So here we can say that

$P_g = \rho g h$

now when liquid is filled to height "h" in base area "A" then gauge pressure of the liquid at the bottom is given as

$P_g = 0.30 atm$

now we put the whole liquid into another cylinder with twice radius of the first cylinder

So area becomes 4 times

now by volume conservation we can say that if area is increased by 4 times then height of liquid will decrease by 4 times

so we have

$h' = \frac{h}{4}$

so gauge pressure is given as

$P_g' = \frac{0.30}{4} = 0.075 atm$

5 0

3 years ago