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gavmur [86]
3 years ago
7

What is the absolute pressure at a depth of 9.05 m below the surface of a deep lake? Assume atmospheric pressure is 1.01 × 10 5

Pa .
Physics
1 answer:
Arte-miy333 [17]3 years ago
8 0

Answer:

1.89 x 10⁵ Pa

Explanation:

given,

depth of the lake = 9.05 m

atmospheric pressure = 1.01 × 10⁵ Pa

absolute pressure = ?

we know,

Absolute pressure = atmospheric pressure + gauge pressure

Absolute pressure = 1.01 × 10⁵ Pa + ρgh

Absolute pressure = 1.01 × 10⁵ Pa + 1000 x 9.8 x 9.05

Absolute pressure = 1.89 x 10⁵ Pa

hence, absolute pressure is equal to 1.89 x 10⁵ Pa

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A rock dropped on the moon will increase it's speed from 0 m/s to 8.15 m/s in about 5 seconds what is the acceleration of the ro
Lunna [17]

Using the formula:


a = (Vf - Vi) / t


Our initial velocity is 0 m/s, and our final velocity is 8.15 m/s, with a time period of 5 seconds:


a = (8.15 - 0.0) / 5

a = 1.63 m/s^2


If you know the acceleration due to gravity on the Moon, you can confirm this answer. The recorded gravitational acceleration on the Moon is 1.62 m/s^2.

5 0
3 years ago
A 2-kg cart, traveling on a horizontal air track with a speed of 3m/s, collides with a stationary 4-kg cart. The carts stick tog
ladessa [460]

Answer:

The impulse exerted by one cart on the other has a magnitude of 4 N.s.

Explanation:

Given;

mass of the first cart, m₁ = 2 kg

initial speed of the first car, u₁ = 3 m/s

mass of the second cart, m₂ = 4 kg

initial speed of the second cart, u₂ = 0

Let the final speed of both carts = v, since they stick together after collision.

Apply the principle of conservation of momentum to determine v

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 3 + 0 = v(2 + 4)

6 = 6v

v = 1 m/s

Impulse is given by;

I = ft = mΔv = m(

The impulse exerted by the first cart on the second cart is given;

I = 2 (3 -1 )

I = 4 N.s

The impulse exerted by the second cart on the first cart is given;

I = 4(0-1)

I = - 4 N.s (equal in magnitude but opposite in direction to the impulse exerted by the first).

Therefore, the impulse exerted by one cart on the other has a magnitude of 4 N.s.

8 0
3 years ago
A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturi
Darina [25.2K]

Answer: P = 36.75W

The additional power needed to account for the loss is 36.75W.

Explanation:

Given;

Mass of the runner m= 60 kg

Height of the centre of gravity h= 0.5m

Acceleration due to gravity g= 9.8m/s

The potential energy of the body for each step is;

P.E = mgh

P.E = 60 × 9.8 × 0.5

PE = 294J

Since the average loss per compression on the leg is 10%.

Energy loss = 10% (P.E)

E = 10% of 294J

E = 29.4J

To calculate the runner's additional power

given that time per stride is = 0.8s

Power P = Energy/time

P = E/t

P = 29.4J/0.8s

P = 36.75W

5 0
3 years ago
This when a satellite orbits in an oval-shaped path around a central object.
Ilya [14]

Elliptical orbit.<<<<<<<<<<



8 0
3 years ago
Consider a parachutist that has reached terminal velocity. which of the following is true? A.) The acceleration of the parachuti
trasher [3.6K]
The correct answer is:
<span>B.) At terminal velocity there is no net force 

In fact, when the parachutist reaches the terminal velocity, his velocity does not change any more. It means that the acceleration acting on the parachutist is zero, and for Newton's second law, this means the net force acting on him is zero:
</span>\sum F = ma = 0
<span>because the acceleration is zero: a=0. 
This also means that the two relevant forces acting on the parachutist (gravity, downward, and air resistance, upward) are balanced to produce a net force equal to zero.</span>
8 0
3 years ago
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