Is this the actual answer??

If a 50-kg Person is running at a rate of 2m/s, the person's momentum is _____ kg• m/s.

sergiy2304 [10]

Answer:

100

Explanation:

Use equation p=mv

p=(50kg)(2m/s)= 100

8 0

3 years ago

Explore how archemides principle is applied in building a ship and submarine

Sladkaya [172]

Answer:

Principle Archimedes is applied in building a ship and submarine using the manipulating that buoyancy, is controlled the ballast tank system.

Explanation:

Submarine is rather had they focused on main parts of the submarine,he is complex and long process implementation,the most submarine design like submarine stability.

Submarine stability is complete and the fundamental Archimedes principle to arrive the weight of submarine is equal to buoyancy force.

Submarine into the parts and components of ballast tank the sequence in diving and surfacing,there two vital parts:- flood parts and air vents

flood parts:- at the bottom position and allow water to enter or leave that tank.

air vents:- air vents at the top of the pressure hall,and that they submarine dive.

this time submarine is most modern system is depth is 300 to 450 meters,high pressure air is 15 bar is tank air valve.

submarine is basic of the effective volume of all the submarine surfaced condition,submarine minus to the free water flood is equal to the fully pressure hull,submarine is the surfaced condition.

3 0

2 years ago

A hydrometer is made of a tube of diameter 2.3cm.The mass of the tube and it's content is 80g. If it floats in a liquid density

iris [78.8K]

Answer:

The depth to which the hydrometer sinks is approximately 24.07 cm

Explanation:

The given parameters are;

The diameter of the hydrometer tube, d = 2.3 cm

The mass of the content of the tube, m = 80 g

The density of the liquid in which the tube floats, ρ = 800 kg/m³

By Archimedes' principle, the up thrust (buoyancy) force acting on the hydrometer = The weight of the displaced liquid

When the hydrometer floats, the up-thrust is equal to the weight of the hydrometer which by Archimedes' principle, is equal to the weight of the volume of the liquid displaced by the hydrometer

Therefore;

The weight of the liquid displaced = The weight of the hydrometer, W = m·g

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ W = 80 g × g

The volume of the liquid that has a mass of 80 g (0.08 kg), V = m/ρ

V = 0.08 kg/(800 kg/m³) = 0.0001 m³ = 0.0001 m³ × 1 × 10⁶ cm³/m³ = 100 cm³

The volume of the liquid displaced = 100 cm³ = The volume of the hydrometer submerged, $V_h$

$V_h$ = A × h

Where;

A = The cross-sectional area of the tube = π·d²/4

h = The depth to which the hydrometer sinks

h = $V_h$ /A

∴ h = 100 cm³/( π × 2.3²/4 cm²) ≈ 24.07 cm

The depth to which the tube sinks, h ≈ 24.07 cm.

3 0

2 years ago

A 15kg dog jumps out of a 40kg canoe. If the dogs velocity is 1.2m/s, what is the velocity of the canoe?

vivado [14]

Answer:

v = -0.45 m/s

Explanation:

Assuming the canoe was initially at rest with momentum L = 0

and that the dog's velocity is in the positive direction

conservation of momentum

0 = 15(1.2) + 40v

v = -0.45 m/s

6 0

2 years ago

Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts

podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height =?

$y_{0}$ : is the initial height = 1 m

$v_{0_{y}$ : is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time

First, we need to find the time by using the following equation:

$t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s$

Now, the height is:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m$

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

$0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}$

$1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0$

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

$x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m$

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!

4 0

3 years ago