3 physical adaptations of a frog

The question relates to Archimedes' principle which states that the buoyant force experienced by an object immersed in a fluid is equal to the weight of (the force of gravity on) the displaced fluid

The given parameters are;

The combined mass of the person and the raft, m = 233 kg

The liquid on which the raft is located = Water

The density of water, $\rho _{water}$ = 1000 kg/m³

Weight = Mass, m × g

Where;

m = The mass of the object

g = The acceleration due to gravity = 9.8 m/s²

Given that the raft is on the surface of the water (floating), the buoyant force is equal to the combined weight of the person and the raft = 233 kg

The combined weight of the person and the raft, $W_{combined}$ = 233 kg × 9.8 m/s² = 2,283.4 N

Therefore;

The buoyant force = 2,283.4 N = The weight of the water displaced

The mass of the water displaced, $m_{water}$ , = 2,283.4 N/(9.8 m/s²) = 233 kg

Density = Mass/Volume

The volume of water displaced by the raft = The mass of the water displaced/(The density of the water) = 233 kg/(1,000 kg/m³) = 0.233 m³.

3 0

2 years ago

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In a uniform circular motion map, what is always true? Check all that apply.

grandymaker [24]

Answer:

Velocity vectors are always perpendicular to the circle.

Acceleration vectors point toward the center of the circle.

Velocity vectors are the same length.

Explanation:

(THESE ARE NOT MY WORDS BTW)

1) Acceleration and velocity are vectorial quantities, which means they have magnitude and direction.

2) In a circular motion velocity direction changes all the time, which means that it is accelerated.

3) In a uniform circular motiion, the velocity changes in a constant value. This is the rate of change of velocity, which is the magnitude of the acceleration, is constant (uniform).

4) The velocity is perpendicular to the path, i.e. the circle. You can see it if you think that if the object stopped changing the direction, then the object would follow a straight path (as per inertia principle). That is why this velocity is called tangential velocity (to differentiate it of the angular velocity).

This is what the option C says "Velocity vectors are always perpendicular to the circle". Then this is true.

5) The constant change of direction in a circular path, means that the object is been pushed, accelerated, toward the center of a circle. This is, all the time the object in motion tries to follow the perpendicular path but a push (a force) directed to the center of the circle changes its direction. Such force accelerates the object toward the center of the circle. So, the acceleration vectors point toward the center of the circle, which is what the option D says. So, this is also true.

6) Since the motion is uniform, the magnitude or length of the velocity vectors are always the same, are constant. So, the option E. is also true.

6 0

2 years ago

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A small bag of sand is released from an ascending hot-air balloon whose upward constant velocity is v0 = 2.45 m/s. knowing that

lakkis [162]

When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:

H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s

t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s

Total time = 0.125 s + 4.495 s = 4.62 seconds

8 0

3 years ago