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ddd [48]
2 years ago
11

A person is standing on a raft; their

Physics
2 answers:
klemol [59]2 years ago
5 0

Answer:

the volume displaced by the raft = 0.233 m3

Explanation:

correct for Acellus

krok68 [10]2 years ago
3 0

Answer:

The volume of water displaced by the raft is 0.233 m³

Explanation:

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.

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A lever allows a 35 N load to be lifted with a force of 7 N. What is the mechanical advantage of the lever?
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Simple question please help! A ball is thrown 19.0 m/s at an angle of 40.0º with the horizontal. Assume the ball is thrown at gr
d1i1m1o1n [39]

a. 1.51 s

In this part of the problem, we are only interested in the horizontal motion of the ball. Along the horizontal direction, the motion of the ball is a uniform motion with constant velocity, which is equal to the horizontal component of the initial velocity:

v_x = v_0 cos \theta = (19.0 m/s)(cos 40^{\circ})=14.6 m/s

So, the ball travels horizontally at a speed of 14.6 m/s; in order to cover the distance of d = 22.0 m that separates it from the wall, the time need is

t=\frac{d}{v_x}=\frac{22.0 m}{14.6 m/s}=1.51 s

b. 7.25 m

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y(t)=v_{0y} t -\frac{1}{2}gt^2

We know that the ball hits the wall at t=1.51 s, so if we substitute this value into the previous formula, we find at what height y the ball hits the wall:

y(1.51 s)=(12.2 m/s)(1.51 s)-\frac{1}{2}(9.8 m/s^2)(1.51 s)^2=7.25 m

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