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2 years ago

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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if you are in Birmingham AL, and you want to use your cell phone to talk to your cousin in Houston, TX, what must occur in order

lesantik [10]

Answer:

For communication to be communicated between you and your cousin who is in Houston,Texas, when a call is made by you, a request is made to the specific phone, and the telephone tower will get the request from the mobile phone. Then a signal is sent via a transmitter underground, by this the satellite communicates with the local receiver in Houston, that is linked to the local tower over there, the tower would request for your cousin's number and connects the two of you, once a link is set.

Explanation:

From the example stated, what is required for such for a far distance, is a communication satellite link.

When a call is made by you, the a connection request is sent to the specified phone.The telephone tower receives the request from The mobile phone. The local tower(Birmingham,Al) is linked to a ground transmitter by the means of a Fiber optical cable.

A signal is sent to satellite via the ground transmitter.The satellite then set's off the local receiver in (Houston,Texas) which on it's end is connected to the local tower there. This tower then ask for your cousin's mobile for a call that will be incoming, a link is set, once he/she receives the call, from there a conversation can be done.

3 0

3 years ago

Calculate the critical angle for light going from Glycerine to air.

Georgia [21]

The refractive index for glycerine is $n_g=1.473$ , while for air it is $n_a = 1.00$ .

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
$\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}$

6 0

2 years ago

Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC

nignag [31]

Answer:

ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

ΔL = α L₀ ΔT

ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

ΔL = 3.8236 10⁻⁴ m

using the criterion of three significant figures

ΔL = 3.82 10⁻⁴ m

5 0

2 years ago

How do you solving kinematic equations for horizontal projectiles?

daser333 [38]

See projectiles are very simple unless you understand its core concepts....projectile is nothing just mixture of upward motion and horizontal motion....
THE KEY IS FORGET THE NAME PROJECTILE...ITS JUST HORIZONTAL MOTION + VERTICAL MOTION