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diamong [38]
3 years ago
8

Why does a rock weigh less in water than it does on land?

Physics
1 answer:
Cloud [144]3 years ago
7 0
Water is more dense then air so it sorta holds the rock as it sinks
You might be interested in
A small sculpture made of brass (rho brass = 8470 kg/m 3 ) is believed to have a secret central cavity. The weight of the sculpt
GaryK [48]

Answer:

Volume of secret cavity = 4 x 10⁻⁶m³

Explanation:

The weight of the sculpture in air is 15.76 N

Mass of the sculpture = 1.61 kg

Mass = Volume x Density

1.61 = V x 8470

Volume of brass =1.90 x 10⁻⁴ m³

When it is submerged in water, the weight is 13.86 N.

That is

  Weight of sculpture - Weight of water displaced = 13.86 N

  15.76 - Weight of water displaced = 13.86

   Weight of water displaced = 1.9 N

   Mass of water displaced = 0.194 kg

 Mass = Volume x Density

  0.194 = V x 1000

  Volume of water displaced =1.94 x 10⁻⁴ m³

Volume of secret cavity =  Volume of water displaced - Volume of brass material

Volume of secret cavity = 1.94 x 10⁻⁴-1.94 x 10⁻⁴ = 0.04x 10⁻⁴ = 4 x 10⁻⁶m³

7 0
4 years ago
When analyzing forces, it is allowable to break the motion and forces into horizontal and vertical components.
jonny [76]

The statement above is TRUE.

When analyzing forces it is permitted to break the motion and forces into horizontal and vertical components. This is especially true for two dimensional projectile motion, in order to simplify the calculation, the forces and the motion has to be broken down into horizontal and vertical components. The vertical component is affected by the force of gravity while the horizontal component is not affected by gravity for short displacements. With greater displacement, the force of gravity comes into play.

8 0
3 years ago
Suppose a 1 Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to Ear
vagabundo [1.1K]

Answer:

The time required to send the data from Earth to Moon will be 1.28s while for a two way communication, to send it back to the earth, it will take double time i.e. <em>RTT = 2.56s </em>

Explanation:

Distance between Earth and Moon = 385,000 km = 3.85 x 10⁸m

Speed of data travel = speed of light ≈ 3 x 10⁸m/s

As, v=d/t

t=d/v

t=\frac{3.85*10^{8} }{3*10^{8}}

t=1.28s

RTT = Double of single way time taken = 2x1.28

<em><u>RTT=2.56s</u></em>

7 0
3 years ago
A gold wire has a cross-sectional area of 1.0 cm^2 and a resistivity of 2.8 × 10^-8 Ω ∙ m at 20°C. How long would it have to be
Karo-lina-s [1.5K]

Answer:

Length, l = 3.57 meters

Explanation:

It is given that,

Area of cross-section of gold wire, A=1\ cm^2=0.0001\ m^2

Resistivity of gold wire, \rho=2.8\times 10^{-8}\ \Omega-m

Resistance, R = 0.001 ohms

Resistance in terms of length and area is given by :

R=\rho \dfrac{l}{A}

l=\dfrac{RA}{\rho}

l=\dfrac{0.001\times 0.0001}{2.8\times 10^{-8}}

l = 3.57 meters

So, the length of the wire is 3.57 meters. Hence, this is the required solution.

4 0
3 years ago
A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion s
loris [4]

To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

m*V_i = m_1V_1+m_2V_2

Here,

m = Total mass (8Kg at this case)

m_1=m_2 = Mass each part

V_i = Initial velocity

V_2 = Final velocity particle 2

V_1 = Final velocity particle 1

The initial kinetic energy would be given by,

KE_i=\frac{1}{2}mv^2

KE_i = \frac{1}{2}8*5^2

KE_i = 100J

In the end the energy increased 100J, that is,

KE_f = KE_i KE_{increased}

KE_f = 100+100 = 200J

By conservation of the moment then,

m*V_i = m_1V_1+m_2V_2

Replacing we have,

(8)*5 = 4*V_1+4*V_2

40 = 4(V_1+V_2)

V_1+V_2 = 10

V_2 = 10-V_1(1)

In the final point the cinematic energy of EACH particle would be given by

KE_f = \frac{1}{2}mv^2

KE_f = \frac{1}{2}4*(V_1^2+V_2^2)

200J=\frac{1}{2}4*(V_1^2+V_2^2)(2)

So we have a system of 2x2 equations

V_2 = 10-V_1

200J=\frac{1}{2}4*(V_1^2+V_2^2)

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: V_1 = 10m/s

Then replacing in (1) we have that

PART B: V_2 = 0m/s

8 0
3 years ago
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