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Bogdan [553]
3 years ago
7

When heather grabbed the glass of ice water, the ice cubes were floating at the top. why were the ice cubes floating in the wate

r?
Physics
1 answer:
tiny-mole [99]3 years ago
8 0
The ice cubes were floating in water because they are less dense than liquid water. When water is frozen, a structure that is crystalline is formed that is held by hydrogen bonding. Due to the orientation of these bonds, the moleules would push far away from each other causing it to have a bigger volume and a lower density.
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The human nervous system can propagate nerve impulses at about 102 m>s. Estimate the time it takes for a nerve impulse to tra
Olin [163]

Answer:

t = 0.196 s

Explanation:

The speed of a pulse is determined by the characteristics of the medium, its density and its resistance to stress, as long as these remain the speed will be constant for which we can use the kinetic expressions of the uniform movement

          v = x / t

          t = x / v

calculate

          t = 2/102

          t = 0.196 s

4 0
3 years ago
An elevator supported by a single cable descends a shaft at a constant speed. The only forces acting on the elevator are the ten
VikaD [51]

Answer:

C. The net work done by the two forces is zero joules.

Explanation:

Work (W) is defined as the product of force F) by the distance (d)the body travels due to this force.  

W= F*d Formula ( 1)

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

The component of the force that performs work must be parallel to the displacement.  

Calculation of the forces

We apply Newton's second law:

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

∑F = m*a  

a = 0 because the elevator moves at constant speed.

∑F=0

T-W = 0

T = W  

Calculation of work done by the forces

W₁ :  work done by the gravity= (+ W*d ) J

W₂ : work done by the tension force=( -T*d) J

Net work done by the two forces(W₁-W₂):

W₁= W₂ .because T = W  

W₁-W₂ =0

3 0
3 years ago
The circumference of a sphere was measured to be
professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

\phi = 76cm

Error (dr) = 0.5cm

The radius then would be

\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm

And

\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr

PART A ) For the Surface Area we have that,

A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)

Maximum error:

dA = 4(\frac{38}{\pi})(0.5)

dA = \frac{76}{\pi}cm^2

The relative error is that between the value of the Area and the maximum error, therefore:

\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}

\frac{dA}{A} = 0.01315 = 1.31\%

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

V = \frac{4}{3} \pi r^3

V = \frac{4}{3} \pi (\frac{38}{\pi})^3

V = \frac{219488}{3\pi^2}

Therefore the Maximum Error would be,

\frac{dV}{dr} = \frac{4}{3} 3\pi r^2

dV = 2r^2 (2\pi dr)

dV = 4r^2 (\pi dr)

Replacing the value for the radius

dV = 4(\frac{38}{\pi})^2(0.5)

dV = \frac{2888}{\pi^2} cm^3

And the relative Error

\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }

\frac{dV}{V} = 0.03947

\frac{dV}{V} = 3.947\%

3 0
3 years ago
In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

3 0
3 years ago
Explain why the types of technology valued can vary.
Elan Coil [88]
Over time, the types of technology can vary and be improved upon so that more advanced techniques become more valued. This could be the situation with mining whereby back in the 1500's in underground mines the rock was broken by fire setting ie lighting a fire below the rock face to heat up the rock and then throwing cold water on it to crack it, so that it could be dug by hand. With the advent of explosives, this all changed so that the rock could be blasted. The increase in advance rates for an underground heading have thus gone from 5-20 feet per month to up to 300meters (984 ft) per month for a 24/7 mining operation, which is a huge improvement.
4 0
3 years ago
Read 2 more answers
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