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Effectus [21]
3 years ago
14

What is the maximum mass that can hang without sinking from a 20-cm diameter Styrofoam sphere in water? Assume the volume of the

mass is negligible compared to that of the sphere.
Physics
1 answer:
alex41 [277]3 years ago
3 0

Answer:

the maximum mass that can hang without sinking is 2.93 kg

Explanation:

Given: details:

sphere diameter  d = 20 cm

so, radius r = 10 cm  = 0.10 m

density of the Styrofoam sphere D = 300 kg/m3

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

                                                   =\frac{4}{3} \pi 0.10^3

                                                   =4.18*10^{-3} m^3

we know that

Density = \frac{Mass}{Volume}

mass  M = Density * Volume

                                  = (300)(4.18*10^{-3} m3)

                                  =1.25 kg

mass of the water displace = volume *density  of water

                                                 = 4.18*10^{-3} m3 * 1000

                                                 = 4.18 kg

The difference between the mass of water and mass of styrofoam is the amount of mass that the sphere can support

=4.18 kg  -1.25 kg

= 2.93 kg

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Which would melt first, germanium with a melting point of 1210 k or gold with a melting point of 1064oc?
iren2701 [21]
<span>Germanium To determine which melts first, convert their melting temperatures so they're both expressed on same scale. It doesn't matter what scale you use, Kelvin, Celsius, of Fahrenheit. Just as long as it's the same scale for everything. Since we already have one substance expressed in Kelvin and since it's easy to convert from Celsius to Kelvin, I'll use Kelvin. So convert the melting point from Celsius to Kelvin for Gold by adding 273.15 1064 + 273.15 = 1337.15 K So Germanium melts at 1210K and Gold melts at 1337.15K. Germanium has the lower melting point, so it melts first.</span>
8 0
3 years ago
9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the
mojhsa [17]

Answer:

Moment about SHOULDER  ∑ τ = 3.17 N / m,

Moment respect to ELBOW   Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

         ∑ τ = I α

   

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

          ∑ τ = I α

           I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

            I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

            I_mass = m L²

we substitute

           ∑ τ = (⅓ m L² + M L²) α

let's calculate

          ∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

           ∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

         Στ = I α

         I = I_ forearm + I_mass

         I = ⅓ m (L-0.03)² + M (L-0.03)²

         

let's calculate

        i = ⅓ 2.3 0.47² + 0.5 0.47²

        I = 0.2798 Kg m²

        Στ = 0.2798 10

        Στ= 2.80 N m

3 0
3 years ago
Light with energy equal to three times the work function of a given metal causes the metal to eject photoelectrons. What is the
Mrac [35]

The ratio of the maximum photoelectron kinetic energy to the work function will be 3:1.

<h3 /><h3>What is the photoelectric effect?</h3>

When a medium receives electromagnetic radiation, electrostatically charged particles are emitted from or inside it.

The emission of ions from a steel plate when light falls on it is a common definition of the effect. The substance could be a solid, liquid, or gas; and the released particles could be protons or electrons.

A particular metal emits photoelectrons when exposed to light with energy three times its work function:

\rm KE=3 \phi

The ratio of the maximum photoelectron kinetic energy to the work function will be;

R=\frac{E}{\phi} \\\\ R=\frac{3 \phi}{\phi} \\\\ R= 3

Hence, the ratio of the maximum photoelectron kinetic energy to the work function will be 3:1.

To learn more about the photoelectric effect refer to the link;

brainly.com/question/9260704

#SPJ1

5 0
1 year ago
1 2 3 4 5 6 7 8 9 10
7nadin3 [17]

Answer:

<h2>C. <u>0.55 m/s towards the right</u></h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

<u>x is the final velocity of the 0.25kg ball</u>

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>

<u></u>

4 0
3 years ago
calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2​
igor_vitrenko [27]

First, balance the reaction:

_ KClO₃   ==>   _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃   ==>   2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃   ==>   2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

7 0
3 years ago
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