Can yall Please fill out them blanks i need help it's due Today.

Drag each label to the correct location on the image

umka2103 [35]

Answer:

you cant really give any answers because i cant tell u were to drag

3 0

2 years ago

The objects listed are placed at the top of a ramp and roll down to the bottom without slipping. Assuming that there is no air r

Thepotemich [5.8K]

Explanation:

For each object, the initial potential energy is converted to rotational energy and translational energy:

PE = RE + KE

mgh = ½ Iω² + ½ mv²

For the marble (a solid sphere), I = ⅖ mr².

For the basketball (a hollow sphere), I = ⅔ mr².

For the manhole cover (a solid cylinder), I = ½ mr².

For the wedding ring (a hollow cylinder), I = mr².

If we say k is the coefficient in each case:

mgh = ½ (kmr²) ω² + ½ mv²

For rolling without slipping, ωr = v:

mgh = ½ kmv² + ½ mv²

gh = ½ kv² + ½ v²

2gh = (k + 1) v²

v² = 2gh / (k + 1)

The smaller the value of k, the higher the velocity. Therefore:

marble > manhole cover > basketball > wedding ring

7 0

3 years ago

Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram

Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

Given the following data:

Force 1 = 500 Newton

Distance 1 = 1.2 meter

Force 2 = 500 Newton

Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

$Work\;done = Force \times distance$

For student 1:

$Work\;done = 500 \times 1.2$

Work done = 600 Joules

For student 2:

$Work\;done = 500 \times 5$

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0

2 years ago

Read 2 more answers

A physical count of merchandise inventory on november 30

Zigmanuir [339]

It depends on the indirect taxes and the share market values

4 0

3 years ago

A 19.12 g mixture of Ca(NO3)2 and KCl is dissolved in 149 g of water. The freezing point of the solution was measured as −5.77 ∘

hichkok12 [17]

Answer:

The mass percentage of calcium nitrate is 31.23%.

Explanation:

Let the the mass of calcium nitrate be x and mass of potassium chloride be y.

Total mass of mixture = 19.12 g

x + y = 19.12 g..(1)

Mass of solvent = 149 g = 0.149 kg

Freezing point of the solution, $T_f$ = -5.77 °C

Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)

The van't Hoff factor contribution by calcium nitrate is 3 and by potassium chloride is 2.So:

i = 3

i' = 2

Freezing point of water = T = 0°C

$\Delta T_f=T-T_f=0^oC-(-5.77^oC)=5.77^oC$

$\Delta T_f=i\times K_f\times m$

$Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}$

$5.77^oC=1.86 ^oC/(mol/kg)\times (\frac{ i\times x}{164 g/mol\times 0.149 kg}+\frac{i'\times y}{74.5 g/mol0.149 kg})$

On solving we get:

$\frac{3x}{164 g/mol}+\frac{2x}{74.5 g/mol}=0.4622 mol$ ....(2)

Solving equation (1)(2) for x and y:

x =5.973 g

y = 13.147 g

Mass percent of $Ca(NO_3)_2$ in the mixture:

$\frac{x}{19.21 g}\times 100=\frac{5.973 g}{19.12 g}=31.23\%$

The mass percentage of calcium nitrate is 31.23%.

5 0

3 years ago