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3 years ago

5

A box that has a mass of 80 kg slides down a ramp with a 30 degree angle. The free-body diagram shows the forces acting on the b

ox.Ignoring friction and air resistance, what is the acceleration of the box, to the nearest tenth?

1 answer:

GaryK [48]3 years ago

3 0

Answer:

4.9

Explanation:

ecad

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The wheel of a stationary exercise bicycle at your gym makes one rotation in 0.670 s. Consider two points on this wheel: Point P

defon

Answer:

0.938 m/s.

Explanation:

Given:

ω = 1 rev in 0.67 s

In rad/s,

1 rev = 2pi rad

ω = 2pi ÷ 0.67

= 9.38 rad/s

Rp = 10 cm

= 0.1 m

V = ω × r

= 9.38 × 0.1

= 0.938 m/s.

5 0

3 years ago

a driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di

daser333 [38]

Answer:

v = 98.75 km/h

Explanation:

Given,

The distance driver travels towards the east, d₁ = 135 km

The time period of the travel, t₁ = 1.5 h

The halting time, tₓ = 46 minutes

The distance driver travels towards the east, d₂ = 215 km

The time period of the travel, t₁ = 2 h

The average speed of the vehicle before stopping

v₁ = d₁/t₁

= 135/1.5

= 90 km/h

The average speed of vehicle after stopping

v₂ = d₂/t₂

= 215/2

= 107.5 km/h

The total average velocity of the driver

v = (v₁ +v₂) /2

= (90 + 107.5)/2

= 98.75 km/h

Hence, the average velocity of the driver, v = 98.75 km/h

7 0

3 years ago

You and your friend live in different states and plant identical hydrangea bulbs in your garden. When they bloom, you send each

Whitepunk [10]

Answer:

sample. rewrite in your own words if you'd like.

Explanation:

Since the hydrangeas are changing color depending on whether the soil is acidic or basic, they must have an indicator. This means that any hydrangea planted in acidic soil will have blue flowers and any hydrangea planted in basic soil will have pink. One common indicator present in many plants is anthocyanin, so maybe hydrangeas contain that chemical.

8 0

3 years ago

Read 2 more answers

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

6 0

3 years ago

Select the correct answer.

andrew-mc [135]

I think the answer would be C

7 0

3 years ago

Read 2 more answers