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2 years ago

How is momentum of an object related to stopping distance

Physics

1 answer:

Rama09 [41]2 years ago

6 0

Answer:

To stop a car must lose its momentum and its kinetic energy. ... If the forces on two cars are equal then the greater the kinetic energy the greater the distance before stopping. But there will be a relationship to the momentum because momentum and mass are both related to the kinetic energy.

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If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

stich3 [128]

I would think that you would have to do 42/2=21Hz, but I'm not sure...

5 0

3 years ago

Solve the question that follows using the equation for the conversion of Celsius to Fahrenheit. F=95(C)+32 On February 9, 1934,

SVEN [57.7K]

Answer:

Buffalo, NY

Explanation:

Temperature in Buffalo, NY = -29°C

In order to compare the temperatures we need to convert them to the same scale.

$F=\dfrac{9}{5}C+32\\\Rightarrow F=\dfrac{9}{5}\times -29+32\\\Rightarrow F=-20.2$

So, the temperature in Buffalo, NY was -20.2°F and the temperature in Anchorage, AL was 19°F.

Hence, it was colder in Buffalo, NY than in Anchorage, AL.

3 0

3 years ago

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

HELPPPP I HAVE A TEST TODAY AND I LITERALLY CAN'T WITH THIS

dsp73

The horizontal component of velocity is

(22 m/s) • cosine(62°).

The vertical component of velocity is

(22 m/s) • sine(62°).

These are the original components, right after the kick. As time goes on, the horizontal one doesn't change. But the vertical one gets bigger and bigger, because gravity is accelerating the ball downward.

That's the complete story of projectile motion.

7 0

2 years ago

* Without heat you would be hot warm none of these cold

Arturiano [62]

Answer:

It depends on the environment you're in but I'd go with warm

Explanation:

4 0

2 years ago

How is momentum of an object related to stopping distance​

How is momentum of an object related to stopping distance