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tatiyna
3 years ago
8

the car starts from rest at s=0 and increases its speed at at=4m/s^2. Determine the time when the magnitude of acceleration beco

mes 20m/s^. At what point s does this occur
Physics
1 answer:
dem82 [27]3 years ago
5 0

Answer:

<em>Time =  5 seconds</em>

<em>Distance = 50 meters</em>

Explanation:

<u>Constantly Accelerated Motion</u>

When the velocity of a moving object changes at a constant rate, called acceleration, the velocity changes in same amounts in the same times. The question has a mistake when asking when the acceleration is 20 m/s. If the acceleration is constant, the only variable that can change to that value is the velocity. The equation to calculate the speed is

v_f=v_o+a.t

And the distance s is

\displaystyle s=v_o.t+\frac{gt^2}{2}

Given the object starts from rest, vo=0 and vf=20 m/s at a=4\ m/s^2. We compute t

\displaystyle t=\frac{v_f-v_o}{a}=\frac{20-0}{4}

\boxed{t=5\ sec}

Now we compute s

\displaystyle s=0+\frac{4\times 5^2}{2}

\boxed{s=50\ m}

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Answer:

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5 0
2 years ago
A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm
Karo-lina-s [1.5K]

Answer:

the thermistor temperature = 325.68 \ ^0 \ C

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K

Resistance of the thermistor R_1 = 20,000 ohms

Material constant \beta = 3650

Resistance of the thermistor R_2 = 500 ohms

Using the equation :

R_1 = R_2  \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

\frac{R_1}{ R_2} =   \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

Taking log of both sides

In \ \frac{R_1}{ R_2} = In \  \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})

\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})

\frac{1}{T_2} =   \frac{1}{T_1}  -          \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}

{T_2} =  \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}

Replacing our values into the above equation :

{T_2} =  \frac{3650*373}{3650 - In (\frac{20000}{500})373}

{T_2} =  \frac{1361450}{3650 - 3.6888*373}

{T_2} =  \frac{1361450}{3650 - 1375.92}

{T_2} =  \frac{1361450}{2274.08}

{T_2} = 598.68 \ K

{T_2} = 325.68 \ ^0 \ C

Thus, the thermistor temperature = 325.68 \ ^0 \ C

4 0
3 years ago
A first order reaction, A -&gt; products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,
tamaranim1 [39]

Answer: a)  The rate constant, k, for this reaction is 0.00516s^{-1}

b) No t_{\frac{1}{2}} does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

A\rightarrow products

Given: Order with respect to A = 1

Thus rate law is:

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k= rate constant

0.00250=k[0.484]^1

k=0.00516s^{-1}

The rate constant, k, for this reaction is 0.00516s^{-1}

b) Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{2.303}{k}\log\frac{100}{50}

t_{\frac{1}{2}}=\frac{0.69}{k}

Thus t_{\frac{1}{2}} does not depend on concentration.

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Answer:

<em>C</em>

Explanation:

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A train travels 600 km in 4 hours. What is the speed of the train?
tekilochka [14]

Answer:

150 km/h

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