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kogti [31]
2 years ago
10

An apple dropped from the branch of a tree hits the ground in 0.5 s. If the acceleration of the apple during its motion is 10 ms

-2 in the downward direction. Calculate ----- Its speed just before it hits the ground. What is its average velocity during 0.5 s? Calculate the height of the branch of the tree from the ground?
Physics
1 answer:
Ipatiy [6.2K]2 years ago
7 0

Given that,

Time = 0.5 s

Acceleration = 10 m/s²

(I). We need to calculate the speed of apple

Using equation of motion

v=u+at

Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

v=0+10\times0.5

v=5\ m/s

(III). We need to calculate the height of the branch of the tree from the ground

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Put the value into the formula

s=0+\dfrac{1}{2}\times10\times(0.5)^2

s=1.25\ m

(II). We need to calculate the average velocity during 0.5 sec

Using formula of average velocity

v_{avg}=\dfrac{\Delta x}{\Delta t}

v_{avg}=\dfrac{x_{f}-x_{i}}{t_{f}-t_{0}}

Where, x_{f}= final position

x_{i} = initial position

Put the value into the formula

v_{avg}=\dfrac{1.25+0}{0.5}

v_{avg}=2.5\ m/s

Hence, (I). The speed of apple is 5 m/s.

(II). The average velocity during 0.5 sec is 2.5 m/s

(III). The height of the branch of the tree from the ground is 1.25 m.

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A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm
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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

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