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zvonat [6]
3 years ago
9

An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 12.7 m/s2 . At t1 the rocket e

ngine is shut down and the sled moves with constant velocity v for another t2 s. The total distance traveled by the sled is 5.85 × 103 m and the total time is 94.6 s. Find t1. Answer in units of s.
Physics
1 answer:
sp2606 [1]3 years ago
6 0

Answer:

The value of t₁ is 5.119 sec.

Explanation:

Given that,

Acceleration = 12.7 m/s²

Total distance D=5.85\times10^{3}\ m

Time = 94.6 sec

We need to calculate the distance

Using equation of motion

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

Put the value into the formula

s_{1}=0+\dfrac{1}{2}12.7t_{1}^2...(I)

We need to calculate the velocity

Using equation of motion

v=u+at

Put the value into the formula

v=0+12.7t_{1}

v=12.7t_{1}

The sled while moves with constant velocity v for time t₂

Using formula of distance

s_{2}=vt_{2}

s_{2}=12.7t_{1}t_{2}

The total distance is

s_{1}+s_{2}=5.85\times10^{3}

Put the value of s₁ and s₂

\dfrac{1}{2}12.7t_{1}^2+12.7t_{1}t_{2}=5.85\times10^{3}

6.35t_{1}+12.7t_{1}t_{2}=5.85\times10^{3}...(II)

The total time is

t_{1}+t_{2}=94.6....(III)

From equation (II) and (III)

After solving

t_{1}=5.119\ sec

Hence, The value of t₁ is 5.119 sec.

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Answer: The value of the celsius temperature of the cube is 472.2°c.

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T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}

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T_{2}=472.2^{\circ}c

Therefore, the value of the celsius temperature of the cube is 472.7°c.    

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