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3 years ago

9

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

a gazelle assumes an acceleration of 4.2 m/s2 for 6.5 s , after which the gazelle continues at a steady speed.

1 answer:

NikAS [45]3 years ago

6 0

Answer:

The gazelles top speed is 27.3 m/s.

Explanation:

Given that,

Acceleration = 4.2 m/s²

Time = 6.5 s

Suppose we need to find the gazelles top speed

The speed is equal to the product of acceleration and time.

We need to calculate the gazelles top speed

Using formula of speed

$v=at$

Where, v = speed

a = acceleration

t = time

Put the value into the formula

$v=4.2\times6.5$

$v=27.3\ m/s$

Hence, The gazelles top speed is 27.3 m/s.

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A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 85.0 m/s2

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Answer:

Maximum height attained by the model rocket is 2172.87 m

Explanation:

Given,

Initial speed of the model rocket = u = 0
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We have to consider the whole motion into two parts

In first part the rocket is moving with an acceleration of a = 85.0 $m/s^2$ for the time t = 2.30 s before the fuel abruptly runs out.

Let $s_1$ be the height attained by the rocket during this time intervel,

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And Final velocity at that point be v

$\therefore v\ =\ u\ +\ at\\\Rightarrow v\ =\ 0\ +\ 85.0\times 2.3\\\Rightarrow v\ =\ 195.5\ m/s.$

Now, in second part, after reaching the altitude of 224.825 m the fuel abruptly runs out. Therefore rocket is moving upward under the effect of gravitational acceleration,

Let ' $s_2$ ' be the altitude attained by the rocket to reach at the maximum point after the rocket's fuel runs out,

At that insitant,

initial velocity of the rocket = v = 195.5 m/s.

a = $-g\ =\ -9.81\ m/s^2$
Final velocity of the rocket at the maximum altitude = $v_f\ =\ 0$

From the kinematics,

$v^2\ =\ u^2\ +\ 2as\\\Rightarrow 0\ =\ u^2\ -\ 2gs_2\\\Rightarrow s_2\ =\ \dfrac{u^2}{2g}\\\Rightarrow s_2\ =\ \dfrac{195.5^2}{2\times 9.81}\\\Rightarrow s_2\ =\ 1948.02\ m$

Hence the maximum altitude attained by the rocket from the ground is

$s\ =\ s_1\ +\ s_2\ =\ 224.85\ +\ 1948.02\ =\ 2172.87\ m$

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3 years ago

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3 years ago

A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 94.8 g of water at 60.4°C with 94.8

madreJ [45]

Answer:

$396.65$ JC⁻¹

Explanation:

$m_{a}$ = mass of water added to calorimeter = 94.8 g

$T_{ai}$ = initial temperature of the water added = 60.4 C

$c_{w}$ = specific heat of water = 4.184 Jg⁻¹C⁻¹

$m_{c}$ = mass of water available to calorimeter = 94.8 g

$T_{ci}$ = initial temperature of the water in calorimeter = 22.3 C

$T_{f}$ = final equilibrium temperature = 35 C

$Q_{C}$ = Heat gained by calorimeter

Using conservation of heat

Heat gained by calorimeter = Heat lost by water added - heat gained by water in calorimeter

$Q_{C} = m_{a} c_{w} (T_{ai} - T_{f}) - m_{c} c_{w} (T_{f} - T_{ci})$

$Q_{C} = (94.8) (4.184) (60.4 - 35) - (94.8) (4.184) (35 - 22.3)$

$Q_{C} = 5037.4$ J

$T$ = Change in temperature of calorimeter

Change in temperature of calorimeter is given as

$T = T_{f} - T_{ci}$

$T = 35 - 22.3$

$T = 12.7$ C

Heat capacity of calorimeter is given as

$c_{cm} = \frac{Q_{C}}{T}$

$c_{cm} = \frac{5037.4}{12.7}$

$c_{cm} = 396.65$ JC⁻¹

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3 years ago