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nikitadnepr [17]

3 years ago

14

You are at a circus and you see a stunt man climb up 29.4 meters into a cannon. He gets fired horizontally out of the cannon wit

h a speed of 57.1 m/s. How long was stunt man in the air for?

1 answer:

klemol [59]3 years ago

8 0

Answer:

t = 2.44 s

Explanation:

Given that,

A stunt man climb up 29.4 meters into a cannon. He gets fired horizontally out of the cannon with a speed of 57.1 m/s.

We need to find the time for which was the stunt man in the air. Let it is t. It can be calculated using second equation of motion as follows :

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

u is initial speed and it is 0

$s=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2s}{g}} \\\\t=\sqrt{\dfrac{2\times 29.4}{9.8}} \\\\t=2.44\ s$

So, the stunt man is in the air for 2.44 seconds.

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velikii [3]

<u>Answer</u>:

The coefficient of static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

<u>To Find:</u>

The coefficient of static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

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A grinding wheel is a uniform cylinder with a radius of 8.65 cm and a mass of 0.400 kg . Calculate its moment of inertia about i

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Answer:

I = 1.5*10⁻³ kg*m²

Explanation:

It can be showed that the moment of inertia (or rotational inertia) for a uniform cylinder of mass m and radius r, respect an longitudinal axis going through its center (parallel to the height of the cylinder) can be written as follows:

$I = \frac{1}{2}*m*r^{2} = \frac{1}{2}*0.400 kg*(0.0865m)^{2} = 1.5e-3 kg*m2$

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Answer:

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Explanation:

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A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

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