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4 years ago

Please help! I especially need help with the second question but help with the first one would be most appreciated!

Physics

1 answer:

lara31 [8.8K]4 years ago

4 0

Answer:

a) Team A will win.

b) The losing team will accelerate towards the middle line with 0.01 m/ $s^{2}$

Explanation:

Given that Team-A pulls with a force , $F_{1} = 50N$

and Team-B pulls with a force , $F_{1} = 45N$

∵ $F_{1} > F_{2}$

The rope will move in the direction of force $F_{1}$ .

∴ Team-A will win.

b) Considering both the teams as one system of total mass , $m = 246+253 = 499 kg$

Net force on the system , $F = F_{1} - F_{2}$ = 50-45 = 5N

Applying Newtons first law to the system ,

F = ma , where 'a' is the acceleration of the system.

Since , both the teams are connected by the same rope , their acceleration would be the same.

∴ 5 = 499×a

∴ a = 0.01 m/ $s^{2}$

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3 years ago

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3 years ago

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3 years ago

Scenario 2: Use the following information to answer questions 3 and 4:

NemiM [27]

Answer:

A. 52 min

.A. 47 watts

Explanation:

Given that;

jim weighs 75 kg

and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.

Using the following relation to determine the amount of calories burned per minute while walking; we have:

$\dfrac{MET*weight (kg)*3.5}{200}$

here;

MET = energy cost of a physical activity for a period of time

Obtaining the data for walking with a speed of 3.3 mph From the standard chart for MET, At 3.3 mph; we have our desired value to be 4.3

However;

the calories burned in a minute = $\dfrac{4.3*75 (kg)*3.5}{200}$

= 5.644

Therefore, for walking for 52 mins; Jim burns approximately 293.475 kcal which is nearest to 300 kcal.

Given that:

mass m = 75 kg

intensity = 6 kcal/min

The eg ergometer work rate = ??

Applying the formula:

$V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7$

where ;

$V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}$

$V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}$

$V_O_2 ( intensity ) = 0.0012$

∴ $0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min$

Converting to watts;

Since; 6.118kg-m/min is = 1 watt

Then 291.66 kgm /min will be equal to 47.67 watts

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3 0

4 years ago

A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m verticall

lorasvet [3.4K]

Explanation:

Given:

t = 20 seconds

x = 3000 m

y = 450 m

a) To find the vertical component of the initial velocity $v_{0y}$ , we can use the equation

$y = v_{0y}t - \frac{1}{2}gt^2$

Solving for $v_{0y}$ ,

$v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}$

$\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}$

$\:\:\:\:\:\:\:=120.5\:\text{m/s}$

b) We can solve for the horizontal component of the velocity $v_{0x}$ as

$x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}$

$v_{0x} = 150\:\text{m/s}$

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3 years ago