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2 years ago

An athlete runs four laps of a 400 m track. What is the athlete's total displacement?

Physics

1 answer:

sasho [114]2 years ago

7 0

Answer:

○ 0 $\checkmark$

Explanation:

Even if the athelete runs four laps of a 400 m track, its displacement will be 0 because the displacement is the shortest distance from its to final position. And, here the final and initial position is same since he comes to its initial position after covering certain distance. So, displacement is 0.

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Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured

Mamont248 [21]

Answer:

74.86°C

Explanation:

P₂ = Vapour pressure of water at sea level = 760 mmHg

P₁ = Pressure at base camp = 296 mmHg

T₂ = Temperature of water = 373 K

ΔH°vap for H2O = 40.7 kJ/mol = 40700 J/mol

R = Gas constant = 8.314 J/mol K

From Claussius Clapeyron equation

$ln\frac{P_2}{P_1}=\frac{\Delta H}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\\\Rightarrow ln\frac{760}{296}=\frac{40700}{8.314}\left(\frac{1}{T_1}-\frac{1}{373}\right)\\\Rightarrow ln\frac{760}{296}\times \frac{8.314}{40700}+\frac{1}{373}=\frac{1}{T_1}\\\Rightarrow 0.0028735=\frac{1}{T_1}\\\Rightarrow T_1=347.996\ K$

T₁ = 347.996 K = 74.86°C

∴Water will boil at 74.86°C

4 0

3 years ago

In a clock, the average speed is maximum for the tip of a) Seconds hand b) Hours hand c) Minutes hand

Simora [160]

A, hope this helped! I didn’t really get it but I think it’s correct?

3 0

3 years ago

Read 2 more answers

If the weight of the ruler is one Newton ,Gc cannot have a value more than 25cm

nevsk [136]

If the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm.

The given parameters:

Weight of the ruler = 1 N

<h3>What is center of gravity (CG)?</h3>

Center of gravity is the point at which the weight of an object is concentrated.

Let the length of the ruler = L

The center of the gravity of the ruler is calculated as follows;

$X_{CG} = \frac{W(L_0) + W(L -X_{CG})}{W} \\\\X_{CG} = \frac{1(0) + 1(L -X_{CG})}{1}\\\\X_{CG} = L - X_{CG}\\\\X_{CG } + X_{CG} = L\\\\2X_{CG} = L\\\\X_{CG} = \frac{L}{2} \\\\when , \ L = 50 \ cm\\\\X_{CG} = \frac{50}{2} \\\\X_{CG} = 25 \ cm$

Thus, if the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm. This may change if the length of the ruler changes because the center of gravity of uniform ruler depends on the length of the ruler.

Learn more about center of gravity here: brainly.com/question/6765179

4 0

1 year ago

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

$\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W$

$\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})$

$h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}$

$2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]$

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0

3 years ago

A particle moves along a straight line with an acceleration of a = 5>(3s 1>3 + s 5>2) m>s2, where s is in meters. De

Len [333]

Answer:

v=1.295

Explanation:

What we are given:

a=5÷(3s^(1/3)+s^(5/2)) m/s^2

Start by using equation a ds = v dv

This problem requires a numeric method of solving. Therefore, you can integrate v ds normally, but you must use a different method for a ds The problem should look like this:

$\int\limits^a_b {x} \, dx$

a=2

b=1

x=5÷(3s^(1/3)+s^(5/2)) m/s^2

dx=dv

Integrate the left side the standard method.

$\int\limits^a_b {x} \, dx$

a=v

b=0

dx=dv

Integrating

=v^2/2

Use Simpson's rule for the right site.

$\int\limits^a_b {x} \, dx$

a=b

b=a

x=f(x)

f(x)=b-a/6*(f(a)+4f(a+b/2)+f(b)

If properly applied. you should now have the following equation:

v^2/2=5[(1/6*(0.25+4(0.162)+(0.106)]

=0.8376

Solve for v.

v=1.295

5 0

2 years ago

An athlete runs four laps of a 400 m track. What is the athlete's total displacement?​

An athlete runs four laps of a 400 m track. What is the athlete's total displacement?