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3 years ago

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A man works on a striaght road from his home to market 2.5km away dwith a speed of 5km/h.Finding the market closed, he instantly

truns and walk back home with a speed of 7.5km/h. The average speed of the man over the interval of time 0.40min.Is equal to?

1 answer:

umka21 [38]3 years ago

7 0

Answer:

5.625km/h

Explanation:

We are given that

Distance between home and market, d=2.5 km

Speed, v1=5km/h

Speed, v2=7.5 km/h

We have to find the average speed of the man over the interval of time 0 to 40min.

Time, $t=\frac{distance}{speed}$

Using the formula

$t_1=\frac{2.5}{5}=0.5 h=0.5\times 60=30 min$

1 hour =60 min

$t_2=\frac{2.5}{7.5}=\frac{1}{3}hour=\frac{60}{3}=20min$

Distance traveled by man in 10 min with speed 7.5 km/h= $\frac{2.5}{20}\times 10=2.5/2km$

Therefore,

Total distance covered=2.5+2.5/2=3.75 km

Time=40 min=40/60=2/3 hour

Average speed= $\frac{total\;distance}{total\;time}$

Average speed= $\frac{3.75}{2/3}=5.625km/h$

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Answer:

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Explanation:

Given that

T₁ = 745 K

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a)

The entropy change of reservoir 745 K

$S_1=-\dfrac{7190}{745}\ J/K$

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c)

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A sportscar has a mass of 1500 kg and accelerates at 5 meters per second squared. What is the magnitude of the force acting on t

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3 years ago

An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl

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We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 0 m/s

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Acceleration, a = 9.81 m/s²

Substituting

s = ut + 0.5 at²

192 = 0 x t + 0.5 x 9.81 x t²

t = 6.26 seconds

Now we need to find horizontal distance traveled in 6.26 seconds by the package.

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 41 m/s

Time, t = 6.26 s

Acceleration, a = 0 m/s²

Substituting

s = ut + 0.5 at²

s = 41 x 6.26 + 0.5 x 0 x 6.26²

s = 256.52 m

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b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

Initial velocity, u = 0 m/s

Time, t = 6.26 s

Acceleration, a = 9.81 m/s²

Substituting

v = u+ at

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Vertical component of velocity = 61.41 m/s

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where

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$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

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The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago