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skad [1K]
2 years ago
12

In Michael Johnson's world-record 400 m sprint, he ran the first 100 m in 11.20 s; then he reached the 200 m mark after a total

time of 21.32 s had elapsed, reached the 300 m mark after 31.76 s,and finished in 43.18 s.
During what 100 m segment was his speed the highest?

a) Between the start and the 100 m mark.

b) Between the 100 m mark and the 200 m mark.

c) Between the 200 m mark and the 300 m mark.

d)Between the 300 m mark and the finish.
Physics
1 answer:
GrogVix [38]2 years ago
6 0

Answer:

b)

Explanation:

Assuming that we are talking about average speed during any segment, we can apply the definition of average speed, as follows:

v(avg) = Δx / Δt = (xf-x₀) / (tfi-t₀)

Using this definition for the 4 segments, we have:

1) v(0-100m) = 100 m / 11.20 sec = 8.93 m/s

2) v(100m-200m) = 100 m / (21.32 s - 11.2 s) = 100 m / 10.12 s = 9.88 m/s

3) v(200m -300m) = 100 m / (31.76 s- 21.32s) = 100 m / 10.44 s = 9.58 m/s

4) v(300m-400m) = 100 m / (43.18 s - 31.76 s) = 100 m / 11.42 s = 8.76 m/s

As we can see, the highest speed was reached between the 100m mark and the 200m mark, so the statement b) is the one that results to be true.

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A power washer is being used to clean the siding of a house. Water enters at 20 C, 1 atm, with a volumetric flow rate of 0.1 lit
VladimirAG [237]

Answer:

W=1259W=1.2Kw

Explanation:

Hello!

The first step to solve is to find the enthalpies at the entrance (state 1) and the exit of the washer, for this we use the thermodynamic tables.

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

state 1

h1(T=20C,P=1atm)=83.93kJ/kg

state 2

h1(T=23C,P=1atm)=96.48kJ/kg

now we find the mass flow remembering that it is the product of the flow rate by the density=

m= mass flow

M=\alpha q

q=flow=0.1l/S=0.0001M^3/s

α=Density=1000kg/m^3

m=0.1kg/s

Now we draw the energy flows in the washer (see attached image) and propose the first law of thermodynamics that states that the energy that enters must be equal to the one that comes out

mh1+W+\frac{v1^2}{2g} =0.1W+mh2+\frac{v2^2}{2g}+mgH

where

m=mass flow

h=entalpy

v=speed

W=power input

g=gravity

H=height

SOLVING FOR W

W=\frac{m2h2-m1h1+m\frac{v2^2-v1^2}{g}+mgH}{1.1}

W=\frac{(0.1)(96480)-(0.1)(83930)+\frac{0.1(50^2-0.2^2)}{2}+0.1(9.81)(5)}{1.1}

W=1259W=1.2Kw

5 0
2 years ago
In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to
Salsk061 [2.6K]

Explanation:

1.2 \mathrm{N} ; 2 \mathrm{N}

2.200 \mathrm{N} ; 200 \mathrm{N}

4.2 \mathrm{N} ; 2 \mathrm{N} ; 4 \mathrm{N}

5.2 \mathrm{N} ; 2 \mathrm{N} ; 2 \mathrm{N}

6.2 \mathrm{N} ; 2 \mathrm{N} ; 3 \mathrm{N}

8.200 \mathrm{N} ; 200 \mathrm{N} ; 5 \mathrm{N}

In only the above cases (i.e 1,2,4,5,6,8 ) the object possibly moves at a constant velocity of 256 \mathrm{m} / \mathrm{s}

You should have noticed that the sets of forces applied to the object are the same asthe ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes nodistinction between the state of re st and the state of moving at a constant velocity(even a high velocity).

In both cases, the net force applied to the object must equal zero.

7 0
2 years ago
Paula is studying two different animals. Both animals are classified within the same genus, but they are different species. Base
Ksivusya [100]

Answer:c

Explanation:

i think

6 0
2 years ago
Read 2 more answers
At time t=0 , a cart is at x=10 m and has a velocity of 3 m/s in the −x -direction. The cart has a constant acceleration in the
DENIUS [597]

Answer:

Explanation:

The minimum magnitude of acceleration = 3 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 3 x 1²

= - 3 + 1.5

= - 1.5 m

position at t = 1 s

= 10 - 1.5

= 8.5 m

The maximum  magnitude of acceleration = 6 m /s²

displacement at t = 1

s = ut + 1 /2 at²

= -3 x 1 + .5 x 6 x 1²

= - 3 + 3

=  0

position at t = 1 s

= 10 +0

= 10  m

So range of position is 8.5 m to 10 m .

7 0
2 years ago
Disturbed by speeding cars outside his workplace, Nobel laureate Arthur Holly Compton designed a speed bump (called the "Holly h
Bezzdna [24]
:<span>  </span><span>30.50 km/h = 30.50^3 m / 3600s = 8.47 m/s 

At the top of the circle the centripetal force (mv²/R) comes from the car's weight (mg) 

So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road .. 

Fn = mg - mv²/R 
Fn = m(g - v²/R) .. .. 1800kg (9.80 - 8.47²/20.20) .. .. .. ►Fn = 11 247 N (upwards) 
(b) 
When the car's speed is such that all the weight is needed for the centripetal force .. then the net downward force (Fn), and the reaction from the road, becomes zero. 

ie .. mg = mv²/R .. .. v² = Rg .. .. 20.20m x 9.80 = 198.0(m/s)² 

►v = √198 = 14.0 m/s</span>
3 0
2 years ago
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