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ANEK [815]
1 year ago
6

Table 2.4 shows how the dispacement of a runner changed during a sprint race. Draw a dispacement-time graph to show this data, a

nd use it to deduce the runner's speed in the middle of the race​
Physics
1 answer:
GalinKa [24]1 year ago
8 0
4. Table 2.4 shows how the displacement of a runner changed
during a sprint race. Draw a displacement–time graph to show
this data, and use it to deduce the runner’s speed in the middle
of the race.
Table 2.4 Data for a sprinter during a race
Displacement
(m)
0 4 10 20 50 80 105
Time (s) 1 2 3 6 9 12
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When you come within 100 ft of the uncontrolled railroad crossing and you cannot see the tracks for 400 ft in both directions, t
Sergio039 [100]
The answer to the blank above is 20. The speed limit given provided that you cannot see the tracks for 400 ft in both directions is always 20 mph. Basically, this 20 mph is decided based upon the traffic laws in order to avoid road casualties. Since it is "uncontrolled railroad crossing", the minimum speed should be implemented to slow down for the purpose of traffic calming measures. Other than this, the 20 mph is also applicable in narrowing roads as well as speed humps.
6 0
2 years ago
A rock is rolling down a hill. At position 1, it’s velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, it’s velo
mr Goodwill [35]

Answer

Hi,

correct answer is {D} 3.5 m/s²

Explanation

Acceleration is the rate of change of velocity with time. Acceleration can occur when a moving body is speeding up, slowing down or changing direction.

Acceleration is calculated by the equation =change in velocity/change in time

a= {velocity final-velocity initial}/(change in time)

a=v-u/Δt

The units for acceleration is meters per second square m/s²

In this example, initial velocity =2.0m/s⇒u

Final velocity=44.0m/s⇒v

Time taken for change in velocity=12 s⇒Δt

a= (44-2)/12  = 42/12

3.5 m/s²

Best Wishes!

5 0
2 years ago
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
yulyashka [42]

Answer:

178200 g mile pounds

Explanation:

Work= Force * Distance= Fh

F=ma=mg where m is mass and g is acceleration due to gravity

Work= 165 pounds *g* 1080 m=  178200 g mile pounds

5 0
2 years ago
A reflecting telescope is built with a 20-cm-diameter mirror having a 1.00 m focal length. it is used with a 10× eyepiece.
kondaur [170]
I really wish I could be helping you. I don't know.
3 0
2 years ago
Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,
balandron [24]

Answer with Explanation:

We are given that

Radius of  solid core wire=r=2.28 mm=2.28\times 10^{-3} m

1mm=10^{-3} m

Radius of each strand  of thin wire=r'=0.456 mm=0.456\times 10^{-3} m

Current density of each wire=J=3750 A/m^2

a.Area =\pi r^2

Where \pi=3.14

Using the formula

Cross section area of copper wire has solid core =3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2

Current density =J=\frac{I}{A}

Using the formula

3750=\frac{I}{16.3\times 10^{-6}}

I=3750\times 16.3\times 10^{-6}=0.061 A

Total number of strands=19

Area of strand wire=A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2

J'=\frac{I'}{A'}

3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}

I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2

I'=0.047 A

b.Resistivity of copper wire=\rho=1.69\times 10^{-8}\Omega-m

Length of each wire =6.25 m

Resistance, R=\frac{\rho l}{A}

Using the formula

Resistance of solid core wire=R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega

Resistance of strand wire=R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega

7 0
2 years ago
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