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yan [13]
3 years ago
15

in a 4 kilometer race, a runner completes the first kilometer in 5.9 minutes, the second kilometer in 6.2 minutes, the third kil

ometer in 6.3 minutes, and the final kilometer in 6 minutes. the average speed of the runner for the race is approximately
Physics
2 answers:
ohaa [14]3 years ago
5 0
To find the average of data collected, add all of the measurements: 5.9+6.2+6.3+6= 12.2

Then, divide the total amount by the number of data collected which is 4: 12.2/4= 3.05

The average speed of the runner of the race is approximately 3.05 km/min

Feel free to ask me any other questions you might have :)
Kruka [31]3 years ago
5 0

Answer:

9.83 km/h

Explanation:

Given: in a 4 kilometer race, a runner completes the first kilometer in 5.9 minutes, the second kilometer in 6.2 minutes, the third kilometer in 6.3 minutes, and the final kilometer in 6 minutes.

To Find:  the average speed of the runner for the race.

Solution:

In a 4 kilometer race,

time required to travel first kilometer = 5.9 \text{minutes}

time required to travel seoond kilometer = 6.2 \text{minutes}

time required to travel third kilometer = 6.3 \text{minutes}

time required to travel final kilometer = 6 \text{minutes}

we know that,

\text{Average speed}=\frac{\text{Total distance traveled}}{\text{Total time taken}}

Total distance traveled = 4  \text{kilometer}

Total time taken = (5.9+6.2+6.3+6)

                           =  24.4 \text{minutes}

                           = \frac{24.4}{60}  \text{hours}

putting values,

\text{Average speed}=\frac{4}{\frac{24.4}{60}}

 \text{Average speed}=\frac{4\times60}{24.4}

                                            =9.83 \text{km}/ \text{h}

the average speed of the runner for the race is approximately 9.83 \text{km}/ \text{h}

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During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between
Whitepunk [10]

Answer:

a. 12.12°

b. 412.04 N

Explanation:

Along vertical axis, the equation can be written as

T_1 sin14 + T_2sinA = mg

T_2sinA = mg - T_1sin12.5           ....................... (a)

Along horizontal axis, the equation can be written as

T_2×cosA = T_1×cos12.5    ......................... (b)

(a)/(b) given us

Tan A = (mg - T_1sin12.5) / T_1 cos12.5

 = (176 - 413sin12.5) / 413×cos12.5

A = 12.12 °

(b) T2 cosA = T1 cos12.5

T2 = 413cos12.5/cos12.12

= 412.04 N

4 0
3 years ago
Read 2 more answers
The atlas stone’s is a strong man competition where athletes have to load 5 stones of masses 100kg, 120kg, 140kg, 160kg and 180k
lyudmila [28]

Answer: 6250 joules

Explanation:

The work needed to lift an object of mass M by a height H is equal to:

w = M*g*H

where h = 10m/s^2

then the total work that he did is equal to the sum of the work for every stone:

W = (100kg*g*H) + (120kg*g*H) + (140kg*g*H) + (160kg*g*H) + (180kg*g*H)

 = (100kg + 120kg + 140kg + 160kg + 180kg)*g*H

= (500kg)*g*H

and now we can repalce g by 10m/s^2 and H by 125cm

But you can notice that we have two different units of distance, so knowing that 100cm = 1m

we can write H =  125cm = (125/100) m = 1.25 m

Then we have:

H = 500kg*10m/s^2*1.25m = 6250 J

3 0
2 years ago
WILL GIVE BRAINLIEST ASAP
Sedaia [141]

Explanation:

1. Height Relatives to reference point, Mass, and strength of the gravitational field it's in

2. Distance in the magnetic field

8 0
2 years ago
A horizontal force of 140 N is needed to pull a 60kg box across the horizontal floor at constant speed . Find the coefficient of
____ [38]

Explanation:

formula for force is:

force=mass × acceleration

but in case of friction

force =coefficient of friction × Normal Reaction

F. = u × R

U = F/R

but when placed horizontally

R= M×G

M=mass=60kg

G=Gravity(10m/s or 9.8m/s)

F=140N

U=140/60×10

U=140/600

U=0.2333333333

approximately to 3 significant figures

U=0.233

if i am correct rate it 5 star

4 0
1 year ago
the total positive charge is QQQ = 1.62×10−6 CC , what is the magnitude of the electric field caused by this charge at point P,
balu736 [363]

Answer:

6.1 × 10^9 Nm-1

Explanation:

The electric field is given by

E= Kq/d^2

Where;

K= Coulombs constant = 9.0 × 10^9 C

q = magnitude of charge = 1.62×10−6 C

d = distance of separation = 1.53 mm = 1.55 × 10^-3 m

E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2

E= 14.58 × 10^3/2.4 × 10^-6

E= 6.1 × 10^9 Nm-1

8 0
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