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lisov135 [29]
3 years ago
11

Someone help please.....

Physics
1 answer:
Westkost [7]3 years ago
8 0

Answer:

0.0928km/min (4dp)

Explanation:

To find the jogger's speed in km per minute, we just need to divide the number of km jogged by the time in minutes it took to jog that distance. This will give us the distance they jogged every minute which is their speed.

4km in 32 minutes:

4/32 = 0.125km/min

2km in 22 minutes:

2/22 = 0.091 (3dp)km/min

1km in 16 minutes:

0.0625km/min

Now to find the average speed of these 3 speeds, we just add them all together and divide by how many values there are (3 values).

Average (mean)  = \frac{0.125+0.091+0.0625}{3}

Average = 0.2785/3

Average speed of jogger = 0.0928 (4dp) km/min

Hope this helped!

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4 years ago
In a lab, the mass of object a is 52 kg. object a weighs
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If you are asking for the weight then the formula is F=mg where f is weight m is mass and g is acceleration due to gravity.m=52kg and g=9.8m/s2(the gravity of earth)
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3 years ago
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A swimmer is swimming to the left with a speed of 1.0 m/s when she starts to speed up with constant acceleration. The swimmer re
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Answer:

Correct answer: t = 2.86 seconds

Explanation:

We first use this formula

V² - V₀² = 2 a d    

where V is the final velocity (speed), V₀ the initial velocity (speed),

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We will calculate the acceleration from this formula

a = (V² - V₀²) / (2 d) = (2.5² - 1²) / (2 · 5) = (6.25 - 1) / 10 = 5.25 / 10

a = 0.525 m/s²

then we use this formula

V = V₀ + a t  => t = (V - V₀) / a = (2.5 - 1) / 0.525 = 1.5 / 0.525 = 2.86 seconds

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God is with you!!!

6 0
4 years ago
In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a
SOVA2 [1]

Answer:0.084 m/s^2

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance d_1=8.13 km

time taken=25 min =1500 s

initial speed v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s

after 8.13 km mark steve started to accelerate

speed after 60 s

v_2=v_1+at

v_2=5.6+a\times 60

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time taken in last part of journey

t_3=1663.6-1560=103.6 s

distance traveled in this time

d_3=v_2\times t_3

d_3=\left ( 5.6+a\times 60\right )103.6

and total distance=d_1+d_2+d_3

10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6

1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6

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4 years ago
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but I think it's a and f

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