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3 years ago

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A jogger ran her first 100 meters in 50 seconds then ran another 150 meters in 50 seconds. What is the jogger’s average speed: W

hat is the jogger’s average speed for the first 100 meters?

1 answer:

Stella [2.4K]3 years ago

6 0

Answer:

Explanation:

Learn vocabulary, terms, and more with flashcards, games, and other study tools. ... A jogger sprints 100 m in 13 seconds. What is her average speed? 7.7 m/s ... Kathryn swam 5 complete laps of a 50 m pool. ... stands still for 4 seconds, then continues to walk for 8 meters moving away from the starting point in 6 seconds.

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Which of the following is a field (action-at-a-distance) force?

Kobotan [32]

Answer:

$F_{grav}$ is a field force, as gravity does not physically apply force and does not require proximity.

Explanation:

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3 years ago

An acorn falls from a branch located 9.8 m above the ground. After 1 s of falling, the acorn's velocity will be 9.8 m/s downward

goldenfox [79]

Answer:

The acorn hasn't hit the ground because it only falsl half of the branch distance from the ground

Explanation:

given information:

h =9.8

t =1 s

g = 9.8

the average speed

v = 1/2 gt²

= 1/2 (9.8) (1)²

= 4.8 m/s

the distance in 1s

h = v t

= 4.8 (1)

= 4.8 m

the acorn hasn't hit the ground because it only falsl half of the branch distance from the ground

=

8 0

2 years ago

What charges form a neutral charge?

brilliants [131]

A negative to a negative charge will make a neutral charge.

4 0

2 years ago

Read 2 more answers

The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a

dimaraw [331]

The absolute pressure is given by the equation,

$P_{abs}=P_{atm}-P_{vac}$

Here, $P_{abs}$ is absolute pressure, $P_{atm}$ is atmospheric pressure and $P_{vac}$ is vacuum pressure.

Therefore,

$P_{abs}=98 kPa-80 kPa=18kPa$

The gage pressure is given by the equation,

$P_{gage}=P_{abs}-P_{atm}$ .

Thus,

$P_{gage}=18kPa-98 kPa=-80 kPa$ .

In kn/m^2,

The absolute pressure,

$P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2$ .

In lbf/in2

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2$

In psi,

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi$ .

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi$

In mm Hg

The absolute pressure,

$P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg$

The gage pressure,

$P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg$

3 0

2 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

$v_{avg} = \frac{6000}{20*60}$

$v_{avg} = 5m/s$

now since effect of wind is in opposite direction then we can say

$V_{net} = v_{bird} - v_{wind}$

$5 = 9 - v_{wind}$

$v_{wind}= 4 m/s$

Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

$time = 7.7 minutes$

Part c)

Total time of round trip when wind is present

$T = t_1 + t_2$

$T = 20 + 7.7 = 27.7 min$

now when there is no wind total time is given by

$T = \frac{6000}{9} + \frac{6000}{9}$

$T = 22.22 min$

So due to wind time will be more

4 0

3 years ago