Which situation will change the direction of the bicycle?

A long-distance runner is running at a constant speed of 5 m/s.

vivado [14]

Answer:

3.33 minutes (3 minutes and 20 seconds)

Explanation:

Speed of the runner = s = 5 m/s

We need to calculate how will it take for runner to complete 1 km. We have the speed, the distance and we need to find the time. Before performing any calculations, we must convert the values to same units.

Speed is in m/s and distance is in kilometers. So we have to either convert speed to km/s or distance into meters. In this case, converting distance into meters would be a convenient option.

1 kilo meters = 1000 meters

The distance, speed and time are related by the equation:

Distance = Speed x Time

So,

Time = Distance/Speed

Using the values, we get:

t = 1000/5

t = 200 seconds

This means, the runner can complete 1 kilometers in 200 seconds. Since, there are 60 seconds in a minute, we can convert this time to minutes, by dividing it by 60. i.e.

$200 \text{ sec} = \frac{200}{60} \text{ min} = 3.33 \text{ min}$

Thus, it will take the runner 3.33 minutes (3 minutes and 20 seconds) to travel 1 km.

3 0

2 years ago

With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release

Dmitry [639]

THIS IS THE COMPLETE QUESTION BELOW

With what minimum speed must you toss a 130 g ball straight up to just touch the 14-m-high roof of the gymnasium if you release the ball 1.1 m above the ground

And what speed does the ball hit the ground? Solve this problem using energy.

Answer

a)minimum speed must you toss a 130 g is 15.9090m/s

b)speed the ball hit the ground is 16.57m/s

Explanation:

a)We know that For any closed/isolated system, the total energy is CONSERVED.

K.E. lost by the ball=The change in P.E of the ball at 1.1m above the ground as well as the P.E. of the ball at 14 m-high roof

This statement can be expressed as the expression below from K.E and P.E energy formula

P.E. = mgh

K.E. = (1/2)mv^2

Therefore,

(mgh1 - mgh2)=(1/2)mv^2

Where h2=the ball height above the ground=1.1m

h1=ball height at roof of the gymnasium= 14m

Then if we substitute we have

[(10) x (0.14) x (9.81)] - [(1.1) x (0.14) x (9.81)] = (1/2)(0.14)(v^2)

16.45137=0.065V^2

V=15.9090m/s

minimum speed must you toss a 130 g is 15.9090m/s

b)To calculate the speed the ball hit the ground?

This is the highest point (14m-high roof),and the type of the energy the ball possesses is Po.tential energy only.

At the lowest point (ground), the energy the ball possesses is K.E. only.

P.E at 10m-high roof = K.E. at ground.

(14) x (0.13) x (9.81) = (1/2) x (0.13) x v^2

17.8542= 0.065V^2

V= 16.57

Therefore,And speed the ball hit the ground is 16.57m/s

6 0

3 years ago

A 3.53-g sample of aluminum completely reacts with oxygen to form 6.67 g of aluminum oxide. find percent composition

katrin2010 [14]

The mass percent composition of aluminum is 52.9% in aluminum oxide.

Mass of the aluminum = 3.53 g

Mass of the aluminum oxide = 6.67 g.

The mass percent of a substance is the mass of the substance divided by the mass of the compound into 100.

Aluminum reacts with oxygen to form aluminum oxide.

The overall balanced equation for the reaction is,

$Aluminum + Oxygen→Aluminum \: oxygen$

$4Al +3O _{2}→2Al _{2}O _{3}$

The mass percent composition of aluminum in the aluminum oxide is,

$Mass \: percent = \frac{ Mass \: of \: the \: substance}{ Mass \: of \: the \: compound} \times 100$

$Mass \: percent = \frac{Mass \: of \: the \: aluminum}{ Mass \: of \: the \: aluminum \: oxide } \times 100$

$Mass \: percent = \frac{3.53}{6.67 } \times 100$

= 52.9 %

Therefore, the mass percent composition of aluminum is 52.9% in aluminum oxide.

To know more about aluminum oxide, refer to the below link:

brainly.com/question/25869623

#SPJ4

7 0

1 year ago

Please help!!!

Dominik [7]

Answer:

i Believe the correct answer is "An open circuit being changed into a closed circuit"

Explanation:

8 0

3 years ago

How know that rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the arkose sandstone

svp [43]

The rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the Arkose sandstone because the Amadeus basin were made up of marine and non-marine sedimentary rocks which are softer compared to quarts which make up mostly the Arkose sandstone.

7 0

3 years ago

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