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Paladinen [302]

3 years ago

14

2 Bikers Rode at a constant speed on a 150 meter track.The data here show each bikers distance for a certain part of the race.Wh

o won the race and by how much

2 answers:

Fittoniya [83]3 years ago

7 0

Biker two won the race by 1 second ( hope this is helpful ).

kykrilka [37]3 years ago

4 0

Answer:

Student 2 won the race by 1 second

Explanation:

Took the quiz :D

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WILL GIVE BRAINLIEST!!!

klemol [59]

A simple machine can make work easier by reduce the amount of energy needed to perform a task, therefore, B. it magnifies the potential energy so that the kinetic energy is greater is the correct answer:)

6 0

4 years ago

What are the answers for numbers 2 & 3?

ivolga24 [154]

The answer is 6 i no this because im in high school this app is for high schoolers now I'll answer more. of ur questions if u make them have more points now u shouldn't play this game because ur prob in elementary but any ways the answer is 6

8 0

3 years ago

A horizontal pipe of inner diameter 2.2 cm carries water with a density of 1000.0 kg/m3 flowing at a rate of 1.5 kg/s. If the pi

EleoNora [17]

The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is, $\rm d_1 = 2.2 \ cm$

initial radius,

$r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm$

The initial crossection area;

$\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2$

The final crossection area;

$\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2$

The initial flow rate is;

R = density ×velocity ×area

$\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1 = 3.947 \ m/sec$

The speed of the water in the wider part will be;

From the continuity equation;

$\rm A_1 V_1 = A_2V_2 \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec$

Hence, the speed of the water in the wider part will be 1.194 m/sec.

To learn more about the speed, refer to the link;

brainly.com/question/7359669

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7 0

2 years ago

Problem 5 You are playing a game where you drop a coin into a water tank and try to land it on a target. You often find this gam

Dvinal [7]

Answer:

Option D: 1.5in in front of the target

Explanation:

The object distance is $y= 6in$ .

Because the surface is flat, the radius of curvature is infinity .

The incident index is $n_i=\frac{4}{3}$ and the transmitted index is $n_t= 1$ .

The single interface equation is $\frac{n_i}{y}+\frac{n_t}{y^i}=\frac{n_t-n_i}{r}$

Substituting the quantities given in the problem,

$\frac{\frac{4}{3}}{6in}+\frac{1}{y^i}= 0$

The image distance is then $y^i=-\frac{18}{4}in =-4.5in$

Therefore, the coin falls $1.5in$ in front of the target

6 0

3 years ago

How many gallons of blood does it pump? (Estimate that the heart pumps 50 cm3 of blood with each beat.)

horrorfan [7]

50 cm^3 = 0.0132086 gallons

6 0

4 years ago

Read 2 more answers