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cestrela7 [59]
3 years ago
5

Look at the graph above. It shows how three runners ran a 100 meter race

Physics
2 answers:
Rzqust [24]3 years ago
8 0

Answer:

1. Albert

2. Charlie

3. 05 Seconds

4. 14 seconds

5. 8.33 m/s

Explanation:

The problems in the given scenarios can be solved just by analyzing the given the graph. Here the x-axis of the graph shows the time in seconds and y-axis shows the distance covered.

1. We can see that Albert finished the race in 12 seconds. This is lesser in comparison with Bob (14 seconds) and Charlie (17 seconds). Thus Albert is the winner of the race.

2. We can see that when Charlie reached at 50 meters, the time is increasing while the distance is same. It implies that he stopped there for some rest.

3. This rest was of 05 seconds.

4. Bob completed the race in 14 seconds.

5. Albert covered 100 m in 12 seconds, so his speed (v) will be,

v = \frac{100}{12}  = 8.33 m/s

anastassius [24]3 years ago
7 0

Answer:

1. Albert

2. Charlie

3. 5 seconds

4. 14 seconds

5. 8.33... meters per second

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What is the net charge of a metal ball if there are 21,749 extra electrons in it?
pickupchik [31]

Answer:

Q=3.47\times 10^{-15}\ C

Explanation:

Given that,

Number of extra electrons, n = 21749

We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is q=1.6\times 10^{-19}\ C

To find the net charge if there are n number of extra electrons is :

Q = n × q

Q=21749\times 1.6\times 10^{-19}\ C

Q=3.47\times 10^{-15}\ C

So, the net charge on the metal ball is 3.47\times 10^{-15}\ C. Hence, this is the required solution.

6 0
3 years ago
An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2. The coil rotates in a uniform magnetic f
Reptile [31]

Answer:

331.75 V

Explanation:

Given:

Number of turns of the coil, N = 40 turns

Area, A = 0.06 m²

Magnetic Field, B = 0.4 T

Frequency, f = 55 Hz

                           Maximum induce emf, E₀ = NABω

but ω = 2πf

                           Maximum induce emf, E₀ = NAB(2πf₀)

                           Maximum induce emf, E₀ = 2πNABf₀

Where;

N is number of turns of the coil

A is area

B is magnetic field

ω is the angular velocity

f is the frequency

                                     E₀ = 2 × π × 40 × 0.06 × 0.4 × 55

                                     E₀ = 342.81 V

The maximum induced emf is 331.75 V

6 0
3 years ago
How does the end point differ from the equivalence <br>point of a titration?​
Gwar [14]

<u>Answer:</u>

<em>Equivalence point and end point are terminologies in pH titrations and they are not the same. </em>

<u>Explanation:</u>

In a <em>titration the substance</em> added slowly to a solution usually through a pippette is called titrante and the solution to which it is added is called titrand. In acid-base titrations acid is added to base or base is added to acid.the strengths of the <em>acid and base titrated</em> determines the nature of the final solution.

At equivalence point the <em>number of moles of the acid</em> will be equal to the number of moles of the base as given in the equation.  The nature of the final solution determines the <em>pH at equivalence point. </em>

<em>A pH less than 7 will be the result if the resultant is acidic and if it is basic the pH will be greater than 7. </em>In a strong base-strong acid and weak base-weak acid titration the pH at the equivalence point will be 7 indicating <em>neutral nature of the solution. </em>

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Rx= 3.5 km

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The answer of this problem is positive 3
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3 years ago
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