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3 years ago

An athlete runs four laps of a 400 m track. What is the athlete's total displacement?

Physics

1 answer:

sasho [114]3 years ago

7 0

Answer:

○ 0 $\checkmark$

Explanation:

Even if the athelete runs four laps of a 400 m track, its displacement will be 0 because the displacement is the shortest distance from its to final position. And, here the final and initial position is same since he comes to its initial position after covering certain distance. So, displacement is 0.

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In a given chemical reaction, the energy of the products is greater than the energy of the reactants. Which statement is true fo

Flura [38]

Energy is released in the reaction

Explanation:

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4 years ago

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An arch carries the thrust of weight to its _____(1)______. With a _____(2)______, the horizontal part of the structure supports

finlep [7]

Option D is correct. An arch carries the thrust of weight to its sides with a post-and-lintel.

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3 years ago

What's the difference between meteoroids, meteorites, and meteors?

SVEN [57.7K]

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4 years ago

a parking lot is going to be 60m wide and 240m long which dimensions could be used please please help

Bezzdna [24]

Answer: The area of the parking lot is 14,400 meters squared.

Explanation:

We have the dimensions of the parking lot.

60m by 240m

The units used here are meters.

Now, if we want to know the area of the parking lot is equal to the product between the length and the width:

A = 60m*240m = 14,400 m^2

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3 years ago

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A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106

LenKa [72]

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

$E_2 = \frac{E_1q_1}{q_2}$

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

$E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C$

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

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3 years ago

An athlete runs four laps of a 400 m track. What is the athlete's total displacement?​

An athlete runs four laps of a 400 m track. What is the athlete's total displacement?