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umka21 [38]
3 years ago
13

The proper length of the first lane of a competitive running track is 400 m (1,312.3 ft). A runner runs track from the starting

line four times, stopping at the starting line.
Which statement is TRUE?

Group of answer choices

The runner's distance is 1600 m, his displacement is 0 m, and his speed is 0 m/s.

The runner's distance is 1600 m, his displacement is 0 m, and his velocity is 0 m/s.

The runner's displacement is 1600 m, his distance is 0 m, and his velocity is 0 m/s.

The runner's distance and displacement are 1600 m and his velocity cannot be calculated given this information.
Physics
2 answers:
charle [14.2K]3 years ago
6 0

Answer:

The runner's distance is 1600 m, his displacement is 0 m, and his velocity is 0 m/s.

Explanation:

It is given that,

Length of the truck, l = 400 m

A runner runs track from the starting line four times, stopping at the starting line. In this case, the distance covered by the runner is, 4 l i.e. 4 × 400 m = 1600 meters.

Distance covered by an object is the overall path travelled during its entire journey. So, runner's distance is 1600 meters.

We know that the displacement of an object is equal to the shortest path covered or it is equal to the difference between final position and initial position. As a result, the displacement of runner is 0 as he is coming back to starting line.

We know that the net displacement divided by total time taken is called velocity of an object. So, velocity will be 0.

Hence, the correct option is (b) "The runner's distance is 1600 m, his displacement is 0 m, and his velocity is 0 m/s".

irakobra [83]3 years ago
4 0

Answer:

The runner's distance is 1600 m, his displacement is 0 m, and his speed is 0 m/s.

Explanation:

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. In this case distance is 400m x 4= 1600m. Displacement is a vector quantity that refers to the distance betwen the first and the last position of an object. In this case is 0 because the runer always track in the same place.

Finally, the velocity is 0 m/s because the runner always tracks a point, and in order to measure the velocity we need a distance between to points and a time.

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Two point charges are placed on the x-axis as follows: charge q1 = 3.95 nC is located at x= 0.198 m , and charge q2 = 4.96 nC is
Yuki888 [10]

Answer:

F = 2.40 × 10^{-6}  N

Explanation:

given data

charge q1 = 3.95 nC

x= 0.198 m

charge q2 = 4.96 nC

x= -0.297 m

solution

force on a point charge kept in electric field F = E × q       ................1

here E is the magnitude of electric field and q is the magnitude of charge

and

first we will get here electric field at origin

So net field at origin is

E = (Kq2÷r2²) - (kq1÷r1²)           ...............2

put here value

E = 9[(4.96÷0.297²)-(3.95÷0.198²)]

E = 400.72 N/C        ( negative x direction )

so that force will be

F = 6 × 10^{-9} × 400.72

F = 2.40 × 10^{-6}  N

7 0
3 years ago
The dentisty of an object relies on both the _________ of the object.
Daniel [21]

Answer:

Density is determined by an object's mass and volume.

Explanation:

If two objects take up the same volume, but one has more mass, then it also has a higher density.

7 0
2 years ago
The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c
maria [59]

To solve this problem we will start using the concepts related to the electric field, from there we will find the load exerted on the body. Through this load it will be possible to make a sum of forces in balance to find the load that a human supports. Finally with these values it will be possible to find the repulsive force. We will proceed as follows,

The electric field is

E= \frac{kQ}{R^2}

Here,

k = Coulomb's Constant

Q = Charge

R = Distance (At this case from the center of mass of the earth to the surface)

Rearranging to find the charge,

Q = \frac{ER^2}{k}

Replacing,

Q = frac{(150)(6.38*10^6)}{8.99*10^9}

Q = 6.79*10^5 C

Since the electric field is directed towards the center of earth, the charge is negative.

PART A) Once the load is found we can proceed to apply the balance of Forces, for which the electrostatic force must be equivalent to the weight, this in order to satisfy the balance, therefore

F_w = F_e

mg = \frac{kQq}{R^2}

Replacing,

(62)(9.8) = \frac{(8.99*10^9)(q)(-6.79*10^5)}{(6.38*10^6)^2}

Solving for q,

q = -4.056C

PART B) Finally using the given distance and the values of the found load we can find the repulsive Force, which is

F =\frac{kq^2}{d^2}

F = \frac{(8.99*10^9)(-4.056)^2}{110^2}

F = 1.22*10^7N

PART C) The answer is no. According to the information found, we can conclude that traveling through an electric field is not viable because there is a repulsive force of great magnitude acting on the body.

3 0
3 years ago
What is the relationship between work and power
Mashcka [7]
Power is the rate work is usually done in.
3 0
3 years ago
Read 2 more answers
17.Explain the different ways that an object can become electrically charged.
Debora [2.8K]

17.

There are three different methods for charging objects:

- Friction: in friction, two objects are rubbed against each other. As a result, electrons can be passed from one object to the other, so one object will gain a net negative charge while the other object will gain a net positive charge due to the lack of electrons.

- Conduction: this occurs when two conductive objects are put in contact with each other, and charges (electrons, usually) are transferred from one object to the other one.

- Induction: this occurs when two objects are brought closer to each other, but not in contact. If one of the two objects has a net charge (different from zero) on its surface, then it will induce a movement of charges in the second object: in particular, in the second object, charges of the opposite polarity will be attracted towards the first object, while charges of same polarity will be repelled further away.

18.

Charged objects produce around themselves an electric field. The strenght of the electric field is given by (assuming the charged objects are spherical)

E=k\frac{q}{r^2}

where k is the Coulomb's constant, q is the magnitude of the charge and r the distance from the centre of the charge. As we see, the strength of the field is inversely proportional to the square of the distance.

Also, the direction of the field is determined by the sign of the charge:

- if the charge is positive, the electric field points away from the charge (this means that other positive charges in the field will be repelled away)

- if the charge is negative, the electric field points towards the charge (this means that other positive charges in the field will be attracted towards it)

19.

Electrical force is given by:

F=k\frac{q_1 q_2}{r^2}

where k is the Coulomb's constant, q1 and q2 are the two charges, and r their separation.

Gravitational force is given by:

F=G\frac{m_1 m_2}{r^2}

where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r their separation.

Similarities between the two forces:

- Both are inversely proportional to the square of the distance between the two objects, r

- Both are non-contact forces (the two objects can experience the forces even if they are not in contact)

- Both forces have infinite range

Differencies between the two forces:

- The electric force can be either attractive or repulsive, while the gravitational force is attractive only

- The electric force is much stronger than the gravitational force, due to the much larger value of the Coulomb's constant k compared to the gravitational constant G

4 0
3 years ago
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