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3 years ago

How fast is the runner moving between 2-3 seconds? Is the other question please help

Physics

1 answer:

Rainbow [258]3 years ago

4 0

Answer:

Concept: Physics analysis

We have a distance vs time graph
That indicates that we are analyzing the distance traveled by either a person or a particle
The runner starts at 4 meters and over the interval of a second runs to 12 meters reaching the 16 meter mark. The runner stood at 16 meter mark for one full second and then runs back 10 meter over the course of one second reaching 6 meter. The runner runs forward again for a second traveling 6 meters, ending at the 12 meter mark.

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Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point. F⃗ 1F→1F_1_vec has a magnitude of 9.20 NN and is directed at an a

BARSIC [14]

Answer:

The x component of the resultant force is -7.27N.

Explanation:

To obtain the x component of the resultant force, first we have to know the x components of the other forces. To do this, we just have to do some trigonometry:

$|F_{1x}|=|F_1|\cos\theta_1=9.20N\cos62.0\°=4.31N \\|F_{2x}|=|F_2|\cos\theta_2=5.00N\cos53.6\°=2.96N$

Since both vectors are in the left side of the y-axis, they have a negative x component. So:

$F_{1x}=-4.31N;\\F_{2x}=-2.96N$

Finally, we sum both components to obtain the component of the resultant force:

$F_{Rx}=-4.31N-2.96N=-7.27N$

In words, the x component of the resultant force is -7.27N.

6 0

3 years ago

Thin Layer Chromatography consists of three parts: The analyte, the stationary phase, and mobile phase. Match each of these term

Margaret [11]

Answer:

Analyte⇒ one of analgesics

stationery phase⇒ silica

mobile phase⇒ solvent

Explanation:

during the thin layer chromatography non volatile mixtures are separated.The technique is performed on the plastic or aluminum foil that is coated with a thin layer.

8 0

3 years ago

The mass luminosity relation L  M 3.5 describes the mathematical relationship between luminosity and mass for main sequence sta

ivanzaharov [21]

Answer:

(a) 11.3 L

(b) 10 M

Explanation:

The mass-luminosity relationship states that:

Luminosity ∝ Mass^3.5

Luminosity = (Constant)(Mass)^3.5

So, in order to find the values of luminosity or mass of different stars, we take the luminosity or mass of sun as reference. Therefore, write the equation for a star and Sun, and divide them to get:

Luminosity of a star/L = (Mass of Star/M)^3.5 ______ eqn(1)

where,

L = Luminosity of Sun

M = mass of Sun

(a)

It is given that:

Mass of Star = 2M

Therefore, eqn (1) implies that:

Luminosity of star/L = (2M/M)^3.5

Luminosity of Star = (2)^3.5 L

Luminosity of Star = 11.3 L

(b)

It is given that:

Luminosity of star = 3160 L

Therefore, eqn (1) implies that:

3160L/L = (Mass of Star/M)^3.5

taking ln on both sides:

ln (3160) = 3.5 ln(Mass of Star/M)

8.0583/3.5 = ln(Mass of Star/M)

Mass of Star/M = e^2.302

Mass of Star = 10 M

3 0

3 years ago

Which of the following is a strength training option?

Vilka [71]

Answer:

D. All of the above

PLZ MARK ME AS BRAINLEIST ;)

5 0

3 years ago

Read 2 more answers

How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest? a.3.41 x 10^5 J b.2.73 x 10^5 J

algol13

Answer:

Work done by the frictional force is $3.41\times 10^5\ J$

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Initial velocity of car, u = 26.1 m/s

Finally, it comes to rest, v = 0

We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.

$W=k_f-k_i$

$W=\dfrac{1}{2}m(v^2-u^2)$

$W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)$

W = −340605 J

$W=3.41\times 10^5\ J$

Hence, the correct option is (a).

6 0

3 years ago