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olya-2409 [2.1K]
3 years ago
6

A world class sprinter can accelerate at 5.0 m/s squared, from the rest in the starting blocks to their top speed in 4.0 seconds

. What distance will such a sprinter cover during that time?
Physics
1 answer:
Airida [17]3 years ago
4 0

Answer:

The distance covered by the sprinter, s = 40 m

Explanation:

Given data,  

The initial velocity of the sprinter, u = 0 m/s

The acceleration of the sprinter, a = 5 m/s²

The time period of acceleration of the sprinter, t = 4 s

Using the II equations of motion

                   s = ut + ½ at²

                       = 0 + ½ (5) (4)²

                       = 40 m

Hence, the distance covered by the sprinter, s = 40 m                    

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Potential energy is the answer
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Answer: The motion of the object will remain the same

Explanation:

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If grade 7 students learn six periods and if one period is 40 minutes then how many seconds they learn per day
Ludmilka [50]

Answer:

C. 14400

Explanation:

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2. A brick is sitting on a building 22 m high. It has a mass of 7.9 kg. What amount of potential
Citrus2011 [14]

Answer:

1703.24J

Explanation:

Given parameters:

Mass of brick  = 7.9kg

Height of building  = 22m

Unknown:

Potential energy of the brick  = ?

Solution:

The potential energy of a body is the energy at rest of the body. Mathematically;

 

         P.E = mgh

m is the mass of the brick

g is the acceleration due to gravity

h is the height of the building

  Insert the given parameters and solve;

        P.E  = 7.9 x 9.8 x 22  = 1703.24J

6 0
3 years ago
A ray of red light in air is incident at an angle of 30. on a
Vlad [161]

Answer:

20 degrees.

Explanation:

From Snell’s law of refraction:

sinθ1•n1 = sinθ2•n2

where θ1 is the incidence angle, θ2 is the refraction angle, n1 is the refraction index of light in medium1, and n2 is the refraction index for virgin olive oil. The incidence angle of the red light is θ1 = 30 degrees.

The red light is in air as medium1, so n1 (air) = 1.00029

So, to find θ2, the refracted angle:

sinθ1•1.00029 = sinθ2•1.464

sin(30)•1.00029 / 1.464 = sinθ2

0.5•1.00029 / 1.464 = sinθ2

sinθ2 = 0.3416291

θ2 = arcsin(0.3416291)

θ2 = 19.976 degrees

To the nearest degree,

θ2 = 20 degrees.

8 0
3 years ago
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