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2 years ago

12. A sprinter has an acceleration of 5m/s” during the first 2 seconds of the race. What

Physics

1 answer:

Gala2k [10]2 years ago

7 0

Answer:

v=10m/s

Explanation:

$a=\frac{vf-vi}{t}\\ 5m/s^{2}=\frac{vf-0}{2}\\ vf=5*2\\vf=10 m/s$

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A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?

irakobra [83]

Answer:

1.2 amps :)

Explanation:

A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?

Known:

R = 10.0 Ω

V = 12.0 V

Unknown:

I = ???

I = V/R

= 12.0 V / 10.0 Ω

= 1.2 amps

6 0

2 years ago

suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain

IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

$K_i + U_i = K_f + U_f$

where :

$K_i = 0$ is the kinetic energy of the car at the top (it starts from rest)

$U_i = mgh$ is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

$U_f = 0$ is the gravitational potential energy of the car at the bottom

Re-arranging the equation, we find

$mgh=\frac{1}{2}mv^2$

and we have:

$g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s$

Solving for h, we find the initial height of the car:

$h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m$

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

5 0

3 years ago

1. Find the charge if the number of electrons is 4 x 10-18

Marianna [84]

22. The answer is 22.

4 0

3 years ago

Plzzz help me will mark brilliant

Setler79 [48]

Answer:

i hope it helped U

stay safe stay happy

3 0

3 years ago

1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s

siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

4 0

2 years ago