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2 years ago

12

a sprinter's whose mass is 70 kg accelerstes from rest to 11 m/s in 6 seconds what is the acceleration what force did this requi

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1 answer:

mr Goodwill [35]2 years ago

6 0

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5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A)

jolli1 [7]

Answer:

(B) 13.9 m

(C) 1.06 s

Explanation:

Given:

v₀ = 5.2 m/s

y₀ = 12.5 m

(A) The acceleration in free fall is -9.8 m/s².

(B) At maximum height, v = 0 m/s.

v² = v₀² + 2aΔy

(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)

y = 13.9 m

(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.

v = at + v₀

-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s

t = 1.06 s

3 0

2 years ago

The weight of a girl with a mass of 40.0 kg is_N

faltersainse [42]

Answer: The weight of a girl with a mass of 40kg is 392.266 Newtons.

Explanation:

5 0

2 years ago

Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observ

Tomtit [17]

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

$\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}$

Put the value into the formula

$525=\sqrt{\dfrac{c+v}{c-v}}\times950$

$\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2$

$\dfrac{c+v}{c-v}=0.305$

$c+v=0.305\times(c-v)$

$v(1+0.305)=c(0.305-1)$

$v=\dfrac{0.305-1}{1+0.305}c$

$v=−0.532c$

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

7 0

2 years ago

Did I do these questions correctly?

Elena L [17]

For number 11, you should say that there is more pollution and fossil fuels being burned.

8 0

2 years ago

A 600‑kg car accelerates at a rate of 3 m/s2. How much net force is acting on the car to cause this acceleration?

finlep [7]

600(kg) x 3(m/s^2) = 1800N (newtons)

4 0

2 years ago