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Andreas93 [3]
3 years ago
15

A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m/s. After he reaches this

maximum speed, his acceleration becomes zero, and then he runs at constant speed. Assume that his acceleration is constant during the first 2.0 s of the race, that he starts from rest, and that he runs in a straight line. (a) How far has the sprinter run when he reaches his maximum speed? (b) What is the magnitude of his average velocity for a race of these lengths: (i) 50.0 m; (ii) 100.0 m; (iii) 200.0 m?
Physics
1 answer:
baherus [9]3 years ago
5 0
20 characters longer later and woah
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Gala2k [10]

Answer:

d. 0.3 N

Explanation:

Force: This is defined as the product of mass of a body and its acceleration.

The S.I unit of Force is Newton (N).

Mathematically Force can be represented as,

F = ma .................. Equation 1

Where F = force, m = mass, a = acceleration.

also

a = (v-u)/t............... Equation 2

Where v = final velocity, u = initial velocity, t = time.

Given: v = 0.60 m/s, u = 0 m/s ( From rest), t= 0.16 s.

Substitute into equation 2

a = (0.60-0)/0.16

a = 3.75 m/s²

Also given: m = 80 g = 0.08 kg.

Substitute into equation 1,

F = 0.08(3.75)

F = 0.3 N.

Hence he average force = 0.3 N

The right option is d. 0.3 N

6 0
3 years ago
What type of circuit is illustrated?
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Answer:

I beleive its B

Explanation:

If not then A but I'm positive its B

6 0
3 years ago
Write the ratio of the following?<br><br>CaCO3<br>C2H6<br>Fe(NO3)3​
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Answer:

C2H6 up the road to be with its own in

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2 years ago
Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistan
Rama09 [41]

Answer:

(a) 0.613 m

(b) 0.385 m

(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s

v = 3.68 m/s², θ = 72.6° below the horizontal

Explanation:

(a)  Take down to be positive.

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 0.350 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²

Δy = 0.613 m

(b) Given in the x direction:

v₀ = 1.10 m/s

a = 0 m/s²

t = 0.350 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²

Δx = 0.385 m

(c) Find: vₓ and vᵧ

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (0.350 s) + 1.10 m/s

vₓ = 1.10 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (10 m/s²) (0.350 s) + 0 m/s

vᵧ = 3.50 m/s

The magnitude is:

v² = vₓ² + vᵧ²

v = 3.68 m/s²

The direction is:

θ = atan(vᵧ / vₓ)

θ = 72.6° below the horizontal

3 0
3 years ago
If an object is projected upward from ground level with an initial velocity of 144144 ft per​ sec, then its height in feet after
Liula [17]

Answer:

4.5 s, 324 ft

Explanation:

The object is projected upward with an initial velocity of

v_0 = 144 ft/s

The equation that describes its height at time t is

s(t) = -16t^2 + 144 t (1)

where t, the time, is measured in seconds.

In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):

v(t) = s'(t) = -32t +144 (2)

At the maximum heigth, the vertical velocity is zero:

v(t) = 0

Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:

0=-32t+144\\t=\frac{144}{32}=4.5 s

And by substituting this value into eq.(1), we also find the maximum height:

s(t) = -16(4.5)^2+144(4.5)=324 ft

3 0
3 years ago
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