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ZanzabumX [31]
2 years ago
5

9a driver has a reaction time of 0.50 s , and the maximum deceleration of her car is 6.0 m/s2 . she is driving at 20 m/s when su

ddenly she sees an obstacle in the road 50 m in front of her.
Physics
1 answer:
dolphi86 [110]2 years ago
7 0
The distance traveled during the reaction time is
d₁ == (20 m/s)*(0.50 s) = 10 m

The deceleration is  6.0 m/s²
The distance, d₂, traveled until the car comes to a complete stop is given by
(20 m/s)²  - 2*(6.0 m/s²)*(d₂ m) = 0
d₂ = 400/12 = 33.333 m

The total distance traveled before the car comes to rest is
d₁ + d₂ = 10 + 33.333 = 43.333 m
Because 43.333 m is less than 50 m, the car will not hit the object.

Answer:
The car comes to a stop before reaching the object. It will not hit the object.

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2 years ago
A jogger runs at a constant rate of 10.0 m every 2.0 seconds. The jogger starts at the origin and runs in the positive direction
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Answer:

(a) 25 m

(b) 75 m

Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

So, the speed of the jogger,

v=\frac{10}{2}=5m/s\;\cdots(i)

Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

So, d=vt

From equation (i)

\Rightarrow d=5t\;\cdots(ii)

As the jogger starts from origin, so, the distance, d, also represents the position of the jogger at the time t s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

d=5\times 5=25 m

So, the jogger is at a distance of 25 m from the origin.

(b) Similarly, for t=15.0 s

d=5\times 15=75 m

So, the jogger is at a distance of 75 m from the origin.

8 0
2 years ago
What is a wave coming into a barrier
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6 0
2 years ago
A force F with arrow applied to an object of mass m1 produces an acceleration of 3.10 m/s2. The same force applied to a second o
Bas_tet [7]

Answer:

(a) The value of the ratio m₁/m₂ is 0.581

(b)  the acceleration of the combined masses is 1.139 m/s²

Explanation:

Given;

The acceleration of force applied to M₁, a₁ = 3.10 m/s²

The same force applied to M₂ has acceleration, a₂ = 1.80 m/s²

Let this force = F

According Newton's second law of motion;

F = ma

(a) the value of the ratio m₁/m₂

since the applied force is same in both cases,  M₁a₁ = M₂a₂

\frac{m_1}{m_2} = \frac{a_2}{a_1} \\\\\frac{m_1}{m_2} = \frac{1.8}{3.1} \\\\\frac{m_1}{m_2} = 0.581

(b) the acceleration of m₁ and m₂ combined as one object under the action force F

F = ma

a = \frac{F}{M} \\\\a =  \frac{F}{m_1 + m_2} \\\\a = \frac{F}{0.581m_2 + m_2}\\\\a = \frac{F}{1.581m_2}

But, m_2 = \frac{F}{a_2} \\\\a = \frac{F}{1.581m_2} =  \frac{F*a_2}{1.581F} \\\\a = \frac{a_2}{1.581} \\\\a = \frac{1.8}{1.581} = 1.139 \ m/s^2

Therefore, the acceleration of the combined masses is 1.139 m/s²

5 0
3 years ago
What is a conclusion?
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