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UNO [17]
3 years ago
11

A car is traveling north at 17.7m/s After 12 s its velocity is 14.1m/s in the same direction. Find the magnitude and direction o

f the car's average acceleration.
a2.7 m/s2, north b0.30 m/s2, south c2.7 m/s2, south d0.30 m/s2, north
Physics
1 answer:
Nat2105 [25]3 years ago
7 0

The car is initially traveling north at 17.7 m/s, and after 12 s, its velocity is 14.1 m/s, still due north. This means that the direction of the car has not changed, so we can already say that the direction of the acceleration is north (if the magnitude of the acceleration is positive) or south (if the magnitude of the acceleration is negative).

To find the magnitude of the average acceleration, we must calculate the ratio between the change in velocity and the time taken:

a=\frac{v_f -v_i}{t}=\frac{14.1 m/s-17.7 m/s}{12 s}=-0.3 m/s^2

Since the acceleration is negative, it means it is in the opposite direction to the motion of the car, therefore south. Therefore, the correct answer is

b) 0.30 m/s2, south

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A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg.
rodikova [14]

Answer:

the moment of inertia of the merry go round is 38.04 kg.m²

Explanation:

We are given;

Initial angular velocity; ω_1 = 37 rpm

Final angular velocity; ω_2 = 19 rpm

mass of child; m = 15.5 kg

distance from the centre; r = 1.55 m

Now, let the moment of inertia of the merry go round be I.

Using the principle of conservation of angular momentum, we have;

I_1 = I_2

Thus,

Iω_1 = I'ω_2

where I' is the moment of inertia of the merry go round and child which is given as I' = mr²

Thus,

I x 37 = ( I + mr²)19

37I = ( I + (15.5 x 1.55²))19

37I = 19I + 684.7125

37I - 19 I = 684.7125

18I = 684.7125

I = 684.7125/18

I = 38.04 kg.m²

Thus, the moment of inertia of the merry go round is 38.04 kg.m²

7 0
4 years ago
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gtnhenbr [62]
What are the following statements? If there's one that mention a description of current action, or motion, that's your answer.
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4 years ago
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Answer:

v=4m/s

Explanation:

The formulas for accelerated motion are:

v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}

We can derive the formula v^2=v_0^2+2ad from them.

We have:

v-v_0=at\\t=\frac{v-v_0}{a}

And substitute:

x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}\\2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2

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We use then our values to calculate the final velocity when starting from rest, traveling a distance 0.002m with acceleration 4000 m/s^2:

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Answer:

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Explanation:

The force (Hooke's Law) is proportional to the deformation. To extend the spring 4 times longer, you will need 4 times the force. In total, 20 N

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3 years ago
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