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g100num [7]
3 years ago
13

A meter stick is held vertically above your hand, with the lower end between your thumb and first finger. On seeing the meter st

ick released, you grab it with these two fingers. You can calculate your reaction time from the distance the meter stick falls, read directly from the point where your fingers grabbed it. If the measured distance is 18.3 cm, what is the reaction time?
Physics
1 answer:
inessss [21]3 years ago
7 0

Answer:

t=0.193s

Explanation:

What is said is that the meter fell d=18.3cm=0.183m under the action of gravity. We can use the formula for accelerated motion:

d=v_0t+\frac{at^2}{2}

Since it departed from rest it will mean that:

d=\frac{at^2}{2}

So our time will be:

t=\sqrt{\frac{2d}{a}}

Which for our values is:

t=\sqrt{\frac{2(0.183m)}{(9.81m/s^2)}}=0.193s

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Answer:

5.4

Explanation:

900/10=90 and 90(60)=5400 meters. 5400/100=5.4

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John and mary are skating at an ice rink. john skates at a constant speed of 6.7 m/s, with respect to the ice surface, directly
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8 0
3 years ago
Particle 1 has mass 4.6 kg and is on the x-axis at x = 5.7 m. Particle 2 has mass 7.2 kg and is on the y-axis at y = 4.2 m. Part
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To solve this problem it is necessary to apply the concepts related to the Gravitational Force, for this purpose it is understood that the gravitational force is described as

F_g = \frac{Gm_1m_2}{r^2}

Where,

G = Gravitational Universal Force

m_i = Mass of each object

To solve this problem it is necessary to divide the gravitational force (x, y) into the required components and then use the tangent to find the angle generated between both components.

Our values are given as,

m_1 =4.6 kg\\m_2 = 7.2 kg\\m_3 = 2.6 kg\\r_1 = 5.7 m\\r_2 = 4.2 m

Applying the previous equation at X-Axis,

F_x = \frac{Gm_1m_3}{R_{1}^2}\\F_x = \frac{6.67*10^{-11}*4.6*2.6}{5.7^2}\\F_x = 2.46*10^{-11}N

Applying the previous equation at Y-Axis,

F_y = \frac{Gm_2m_3}{R_2^2}\\F_y = \frac{6.67*10^{-11}*7.2*2.6}{4.2^2}\\F_y = 7.08*10^{-11} N

Therefore the angle can be calculated as,

tan\theta = \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{7.08*10^{-11}}{2.46*10^{-11}}\\\theta = 71\°

Then in the measure contrary to the hands of the clock the Force in the particle 3 is in between the positive direction of the X and the negative direction of the Y at 71 ° from the positive x-axis.

5 0
4 years ago
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