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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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Use the form of equation:

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This force involves the attraction between objects with mass. (2 points) i need it asap
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Gravitational Mass

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3 years ago
5. Which unit of electricity does the work in a circuit?
Pachacha [2.7K]

Answer:

Volt

Explanation:

Voltage is what makes electric charges move. ... Voltage is also called, in certain circumstances, electromotive force (EMF). Voltage is an electrical potential difference, the difference in electric potential between two places. The unit for electrical potential difference, or voltage, is the volt.

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8 0
3 years ago
A 52.5-turn circular coil of radius 5.35 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0
krek1111 [17]

Answer:

5.43 x 10^-3 Nm

Explanation:

N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A

Torque = N I A B Sin theta

Here, theta = 90 degree

Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455

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3 years ago
How does the magnitude of the electrical force between a pair of charged particles change when they are brought to half their or
Ostrovityanka [42]

Answer:

5. Quadruple

Explanation:

The electrostatic force between two charged particles is given by:

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

If the distance between the charges is reduced to half,

r' = \frac{r}{2}

So the new force will be

F'=k\frac{q_1 q_2}{(r/2)^2}=4(k\frac{q_1 q_2}{r^2})=4F

So, the force will quadruple.

4 0
3 years ago
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