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2 years ago

I need help please...

Physics

2 answers:

Natalka [10]2 years ago

6 0

No cluuuueee :/ sowwwwyyy but good luck

fomenos2 years ago

4 0

No wayyyyyyyyyyyyyyyyyy idk

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What is cohesion force

olchik [2.2K]

Answer:

The force of attraction between the molecules of the same substance

Explanation:

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2 years ago

At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco

Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

$A_1*m=M*A_2$

$A_i =Area$

M,m = Counts per second

Our radios are given by

$r_1 = 11cm$

$R_2 = 20cm$

$m = 65cps$

Therefore replacing we have that,

$A_1*m=M*A_2$

$4\pi r_1^2*m = M * 4\pi R_2^2 M$

$r^2*m=MR^2$

$M = \frac{m*r^2}{R^2}$

$M = \frac{65*11^2}{20^2}$

$M = 19.6625cps$

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

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3 years ago

A child is riding a merry-go-round that has an instantaneous angular speed of 12 rpm. If a constant friction torque of 12.5 Nm i

sammy [17]

Answer:

$-0.25 rad/s^2$

Explanation:

The equivalent of Newton's second law for rotational motions is:

$\tau = I \alpha$

where

$\tau$ is the net torque applied to the object

I is the moment of inertia

$\alpha$ is the angular acceleration

In this problem we have:

$\tau = -12.5 Nm$ (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

$I=50.0 kg m^2$ is the moment of inertia

Solving for $\alpha$ , we find the angular acceleration:

$\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2$

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2 years ago

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ruslelena [56]

Correct your answer to the amount of significant figures the questions wants

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2 years ago