In such a way, human beings are pulled off course by the invisible forces of their unconscious minds. The unconscious minds of people control the thoughts of people.

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3 0

2 years ago

In an arcade game a 0.099 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and releas

Sonja [21]

Answer:

The speed of disk is 1.98 $\frac{m}{s}$

Explanation:

Given:

Mass of $m = 0.099$ kg

Spring constant $k = 244 \frac{N}{m}$

Compression of spring $x = 4 \times 10^{-2}$ m

From energy conservation theorem,

Spring potential energy converted into kinetic energy,

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$

$v = \sqrt{\frac{k x^{2} }{m} }$

$v = \sqrt{\frac{244 \times 16 \times 10^{-4} }{0.099} }$

$v = 1.98 \frac{m}{s}$

Therefore, the speed of disk is 1.98 $\frac{m}{s}$

8 0

3 years ago

Two balloons (m = 0.021 kg) are separated by a distance of d = 16 m. They are released from rest and observed to have an instant

evablogger [386]

(a) $2.56\cdot 10^{-5} C$

According to Newton's second law, the force experienced by each balloon is given by:

F = ma

where

m = 0.021 kg is the mass

a = 1.1 m/s^2 is the acceleration

Substituting, we found:

$F=(0.021)(1.1)=0.0231 N$

The electrostatic force between the two balloons can be also written as

$F=k\frac{Q^2}{r^2}$

where

k is the Coulomb's constant

Q is the charge on each balloon

r = 16 m is their separation

Since we know the value of F, we can find Q, the magnitude of the charge on each balloon:

$Q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(0.0231)(16)^2}{9\cdot 10^9}}=2.56\cdot 10^{-5} C$

(b) $1.6\cdot 10^{14}$ electrons

The magnitude of the charge of one electron is

$e=1.6\cdot 10^{-19}C$

While the magnitude of the charge on one balloon is

$Q=2.56\cdot 10^{-5} C$

This charge can be written as

$Q=Ne$

where N is the number of electrons that are responsible for this charge. Solving for N, we find:

$N=\frac{Q}{e}=\frac{2.56\cdot 10^{-5}}{1.6\cdot 10^{-19}}=1.6\cdot 10^{14}$

5 0

3 years ago

N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e

Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

$KE_i+PE_i=KE_f+PE_f$

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

from the law of conservation of the linear momentum

$m_1v_1=m_2v_2$

Therefore,

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

$=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]$

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

Substitute the values in the above result

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

$=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s$