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3 years ago

Gravity acts on any two objects with mass anywhere in the universe

2 answers:

8 0

- "Gravity is a universal attraction between any two objects with mass. Every object in the universe is affected by the force of gravity. The two factors that affect the magnitude of gravity are the mass of the attracted objects and the distance between them. The greater the masses the greater the gravitational force between two objects."

klemol [59]3 years ago

4 0

Answer: True

Explanation:

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For the long life cells we have to connect them in ____ combination

NISA [10]

Answer:

Parallel combination.

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3 years ago

Read 2 more answers

How would radiation occur in space?

Marizza181 [45]

Answer:

Space radiation is made up of three kinds of radiation: particles trapped in the Earth's magnetic field; particles shot into space during solar flares (solar particle events); and galactic cosmic rays, which are high-energy protons and heavy ions from outside our solar system.

Explanation:

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3 years ago

Given: Saturated air changes temperature by 0.5°C/100 m. The air is completely saturated at the dew point. The dew point has bee

erma4kov [3.2K]

Answer:

Explanation:

Given

saturated air temperature by $0.5^{\circ}C/100 m$

Dew point temperature is given by $t=2^{\circ}C$

Dew point is defined as the temperature after which air no longer to uphold the water vapor fuse with it and some water vapor may condense to a liquid.

air continues to rise for 1400 m

i.e. change in temperature would be $\Delta t =\frac{0.5}{100}\times 1400=7^{\circ}C$

Final temperature $t_f$

$t_f+\Delta t=t$

$t_f=2-7=-5^{\circ}C$

3 0

4 years ago

Two particles are fixed to an x axis: particle 1 of charge −1.50 ✕ 10−7 c at x = 6.00 cm, and particle 2 of charge +1.50 ✕ 10−7

sleet_krkn [62]

Answer : $\underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Explanation :

Given that,

Charge of particle 1 = $-1.50\times10^{-7} c$

Distance x = 6 cm

Charge of particle 2 = $1.50\times10^{-7} c$

Distance x = 27 cm

Total distance = $\dfrac{6+27}{2}$

$r = 16.5\ cm$

Particle 1 is at (6,0) and particle 2 is at (27,0) .

Therefore, midway (16.5, 0)

Now, $r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm$

Formula of electric field

$E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}$

Now, the the electric field due to particle 1

$\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Similarly, the electric field due to particle 2

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Resultant Electric field

$\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}$

$\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Hence, this is the required answer.

3 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

Iteru [2.4K]

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