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2 years ago

What is the square root of 25x10^12

Physics

1 answer:

kirill115 [55]2 years ago

7 0

$\sqrt{25 \times 10^{12} }$

In order to find it's square root, we could make it into two square roots.

$\sqrt{25 \times 10^{12} } = \sqrt{25} \times \sqrt{10^{12}}$

Let us find the square roots of both radicals seprately.

$\sqrt{25} =\sqrt{5*5}=5$

Each pair of a number inside square root gives a number out .

$\sqrt{10^{12}} = \sqrt{10*10*10*10*10*10*10*10*10*10*10*10} \ \ ( \ makes \ 6 \ pairs \ of \ 10 )$

$\sqrt{10*10*10*10*10*10*10*10*10*10*10*10} = 10*10*10*10*10*10$

$= 10^6.$

Therefore,

$\sqrt{25 \times 10^{12} }=\sqrt{25} \times \sqrt{10^{12}}={5 \times 10^6}$

$\sqrt{25 \times 10^{12} } = {5 \times 10^6}$

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Part b suppose the magnitude of the gravitational force between two spherical objects is 2000 n when they are 100 km apart. what

kobusy [5.1K]

<span>b) The force with a distance of 150 km is 889 N c) The force with a distance of 50 km is 8000 N This question looks like a mixture of a question and a critique of a previous answer. I'll attempt to address the original question. Since the radius of the spherical objects isn't mentioned anywhere, I will assume that the distance from the center of each spherical object is what's being given. The gravitational force between two masses is given as F = (G M1 M2)/r^2 where F = Force G = gravitational constant M1 = Mass 1 M2 = Mass 2 r = distance between center of masses for the two masses. So with a r value of 100 km, we have a force of 2000 Newtons. If we change the distance to 150 km, that increases the distance by a factor of 1.5 and since the force varies with the inverse square, we get the original force divided by 2.25. And 2000 / 2.25 = 888.88888.... when rounded to 3 digits gives us 889. Looking at what looks like an answer of 890 in the question is explainable as someone rounding incorrectly to 2 significant digits. If the distance is changed to 50 km from the original 100 km, then you have half the distance (50/100 = 0.5) and the squaring will give you a new divisor of 0.25, and 2000 / 0.25 = 8000. So the force increases to 8000 Newtons.</span>

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2 years ago

Read 2 more answers

The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict

nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ = 400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction, μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

$v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f$