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Paladinen [302]

3 years ago

6

What happens to the electrostatic force between two charged particles if the distance between them is doubled?

1 answer:

Stells [14]3 years ago

5 0

According to Coulomb's Law , The size of the force varies inversely as the square of the distance between the two charges. So ,if the distance between the two charges is doubled, the electrostatic force will become weak by one fourth of the original force.

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Thandy is looking at two cells under the microscope.One is a human cheek cell and the other is a leaf mesophyll cell from a plan

Vanyuwa [196]

Cell are small and us human can't see them with our own eyes. it is in possible to see cell without a microscope

7 0

3 years ago

Creativity is always discouraged in scientific endeavors.

shepuryov [24]

Absolutely false!! Creativity, curiosity encouraged in Scientific endeavors. Science is all about exploring knowledge, it just promotes intelligency!

In Short, Your Answer would be "False"

Hope this helps!

3 0

3 years ago

Read 2 more answers

Suppose that a particular artillery piece has a range R = 9880 yards . Find its range in miles. Use the facts that 1mile=5280ft

Solnce55 [7]

Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

If you do it in steps

R = 9880 yd * 3 ft/yd = 29640 ft

R = 29640 ft / 5280 ft/mi = 5.61 mi

6 0

3 years ago

Compute the dot product of the vectors u and v, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and

OLga [1]

Answer:

$\theta = 106.3 degree$

Explanation:

As we know that

$\vec w = -\hat i + 7\hat j$

$\vec v = 7\hat i - \hat j$

also we know that

$\vec v. \vec w = -14$

it is given as

$\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)$

$\vec v. \vec w = - 7 - 7 = -14$

also we can find the magnitude of two vectors as

$|v| = \sqrt{(-1)^2 + (7)^2}$

$|v| = \sqrt{50}$

similarly we have

$|w| = \sqrt{(7^2) + (-1)^2}$

$|w| = \sqrt{50}$

now we know the formula of dot product as

$\vec v. \vec w = |v||w| cos\theta$

$-14 = (\sqrt{50})^2cos\theta$

$\theta = cos^{-1}(\frac{-14}{50})$

$\theta = 106.3 degree$

3 0

3 years ago

A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal in

erica [24]

Explanation:

Given that,

Mass of a freight car, $m_1=30,000-kg$

Speed of a freight car, $u_1=0.85\ m/s$

Mass of a scrap metal, $m_2=110,000\ kg$

(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :

$30000\times 0.85=(30000+110000)V\\\\V=\dfrac{30000\times 0.85}{30000+110000}\\\\=0.182\ m/s$

So, the final velocity of the loaded freight car is 0.182 m/s.

(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy

$\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J$

Lost in kinetic energy is 8518.82. Negative sign shows loss.

6 0

3 years ago