Lonic Naming Practice<br> Write the name of the following chemical compound:<br> Na3Cl1

What effect would decreasing the distance between objects have on their gravitational attraction to each other?

Lady_Fox [76]

Decreasing the distance between two objects having a considerable mass would increase the attraction on gravitation. The reverse is true that if you separate or inrease the objects distance would substantially decrease their gravitational attraction. Most object in our planet is held by its gravitational force towards it's center.

5 0

4 years ago

Read 2 more answers

An object moves in a circle at a constant speed of 1.0 m/s. The radius of the circle is 1.0 m. If a force of 1.0 N acts toward t

Vlad1618 [11]

Answer:5

Explanation:

Given

speed of object $v=1\ m/s$

radius of circle $r=1\ m$

Force towards the center $F=1\ N$

Work done is given by the dot product of Force and displacement

and we know know displacement of the object is along the circle which is perpendicular to the force acting therefore Work done will be zero

$W=F\cdot s\cos 90$

$W=0$

4 0

4 years ago

System A has masses m and m separated by a distance r; system B has masses m and 2m separated by a distance 2r; system C has mas

Anna [14]

Answer:

System D --> System C --> System A --> System B

Explanation:

The gravitational force between two masses m1, m2 separated by a distance r is given by:

$F=G \frac{m_1 m_2}{r^2}$

where G is the gravitational constant. Let's apply this formula to each case now to calculate the relative force for each system:

System A has masses m and m separated by a distance r:

$F=G\frac{m \cdot m}{r^2}=G \frac{m^2}{r^2}$

system B has masses m and 2m separated by a distance 2r:

$F=G\frac{m \cdot 2m}{(2r)^2}=G \frac{2m^2}{4r^2}=\frac{1}{2} G \frac{m^2}{r^2}$

system C has masses 2m and 3m separated by a distance 2r:

$F=G\frac{2m \cdot 3m}{(2r)^2}=G \frac{6m^2}{4r^2}=\frac{3}{2} G \frac{m^2}{r^2}$

system D has masses 4m and 5m separated by a distance 3r:

$F=G\frac{4m \cdot 5m}{(3r)^2}=G \frac{20m^2}{9r^2}=\frac{20}{9} G \frac{m^2}{r^2}$

Now, by looking at the 4 different forces, we can rank them from the greatest to the smallest force, and we find:

System D --> System C --> System A --> System B

5 0

4 years ago

Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia

Crank

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

3 0

3 years ago

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both

Nuetrik [128]

Answer:

Part a)

Width of the slit is

$a = 580 nm$

Part b)

Ratio of intensity is given as

$\frac{I}{I_o} = 0.81$

Explanation:

Part a)

As we know by the formula of diffraction we will have

$a sin\theta = \lambda$

so we have

$\theta = 90$

$\lambda = 580 nm$

so we will have

$a sin90 = 580 nm$

$a = 580 nm$

Part b)

As we know that the intensity in diffraction pattern is given as

$I = I_o (\frac{sin\theta}{\theta})^2$

$\frac{I}{I_o} = (\frac{sin\theta}{\theta})^2$

so for angle 45 degree

$\frac{I}{I_o} = (\frac{sin45}{\pi/4})^2$

$\frac{I}{I_o} = 0.81$

7 0

4 years ago

Lonic Naming Practice Write the name of the following chemical compound: Na3Cl1