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Olin [163]
2 years ago
9

A compound held together by ionic bonds is called a

Physics
1 answer:
kobusy [5.1K]2 years ago
5 0

Answer:

its probably covalent compound

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A lens is used to make an image of magnification -1.50. The image is 30.0 cm from the lens. What is the object’s distance (unit=
jeka57 [31]

Answer: 20cm

Explanation:

Given the following data:

Magnification of image = - 1.50

Distance of image from the lens = 30.0cm

The Magnification of an image is the ratio of the image distance from the lens and the object distance from the lens.

Mathematically,

Magnification(M) = - (image distance (di) / object distance (do))

M = - (di/do)

-1.50 = - (30/do)

do = 30 / 1.5

do = 20 cm

Object is placed at a distance of 20cm from the lens.

5 0
3 years ago
A 120 kg quarterback is stationary in the "pocket" and hit by a defensive tackle with 4000 Newtons of force. What acceleration d
Leni [432]

Answer: 33.3 m/s^2

Explanation:

Here we need to use the second Newton's law.

F = m*a

Force equals mass times the acceleration.

Here we know that the force is F = 4000 N, and the mass is 120kg, then the acceleration must be:

4000 N = 120kg* a

a = (4000/120) m/s^2 = 33.3 m/s^2

6 0
3 years ago
I need a short summary of what a galixy is and does, 30+ points
frosja888 [35]
A galaxy consist of billions of stars that contains gases and dust held together by gravity.
7 0
4 years ago
Who discovered the basic laws of planetary orbits?.
Phoenix [80]

Answer:

kepler

Explanation:

While Copernicus rightly observed that the planets revolve around the Sun, it was Kepler who correctly defined their orbits. At the age of 27, Kepler became the assistant of a wealthy astronomer, Tycho Brahe, who asked him to define the orbital mars

6 0
2 years ago
A positively charged particle Q1 = +15nC is held fixed at the origin. A second charge Q2 of mass m = 9.5g is floating a distance
Paladinen [302]

Answer:

Q_2=4.4293\times 10^{22}\ nC

Explanation:

Given

charge on the fixed particle, Q_1=15\times 10^{-9}\ C

mass of the second charge, m_2=9.5\times 10^{-3}\ kg

distance between the fixed charge and the floating charge on the top, d=0.25\ m

  • According to the question the second charge is floating just above the fixed positive charge despite of gravity this means that the floating charge is also positive in nature and hence feels the repulsion from the fixed charge which is equal in magnitude to the gravitational force on the charge.

m_2.g=\frac{1}{4\pi\epsilon_0} \times \frac{Q_1.Q_2}{d^2}

where:

\epsilon_0= permittivity of free space

g = acceleration due to gravity

9.5\times 10^{-3}\times 9.8=8.85\times 10^{-9}\times \frac{15\times 10^{-9}\times Q_2}{0.25^2}

Q_2=4.4293\times 10^{22}\ nC

7 0
4 years ago
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