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ycow [4]
3 years ago
10

The diagrams show objects’ gravitational pull toward each other. Which statement describes the relationship between diagram X an

d Y?
A. Gravity attracts only larger objects toward one another.
B. Gravity attracts larger objects only if they are close to one another.
C.If the masses of the objects increase, then the force between them also increases.
D.If distance between the objects increases, then the amount of force also increases.
Physics
2 answers:
creativ13 [48]3 years ago
8 0

' C ' is the only correct statement on the list.  We don't know anything about diagram-x or diagram-y because we can't see them.

Nookie1986 [14]3 years ago
7 0

The answer is c on edugentiy hope this helped

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A teacher used a Geiger-Muller(GM)tube to measure the background radiation in her laboratory.
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It changes because radiation always differs in the atmosphere because of the nuclear plants and of pollution around us. Even though it is indoors.
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3 years ago
Explain why atmospheric pressure changes as atmospheric depth changes.
brilliants [131]
The lower you go <span>in relation to the top of the atmosphere the larger the column of air is that is pressing down on you. </span>
4 0
3 years ago
Two particles with charges of 2nC and 5nC are separated by a distance of 3m. The charge 2nC is placed on the left. Find the forc
Vanyuwa [196]

Answer:

1\cdot 10^{-8} N to the left

Explanation:

The magnitude of the electrostatic force between two charges is given by the following equation:

F=k\frac{q_1 q_2}{r^2}

where:

k=9\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the magnitude of the two charges

r is the distance between the two charges

Moreover, the force is:

- Attractive if the charges have opposite sign

- Repulsive if the charges have same sign

In this problem, we have:

q_1=2nC=2\cdot 10^{-9}C is the magnitude of charge 1

q_2=5nC =5\cdot 10^{-9}C is the magnitude of charge 2

r = 3 m is the distance between the two charges

Substituting, we find the force on both charges:

F=\frac{(9\cdot 10^9)(5\cdot 10^{-9})(2\cdot 10^{-9}}{3^2}=1\cdot 10^{-8} N

Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).

5 0
3 years ago
Find a numerical value for ρearth, the average density of the earth in kilograms per cubic meter. use 6378km for the radius of t
Andre45 [30]

Answer:

5501 kg/m^3

Explanation:

The value of g at the Earth's surface is

g=\frac{GM}{R^2}=9.70 m/s^2

where G is the gravitational constant

M is the Earth's mass

R=6378km = 6.378 \cdot 10^6 m is the Earth's radius

Solving the formula for M, we find the value of the Earth's mass:

M=\frac{gR^2}{G}=\frac{(9.81 m/s^2)(6.378\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=5.98\cdot 10^{24}kg

The Earth's volume is (approximating the Earth to a perfect sphere)

V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (6.378\cdot 10^6 m)^3=1.087\cdot 10^{21} m^3

So, the average density of the Earth is

\rho = \frac{M}{V}=\frac{5.98\cdot 10^{24} kg}{1.087\cdot 10^{21} m^3}=5501 kg/m^3

4 0
4 years ago
Read 2 more answers
On an air track, a 400.5 g glider moving to the right at 2.30 m/s collides elastically with a 500.0 g glider moving in the oppos
Yanka [14]

Answer:

3.53 m/s towards the left

Explanation:

m_1 = Mass of first glider = 400.5 g

m_2 = Mass of Second glider = 500 g

u_1 = Initial Velocity of first object = 2.3 m/s

u_2 = Initial Velocity of second object = -2.95 m/s

As momentum and Energy is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.4005-0.5}{0.4005+0.5}\times 2.3+\frac{2\times 0.5}{0.4005+0.5}\times -2.95\\\Rightarrow v_1=-3.53\ m/s

The velocity of first glider after collision is 3.53 m/s towards the left

4 0
3 years ago
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