Answer: It would increase.
Explanation:
The equation for determining the force of the gravitational pull between any two objects is:
Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.
Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.
Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.
Answer:
the first one answer is no
Answer:
1.6 m/s
Explanation:
First you need to find the momentums of each disc by multiplying their velocities with mass.
disc 1: 7*1= 7 kg m/s
disc 2: 1*9= 9 kg m/s
Second, you need to find the total momentum of the system by adding the momentums of each sphere.
9+7= 16 kg m/s
Because momentum is conserved, this is equal to the momentum of the composite body.
Finally, to find the composite body's velocity, divide its total momentum by its mass. This is because mass*velocity=momentum
16/10=1.6
The velocity of the composite body is 1.6 m/s.
Explanation:
1.Pick up litter and throw it away in a garbage can.
2.Blow or sweep fertilizer back onto the grass if it gets onto paved areas. ...
3.Mulch or compost grass or yard waste. ...
4.Wash your car or outdoor equipment where it can flow to a gravel or grassed area instead of a street.
5.Don't pour your motor oil down the storm drain.
Answer:
The correct answer is B
Explanation:
Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity
Φ
= ∫ E. dA = 
/ ε₀
For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product
∫ E dA = / ε₀
The area of a sphere is
A = 4π r²
E 4π r² = / ε₀
E = (1 /4πε₀
) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀
) Q / (1.25R)²
E = (1 /4πε₀
) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀
) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B
