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nikitadnepr [17]
3 years ago
9

The force of gravity is (4 points) 1 a contact force 2 a distance force 3 an electrical force 4 a magnetic force

Physics
1 answer:
d1i1m1o1n [39]3 years ago
4 0
It is a distance force
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Apply the concept of density to explain why oil floats on water.
Sauron [17]
Anything less dense than water will float, like oil. Anything more dense than water will sink, like rock.
6 0
3 years ago
Form a hypothesis about how deeply water could erode and about how deeply glaciers could erode
Softa [21]
A hypothesis is an educated guess. It's your own opinion!
8 0
3 years ago
Two 25.0N weights are suspended at opposite ends of a rope that passes over a frictionless pulley. What is the tension in the ro
goldenfox [79]

Answer:

tension in rope = 25.0 N

Explanation:

  • Two forces act on the suspended weight. The force coming down is the gravitational force and the upward force by the tension in the rope.
  • Since the suspended weight is not accelerating so that the net force will be zero. Therefore the tension in the rope should be 25 N.

       ∑F = F - W = 0

       so

       F = W

       so tension in rope = F  = T  = 25 N

8 0
3 years ago
Choose the correct statement regarding the sign conventions for lenses.
Savatey [412]

Answer:

a) false

b) True

c) True

d) False

e) False

Explanation:

a) False

For a diverging lens, the focal length is negative while it is positive for a converging lens

b) True

Image distances  for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged

c) True

d) False

For a diverging lens, the focal length is negative while it is positive for a converging lens

e) False

Image distances  for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged

3 0
3 years ago
wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the
lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)

The electric field due to charge QB at P is EB leftwards

E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0
3 years ago
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