-- heat the magnet red-hot in a flame
-- drop it on the floor several times, or beat it with a hammer
-- place it in a strong, rapidly alternating external magnetic field
Answer:
<h2>154.73N</h2>
Explanation:
The question is incomplete. Here is the complete question.
Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.
Check the diagram related to the question in the attachment below for better understanding.
The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.
The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).
Ty = 15sin31°
Ty = 7.73N
W = mass * acceleration due to gravity
W = 15.0*9.8
W = 147N
The normal force is therefore expressed as;
N = Ty + W
N = 7.73 + 147
N = 154.73N
The radial velocity method preferentially detects large planets close to the central star
- what is the Radial velocity:
The radial velocity technique is able to detect planets around low-mass stars, such as M-type (red dwarf) stars.
This is due to the fact that low mass stars are more affected by the gravitational tug of planets.
When a planet orbits around a star, the star wobbles a little.
From this, we can determine the mass of the planet and its distance from the star.
hence we can say that,
option D is correct.
The radial velocity method preferentially detects large planets close to the central star
Learn more about radial velocity here:
<u>brainly.com/question/13117597</u>
#SPJ4
Answer:
It is important because it carries useful energy through your house that you can use for a variety of tasks.
Explanation:
Hope this helped !
Answer:
Charge on each is 2 x 10⁻¹⁰.
Explanation:
We know that Force between two point charges is given b the Coulomb's law as:
F = kq₁q₂/r^2
k = 9 x 10^9
r = 3.00 cm
= 0.03 m
q₁ = q₂
F = 4.00 x 10^-7
Rearranging the formula, we get:
F = k q²/r²
q² = Fr²/k
q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)
q² = 4 x 10⁻²⁰
q = 2 x 10⁻¹⁰
As there is force of repulsion between the charges, the charges must be both positive or both negative.
