Answer:
<h2>3.36J</h2>
Explanation:
Step one:
given data
mass m= 1.3kg
distance moved s= 2.8m
opposing frictional force= 0.34N
assume g= 9.81m/s^2
we know that work done= force *distance moved
1. work done to push the book= 1.55*2.8=4.34J
2. Work against friction = force of friction x distance
= 0.34*2.8=0.952J
Step two:
the work done on the book is the net work, which is
Network done= work done to push the book- Work against friction
Network done= 4.32-0.952=3.36J
<u>Therefore the work of the 1.55N 3.36J</u>
Answer:
8.94*10^22 kg
Explanation:
Given that
Mass of Lo, M = ?
Radius of Lo, r = 1.82*10^6 m
Acceleration on Lo, g = 1.80 m/s²
Gravitational constant, G = 6.67*10^-11
Using the formula
g = GM/r²
Solution is attached below
Answer is 8.94*10^22 kg
B represents the direction of the magnetic field around the wire
Explanation:
A wire carrying an electric current always produces a magnetic field around itself. The lines of the magnetic field produced by a current-carrying wires are concentric circles around the wire. The magnitude of the field is given by the formula:
where
is the vacuum permeability
I is the current in the wire
r is the distance from the wire
The direction of the field lines is given by the so-called right hand rule, shown in the figure. Basically, the thumb of the right hand is placed in the direction of the electric current, while the other fingers are "wrapped" around the thumb: the direction of the other fingers give the direction of the magnetic field lines.
Learn more about magnetic field:
brainly.com/question/3874443
brainly.com/question/4240735
#LearnwithBrainly
We need to see what forces act on the box:
In the x direction:
Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.
In the y direction:
N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.
From N-Gcosα=0 we get:
N=Gcosα, we will need this for the force of friction.
Now to solve for Fh:
Fh=ma + Ff + Gsinα,
Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²
Fh=ma + μmgcosα+mgsinα
Now we plug in the numbers and get:
Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N
The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
