Newton's second law allows us to find that the correct answer is:
- Less than weight of the skater
Newton's second law states that the net force and is propositional to the mass and acceleration of the body
For these problems it is essential to set a reference system, with respect to which to carry out the measurements, in this case we set a coordinate system with the x axis parallel to the plane and positive in the direction of movement and the y axis perpendicular to the plane.
In the attachment we can see a free-body diagram of the problem, let's work each axis separately
x-axis
Wₓ -fr = m a
y-axis
N -
= 0
N =
Where Wₓ and W_y are the components of the weight, fr the friction force that opposes the movement, m the mass and the acceleration of the body
let's use trigonometry to find the components of the weight
cos θ =
sin θ =
= W cos θ
Wₓ = W sin θ
we substitute
N = mg cos θ
From this equation we can see that the normal is less than the weight of the body.
In conclusion using Newton's second law we find that the correct answer is:
- Less than th weight of the skater
Learn more about Newton's second law here:
brainly.com/question/13685393
Answer:
Both A and C
Explanation:
I just got it correct on Edg
Yes it is an ion as it has a positive charge (i.e more protons than electrons, in this case there is no electrons). It's also an ion of an isotope of a hydrogen atom.
Hope this helps :).
newton 1st law: w no external force, a body will stay at rest or in constant uniform motion.
a motor raises the car up the tower. cuz it moves w/ constant vel, there is no external force: the motor force is the same n in opposity direction as the gravity force on car.
Answer:
a-
V= IR
9V = I ×( 12+6)
I = 9/ 18 A = 0.5 A
b
V=IR
240 = 6 A ×( 20 + R)
40 = 20 + R
R = 20 ohm
c
resultant resistance of the 2 parallel resistances= Ro
1/Ro = 1/ 5 + 1/ 20
1/Ro =( 20+5)/100
= 1/Ro = 1/4
Ro= 4 ohm
V=IR
V = 2A × ( 1+ 4 OHM)
V = 10V
d
equivalent resistance = Ro
1/Ro = 1/(2+8) + 1/(5+5)
1/Ro = 1/10 +1/10
2/10 = 1/ Ro
Ro= 10/2 = 5 ohm
V = IR
12V = I × 5Ohm
I=2.4 A