Answer:
The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N
Explanation:
For this exercise we will use Newton's second law, let's set a reference system where the x axis is parallel to the plane, in the adjoint we can see the forces in the system.
sin 35 = Wₓ / W
cos 35 = W_y / W
Wₓ = W sin 35
W_y = W cos 35
Wₓ = 2500 9.8 sin 35
Wₓ = 14052.6 N
let's write the equations for each axis
and
Y axis
N-W_y = 0
N = W_y
X axis
F -Wₓ = m a
F = Wₓ + m a = mg sin 35 + m a
F = m (a + g sin 35)
let's calculate
F = 2500 (5 + 9.8 sin 35)
F = 26552.6 N
The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N
Answer:
work being done on an object.
Answer. A. A man pushes a couch across the room
diagram identifies an axle:
Answer: B. Z
simple machine described as a shaft at the center of a wheel:
Answer: D. Wheel and axle
Type of lever:
A. a catapult
Answer:
a)n= 3.125 x 
electrons.
b)J= 1.515 x 
A/m²
c)
=1.114 x 
m/s
d) see explanation
Explanation:
Current 'I' = 5A =>5C/s
diameter 'd'= 2.05 x 
m
radius 'r' = d/2 => 1.025 x m
no. of electrons 'n'= 8.5 x 
a) the amount of electrons pass through the light bulb each second can be determined by:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
As we know that: Q= ne
where e is the charge of electron i.e 1.6 x 
C
n= Q/e => 5/ 1.6 x
n= 3.125 x electrons.
b) the current density 'J' in the wire is given by
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x )²)
J= 1.515 x A/m²
c) The typical speed'' of an electron is given by:
= 
=1.515 x / 8.5 x x |-1.6 x |
=1.114 x m/s
d) According to these equations,
J= I/A
= =
If you were to use wire of twice the diameter, the current density and drift speed will change
Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.
Also drift velocity will decrease as it is inversely proportional to the area
Answer is A of course lol Fire needs oxygen as an essential fuel to burn.
Speed (ex: meters/second, miles/hour)
