I think is 3.5 pretty sure
Answer:
How do you find the poles of magnet?
<em>Unless they came marked with “N” or “S,” the poles of a magnet look the same. One easy way to tell which pole is north and which is <u>south is to set your magnet near a compass</u>. The needle on the compass that normally points toward the north pole of the Earth will move toward the magnet's south pole.</em>
Explanation:
<em>I </em><em>hope</em><em> it</em><em> will</em><em> help</em><em> you</em><em>.</em><em>.</em><em>.</em><em>.</em>
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<em>#</em><em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em>o</em><em>n</em><em>l</em><em>e</em><em>r</em><em>a</em><em>n</em><em>i</em><em>n</em><em>g</em>
Answer:
(a). The work done is 7001 MeV.
(b). The momentum of this proton is .
Explanation:
Given that,
Speed = 0.993 c
We need to calculate the work done
Using work energy theorem
The work done is equal to the kinetic energy relative to the proton
Put the value into the formula
(b). We need to calculate the momentum of this proton
Using formula of momentum
Put the value into the formula
Hence, (a). The work done is 7001 MeV.
(b). The momentum of this proton is .
Answer:
F= 4788 N
Explanation:
Because the car moves with uniformly accelerated movement we apply the following formula:
vf²=v₀²+2*a*d Formula (1)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
d=36.9 m
v₀=14.0 m/s m/s
vf= 0
Calculating of the acceleration of the car
We replace dta in the formula (1)
vf²=v₀²+2*a*d
(0)²=(14)²+2*a*(36.9)
-(14)²= (73.8) *a
a= - (196) / (73.8)
a= - 2.66 m/s²
Newton's second law of the car in direction horizontal (x):
∑Fx = m*ax Formula (2)
∑F : algebraic sum of the forces in direction x-axis (N)
m : mass (kg)
a : acceleration (m/s²)
Data
m=1800 Fkg
a= - 2.66 m/s²
Magnitude of the horizontal net force (F) that is required to bring the car to a halt in a distance of 36.9 m :
We replace data in the formula (2)
-F= (1800 kg) * ( -2.66 m/s²
)
F= 4788 N