1) Length of the wire.
2) Thickness of the wire.
3) Temperature.
4) Type of metal.
Hope this helps!
-Payshence
Answer:
the ball travelled approximately 60 m towards north before stopping
Explanation:
Given the data in the question;
First course :
= 0.75 m/s²,
= 20 m,
= 10 m/s
now, form the third equation of motion;
v² = u² + 2as
we substitute
² = (10)² + (2 × 0.75 × 20)
² = 100 + 30
² = 130
= √130
= 11.4 m/s
for the Second Course:
= 11.4 m/s,
= -1.15 m/s²,
= 0
Also, form the third equation of motion;
v² = u² + 2as
we substitute
0² = (11.4)² + (2 × (-1.15) ×
)
0 = 129.96 - 2.3
2.3
= 129.96
= 129.96 / 2.3
= 56.5 m
so;
|d| = √(
² +
² )
we substitute
|d| = √( (20)² + (56.5)² )
|d| = √( 400 + 3192.25 )
|d| = √( 3592.25 )
|d| = 59.9 m ≈ 60 m
Therefore, the ball travelled approximately 60 m towards north before stopping
Answer:
Explanation:
Part A) Using
light intensity I= P/A
A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2
Radius= Diameter/2
P= power= 10*10^-3=0.01 W
light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2
Part B) Using
I=c*ε*E^2/2
rearrange to solve for E=
((I*2)/(c*ε))
c is the speed of light which is 3*10^8 m/s^2
ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1
I= the already solved light intensity= 8.85*10^10 W/m^2
amplitude of the electric field E=
(9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)
---> E=
(1.8*10^11) / (2.66*10^-3) =
(6.8*10^13) = 8.25*10^6 V/m
As it is given that the air bag deploy in time

total distance moved by the front face of the bag

Now we will use kinematics to find the acceleration




now as we know that

so we have

so the acceleration is 400g for the front surface of balloon
Answer:
1.6 m/s²
Explanation:
Weight equals mass times acceleration due to gravity.
F = mg
14.4 N = (9 kg) g
g = 1.6 m/s²