You’re correct, it’s gravity
Answer:
Explanation:
a ) Let the distance required in former case be d₁ .
initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 7.00 m /s²
v² = u² - 2 a s
0 = 30² - 2 x 7 x d₁
d₁ = 64.28 m
b) initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 5.00 m /s²
v² = u² - 2 a s
0 = 30² - 2 x 5 x d₂
d₂ = 90 m
c)
t = .5 s
s₁ = ut - .5 at²
= 30 x .5 - .5 x 7 x .5²
= 15 - .875
= 14.125 m
t = .5 s
s₂ = ut - .5 at²
= 30 x .5 - .5 x 5 x .5²
= 15 - .625
= 14.375 m
Planck's constant. A physical constant adopted in 2011 by the CGPM.
Answer:![\vec{v_R}=\hat{i}[-329.11]+\hat{j}[516.18]](https://tex.z-dn.net/?f=%5Cvec%7Bv_R%7D%3D%5Chat%7Bi%7D%5B-329.11%5D%2B%5Chat%7Bj%7D%5B516.18%5D)
Explanation:
Given
Plane is initially flying with velocity of magnitude 
at angle of
with North towards west
Velocity of plane airplane can be written as

Now wind is encountered with speed of
at angle of 

resultant velocity


![\vec{v_R}=\hat{i}[-385.67+56.56]+\hat{j}[459.62+56.56]](https://tex.z-dn.net/?f=%5Cvec%7Bv_R%7D%3D%5Chat%7Bi%7D%5B-385.67%2B56.56%5D%2B%5Chat%7Bj%7D%5B459.62%2B56.56%5D)
![\vec{v_R}=\hat{i}[-329.11]+\hat{j}[516.18]](https://tex.z-dn.net/?f=%5Cvec%7Bv_R%7D%3D%5Chat%7Bi%7D%5B-329.11%5D%2B%5Chat%7Bj%7D%5B516.18%5D)
for direction 

west of North
The options are missing and they are;
A) the electric force increases because the balloon loses its charge.
B) the electric force increases because the distance increases.
C) the electric force decreases because the distance increases.
D) the electric force decreases because his hair loses its charge.
Answer:
Correct answer is option C - the electric force decreases because the distance increases.
Explanation:
The formula for electric force is;
F = k•q1•q2/r²
Where;
K is coulombs constant
q1 and q2 are particle charges
r is distance
So,looking at the formula given earlier, if we increase the distance, the denominator will increase and thus the Force will decrease.
So the correct option is option C