Which way is it being pulled?
Answer:
vf = 11.2 m/s
Explanation:
m = 10 Kg
F = 2*10² N
x = 4.00 m
μ = 0.44
vi = 0 m/s
vf = ?
We can apply Newton's 2nd Law
∑ Fx = m*a (→)
F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m
⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²
then , we use the equation
vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)
⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s
Answer:
F = 3.98 kN
Explanation:
GIVEN DATA:
sides of box = 17 cm
pressure = 1 atm = 101325 N/m2
T2 = 378K
T1 = 278 K
final pressure can be calculate by using below relation
we know that
force = pressure * area
therefore force is
F = 3.98 kN
<span> If You Have Nothing to Hide, You Have Nothing to Fear</span>
final velocity = initial
velocity + (acceleration x time) <span>
3.9 m/s = 0 m/s + (acceleration x 0.11 s)
3.9 m/s / 0.11 s = acceleration
30.45 m/s^2 = acceleration
distance = (initial velocity x time) +
1/2(acceleration)(time^2)
distance (0 m/s x 0.11 s) + 1/2(30.45 m/s^2)(0.11s ^2)
<span>distance = 0.18 m</span></span>
