Answer:
A = [m/s]
B = [m/s²]
Explanation:
Assuming that V has SI units of m/s, then A and BT must also have units of m/s.
A = [m/s]
BT = [m/s]
Since T has SI units of s:
B [s] = [m/s]
B = [m/s²]
By definition, the impulse is given by:
Where,
- <em>F: refers to the force that I feel applied
</em>
- <em>Dt: It is the time differential in which force is applied.
</em>
Therefore, we observe that the distinction between impulse and force is that the impulse is the force that is applied during a certain time.
Answer:
The distinction between impulse and force involves the time during which force is applied.
Answer:
a) S = v₀² / 4 g sin θ
Explanation:
Let's apply Newton's second law, let's take a coordinate system with an axis parallel to the plane and the other perpendicular, in this case the only force that we have to decompose the weight (W)
Wx = W sin θ
Wy = W cos θ
First case. Body slides down
X axis
Wx-fr = 0
Axis y
N -Wy = 0
N = Wy
fr = Wx = W sint θ
Miu N = W sint θ
Miu W cos θ = Wsin θ
Miu = tan θ
Second case. Body raises the plane
X axis
Wx + Fr = m a
Axis y
N-Wy = 0
let's find the acceleration of the body going up
a = (Wx + fr) / m
fr = μ N = μ Wy
fr = μ mg cos θ
a = (mg sin θ + μ mg cos θ) / m
a = g (sin θ + μ cos θ)
a = g (sin θ + tan θ cos θ)
a = g (sin θ + sin θ)
a = g 2 sin 2
With the kinematic equation we find the distance that goes up, at the highest point the zero speed (vf = 0)
Vf² = v₀² - 2 a t S
0 = v₀² -2a S
S = v₀² / 2 a
S = v₀² / 2 (g 2sin θ)
S = v₀² / 4 g sin θ
b) in this case the block tries to slide down whereby the friction force opposes this movement
Wx- fr =, m a
mg sin θ - μ mg cos θ = m a
g (sin θ - μ cos θ) = a
a = g 2 sin θ
so that the body slides depends on the angle T for angles close to zero the body does not slide
Answer:
Let N0 be the initial atoms of Be11
N0 / 2^1 = N0 / 2 Number of Be11 after 1 half-life
N0 / 2^2 = N0/ 4 Number of Be11 after 2 half-lives
30/13.81 = 2.17 half lives
N0 / 2^2.17 = N0 / 4.51 = .222 N0 or 22.2 % of atoms remain
Answer
given,
mass of the steel ball = 35 g = 0.035 kg
the horizontal distance is
( v cos θ ) t = 4.8..........(1)
the vertical distance is
...(2)
since,
..................(3)
using equation(2) and (3)
(v t)² (cos²θ+ sin²θ) = 4.8² + 4²
(v t)² = 39.04
t = 0.76 s
the speed is
vt = 6.25
v × 0.76 = 6.25
v = 8.22 m/s