Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled
Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²
Answer: 1.95 x 10¹⁴ m/s²
Answer:
ΔP = 14.5 Ns
I = 14.5 Ns
ΔF = 5.8 x 10³ N = 5.8 KN
Explanation:
The mass of the ball is given as 0.145 kg in the complete question. So, the change in momentum will be:
ΔP = mv₂ - mv₁
ΔP = m(v₂ - v₁)
where,
ΔP = Change in Momentum = ?
m = mass of ball = 0.145 kg
v₂ = velocity of batted ball = 55.5 m/s
v₁ = velocity of pitched ball = - 44.5 m/s (due to opposite direction)
Therefore,
ΔP = (0.145 kg)(55.5 m/s + 44.5 m/s)
<u>ΔP = 14.5 Ns</u>
The impulse applied to a body is equal to the change in its momentum. Therefore,
Impulse = I = ΔP
<u>I = 14.5 Ns</u>
the average force can be found as:
I = ΔF*t
ΔF = I/t
where,
ΔF = Average Force = ?
t = time of contact = 2.5 ms = 2.5 x 10⁻³ s
Therefore,
ΔF = 14.5 N.s/(2.5 x 10⁻³ s)
<u>ΔF = 5.8 x 10³ N = 5.8 KN</u>
Answer:
Acceleration=3.95
Explanation:Use the formula a=m/f
a=128.6/32.5
a=3.95
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
EQUATOR IS THE LINE THAT DIVIDES EARTH INTO TWO HEMISPHERES .
