Torque is equal position vector times (r) times force vector
(F). Since F= 10 N and r = 0.1 m, so the
torque is equal to (10 N) x ( 0.1 m) = 1Nm. The direction of the torque would
be into the screen, clockwise rotation.
Answer:
Explanation:
We shall apply the formula for velocity in case of elastic collision which is given below
v₁ = (m₁ - m₂)u₁ / (m₁ + m₂) + 2m₂u₂ / (m₁ + m₂)
m₁ and u₁ is mass and velocity of first object , m₂ and u₂ is mass and velocity of second object before collision and v₁ is velocity of first velocity after collision.
Here u₁ = 22 cm /s , u₂ = - 14 cm /s . m₁ = 7.7 gm , m₂ = 18 gm
v₁ = ( 7.7 - 18 ) x 22 / ( 7.7 + 18 ) + 2 x 18 x - 14 / ( 7.7 + 18 )
= - 8.817 - 19.6
= - 28.4 cm / s
Answer:
0.358Kg
Explanation:
The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system
0.5Ke^2 = 0.5Mv^2
0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2
Using conservation of momentum between the bullet and the block
0.0115(265) = (M + 0.0115)v
3.0475 = (M + 0.0115)v
v = 3.0475/(M + 0.0115)
plugging into Energy equation
12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2
12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )
12.56 = 0.5 × 9.2872/ M + 0.0115
12.56 = 4.6436/ M + 0.0115
12.56 ( M + 0.0115 ) = 4.6436
12.56M + 0.1444 = 4.6436
12.56M = 4.6436 - 0.1444
12.56 M = 4.4992
M = 4.4992÷12.56
M = 0.358 Kg
Answer:
²₁H + ³₂He —> ⁴₂He + ¹₁H
Explanation:
From the question given above,
²₁H + ³₂He —> __ + ¹₁H
Let ⁿₐX be the unknown.
Thus the equation becomes:
²₁H + ³₂He —> ⁿₐX + ¹₁H
We shall determine, n, a and X. This can be obtained as follow:
For n:
2 + 3 = n + 1
5 = n + 1
Collect like terms
n = 5 – 1
n = 4
For a:
1 + 2 = a + 1
3 = a + 1
Collect like terms
a = 3 – 1
a = 2
For X:
n = 4
a = 2
X =?
ⁿₐX => ⁴₂X => ⁴₂He
Thus, the balanced equation is
²₁H + ³₂He —> ⁴₂He + ¹₁H
