Answer:
27.22 m/s
Explanation:
Let the speed of clay before impact is u.
the speed of clay and target is v after impact.
use conservation of momentum
momentum before impact momentum after impact
mass of clay x u = (mass of clay + mass of target) x v
100 x u = (100 + 500) x v
u = 6 v .....(1)
distance, s = 2.1 m
μ = 0.5
final velocity is zero. use third equation of motion
v'² = v² + 2as
0 = v² - 2 x μ x g x s
v² = 2 x 0.5 x 9.8 x 2.1 = 20.58
v = 4.54 m/s
so by equation (1)
u = 6 x 4.54 = 27.22 m/s
thus, the speed of clay before impact is 27.22 m/s.
Answer:
2.5kN.m
Explanation:
Torque is directly proportional to pitch diameter
= Ta/Tb= Da/Db
=120/Tb= 0.25/0.5
Tb= 2.469kN.m approx 2.5kN.m
Answer: 5 m/s^2
Explanation: In order to solve this question we have to use the kinematic equation given by:
Vf= Vo+a*t where V0 is zero.
we know that it takes Vf( 30 m/s) in 6 seconds
so
a=(30 m/s)/6 s= 5 m/s^2
Convection currents in the mantle.
Answer:
6.20×10⁴ V/m
Explanation:
The magnitude of electric field is:
E = √(Eₓ² + Eᵧ²)
where Eₓ = ∂φ/∂x and Eᵧ = ∂φ/∂y.
φ = 1.11 (x² + y²)^-½ − 429x
Eₓ = -0.555 (x² + y²)^-(³/₂) (2x) − 429
Eᵧ = -0.555 (x² + y²)^-(³/₂) (2y)
Evaluating at (0.003, 0.003):
Eₓ = -44034 V/m
Eᵧ = -43605 V/m
The magnitude is:
E = 61971 V/m
Rounded to three significant figures, the strength of the electric field is 6.20×10⁴ V/m.
