Answer:
3.08 Nm
Explanation:
N = 200, diameter = 6 cm, radius = 3 cm, I = 7 A, B = 0.90 T, Angle = 30 degree
The angle made with the normal of the coil, theta = 90 - 30 = 60 degree
Torque = N I A B Sin Theta
Torque = 200 x 7 x 3.14 x 0.03 x 0.03 x 0.90 x Sin 60
Torque = 3.08 Nm
Answer:
a) 4 289.8 J
b) 4 289.8 J
c) 6 620.1 N
d) 411 186.3 m/s^2
e) 6 620.1 N
Explanation:
Hi:
a)
The kinetic energy of the bullet is given by the following formula:
K = (1/2) m * v^2
With
m = 16.1 g = 1.61 x 10^-2 kg
v = 730 m/s
K = 4 289.8 J
b)
the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest) Ki = 0, therefore:
W = ΔK = Kf - Ki = 4 289.8 J
c)
The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).
If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:
W = F * L
Where F is the net force and L is the length of the barrel, that is:
F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N
d)
The acceleration can be found dividing the force by the mass:
a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2
e)
The force will have a magnitude equal to c) and direction along the barrel towards the exit
Answer:
we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level
for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.
Explanation:
A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?
It decreases in speed on its way down and increases in speed on its way down.
it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center
.It increases in speed on his way down because its under the influence of gravity
from newton's equation of motion we can check by
using V^2=u^2+2as
we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level
for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.
Answer:
432.78 Kg
Explanation:
From the question given above, the following data were obtained:
Distance apart (r) = 6.8 m
Force of attraction (F) = 5.4×10¯⁸ N
Mass of Daffy Duck (M₁) = 86.5 kg
Mass of Minnie Duck (M₂) =?
NOTE: Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
The mass of Minnie Duck can be obtained as follow:
F = GM₁M₂ / r²
5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 6.8²
5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 46.24
Cross multiply
6.67×10¯¹¹ × 86.5 × M₂ =5.4×10¯⁸ × 46.24
Divide both side by 6.67×10¯¹¹ × 86.5
M₂ = 5.4×10¯⁸ × 46.24 / 6.67×10¯¹¹ × 86.5
M₂ = 432.78 Kg
Therefore, the mass of Minnie Duck is 432.78 Kg
Answer: 50.7 J
Explanation:
Given
mass of lion is 
The initial speed of the lion is 
increased speed of lion is 
Initially, its kinetic energy is 
Final kinetic energy 
work did by lion after speed up is 
![\Rightarrow W=\dfrac{1}{2}\times 2.6[8^2-5^2]\\\\\Rightarrow W=1.3\times [39]=50.7\ J](https://tex.z-dn.net/?f=%5CRightarrow%20W%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%202.6%5B8%5E2-5%5E2%5D%5C%5C%5C%5C%5CRightarrow%20W%3D1.3%5Ctimes%20%5B39%5D%3D50.7%5C%20J)
