Answer:
1) Given acceleration due to gravity as 10N/kg
N/kg is the same as m/s². From W=mg
making g subject ... you have g = W/m
where m is in kg and W is in N
hence we have N/kg
So g=10ms-²
Gravitational Potential Energy=mgh
At height 4m
PE=6 x 10 x 4
=240J.
At height 6m
PE=6 x 10 x 6
=360J.
2) KE=1/2mv²
=1/2 x 6 x 5² =75J
When the speed is doubled... it becomes 2x
so V = 2x5=10ms-¹
KE=1/2 x 6 x 10² =300J.
3) KE=1/2mv²
KE is given as 100J
mass=0.5kg
Making V the subject
you have v² =2KE/m
v²=2 x 100/0.5
v²=400
taking square root
v=20ms-¹.
Have a great day✌
Answer:
a

b

Explanation:
From the question we are told that
The speed of the airplane is 
The angle is 
The altitude of the plane is 
Generally the y-component of the airplanes velocity is

=> 
=>
Generally the displacement traveled by the package in the vertical direction is

=> 
Here the negative sign for the distance show that the direction is along the negative y-axis
=> 
Solving this using quadratic formula we obtain that

Generally the x-component of the velocity is

=> 
=>
Generally the distance travel in the horizontal direction is

=> 
=> 
Generally the angle of the velocity vector relative to the ground is mathematically represented as
![\beta = tan ^{-1}[\frac{v_y}{v_x } ]](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%20tan%20%5E%7B-1%7D%5B%5Cfrac%7Bv_y%7D%7Bv_x%20%7D%20%5D)
Here
is the final velocity of the package along the vertical axis and this is mathematically represented as

=>
=>
and v_x is the final velocity of the package which is equivalent to the initial velocity 
So
![\beta = tan ^{-1}[-130.05}{57.96 } ]](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%20tan%20%5E%7B-1%7D%5B-130.05%7D%7B57.96%20%7D%20%5D)

The negative direction show that it is moving towards the south east direction
Explanation:
C is correct.
Newton second law states that force is directly proportional to acceleration with m being the constant of variation.

So


A is wrong, the constant g only happens in free fall or in vertical direction
B and D are wrong due to the mathematical error or equation error
Answer:
Work done= Energy transferred
Explanation:
Work is the transfer of energy. In physics we say that work is done on an object when you transfer energy to that object. If you put energy into an object, then you do work on that object (mass).
Answer
given,
L(t) = 10 - 3.5 t
mass of particle = 2 Kg
radius of the circle = 3.1 m
a) torque
τ = 
τ = 
τ = -3.5 N.m
Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.
L decreases the angular acceleration upward. so, net torque is upward.
b) Moment of inertia of the particle
I = m R^2
I = 2 x 3.1²
I = 19.22 kg.m²
L = I ω
ω = 
ω = 
ω = 
A = 0.52 rad/s B = -0.182 rad/s²