potential is equal to mgh
we know that mass = 1220 kg
g=9.8 0r 10
400=1220*9.8*x
400=11956x
x=400/11956
Answer:
7m/s²
Explanation:
Given parameters:
Velocity at the top = 1m/s
Velocity at the bottom = 36m/s
Time = 5s
Unknown:
Average acceleration = ?
Solution;
Acceleration is the rate of change of velocity with time. It is expressed as;
A = 
v is the velocity at the top
u is the velocity at the bottom
t is the time taken
Now, insert the parameters and solve;
A = 
= 7m/s²
Answer:
please read the answer below
Explanation:
To find the electric field you can consider the Gaussian law for a cylindrical surface inside the slab.
where Qint is the charge inside the Gaussian surface, AG is the area of the surface and rho is the charge density of the slab.
By using the formula for the volume of a cylinder you obtain:
where h is the height. If you assume that the slab is in the interval (-zo<z<z0) you can write VG:
Finally, by replacing in the expression for E you get:
hence, for z>0 you obtain E=pz/eo > 0
for z<0 -> E=pz/eo < 0
Answer:
because they are same and their properties
Answer:
F=(-4.8*10^22,0,0) N
Explanation:
<u>Given :</u>
We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°
Solution :
We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.
The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next
F=|F|cosФp (1)
Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet
F=|F|cosФp
=-4.8*10^22 N*p
<em>As this force is in one direction, we could get its vector as next </em>
F=(-4.8*10^22,0,0) N
F=(0,-4.8*10^22,0) N
F=(0,0-4.8*10^22) N
The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.
