A: 132.9w because 2525\19 is how much energy transferred per second which is also known as the power
Answer:
Answer: <u>Height</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>2</u><u>0</u><u>4</u><u> </u><u>m</u>
Explanation:
At the highest point, it is called the maximum height.
• From third newton's equation of motion:
• At maximum height, v is zero
• u is initial speed
• g is -9.8 m/s²
• s is the height
Answer
given,
mass of block (m)= 6.4 Kg
spring is stretched to distance, x = 0.28 m
initial velocity = 5.1 m/s
a) computing weight of spring
k x = m g
k = 224 N/m
b)
c)
d)
e)
A = 0.682 m
Force =
=
F = 94.20 N
-- find the horizontal and vertical components of F1.
-- find the horizontal and vertical components of F2.
-- find the horizontal and vertical components of F3.
-- add up the 3 horizontal components; their sum is the horizontal component of the resultant.
-- add up the 3 vertical components; their sum is the vertical component of the resultant.
-- the magnitude of the resultant is the square root of (vertical component^2 + horizontal component^2)
-- the direction of the resultant is the angle whose tangent is (vertical component/horizontal component), starting from the positive x-direction.
Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:
ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:
r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:
Finally, you obtain for E:
hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C