The relevant equation we can use in this problem is:
h = v0 t + 0.5 g t^2
where h is height, v0 is initial velocity, t is time, g is
gravity
Since it was stated that the rock was drop, so it was free
fall and v0 = 0, therefore:
h = 0 + 0.5 * 9.81 m/s^2 * (4.9 s)^2
<span>h = 117.77 m</span>
Answer:
Explanation:
The region around a charged particle where another charged particle experiences a force of attraction or repulsion is called electric field.
The strength of electric field is defined as the force experienced by the unit positive test charge.
E = F / q
Electric field strength is a vector quantity and it is measured in newton per coulomb.
Where, F is the force of attraction or repulsion between the two charges and q is the test charge on which the electric field strength is to be calculated.
The strength of electric field is more if the field is strong. It means more be the electric field strength at a point more be the electric field.
Answer:
20.4m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Distance = 53m
Time = 5.2s
Unknown:
Acceleration = ?
Solution:
This is a linear motion and we use the right motion equation;
S = ut +
at²
S is the distance
u is the initial velocity
a is the acceleration
t is the time
Insert the parameters and solve;
53 = (0x 5.2) +
x a x 5.2
53 = 2.6a
a =
= 20.4m/s²
Answer:
The length at the final temperature is 11.7 cm.
Explanation:
We need to use the thermal expansion equation:

Where:
- L(0) is the initial length
- ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
- ΔL is the final length minus the initial length (L(f)-L(0))
- α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)
So, we have:



Therefore, the length at the final temperature is 11.7 cm.
I hope it helps you!
Answer:
5.31143691523 m/s²
Explanation:
m = Mass = 280 g
x = Displacement of spring = 21.7 cm
Time period

Angular velocity is given by


From Hooke's law

The acceleration due to gravity on the planet is 5.31143691523 m/s²
Yes, I have been able to satisfy my curiosity.