A sort of electricity is a light bulb or a phone / computer charger. plants food water. the sun and rain . that’s what i’m guessing!
Complete question is;
An experiment is carried out to measure the extension of a rubber band for different loads.
The results are shown in the image attached.
What figure is missing from the table?
Answer:
17.3 cm
Explanation:
The image attached showed values for load, extension and initial length.
Now, the first length there is 15.2 cm and as such it's corresponding extension is 0 because it has no preceding measured length.
The second measured length is 16.2 cm. Since it's initial measured length is 15.2 cm, then the extension has a formula; final length - initial length.
This gives: 16.2 - 15.2 = 1 cm
This corresponds to what is given in the table.
For the next measured length, it is blank but we are given the extension to be 2.1 cm. Now, since the initial measured length is 15.2 cm.
Thus;
2.1 cm = Final length - 15.2 cm
Final length = 15.2 + 2.1
Final length = 17.3 cm
Frictional force between the object and the floor=5 N
Explanation:
power= 50 W
velocity= 10 m/s
power= force * velocity
50=F * 10
F=50/10
F=5 N
Thus the force of friction= 5 N
Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s
