Ya it looks great to be in a story
Answer:
28.5 m/s
18.22 m/s
Explanation:
h = 20 m, R = 20 m, theta = 53 degree
Let the speed of throwing is u and the speed with which it strikes the ground is v.
Horizontal distance, R = horizontal velocity x time
Let t be the time taken
20 = u Cos 53 x t
u t = 20/0.6 = 33.33 ..... (1)
Now use second equation of motion in vertical direction
h = u Sin 53 t - 1/2 g t^2
20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)
t = 1.17 s
Put in equation (1)
u = 33.33 / 1.17 = 28.5 m/s
Let v be the velocity just before striking the ground
vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s
vy = uSin 53 - 9.8 x 1.17
vy = 28.5 x 0.8 - 16.66
vy = 6.14 m/s
v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2
v = 18.22 m/s
Answer:
Local action is removed by amalgamating zinc rod.
Explanation:
This prevents chemical reaction to occur because impurities in the zinc cannot get into contact with an electrolyte.
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
Answer:B 20 newtons opposite to the direction of the applied force
Explanation:
