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
The gravitational pull of Earth is stronger in satellite A
Answer:
Distance is path length covered by particle. When particle moves along half circle, it covers half the circumference therefore distance covered is (2×pi×r)/2 = pi× r. ... Hence displacement is equal to diameter or 2 times the radius of circle.
Answer:
A
Explanation:
I only think its A because of the gravity part...sorry im not good at explaining
Answer:
Length = 2.32 m
Explanation:
Let the length required be 'L'.
Given:
Resistance of the resistor (R) = 3.7 Ω
Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]
Resistivity of the material of rod (ρ) = 
First, let us find the area of the circular rod.
Area is given as:
Now, the resistance of the material is given by the formula:
Express this in terms of 'L'. This gives,
Now, plug in the given values and solve for length 'L'. This gives,
Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.
