Answer:
2.605m
Explanation:
Using the formula for calculating Range (distance travelled in horizontal direction)
Range R = U√2H/g
U is the speed = 4.8m/s
H is the maximum height = ?
g is the acc due to gravity = 9.8m/s²
R = 3.5m
Substitute into the formula and get H
3.5 = 4.8√2H/9.8
3.5/4.8 = √2H/9.8
0.7292 = √2H/9.8
square both sides
0.7292² = 2H/9.8
2H = 0.7292² * 9.8
2H = 5.21
H = 5.21/2
H = 2.605m
Hence the height of the ball from the ground is 2.605m
1 g = 1 ÷ 1000 kg
= 0.001 kg
1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³
1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³
The density is 1000 kg/m³.
Answer:
F = 4212 N
Explanation:
Given that,
Mass of a car, m = 1300 kg
Speed of car on the road is 9 m/s
Radius of curve, r = 25 m
We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :

So, the force has a magnitude of 4212 N
Answer:
a) E = 0
b) 
Explanation:
The electric field for all points outside the spherical shell is given as follows;
a) 
From which we have;

E = 0/A = 0
E = 0
b) 


By Gauss theorem, we have;

Therefore, we get;

The electrical field outside the spherical shell


Therefore, we have;

Answer:
(a) The angle of projection is 63 degree.
(b) The velocity of projection is 24.5 m/s.
Explanation:
Height, h = 1 m
horizontal distance, d = 50 m
time, t = 4.5 s
Let the initial velocity is u and the angle is A.
(a) Horizontal distance = horizontal velocity x time
50 = u cos A x 4.5
u cos A = 11.1 .....(1)
Use second equation of motion in vertical direction

Divide (2) by (1)
tan A = 1.97
A = 63 degree
(b) Substitute the value of A in equation (2)
u x sin 63 = 21.8
u = 24.5 m/s