Answer:21.45 m/s
Explanation:
Given
Mass of sport car=920 kg
Mass of SUV=2300 kg
distance to which both car skid is 2.4 m
coefficient of friction ()=0.8
Let u be the initial velocity of both car at the starting of skidding
and they finally come to zero velocity
s=2.4 m
u=6.13 m/s
so before colliding sport car must be travelling at a speed of
(conserving momentum)
v=21.45 m/s
Answer:
t = 0.714 s and x = 5.0 m
Explanation:
This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed
vₓ = 7.0 m / s
Let's find the time it takes to get to the river
y = y₀ + v_{oy} t - ½ g t²
the initial vertical speed is zero and when it reaches the river its height is zero
0 = y₀ + 0 - ½ g t²
t =
t = ra 2 2.5 / 9.8
t = 0.714 s
the distance traveled is
x = vₓ t
x = 7.0 0.714
x = 5.0 m
Answer:
f = 12 cm
Explanation:
<u>Center of Curvature</u>:
The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of mirror.
<u>The Radius of Curvature</u>:
The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of mirror. It is the distance from pole to the center of curvature.
<u>Focal Length</u>:
The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’.
The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:
where,
f = focal length = ?
R = Radius of curvature = 24 cm
Therefore,
<u>f = 12 cm</u>