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:
Explanation:
During rescue missions, different types of energy can be devices for flashlight, this could be human powered energy such as squeezing or compressing. In flashlight electrical energy is converted to light and thermal energy.
A squeezing or compressing to get energy for flashlight can be regarded as "DYNAMO PROCESS" it involves spinning of "fly wheels" into the flashlight through consistent squeezing ,which is connected to a dynamo(Dynamo supply electrical current). Hence the needed light is seen on the bulb of the flashlight.
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
Your answer is infrared, visible, ultraviolet.
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>