Answer:
(D) 0.99 cm
Explanation:
Given that the radius of curvature of the mirror is 25 cm.
And another car is following which is behind the mirror of 20 m.
Focal length is half of the radius of curvature and it is negative for convex lens.
Now the mirror formula.
So,
Now
Magnification is,
So,
So, Height of the image
Therefore, the image height is 0.99 cm.
Answer:
(a) The speed of the target proton after the collision is:, and (b) the speed of the projectile proton after the collision is: .
Explanation:
We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:, and y axle:. Now replacing the value given as: , for the projectile proton and according to the problem are perpendicular so , and assuming that , we get for x axle: and y axle: , then solving for , we get: and replacing at the first equation we get:, now solving for , we can find the speed of the projectile proton after the collision as: and , that is the speed of the target proton after the collision.
Answer: Part(a)=0.041 secs, Part(b)=0.041 secs
Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction
now we know that
a=-9.81 m/s^2 ( negative because it is pulling the player downwards)
we also know that
s=76 cm= 0.76 m ( maximum s)
using kinetic equation
where v is final velocity which is zero at max height and u is it initial
hence
now we can find time in the 15 cm ascent
using quadratic formula
t=0.0409 sec
the answer for the part b will be the same
To find the answer for the part b we can find the velocity at 15 cm height similarly using
where s=0.76-0.15
as the player has traveled the above distance to reach 15cm to the bottom
when the player reaches the bottom it has the same velocity with which it started which is 3.861
hence the time required to reach the bottom 15cm is
t=0.0409
Answer:
12.7m/s
Explanation:
Given parameters:
Mass of the diver = 77kg
Height = 8.18m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we use one of the motion equations.
v² = u² + 2gh
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
h is the height
v² = 0² + (2 x 9.8 x 8.18)
v² = 160.3
v = 12.7m/s