the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Explanation:

When the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Changes can occur that the gliders unite and move with a cosecant speed less than the initial one.

The whole process must be analyzed using conservation of the moment.

p₀ = m v₀

celestines que clash case

p_f = (m + M) v

po = pf

m v₀ = (n + M) v

v = $\frac{m}{m+M}$

calculemos

v= $\frac{0.200}{0.200+M} 0.450$

v= 0.09 m/s

elastic shock case

p₀ = m v₀

p_f = m v₁ +M v₂

p₀ = p_f

m v₀ = m v₁ + m v₂

6 0

3 years ago

Question 1 of 10

ss7ja [257]

Answer: ME= E total - E thermal

5 0

2 years ago

Calculate the number of images formed by two plane mirrors inclined at angel of 18 to each other

allochka39001 [22]

Answer:

please give me brainlist and follow

Explanation:

Formula for number of images formed by two plane mirrors incident at an angle θ is n = 360∘θ. If n is even, the number of images is n-1, if n is an odd number of images.

4 0

2 years ago

In a particular case of an object in front of a spherical mirror with a focal length of +12.0 cm, the magnification is +4.00.(a)

salantis [7]

Answer:

9 cm

-36 cm

Explanation:

u = Object distance

v = Image distance

f = Focal length = 12

m = Magnification = 4

$m=-\frac{v}{u}\\\Rightarrow 4=-\frac{v}{u}\\\Rightarrow v=-4u$

Lens equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{12}=\frac{1}{u}+\frac{1}{-4u}\\\Rightarrow \frac{1}{12}=\frac{3}{4u}\\\Rightarrow u=9\ cm$

Object distance is 9 cm

$v=-4\times 9=-36\ cm$

Image distance is -36 cm (other side of object)

7 0

3 years ago