11

Two cars are travelling in the same direction on a road. The blue car is travelling at 50 m/s in front of the red car, which is

Anestetic [448]

Explanation:

The speed of the red car, relative to the blue car, is:

v = 75 m/s − 50 m/s

v = 25 m/s

6 0

3 years ago

Can someone please help

ki77a [65]

Answer:

Acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

Given:

initial speed of hammer = 0 $\frac{m}{s}$

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

s = ut + $\frac{1}{2} a t^{2}$

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

s = ut + $\frac{1}{2} a t^{2}$

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + $\frac{1}{2} g 1^{2}$

g = 15 × 2

g = 30 $\frac{m}{s^{2} }$

Thus, acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

8 0

3 years ago

When a person stands on a rotating merry-go-round the frictional force?

wel

Answer. dry friction- <span> resists relative lateral motion of two solid surfaces in contact.</span>

4 0

3 years ago

What happens to molecules when their kinetic energy decreases?

Illusion [34]

Answer:

The speed of molecule decreases and temperature also decreases

Explanation:

Kinetic energy of the molecules of a subsance is directly proportional to the temperature of molecule So as the kinetic energy decrease, temperature also decreases. decreses their speed.

6 0

3 years ago

Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep

SIZIF [17.4K]

Answer:

1) iii i= 1m, 2) iii and iv, 3) i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

f = r / 2

f = 2/2

f = 1 m

The builder's equation

1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

1 / f = 0 + 1 / i

i = f

i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

R₂ = infinity

R₁ > 0

Cas2 Flat-concave (divergent) lens

R₂ = infinity

R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

1 / f₂ = 1 / (L-f₁) + 1 / i

1 / i = 1 / f₂ - 1 / (L-f₁)

1 / i = (L-f₁-f₂) / f₂ (L-f₁)

i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

6 0

4 years ago