When not in a vacuum, what force causes objects in freefall to fall at different speeds??

A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

$\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

6 0

3 years ago

Read 2 more answers

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

3 years ago

When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo

Sliva [168]

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s first law, the net force on gymnast,

$F_{net} =F-W=ma$

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast ( $9\times g$ acceleration due to gravity)

Therefore,

$F= ma+W$ OR $F=ma+mg=m(g+a)$

Given $m = 30 kg$ and $a=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}$

Substituting these values in above formula and calculate the force exerted by the gymnast,

$F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )$

$F=3.537\times10^{3}N$

6 0

3 years ago

What is the force exerted on a moving charge of –2.0 μC at a 20° angle through a magnetic field of 3.0 × 10–4 T with a velocity

san4es73 [151]

1.0 x 10-3 N is the correct answer!

6 0

3 years ago

Read 2 more answers

What is the role of Langerhans cells?

boyakko [2]

Langerhans cells exist on the surface of the skin. They help to protect you by keeping dangerous microbes from entering your body. Langerhans cells do this in two very different ways: by protecting your skin from infection and also by stimulating allergic reactions.

4 0

2 years ago