Prove f=ma <br>i will give

A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.

kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

$\frac{1}{p}+\frac{1}{q}=\frac{1}{f}$

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm$

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

$y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm$

And the positive sign means the image is upright.

6 0

3 years ago

A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between

SashulF [63]

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

$\Delta W = \Delta KE$

$\Delta W = \frac{1}{2} mv^2$

Here,

m = mass

v = Velocity

Our values are given as,

$m = 79.7kg$

$v = 4.77m/s$

Replacing,

$\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2$

$\Delta W = 907J$

Therefore the mechanical energy lost due to friction acting on the runner is 907J

6 0

3 years ago

Acceleration refers to any changes in

Reptile [31]

Answer:

speed

Explanation:

6 0

3 years ago

Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed v

d1i1m1o1n [39]

1) A) Object 1 has the greater momentum

The magnitude of the momentum of an object is given by:

$p=mv$

where

m is the mass of the object

v is its speed

Object 1 has a mass of $m_1 = 2m$ and a speed of $v_1 = v$ , so its momentum is

$p_1 = m_1 v_1 = (2m)(v)=2mv$

Object 2 has a mass of $m_2 = m$ and a speed of $v_2 = \sqrt{2} v$ , so its momentum is

$p_2 = m_2 v_2 = (m)(\sqrt{2} v)=\sqrt{2}mv$

So we see that $p_1 > p_2$ , so object 1 has the greater momentum.

2) The objects have the same kinetic energy.

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

Object 1 has a mass of $m_1 = 2m$ and a speed of $v_1 = v$ , so its kinetic energy is

$K_1 = \frac{1}{2}m_1 v_1^2 = \frac{1}{2}(2m)(v)^2=mv^2$

Object 2 has a mass of $m_2 = m$ and a speed of $v_2 = \sqrt{2} v$ , so its kinetic energy is

$K_2 = \frac{1}{2}m_2 v_2^2 = \frac{1}{2}(m)(\sqrt{2} v)^2=mv^2$

So we see that $K_1 =K_2$ , so the objects have same kinetic energy

5 0

3 years ago

What produces the distinctive flame color of different substances when they are ignited?

shepuryov [24]

Answer:

Electrons in different metals

Explanation:

When you heat an atom, some of its electrons are "excited* to higher energy levels. When an electron drops from one level to a lower energy level, it emits a quantum of energy. The different mix of energy differences for each atom produces different colours. Each metal gives a characteristic flame emission spectrum.

4 0

3 years ago

Prove f=ma i will give ​

Prove f=ma i will give