What are density conversions

A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal

slamgirl [31]

Answer:

18.89cm

Explanation:

As we know that the person is standing 5m in front of the camera

$d_0=5m=500cm$

The focal length of the lens =50cm

f=50 cm

By Lens formula we have:

$\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm$

By the formula of magnification

$\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm$

The height of the image formed is 18.89cm.

5 0

3 years ago

How does the particle model indicate positive acceleration and how does it indicate negative acceleration? An example of particl

MrMuchimi

Answer:

Explanation:

The diagram has a fairly simple explanation. In the top diagram, the space between the particle is increasing. That means that acceleration is increasing. The bottom diagram shows just the opposite. The particle starts off making large "distances" between where the particle is recorded and then the distances between recordings lessens and the particle is slowing down.

Rule: the greater the "distance" between dot positions, the greater the acceleration, because the speed is large.

Top diagram: increasing distance between dots = larger speed. The distance becomes greater as the particle moves to the right.

Bottom diagram: starts off large and decreases as we move from left to right = - acceleration.

5 0

3 years ago

A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff

Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0

1 year ago

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 33

Anuta_ua [19.1K]

Answer:

$\frac{v_{2}}{v_{1}}=2$ .

Explanation:

The average kinetic energy per molecule of a ideal gas is given by:

$\bar{K}=\frac{3k_{B}T}{2}$

Now, we know that $\bar{K} = (1/2)m\bar{v}^{2}$

Before the absorption we have:

$(1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2}$ (1)

After the absorption,

$(1/2)m\bar{v_{2}}^{2}=\frac{3k_{B}T_{2}}{2}$ (2)

If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)

$\frac{v_{2}^{2}}{v_{1}^{2}}=\frac{T_{2}}{T_{1}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}$

$\frac{v_{2}}{v_{1}}=\sqrt{4}$

Therefore the ratio will be $\frac{v_{2}}{v_{1}}=2$

I hope it helps you!

4 0

3 years ago

Read 2 more answers

A 5.00 kg pendulum swings back and forth. At the top of its arc it reaches a height of 0.36 m. What is the velocity of the pendu

azamat

Answer:

Gravitational potential energy = (mass) x (gravity) x (height)

Kinetic energy (of a moving object) = (1/2) (mass) x (speed)²

When the pendulum is at the top of its swing, its potential energy is

(mass) x (gravity) x (height)

= (5 kg) x (9.8 m/s²) x (0.36 m)

= (5 x 9.8 x 0.36) joules

17.64 joules .

Energy is conserved ... it doesn't appear or disappear ... so that number is exactly the kinetic energy the pendulum has at the bottom of the swing, only now, it's kinetic energy:

17.64 joules = (1/2) x (mass) x (speed)²

17.64 joules = (1/2) x (5 kg) x (speed)²

Divide each side by 2.5 kg:

17.64 joules / 2.5 kg = speed²

Write out the units of joules:

17.64 kg-m²/s² / 2.5 kg = speed²

(17.64 / 2.5) (m²/s²) = speed²

7.056 m²/s² = speed²

Take the square root of each side: Speed = √(7.056 m²/s²) = 2.656 m/s.

Looking through the choices, we're overjoyed to see that one if them is ' 2.7 m/s '. Surely that's IT !

Hope this will help you ...

8 0

3 years ago