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tensa zangetsu [6.8K]
3 years ago
7

A converging lens of focal length 20 cm is placed in contact with a diverging lens of focal length 30 cm. The focal length of th

is combination is
Physics
1 answer:
kramer3 years ago
7 0

Answer:

The focal lenth (F) =+10.0cm

Explanation:

The formular for combined focal length (F) is given as;

1/f = 1/f_{1} + 1/f_{2}

In this  question,

F1 = 20cm

F2  = -30cm

Plugging the values into the formuar above,

1/f = 1/20 + 1/(-30)

1/f = 0.05 - 0.033[tex]1/f = -0.017f = [tex]1/ -0.017

f = 58.82cm

i.e. the combination behaves as a converging lens (because of the postive sign) of focal length 58.82cm .

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Heptane and water do not mix, and heptane has a lower density (0.684 g/mL) than water (1.00 g/mL). A graduated cylinder contains
lakkis [162]

Given that the density of heptane is

d_h=\frac{0.684g}{mL}

The mass of heptane is

m_h=31\text{ g}

The density of water is

d_w=\frac{1g}{mL}

The mass of water is

m_w=37\text{ g}

The volume of heptane will be

\begin{gathered} V_h=\frac{m_h}{d_h} \\ =\frac{31}{0.684} \\ =45.32\text{ mL} \end{gathered}

The volume of water will be

\begin{gathered} V_w=\frac{m_w}{d_w} \\ =\frac{37}{1} \\ =37\text{ mL} \end{gathered}

Thus, the volume of heptane is 45.32 mL and the volume of water is 37 mL.

The total volume of liquid in the cylinder will be

\begin{gathered} V=V_h+V_w \\ =45.32+37 \\ =82.32\text{ mL} \end{gathered}

The total volume of liquid in the cylinder will be 82.32 mL.

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1 year ago
What happens when an object with a lower density is placed in a container with an
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The only thing that definitely happens in every such case is:
The container becomes heavier.
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3 years ago
Help :((((((((((((((((((((
Ann [662]

3-6 seconds time interval is the object slowing down.

The correct option is C.

<h3>What is a time interval?</h3>

The time interval is the span of time among two specified times. To put it another way, it is the amount of time that has passed between the event's start and finish.

<h3>What are different time intervals?</h3>

The time interval is the length of time that the aim uses to gather data and determine values. The critical overview can be one or more seconds, minutes, hours, days, weeks, or months. The period must be greater than zero and positive. When providing minutes, the amount of minutes must divide evenly by 60.

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The complete question is -

During which time interval is the object slowing down ?

a- 8-10 seconds

b- 6-8 seconds

c- 3-6 seconds

d- 0-3 seconds

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1 year ago
A rock is dropped from a garage roof from rest. the roof is 6.0 m from the ground. determine the velocity of the rock as it hits
Dmitrij [34]

From rest, a rock is dropped from a garage roof. The roof is 6.0 meters above ground level. The rock will reach the earth at a speed of 10.849 meters per second.

<h3>What is velocity?</h3>

The change of displacement with respect to time is defined as the velocity.  Velocity is a vector quantity.

it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.

Given data:

V(Final velocity)=? (m/sec)

h(height)= 6.0 m

u(Initial velocity)=0 m/sec

g(gravitational acceleration)=9.81 m/s²

Newton's third equation of motion:

\rm v_y^2 = u_y^2+ 2gh \\\\\rm v_y^2 = 0+ 2gh\\\\\  v_y= \sqrt{2\times 9.81 \ (m/s^2)\times 6.0 (m)} \\\\ v_y=10.849 \ m/sec

Hence, the velocity of the rock as it hits the ground will be 10.849 m/sec.

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7 0
2 years ago
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
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