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Alexxx [7]
2 years ago
6

A certain slide projector has a 100 mm focal length lens. How far away is the screen, if a slide is placed 103 mm from the lens

and produces a sharp image?
Physics
1 answer:
Vlad [161]2 years ago
4 0

Answer:

3.43 m

Explanation:

f = 100 mm

u = - 103 mm

Let v be the distance between the screen and the lens of the projector.

Use lens equation

1 / f = 1 / v - 1 / u

1 / 100 = 1 / v + 1 / 103

1 / v = 1 / 100 - 1 / 103

1 / v = (103 - 100) / (100 x 103)

1 / v = 3 / 10300

v = 3433.33 mm = 3.43 m

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If no friction acts on a diver during a dive, then which of the following statements is true? A) The total mechanical energy of
EleoNora [17]
If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.

Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).

The initial PE = mgh₁ and the initial KE  = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
 
Therefore
(KE + PE) ₁ = (KE + PE)₂

Evaluate the given answers.
A) The total mechanical energy of the system increases.
     FALSE

B) Potential energy can be converted into kinetic energy but not vice versa.
     TRUE

C) (KE + PE)beginning = (KE + PE) end.
     TRUE

D) All of the above.
     FALSE

4 0
2 years ago
Read 2 more answers
Because the soles of your shoes have cleats, you can exert a forward force of 100 N even on slippery ice. A 10-kg picnic cooler
Brilliant_brown [7]

Answer:

you must throw 3 snowballs

Explanation:

We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment

Initial. Before bumping that refrigerator

          p₀ = n m v₀

Where n is the snowball number

Final. When the refrigerator moves

         pf = (n m + M) v

The moment is preserved because the forces during the crash are internal

        n m v₀ = (n m + M) v

        n m (v₀ - v) = M v

        n = M/m    v/(vo-v)

Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton

          F = m a

          a = F / m

The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm

          v₁² = v₀² + 2 a x

          v₁² = 0+ 2 (100/1) 1

          v₁ = 14.14 m / s

This is the initial speed for the crash

         v₀ = v = 14.14 m / s

  Let's calculate

           n = M/m   v/ (v₀-v)

           n = 10/1   3 / (14.14 -3)

          n = 2.7 balls

you must throw 3 snowballs

7 0
2 years ago
A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one
forsale [732]

Answer:

C. Both reach the bottom at the same time.

Explanation:

For a rolling object down an inclined plane , the acceleration is given below

a = g sinθ / (1 + k² / r² )

θ is angle of inclination , k is radius of gyration , r is radius of the cylinder

For cylindrical object

k² / r² = 1/2

acceleration =  g sinθ  /( 1 + 1/2 )

= 2 g sinθ / 3  

Since it does not depend upon either mass or radius , acceleration of both the cylinder will be equal . Hence they will reach the bottom simultaneously.

6 0
2 years ago
A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 cans, 0.355 - l each,
Alona [7]
Flow rate = 220*0.355 l/m = 78.1 l/min = 1.3 l/s = 0.0013 m^3/s

Point 2:
A2= 8 cm^2 = 0.0008 m^2
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa

Point 1:
A1 = 2 cm^2 = 0.0002 m^2
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 = ?
Height = 1.35 m

Applying Bernoulli principle;
P2+1/2*V2^2/density = P1+1/2*V1^2/density +density*gravitational acceleration*height
=>152000+0.5*1.625^2*1000=P1+0.5*6.5^2*1000+1000*9.81*1.35
=> 153320.31 = P1 + 34368.5
=> P1 = 1533210.31-34368.5 = 118951.81 Pa = 118.95 kPa
3 0
2 years ago
Read 2 more answers
Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba
lions [1.4K]

Answer:

a) Probability mass function of x

x P(X=x)

0 0.0602

1 0.0908

2 0.1700

3 0.2050

4 0.1800

5 0.1550

6 0.0843

7 0.0390

8 0.0147

b) Cumulative Distribution function of X

x F(x)

0 0.0602

1 0.1510

2 0.3210

3 0.5260

4 0.7060

5 0.8610

6 0.9453

7 0.9843

8 1.0000

The cumulative distribution function gives 1.0000 as it should.

Explanation:

Probability of arriving late = 0.43

Probability of coming late = 0.57

Let's start with the probability P(X=0) that exactly 0 people arrive late, the probability P(X=1) that exactly 1 person arrives late, the probability P(X=2) that exactly 2 people arrive late, and so on up to the probability P(X=8) that 8 people arrive late.

Interpretation(s) of P(X=0)

The two singles must arrive on time, and the three couples also must. It follows that P(X=0) = (0.57)⁵ = 0.0602

Interpretation(s) of P(X=1)

Exactly 1 person, a single, must arrive late, and all the rest must arrive on time. The late single can be chosen in 2 ways. The probabiliy that (s)he arrives late is 0.43.

The probability that the other single and the three couples arrive on time is (0.57)⁴

It follows that

P(X=1) = (2)(0.43)(0.57)⁴ = 0.0908

Interpretation(s) of P(X=2)

Two late can happen in two different ways. Either (i) the two singles are late, and the couples are on time or (ii) the singles are on time but one couple is late.

(i) The probability that the two singles are late, but the couples are not is (0.43)²(0.57)³

(ii) The probability that the two singles are on time is (0.57)²

Given that the singles are on time, the late couple can be chosen in 3 ways. The probability that it is late is 0.43 and the probability the other two couples are on time is (0.57)².

So the probability of (ii) is (0.57)²(3)(0.43)(0.57)² which looks better as (3)(0.43)(0.57)⁴ It follows that

P(X=2) = (0.43)²(0.57)³ + (3)(0.43)(0.57)⁴ = 0.0342 + 0.136 = 0.1700

Interpretations of P(X=3).

Here a single must arrive late, and also a couple. The late single can be chosen in 2 ways. The probability the person is late but the other single is not is (0.43)(0.57).

The late couple can be chosen in 3 ways. The probability one couple is late and the other two couples are not is (0.43)(0.57)². Putting things together, we find that

P(X=3) = (2)(3)(0.43)²(0.57)³ = 0.2050

Interpretation(s) P(X=4)

Since we either (i) have the two singles and one couple late, or (ii) two couples late. So the calculation will break up into two cases.

(i) Two singles and one couple late

Two singles' probability of being late = (0.43)² and One couple being late can be done in 3 ways, so its probability = 3(0.43)(0.57)²

(ii) Two couples late, one couple and two singles early

This can be done in only 3 ways, and its probability is 2(0.57)³(0.43)²

P(X=4) = (3)(0.43)³(0.57)² + (3)(0.57)³(0.43)² = 0.0775 + 0.103 = 0.1800

Interpretations of P(X=5)

For 5 people to be late, it has to be two couples and 1 single person.

For couples, The two late couples can be picked in 3 ways. Probability is 3(0.43)²(0.57)

The late single person can be picked in two ways too, 2(0.43)(0.57)

P(X=5) = 2(3)(0.43)³(0.57)² = 0.1550

Interpretations of P(X=6)

For 6 people to be late, we have either (i) the three couples are late or (ii) two couples and the two singles.

(i) Three couples late with two singles on time = (0.43)³(0.57)²

(ii) Two couples and two singles late

Two couples can be selected in 3 ways, so probability = 3(0.43)²(0.57)(0.43)²

P(X=6) = (0.43)³(0.57)² + 3(0.43)⁴(0.57) = 0.0258 + 0.0585 = 0.0843

Interpretation(s) of P(X=7)

For 7 people to be late, it has to be all three couples and only one single (which can be picked in two ways)

P(X=7) = 2(0.57)(0.43)⁴ = 0.0390

Interpretations of P(X=8)

Everybody had to be late

P(X=8) = (0.43)⁵ = 0.0147

6 0
2 years ago
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