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Ede4ka [16]
2 years ago
9

A net force of 15 N is exerted on an encyclopedia to cause it to accelerate at a rate of 5 m/s2. Determine the mass of the encyc

lopedia.
Physics
1 answer:
lina2011 [118]2 years ago
6 0

Answer:

<h2>3 kg</h2>

Explanation:

The mass of the encyclopedia can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{15}{5} \\

We have the final answer as

<h2>3 kg</h2>

Hope this helps you

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Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l
Masja [62]

Answer:

a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

Explanation:

The builder's equation is

          1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

      I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0
3 years ago
If a data point is way off the trend line which of the following will not help resolve the problem
S_A_V [24]
Need more details to answer
3 0
3 years ago
A 0.50-kg red cart is moving rightward with a speed of 40 cm/s when it collides
lukranit [14]

The momentum of the red cart before the collision is 0.2 kgm/s and the blue cart is 0.

The momentum of the red cart after the collision is 0.05 kgm/s and the blue cart is 0.15 kgm/s.

The change in momentum of the system of the carts is 0.

<h3>Initial momentum of the carts before collision</h3>

The momentum of the carts before the collision is calculated as follows;

P(red) = 0.5 kg x 0.4 m/s = 0.2 kgm/s

P(blue) = 1.5 x 0 = 0

<h3>Momentum of the carts after collision</h3>

The momentum of the carts after the collision is calculated as follows;

P(red) = 0.5 x 0.1 = 0.05 kgm/s

P(blue) = 1.5 0.1 = 0.15 kgm/s

<h3>Change in momentum of the carts</h3>

\Delta P = P_f - P_i

ΔP = (0.05 + 0.15) - (0.2)

ΔP = 0

Learn more about momentum here: brainly.com/question/7538238

8 0
2 years ago
1.Suppose someone pulls a cart up a ramp a distance of 85cm along the ramp with a force of 15N.
Drupady [299]

1. 12.75 J

Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

W=Fd

where

F = 15 N is the magnitude of the force

d = 85 cm = 0.85 m is the displacement of the cart

Substituting in the formula, we get

W=(15 N)(0.85 m)=12.75 J


2. 10.6 N

In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).

Therefore, the work done is

W=Fd=12.75 J

However, in this case the displacement is

d = 120 cm = 1.20 m

Therefore, the magnitude of the force in this case is

F=\frac{W}{d}=\frac{12.75 J}{1.20 m}=10.6 N

3 0
3 years ago
seagull is flying at a rate of 20 miles per hour south, it encounters wind blowing 20 miles per hour north. What is the resultan
KiRa [710]

Answer:

the winds will make the bird stop

Explanation:

is basically 20 - 20

8 0
3 years ago
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