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julia-pushkina [17]
2 years ago
8

A vehicle accelerates from 0 to 30 m/s in 10 seconds on a straight road

Physics
2 answers:
Solnce55 [7]2 years ago
6 0
What's the question?

Tresset [83]2 years ago
3 0

Answer:

a = 3 m/s/s

Explanation:

Acceleration is defined as the rate of change in velocity

so we will have

a = \frac{v_f - v_i}{t}

so here we have

v_f = 30 m/s

v_i = 0

\Delta t = 10 s

now we have

a = \frac{30 - 0}{10}

a = 3 m/s^2

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A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initiall
Svetlanka [38]

Answer: (d)

Explanation:

Given

Mass of object m=2\ kg

Speed of object u=5\ m/s

Mass of object at rest M=3\ kg

Suppose after collision, speed is v

conserving momentum

\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s

Initial kinetic energy

k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J

Final kinetic energy

k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J

So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.

4 0
2 years ago
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
A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work
kifflom [539]

Answer: Option D: 5.5×10²Joules

Explanation:

Work done is the product of applied force and displacement of the object in the direction of force.

W = F.s = F s cosθ

It is given that the force applied is, F = 55 N

The displacement in the direction of force, s = 10 m

The angle between force and displacement, θ = 0°

Thus, work done on the object:

W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J

Hence, the correct option is D.

3 0
3 years ago
An electric motor moves a chain that pulls the roller coaster car 180 meters to the top of the first hill. The chain exerts a fo
MrRa [10]
The energy added here is potential energy since it is moving upward 180 meters in a gravitational field.  This is then turned into KE when it rolls down.  2524N x 180m = 454,320J
3 0
3 years ago
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How would you describe the behavior of particles in a solid?
4 0
3 years ago
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