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dezoksy [38]
3 years ago
13

Which of think following materials is the best is the best choice for an insulating material?

Physics
1 answer:
RSB [31]3 years ago
7 0
You are correct, the answer is C. 

Good insulators are materials that do not allow heat, electricity, light or sound pass through easily. Materials like steel and silver (which are metals) are not good insulators because they make good conductors. They easily allow heat, electricity and the like pass through them. 

Thus, the answer is letter C. Rubber. 
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Jack has two boxes: one is 148g and one is 78g. if jack pushes both boxes with the same amount of force which will accelerate fa
kifflom [539]
1) A The 78g
2) C Push on the wagon in the opposite direction as Jack with a force that is the same as Jack is applying.
5 0
3 years ago
Visible light includes all of the
krek1111 [17]

Answer:Visible light is a small part of the electromagnetic spectrum. The spectrum covers everything from gamma rays, x-rays, visible light, infrared, microwave and radio waves. Each part of the spectrum, including the different colors of visible light, have different wavelengths (the space between each wave).

Explanation:

6 0
2 years ago
A) an electron has an initial speed of 226000 m/s. if it undergoes an acceleration of 4.0 x 1014 m/s2, how long will it take to
KIM [24]

initial speed of 226000 m/s

acceleration of 4.0 x 1014 m/s2,

speed of 781000 m/s

What is Acceleration?

  • Acceleration is a rate of change of velocity with respect to time with respect to direction and speed.
  • A point or an object moving in a straight line is accelerated if it speeds up or slows down.
  • Acceleration formula can be written as,

                    a = (v - u ) / t m/s²

As we have to find the time taken, the formula can be altered as,

t = \frac{v-u}{a}

where, t - time taken to reach a final speed

v - final velocity

u - initial velocity

a - acceleration.

Substituting all the given values,

t =\frac{781000 - 226000} {4* 1014}

= 1.3875 × 10⁻⁹ seconds.

So, taken to reach the final speed is found to be 1.3 × 10⁻⁹ 8iH..

7 0
1 year ago
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
3 years ago
A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The
lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

E_{i} = E_{f}

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}                              

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s      

 

3) When the object is at the equilibrium position, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s

I hope it helps you!                                                                                        

8 0
3 years ago
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