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Serjik [45]
3 years ago
14

Which kind of electromagnetic waves has a wavelengths greater than 10 cm?

Physics
1 answer:
bonufazy [111]3 years ago
3 0
Radio waves
Hope this helps
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Determine the mechanical energy of this object a 1-kg ball rolls on the ground at <br> m/s
dedylja [7]
Mechanical energy = potential energy + kinetic energy
The ball is on the ground so it has no potential energy. that's all i know.
8 0
3 years ago
Which types of waves requires matter to carry energy? electromagnetic waves only mechanical waves only electromagnetic and mecha
gogolik [260]

Answer: Mechanical waves

Explanation:

Mechanical waves require a medium in order to transport their energy from one location to another. A sound wave is an example of a mechanical wave. Sound waves are incapable of traveling through a vacuum.

7 0
3 years ago
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Inertia law what happens when You are standing on the bus and the bus stops abruptly:
I am Lyosha [343]

Answer:

You will fly forward in the bus until you hit something.

Explanation:

While standing there on the bus, you are traveling at the same speed as the bus. If the bus suddenly stops, you will still be traveling at the same speed you started with. That is until you hit something hard enough or big enough to stop you.

4 0
4 years ago
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A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor sto
Fed [463]

Answer:

a)   a = 34.375 m / s²,  b)    v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

             v_f = \frac{x-x_1}{t}

we substitute the values

             v_f = \frac{ 6600 -x_1}{4}  

The initial part of the movement is carried out with acceleration

             v_f = v₀ + a t

             x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

             x₁ = ½ a t²

             v_f = a t

we substitute the values

              x₁ = 1/2  a 16²

              x₁ = 128 a

              v_f = 16 a

let's write our system of equations  

               v_f = \frac{6600 - x_1}{4}

               x₁ = 128 a

               v_f = 16 a

we substitute in the first equation  

               16 a = \frac{6600 -128 a}{4}

               16 4 a = 6600 - 128 a

                a (64 + 128) = 6600

                a = 6600/192

                 a = 34.375 m / s²

b) let's find the time to reach this height

                x = ½ to t²

                t² = 2y / a

                t² = 2 5100 / 34.375

                t² = 296.72

                t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

               v_f = 16 a

               v_f = 16 34.375

               v_f = 550 m / s

8 0
3 years ago
A 2.55 kg block starts from rest on a rough inclined plane that makes an angle of 60 degree with the horizontal. the coefficient
lorasvet [3.4K]

Answer:

Change in mechanical energy = work done by friction

so it is equal to

W = -8.16 J

Explanation:

As we know that change in mechanical energy must be equal to the work done by non conservative forces only

So here when block moves down the inclined plane then the work done by friction force is given as

W = F.d

here we have

F = \mu F_n

here we know that

F_n = mg cos\theta

so we have

F_n = 2.55(9.81)(cos60)

F_n = 12.5 N

Now the friction force on the block is given as

F_f = \mu F_n

F_f = 0.25 \times 12.5

F_f = 3.13 N

now work done by the friction is given as

W = -(3.13)(2.61)

W = -8.16 J

7 0
4 years ago
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