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Firdavs [7]
3 years ago
15

Radio waves travel 300,000,000 m/s. The frequency is 101,700,000. ehats the wavelength​

Physics
1 answer:
Nesterboy [21]3 years ago
8 0

Answer:

we have formula of frequency :

frequency(f)= speed of sound(c)/wavelength(λ)

for wavelength we swipe it with frequency as follows

λ=c/f

λ=300,000,000/101,700,000

λ=2.949

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A projectile is fired with initial speed vo at an angle of 45o above the horizontal. Assume no air resistance.
katrin2010 [14]

Answer:

The correct answer is a

Explanation:

At projectile launch speeds are

X axis     vₓ = v₀ = cte

Y axis     v_{y} = v_{oy} –gt

The moment is defined as

         p = mv

For the x axis

         pₓ = mvₓ = m v₀ₓ

As the speed is constant the moment is constant

For the y axis

        p_{y} = m v_{y} = m (v_{oy} –gt) = m v_{oy} - m (gt)

Speed ​​changes over time, so the moment also changes over time

Let's examine the answer

i   True

ii False.  The moment changes with time

The correct answer is a

7 0
3 years ago
An object 1,000 km above earths surface would accelerate
Verizon [17]

The acceleration which is gained by an object because of the gravitational force is called its acceleration due to gravity. Its SI unit is m/s2. Acceleration due to gravity is a vector, which means it has both a magnitude and a direction. The formula is ‘the change in velocity= gravity x time’ The acceleration due to gravity at the surface of Earth is represented as g. It has a standard value defined as 9.80665 m/s2.[1]

5 0
3 years ago
A skydiver has a mass of 65 kg.<br> Calculate the weight of the skydiver.
Vinil7 [7]
F=m*a
F=65 kg *9.8 m/s^2
F=637 N (Newtons) — this is the weight
3 0
3 years ago
Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa
lyudmila [28]

Answer:

  • No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field \vec{E} is related to the gradient of the electric potential V with :

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})

This means that for constant electric potential the electric field must be zero:

V(\vec{r}) = k

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k

\vec{E} (\vec{r}) = -  (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k

\vec{E} (\vec{r}) = -  (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})

\vec{E} (\vec{r}) = -  (0,0,0)

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4

give an electric field of zero at point (0,0,0)

8 0
3 years ago
What is the primary difference between an ideal emf device and a real emf device?.
EleoNora [17]

Answer:

A real emf device has an internal resistance, but an ideal emf device does not.

3 0
2 years ago
Read 2 more answers
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