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azamat
3 years ago
9

Which of the following frequencies falls in the range of RF waves used by commercial radio broadcasting stations?

Physics
2 answers:
dybincka [34]3 years ago
8 0
<span>  A. 6,000,000 Hz =3 they were correct</span>
shusha [124]3 years ago
8 0

It's D. 600,000 Hz

I just took the test :)

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What is 74000000 in scientific notation
bekas [8.4K]
It is 7.4x(10 to the power of 7)
3 0
3 years ago
A person pushing a bin with 45 N of force slides the 3 kg plastic bin on a rough surface with friction. The plastic bin is movin
Firdavs [7]

Answer:

F = 0

Explanation:

Given that,

Force acting to push a bin = 45 N

Mass of the plastic bin, m = 3 kg

The plastic bin is moving with a constant velocity.

We need to find the net force acting on the box. Constant velocity means the change in velocity is equal to 0. It means acceleration will be 0.

As a result, the force acting on the box is equal to 0.

7 0
3 years ago
A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a
vredina [299]

(a) 4.0 s

The acceleration of the car is given by

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

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

a=-5.0 m/s^2 is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s

(b) 40 m

The stopping distance of the car can be calculated by using the equation

v^2 - u^2 = 2ad

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m

(c) -5.0 m/s^2

The deceleration is given by the problem, and its value is -5.0 m/s^2.

(d) 5000 N

The net force applied on the car is given by

F=ma

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

a=5.0 m/s^2 is the magnitude of the acceleration

Solving the formula, we find

F=(1000 kg)(5.0 m/s^2)=5000 N

(e) 2.0\cdot 10^5 J

The work done by the force applied by the car is

W=Fd

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

W=(5000 N)(40 m)=2.0\cdot 10^5 J

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0
4 years ago
8x^3/27y^8×9y^3/12x^2
konstantin123 [22]
2x/ 9y^5  this is your answer

4 0
4 years ago
The force acting between two charged particles a and b is 5.2 x 10^-5 newtons. Charges a and b are 2.4 x 10^-2 apart. If the cha
Aneli [31]
4.62x10^-11
hope this helped!
5 0
4 years ago
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