1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
FromTheMoon [43]
3 years ago
10

To receive a different radio channel, you must change the

Physics
2 answers:
NikAS [45]3 years ago
8 0
<span>To receive a different radio channel, you must change the </span>A. wave frequency being received
forsale [732]3 years ago
6 0
I Believe the answer Is A wave frequency Being recieved


You might be interested in
A 65-kg skier grips a moving rope that is powered by an engine and is pulled at a constant speed to the top of a 230 hill. The s
Sveta_85 [38]

Answer:

The required power by the engine is 33.0 hp

Explanation:

Solution

Newton's second law says that, the net force Fnet on an object of mass m will accelerates the object

Where

Fnet = ma

a = acceleration

θ = angle of incline,

m = mass of the 30 skiers,

f = frictional force

N = normal force

mg sinθ, mg cos θ are components of weight skier

F = the force applied by engine

Now,

The skier mass  is 65 kg

We calculate the mass of the 30 skier

m = 30 (65kg) = 1950 kg

Calculate the net force acting on the skiers along the x-axis

Fnet, x=ma

Now,

F-mg sin θ - f = 0

F= mg sin θ + f -----(1)

The kinetic frictional force is denoted by

f = μk N ------(2)

μk = The coefficient of the kinetic friction

We now, calculate the net force acting on the skiers along y axis

Fnet, y = ma

N- mg cos θ = 0

so,

N = mg cos θ

This value is  substituted in equation (2)

f = μk mg cos θ

we substitute the value for equation (1)

F = mg sin θ + μk mg cos θ

mg =  sinθ + μk cos θ)-----(3)

The next step is to calculate the work done by the engine in pulling the skiers, the incline top by applying the equation 3

W = Fx

= mg ( sinθ + μk cos θ)x

x = the displacement

we now substitute 1950 kg for m, 23° for θ, 0.10 for μk and 320m for x

so,

W = mg ( sinθ + μk cos θ)x

= (1950 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (320 m)

= 2.99 * 10 ^6 J

Then,

The time from minute to s is converted

t =(2.0min) ( 60sec/1.0min) = 120 sec

Now we calculate the power needed by the engine to pull the skiers at the incline top

Thus,

P = W/t

we substitute  2.955 * 10 ^6 J for W and  120 s for t

we have,

P = 2.955 * 10 ^ 6 J/ 120 s

= ( 2.4625 * 10 ^ 4 W) (1.0 hp/746 W)

= 33.0 hp

In conclusion the required power by the engine is 33.0 hp

3 0
3 years ago
A bag of sugar weighs 5.00 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixt
gizmo_the_mogwai [7]

Answer:

Earth: 22.246 N

Moon: 3.71 N

Jupiter: 58.72 N

Explanation:

The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.

To calculate anything, we need to convert to standard measurements.

5.00 lbs = 2.27 kg

On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N

Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N

Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N

3 0
3 years ago
Unpolarized light with an intensity of 22.4 ????ux passes through a polarizer whose transmission axis is vertically oriented. (a
irina1246 [14]

Answer:

a)      I₁ = 11.2 Lux , vertical direction , b)      I₂ = 1.44 Lux

Explanation:

a) A polarized is a system that absorbs light that is not polarized in the direction of its axis, therefore half of the non-polarized light must be absorbed

consequently the above the processed light has half of the incident intensity and the directional of the polarized

          I₁ = I₀ / 2

          I₁ = 22.4 / 2

          I₁ = 11.2 Lux

is polarized in the vertical direction

b) The polarized light falls on a second polarizer, therefore it must comply with the law of Malus

         I₂ = I₁ cos² θ

         I₂ = 11.2 cos² 69

         I₂ = 1.44 Lux

8 0
3 years ago
For a constant voltage, how is the resistance related to the current?
Aleksandr-060686 [28]

Answer:

Resistance is inversely proportional to current, so when the resistance doubles, the current is cut in half. Resistance is directly proportional to current, so when the resistance doubles, the current is cut in half.

8 0
1 year ago
Which factor is the most imortant in performing the activities of daily living
wariber [46]

Answer:

Muscular endurance is the most important factor in performing the activities of daily living.

7 0
3 years ago
Other questions:
  • The life cycle of stars may be boiled down into a tug of war, lasting billions of years, between two basic forces. In your own w
    10·1 answer
  • What location on earth experiences the most change the number of daylight hours?
    8·1 answer
  • How can I make the atomic model of boron?
    9·1 answer
  • A ferris wheel of radius 100 feet is rotating at a constant angular speed Ï rad/sec counterclockwise. using a stopwatch, the rid
    7·1 answer
  • A charge alters the space around it. What is this alteration of space called?
    10·1 answer
  • Find the magnitude of the average induced emf in the coil when the magnet is turned off and the field decreases to 0 T in 2.8 s
    11·1 answer
  • A deuteron (a nucleus that consists of one proton and one neutron) is accelerated through a 4.01 kV potential difference. How mu
    5·1 answer
  • When a rope is wrapped around a wheel (or pulley) and pulled, the lever arm is the ________ Blank 1 of the wheel and the angle b
    8·1 answer
  • What is magnetic resonance ???​
    9·2 answers
  • How do you find the missing terms in a number sentence?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!