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iragen [17]
3 years ago
10

A wave transfers _______ as it moves away from the source.

Physics
1 answer:
stiks02 [169]3 years ago
7 0

Answer:

A wave transfers energy as it moves away from the source.

Explanation:

A wave transfers energy as it moves away from the source. A wave is known to be a disturbance that travels through a medium and transfer energy from a point to the other without causing any permanent displacement of the medium itself.

Let's take water as an example of a medium through which wave is transfered. As the water moves away from its source, the wave generates energy. The wave produces a motion which leads to transfer of kinetic energy as the medium moves from its source since kinetic energy is the energy acquired by a body by virtue of its motion.

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A change in which of the following effects the weight of an object?
Masja [62]
Acceleration due to gravity 
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3 years ago
Is NaCH an element? Why or why not?
amm1812
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3 0
3 years ago
What is the weight of an astronaut who has a mass of 90 kg on the moon? (Note: acceleration due to gravity of the moon is 1.62 N
motikmotik

Answer:

W = 145.8 [N]

Explanation:

To solve this problem we must remember that weight is defined as the product of mass by gravity, in this case lunar gravity.

W = m*g

where:

m = mass = 90 [kg]

g = gravity acceleration = 1.62 [kg/m²]

W = 90*1.62

W = 145.8 [N]

8 0
3 years ago
The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro
qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So,  Q =  ∫∫∫ρdV

Q =  ∫∫∫ρr²sinθdθdrdΦ

Q =  ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q =  ∫∫0.2r⁴[-cosθ]drdΦ

Q =  ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q =  ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q =  ∫∫0.2r⁴-[- 2]drdΦ

Q =  ∫∫0.2r⁴(2)drdΦ

Q =  ∫∫0.4r⁴drdΦ

Q =  ∫0.4r⁴dr∫dΦ

Q =  ∫0.4r⁴dr[Φ]

Q =  ∫0.4r⁴dr[2π - 0]

Q =  ∫0.4r⁴dr[2π]

Q =  ∫0.8πr⁴dr

Q =  0.8π∫r⁴dr

Q =  0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0
3 years ago
(11%) Problem 5: A uniform stationary ladder of length L = 2.7 m and mass M = 11 kg leans against a smooth vertical wall, while
jeka94

Answer:

970.2 N

Explanation:

We are given that

Length of ladder=2.7 m

Mass,M=11 kg

Coefficient of friction=\mu=0.45

\theta=51^{\circ}

Mass of painter=8M

Distance from base=d

We have to find the magnitude of the normal force exerted by the floor on the ladder.

Normal force exerted by floor on the ladder=mg+8mg=9mg

Where g=9.8m/s^2

Normal force exerted by floor on the ladder=9\times 11\times 9.8=970.2N

7 0
3 years ago
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