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Leokris [45]
2 years ago
13

PLEASE HELP WILL GIVE BRAINLIEST

Physics
1 answer:
Eduardwww [97]2 years ago
7 0

Multiply the mass by the gravity.

1200kg x -9.8= -11760 N

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The French high-speed train travels at 300 km/h. How long
Agata [3.3K]

Answer:

given , v = 300 km/hr; distance d = 1500 km; then time t = d/v = 1500/300 = 5 hrs

Explanation:

4 0
3 years ago
The alpha line in the balmer series of the hydrogen spectrum consists of light having a wavelength of 6.56. calculate the freque
guajiro [1.7K]
The alpha line in the Balmer series is the transition from n=3 to n=2 and with the wavelength of λ=656 nm = 6.56*10^-7 m. To get the frequency we need the formula: v=λ*f where v is the speed of light, λ is the wavelength and f is the frequency, or c=λ*f. c=3*10^8 m/s. To get the frequency: f=c/λ. Now we input the numbers: f=(3*10^8)/(6.56*10^-7)=4.57*10^14 Hz. So the frequency of the light from alpha line is f= 4.57*10^14 Hz. 
5 0
3 years ago
You turn on your car's headlights while driving at night. What transformation is taking place?
Galina-37 [17]

Answer: C.

Explanation:

7 0
3 years ago
Read 2 more answers
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
2 years ago
A tree moves at 30 km/s relative to the Sun. A squirrel jumps from a branch on
tresset_1 [31]

The reasoning is wrong if we look into Newton's Law of gravitation.

Newton's law of gravitation states  that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

The law is written as follows;

F = \frac{Gm_1m_2}{r^2}

r = \sqrt{\frac{Gm_1m_2}{F} }

The distance between the two particles, is a function of force and their masses not necessarily time of motion.

In the given problem only time of motion was considered which is wrong.

Thus, the reasoning is wrong if we look into Newton's Law of gravitation.

Learn more here: brainly.com/question/19680441

7 0
2 years ago
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