Answer:
The eight Moon phases:
Waxing Crescent: In the Northern Hemisphere, we see the waxing crescent phase as a thin crescent of light on the right. First Quarter: We see the first quarter phase as a half moon. Waxing Gibbous: The waxing gibbous phase is between a half moon and full moon.
The phases of the Moon are the different ways the Moon looks from Earth over about a month. As the Moon orbits around the Earth, the half of the Moon that faces the Sun will be lit up. The different shapes of the lit portion of the Moon that can be seen from Earth are known as phases of the Moon.
<h2>The 8 phases (in order) are:</h2>
- New moon.
- Waxing Crescent.
- First Quarter.
- Waxing Gibbous.
- Full moon.
- Waning Gibbous.
- Third Quarter.
- Waning Crescent.
Explanation:
Hope it is helpful....
Answer:
The tension increases to four times its original value.
Explanation:
v = Velocity
r = Radius
m = Mass of stone
The centripetal force is
The tension will balance the centripetal force
The new tension will be 4 times the old tension
Answer:
a) 27.2 V
b)27.2 V
Explanation:
Charge of the electron =charge of the proton = q = 1.6 × 10⁻¹⁹ C
Radius = r = 0.53×10⁻¹⁰ m
Electric Potential = V = k q/r
k = 9 ×10⁹ N m²/C² = Coulomb's constant.
V = (9 ×10⁹)(1.6 × 10⁻¹⁹)/( 0.53×10⁻¹⁰) = 27.2 V
b) Potential Energy of the electron = k q × q / r
= [(9 ×10⁹)(1.6 × 10⁻¹⁹)(1.6 × 10⁻¹⁹) / (0.53×10⁻¹⁰)] / (1.6 × 10⁻¹⁹) eV,
since 1 electron volt = (1.6 × 10⁻¹⁹)joules
= 27.2 eV
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
The final velocity of the ball that is dropped from a height of 200m is v = 44.73 m/s .
<h3>What is velocity with example?</h3>
The rate at which an object is travelling in one direction is referred to as its velocity. an automobile traveling north on a highway, or a rocket taking off. Its velocity vector's absolute value always is equal to the motion's speed because it is a scalar.
<h3>Briefing:</h3>
Given the initial velocity of the ball (u) = 0
Distance travelled by the ball (s) = 200m
Acceleration (a) = 10 m/s²
As we know:
v² = u² + 2as
Putting values:
v² = 0+2 × (10 m/s²) × (200 m)
v = 44.73 m/s.
To know more about Velocity visit:
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