We can use the equation for Newton's Law of Gravitation
Fg = (Gm₁m₂)/r²
Where gravitational constant = G = 6.674 x 10⁻¹¹ N · m²/kg²
mass m₁ = 0.145 kg
mass m₂ = 6.8 kg
distance between centers of masses = r = 0.5 m
Substitute these values into...
Fg = (Gm₁m₂)/r²
Fg = ((6.674 x 10⁻¹¹)(0.145)(6.8)) / (0.5)²
Fg = 2.63 x 10⁻¹⁰ N
Therefore, your answer should be <span>2.6 × 10–10</span>
Answer:
<u>B. the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animal - like life.</u>
Explanation:
The appropriate spectral range for habitable stars is considered to be "late F" or "G", to "mid-K" or even late "A". <em>This corresponds to temperatures of a little more than 7,000 K down to a little less than 4,000 K</em> (6,700 °C to 3,700 °C); the Sun, a G2 star at 5,777 K, is well within these bounds. "Middle-class" stars (late A, late F, G , mid K )of this sort have a number of characteristics considered important to planetary habitability:
• They live at least a few billion years, allowing life a chance to evolve. <em>More luminous main-sequence stars of the "O", "B", and "A" classes usually live less than a billion years and in exceptional cases less than 10 million.</em>
• They emit enough high-frequency ultraviolet radiation to trigger important atmospheric dynamics such as ozone formation, but not so much that ionisation destroys incipient life.
• They emit sufficient radiation at wavelengths conducive to photosynthesis.
• Liquid water may exist on the surface of planets orbiting them at a distance that does not induce tidal locking.
<u><em>Thus , the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animak - like life.</em></u>
Answer:
option E
Explanation:
given,
diameter = 4 mm
shutter speed = 1/1000 s
diameter of aperture = ?
shutter speed = 1/250 s
exposure time to the shutter time
N is the diameter of the aperture and t is the time of exposure
now,
inserting all the values
N₂² = 4
N₂ = 2 mm
hence , the correct answer is option E
Answer:
Explanation:
Height covered = 12m
time to fall by 12 m
s = 1/2 gt²
12 = 1/2 g t²
t = 1.565 s
Horizontal distance of throw
= 8.5 x 1.565
= 13.3 m
This distance is to be covered by dog during the time ball falls ie 1.565 s
Speed of dog required = 13.3 / 1.565
= 8.5 m /s
b ) dog will catch the ball at a distance of 13.3 m .
Answer:
C. 3.00 s
Explanation:
Given:
Δy = 1.80 m − 46.0 m = -44.2 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
-44.2 m = (0 m/s) t + ½ (-9.8 m/s²) t²
t = 3.00 s
