Ionic bonds have...
- High melting points
- Conductive properties
- Are good insulators
(Extra: Form crystal-like structures over just plain molecules!)
Answer:
Trial 1 is the largest, trial 3 is the smallest
Explanation:
Given:
<em>Trial 1</em>
M₁ = 6·10²² kg
d₁ = 3 500 km = 3.5·10⁶ м
<em>Trial 2</em>
M₂ = 6·10²² kg
d₂ = 7 000 km = 7·10⁶ м
<em>Trial 3</em>
M₃ = 3·10²² kg
d₃ = 7 000 km = 7·10⁶ м
___________
F - ?
Gravitational force:
F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)
F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)
F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)
Trial 1 is the largest, trial 3 is the smallest
The most accurate weather for the next two days would be dry warm weather or severe thunderstorms if there is an occluded front over the area.
<u>Explanation:</u>
The weather front that is created during the cyclogenesis process is an occluded front in meteorology. Cyclogenesis process is the development of extra-tropical cyclone and its intensification.
During the occurrence of this, the warm air is occluded (separated) from the center of cyclone at the surface of the earth.
The cold front rotates the storm as it intensifies and holds up the warm front. This develops an occluded front, that is the boundary which separates the new cold air mass and the older cool air mass that is already in warm front's north.
Answer:
a. 2.1 s
b.0.48 Hz
c. A=24cm
d. 72cm/s
Explanation:
An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider completes 15.0 oscillations in 31.0 s.What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?
What are the period,
period is the time taken for a wave particle to make one complete oscillation
a) 31 / 15 = 2.066 seconds
= 2.1 s
(b) frequency
: this the number of oscillation made in one seconds.
it is also the inverse of the period.
= oscillations / time
= 15/31= 0.48 Hz
(c) amplitude
: maximum displacement from the origin
amplitude = 1/2 of the difference of oscillation marks
= 1/2(57-10) = 47/2cm
23.5cm
A=24cm
(d) maximum speed of the glider?
V=ωA
angular frequency *Amplitude
V=a*pi*f*amplitude
2π x frequency x amplitude = maximum speed
= 2π x .48 x 24
=72.38 cm/s
72cm/s
Answer:
The speed of the roller coaster at this point is 18.74 m/s.
Explanation:
Given that,
Weight of the student, W = 655 kg
Weight of the roller coaster, 
Radius of the roller coaster, r = 18 m
At the bottom of the loop, the weight of the roller coaster us given by :
If m is the mass of the roller coaster,
m = 66.83 kg
So,
v = 18.74 m/s
So, the speed of the roller coaster at this point is 18.74 m/s. Hence, this is the required solution.
