Answer:
b.) All these technologies use radio waves, including high-frequency microwaves
d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.
e.) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.
f.) All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m.
Explanation:
For option D. The frequency range of microwave ovens is 2450 MHz = 2.4 GHz, which intersects with wireless internet technology with range of 2.4 to 2.6 GHz.
For E, wavelenght and frequency are inversely proportional. Wireless internet service has the greatest frequency band and hence the shortest wavelenght band.
For F, in all these radiations, the highest Freq is 2.6 GHz and the lowest is 40 MHz. Wavelenght is speed of light (3x10^8 m/s) divided by the frequency.
2.6 GHz = 2.6x10^9 Hz
Wavelenght = 3x10^8 ÷ 2.6x10^9 = 0.1 m
40 MHz = 40x10^6
Wavelenght = 3x10^8 ÷ 40x10^6 = 7.5 m
60 Miles per hour 60 times 2 is 120
Answer:
A bear normally has a short, thick neck, a rounded head, a pointed muzzle, short ears, and small eyes. Some species have round faces. Bears have poor eyesight, and most have only fair hearing.
Explanation:
Modern Bears are characterized with large body and stocky legs, a long snout, shaggy hair, plantigrade paws with five non-retractile claws and a short tail.
Grizzly bears (Ursus arctos horribilis) have concave faces, a distinctive hump on their shoulders, and long claws about two to four inches long. Both the hump and the claws are traits associated with a grizzly bear's exceptional digging ability. Grizzlies are often dark brown, but can vary from blonde to nearly black.
The brown bear has a slight hump above its shoulder, round ears, a long snout and big paws with long, curved claws that it uses for digging. Unlike the black bear, it can't climb trees. It can weigh between 350-1,500 pounds. When standing on its hind legs it can be up to 5 feet tall.
Hope this helps :)
(I didn't know which type of bear so i did brown bear and grizzly bear)
Answer:
Explanation:
mass of pulley, m3 = 1.5 kg
Radius of pulley, R = 0.09 m
mass of monkey, m2 = 4.5 kg
mass of banana bunch, m1 = 3 kg
Let a is teh acceleration ans T1 and T2 be the tension in the rope.
The moment of inertia of the pulley
I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²
According to Newton's second law
T1 - m1 g = m1 x a .... (1)
m2 g - T2 = m2 x a ..... (2)
(T2 - T1 ) x R = I x α
where, α is the angular acceleration
α = a / R
(T2 - T1)R = 0.5 x m3 x R² x a / R
T2 - T1 = 0.5 x m3 x a ..... (3)
from (1), (2) and (3)
a = 1.78 m/s²
from equation (1)
T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N
from equation (2)
T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N
Answer:
Explanation:
<u>Frictional Force
</u>
When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:
The centripetal acceleration a_c is computed as
Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one
For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as
The normal force N is equal to the weight of the car, thus
Equating both forces
Simplifying
Substituting the values
