Answer:
D) Joe need to walk 481 m in a direction 40.9° north of east
Explanation:
Given:
1. Walk 300 m north
2. Walk 400 m northwest
3. Walk 700 m east-southeast and the treasure is buried there
Taking north as positive y axis and east as positive x axis.
Resolving their positions into x and y vector components.
1. d1 = 300j
2. d2 = -400cos45i + 400sin45j
3 d3 = 700cos22.5i - 700sin22.5j
Resultant position.
d = d1+d2+d3
d = (-400cos45 + 700cos22.5)i + (300+400sin45-700sin22.5)j
d = (363.873)i + (314.964)j
D = √( (363.873)^2 + (314.964)^2)
D =481.25m ~= 481m
Angle = taninverse (314.964/363.873)
Angle = 40.9°
Since the x and y components are both positive, it implies that their position is north of east
Therefore, Joe need to walk 481 m in a direction 40.9° north of east
Answer:
Question A:17.92 m/s
Question B:0.6 seconds
Explanation:
Question A:
Initial velocity(u)=24m/s
Height(h)=13m
acceleration due to gravity(g)=9.8m/s^2
Final velocity=v
v^2=u^2-2xgxh
v^2=24^2-2x9.8x13
v^2=24x24-2x9.8x13
v^2=576-254.8
v^2=321.2
Take them square root of both sides
v=√(321.2)
v=17.92m/s
Question B:
velocity(v)=17.92m/s
Acceleration due to gravity(g)=9.8m/s^2
Initial velocity(u)=24m/s
Time=t
v=u-gxt
17.92=24-9.8xt
Collect like terms
9.8t=24-17.92
9.8t=6.08
Divide both sides by 9.8
9.8t/9.8=6.08/9.8
t=0.6 approximately
The answer should be c....
Answer:
370.6 nm
Explanation:
wavelength in vacuum = 494 nm
refractive index of water with respect to air = 1.333
Let the wavelength of light in water is λ.
The frequency of the light remains same but the speed and the wavelength is changed as the light passes from one medium to another.
By using the definition of refractive index

where, n be the refractive index of water with respect to air
By substituting the values, we get

λ = 370.6 nm
Thus, the wavelength of light in water is 370.6 nm.
The least value varies depends on ammeter range. In the given question, ammeter range is not mentioned. So, the least value of an ammeter is 0.1 or 0.5 (depends on ammeter range).
<u>Explanation:</u>
The least value of an ammeter is the measure of the smallest number which can be observed in ammeter. So the least value will be varying depending upon its range. If we consider the range of ammeter is 30 A and the scale readings are 10 numbers, then the least value will be 30/10 = 3 A per scale.
So the least value is determined as 
So among the given options 0.1 is most suitable for an ammeter with range of 3 A with 30 divisions.