Answer:
d’= (0.561 i ^ - 0.634 j ^) m
, d’= 0.847 m
, 48.5 south east
Explanation:
This is a displacement exercise, one of the easiest methods to solve it is to decompose the displacements in a coordinate system. Let's start with beetle 1
Let's use trigonometry to break down your second displacement
d₂ = 0.89 m θ = 32 north east
sin θ = / d₂
d_{2y} = d2 sin 32
d_{2y} = 0.89 sin 32
d_{2y} = 0.472 m
cos 32 = d₂ₓ / d₂
d₂ₓ = d₂ cos 32
d₂ₓ = 0.89 cos 32
d₂ₓ = 0.755 m
We found the total displacement of the beetle 1
X axis
d₁ = 0.58 i ^
Dₓ = d₁ + d₂ₓ
Dₓ = 0.58 + 0.755
Dₓ = 1,335 m
Axis y
D_{y} = d_{2y}
D_{y} = 0.472 m
Now let's analyze the second beetle
d₃ = 1.37 m θ = 35 north east
Sin (90-35) = d_{3y} / d₃
d_{3y} = d₃ sin 55
d_{3y} = 1.35 sin 55
d_{3y} = 1,106 m
cos 55 = d₃ₓ / d₃
d₃ₓ = d₃ cos 55
d₃ₓ = 1.35 cos 55
d₃ₓ = 0.774 m
They ask us what the second displacement should be to have the same location as the beetle 1
Dₓ = d₃ₓ + dx’
D_{y} = d_{3y} + dy’
dx’= Dₓ - d₃ₓ
dx’= 1.335 - 0.774
dx’= 0.561 m
dy’= D_{y} - d_{3y}
dy’= 0.472 - 1,106
dy’= -0.634 m
We can give the result in two ways
d’= (0.561 i ^ - 0.634 j ^) m
Or in the form of module and address
d’= √ (dx’² + dy’²)
d’= √ (0.561² + 0.634²)
d’= 0.847 m
tan θ = dy’/ dx’
θ = tan⁻¹ dy ’/ dx’
θ = tan⁻¹ (-0.634 / 0.561)
θ = -48.5
º
This is 48.5 south east
Answer:
30 neutrons
Explanation:
A neutral iron atom contains 26 protons and 30 neutrons plus 26 electrons in four different shells around the nucleus. As with other transition metals, a variable number of electrons from iron's two outermost shells are available to combine with other elements.
Answer:
Maximum speed ( v ) = 10.4 m/s (Approx)
Explanation:
Given:
Amplitude A = 15.0 cm = 0.15 m
Frequency f = 11.0 cycles/s (Hz)
Find:
Maximum speed ( v )
Computation:
Angular frequency = 2πf
Angular frequency = 2π(11)
Angular frequency = 69.14
Maximum speed ( v ) = WA
Maximum speed ( v ) = 69.14 x 0.15
Maximum speed ( v ) = 10.371
Maximum speed ( v ) = 10.4 m/s (Approx)
Answer:
The heat is 115478.4 J.
Explanation:
Given that,
Mass of water = 0.400 kg
Power = 200 W
Suppose, we determine how much heat must be added to the water to raise its temperature from 20.0°C to 89.0°C?
We need to calculate the heat
Using formula of heat
Where, m = mass of water
c = specific heat
Put the value into the formula
Hence, The heat is 115478.4 J.
Answer:
The horizontal component of its velocity remains constant and the vertical component of its acceleration is equal to -g.
Explanation:
This is because, the projectile has both vertical and horizontal components of velocity. But, its vertical component of velocity changes as the object moves whereas, its horizontal component of velocity remains constant.
Also, the projectile has only vertical component of acceleration and no horizontal component of acceleration since, its horizontal component of velocity remains constant. Thus, no change in the horizontal component of velocity.
The vertical component of acceleration is equal to -g since, the weight is the only vertical force acting on it.
So, <u>the horizontal component of its velocity remains constant and the vertical component of its acceleration is equal to -g.</u>