Rate of change of momentum = impact force
(m*v-m*u)/t = F
4000*20/t = 80000 (note: v is zero as it stopped)
<span>soo, t = 1 sec</span>
The displacement of Edward in the westerly direction is determined as 338.32 km.
<h3>What is displacement of Edward?</h3>
The displacement of Edward can be determined from different methods of vector addition. The method applied here is triangular method.
The angle between the 200 km north west and 150 km west = 60 + 90 = 150⁰
The displacement is the side of the triangle facing 150⁰ = R
R² = a² + b² - 2abcosR
R² = 150² + 200² - (2x 150 x 200)xcos(150)
R² = 62,500 - (-51,961.52)
R² = 114,461.52
R = 338.32 km
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Astronomers find white dwarfs that distinguish them from main sequence stars because white dwarfs get really hot, we can search for their ultraviolet radiation.
<h3>What is a white dwarf?</h3>
A white dwarf is a very hot star that radiated much energy in the form of ultraviolet radiation.
This UV radiation is initially very bright and then these stars become red with time.
In conclusion, Astronomers find white dwarfs they can search for their ultraviolet radiation.
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Answer:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s
Answer:
a) U = 735 J
, b) U = 125.7 J
, c) U = 0 J
Explanation:
The gravitational power energy is
U = mg y - mg y₀
The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part
a) Rope is horizontal
The height in this case is the same length of the rope
y = 2.10 m
w = mg = 350 N
U = 350 2.10
U = 735 J
b) when the angle is 34º
y = L - L cos 34
y = L (1- cos34)
y = 2.10 (1- cos 34)
y = 0.359 m
U = 350 0.359
U = 125.7 J
c) in this case this point coincides with the reference system
y = 0
U = 0 J
