<span> Maths delivers! Braking distance ... If the </span>car<span> is initially travelling at u</span>m<span>/s, then the stopping distance d </span>m<span> ... the </span>speed<span> of the </span>car<span> at the </span>instant<span> the </span>brakes<span> are applied. ... An object with </span>constant acceleration<span> travels the </span>same<span> distance as it would ... We </span>start<span> with the second equation of motion:.</span>
Answer:
उव्ग्वुव ह्व्झ एउएइहे एइएइएइएएइ सिसुब्स्सी बीस सिस इस्ब एइब
Explanation:
?उग्व्ब्वु विब्सिए इसिग्व विद्बिअब्द सिह्व्व इस्ब्व दिव्ब्स विह्द ऐद्जिइ सुउगव्दी सिइगैगे क्ज्गैइव अजिव्व्ज्व्स कैह्द अजि ह्ज्फ्ज इअह इकुगै ईग इअबे अजिव्ब जैइअब इऐहे ऐइहे ऐइग्गे अत्व्ब ओप्झब रोज दिधिए ऊइफ्ब इसुहद ईउहे सिउउअ दिइब्द स्सिउए ऐइहे सिएय्व एउविये एइव्वे
Answer:
The natural frequency = 50 rad/s = 7.96 Hz
Damping ratio = 0.5
Explanation:
The natural frequency is calculated in this manner
w = √(k/m)
k = spring constant = 5 N/m
m = mass = 2 g = 0.002 kg
w = √(5/0.002) = 50 rad/s
w = 2πf
50 = 2πf
f = 50/(2π) = 7.96 Hz
Damping ratio = c/[2√(mk)] = 0.1/(2 × √(5 × 0.002)) = 0.5
If an atom gains electrons, it develops a negative charge equal to the number of electrons gained.
So the net charge on the copper atom which gained 2 electrons will be -2.
