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dalvyx [7]
3 years ago
12

I need an answer for this plzz!!number 2 anybody can help ??

Physics
1 answer:
Anuta_ua [19.1K]3 years ago
8 0
2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a \:  =  \:  \frac{dv}{dt}  \:  =  \:  \frac{25 \:  \frac{m}{s} }{50 \: s} \:  =  \: 0.5 \:  \frac{m}{ {s}^{2} }
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2ax \:  =  \:  {v}^{2}  \:  -  \:  {u}^{2}
x \:  =  \:  \frac{ {v}^{2}  \:   -  \:  {u}^{2} }{2a}  \:  =  \:   \frac{{(25 \:  \frac{m}{s})}^{2}  \:  -  \:  {(0 \:  \frac{m}{s} )}^{2} }{2 \:  \times  \: 0.5 \:  \frac{m}{ {s}^{2} } } \:  =  \: 625 \: m
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
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if a 60 kg person was standing on a platform at the surface of saturn and they jumped, they would have to push with a force grea
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2 years ago
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Answer:

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\therefore w_f\ =\ w_0\ +\ \alpha t\\\Rightarrow \alpha\ =\ -\dfrac{w_0}{t}\\\Rightarrow \alpha\ =\ -\dfrac{50}{20}\\\Rightarrow \alpha\ =\ -2.5\ rad/s^2

part (b)

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\therefore \theta\ =\ w_0t\ +\ \dfrac{1}{2}\alphat^2\\\Rightarrow \theta\ =\ 50\times 20\ -\ 0.5\times 2.5\times 20^2\\\Rightarrow \theta\ =\ 500\ rad

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3 years ago
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