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borishaifa [10]
3 years ago
12

A train starts from rest and accelerates uniformly, until it has traveled 3.7 km and acquired a velocity of 26 m/s. What is the

acceleration of the train during this time?
A. 0.12 m/s^2
B. 0.082 m/s^2
C. 0.11 m/s^2
D. 0.091 m/s^2
E. 0.10 m/s^2
Physics
1 answer:
bogdanovich [222]3 years ago
7 0
We have:
s=3,7km (3700m)
u=0
v=26m/s
a=
t=
We look up for a formula to solve for a:
v²=u²+2as u²=0 so
v²=2as
(26m/s)²=2a3700m
676/7400=a
a=0,09m/s²

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473.15

Explanation:

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The answer would be Thermal Energy.
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A student of weight 659 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude
Misha Larkins [42]

Answer:

a) The student feel light

b) Nbottom = 758 N

c) N'top= 236 N

d) N'bottom= 1055 N

Explanation:

a) W= 659N , Ntop= 560N

W > Ntop ---> Student feel less weight

b)   Top:

∑F= W - Ntop = m.v²/R

m.v²/R = 659N - 560 N = 99 N

Bottom:

∑F= Nbottom- W = m.v²/R

Nbottom= W + m.v²/R = 659N + 99 N = 758N

c) W= 659 N , Ntop= 560 N , v'=2.v

N'top= ?

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N'top = 659 N - 4. 99 N = 263 N

d)   N'bottom = ?

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4 0
2 years ago
I need to choose a theme for my physics assignment My experiment is finding g
Kobotan [32]
<h3>Question:</h3>

How to find g (acceleration due to gravity)

<h3>Solution:</h3>

We know,

Acceleration due to gravity (g)

=  \frac{GM}{ {R}^{2} }

where, G = Gravitational constant

= 6.67 \times  {10}^{11} N {m}^{2}/k {g}^{2}  \\

M = Mass of the earth

= 6 \times  {10}^{24} \:  kg

R = Radius of the earth

= 6.4 \times  {10}^{6} m

Putting these values of G, M and R in the above formula, we get

g \:  =  \:  \frac{6.67 \times  {10}^{11} N {m}^{2}/k {g}^{2}   \times \: 6 \times  {10}^{24} \:  kg }{(6.4 \times  {10}^{6}m {)}^{2}  }  \\  = 9.8m/ {s}^{2}

So, the value of acceleration due to gravity is

9.8m/s ^{2}

Hope it helps.

Do comment if you have any query.

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3 years ago
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lesantik [10]

Answer:

240 N

Explanation:

6 0
3 years ago
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