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Maslowich
3 years ago
9

Can an object have increasing speed while its acceleration is decreasing?

Physics
1 answer:
alexdok [17]3 years ago
6 0
The best option is C. This is due to friction.
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When two hydrogen nuclei combine to form one helium nucleus, nuclear fusion has taken place. Please select the best answer from
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A particle with charge -40.0nC is on the x axis at the point with coordinate x=0 . A second particle, with charge -20.0 nC , is
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A particle with charge -40.0nC is on the x axis at the point with coordinate x=0 . A second particle, with charge -20.0 nC, is on the x axis at x=0.500 m.

No, there is no point at a finite distance where the electric potential is zero.

Hence, Option D) is correct.

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1 year ago
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1 year ago
An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f
IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}

Similarly, The equation of mption:

x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})

replacing our assumed values:

where v_=0 \ and \ g= 9.81

x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})

x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

1000= 0.981t-0.981(1-e^{-(10)t}) \ m

By diregarding e^{-(10)t} \ m

1000= 0.981t-0.981

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0
3 years ago
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