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3 years ago

What is the mass of a ball being rolled with a force of 200N at 3.5 m/s2?

Physics

1 answer:

aalyn [17]3 years ago

6 0

F=ma (c) Newton, 2d Law
m=F/a=200/3.5=57kg

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The temperature coefficient of a certain conducting material is 5.74 × 10-3 (°C)-1. (a) At what temperature would the resistance

Alex777 [14]

Answer:

Temperature at which the resistance is twice the resistance at $20^{\circ}C$ is $194.216^{\circ}C$

Solution:

As per the question:

Temperature coefficient, $\alpha = 5.74\times 10^{- 3}^{\circ}C$

Reference temperature, $T_{o} = 20^{\circ}C$

Resistance, $R_{t} = 2R_{o}$

Now, using the formula:

$R_{t} = R_{o}(1 + \alpha \Delta T)$

$2R_{o} = R_{o}(1 + \alpha \times (T _ T_{o}))$

$2 = 1 + 5.74\times 10^{- 3}\times (T - 20^{\circ})$

$\frac{1}{5.74\times 10^{- 3}} = T - 20^{\circ}$

$T = 174.216 + 20 = 194.216^{\circ}$

Yes, this temperature holds for all all the conductors of copper, irrespective of the size and shape of the conductor.

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3 years ago

A football punter accelerates a football from rest to a speed of 15 m/s during the time in which his toe is in contact with the

KengaRu [80]

Use F=ma formula
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answer should be around 30N .

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4 years ago

Which tissue provides medical strength plant

Tasya [4]

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3 years ago

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A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,

alexgriva [62]

The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
$S= \frac{1}{2} a t^2$
where a is the acceleration. Using the data of the problem, we can find a:
$a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2$
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J$

and so the power output is
$P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W$

b) power output at t=4.0s
The velocity at t=4.0 s is
$v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J$

and so the power output is
$P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W$

c) Power output at t=6.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J$

and so the power output is
$P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W$

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3 years ago

Which metal is liquid at room temperature?

White raven [17]

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3 years ago

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