2 years ago

Why is copper NOT used as a filament?

Physics

1 answer:

chubhunter [2.5K]2 years ago

6 0

Answer:

Short answer, because copper wire does not have high resistance.

Explanation:

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A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement do

Lubov Fominskaja [6]

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

$\theta=90-40=50^o$

From the triangle OAB

$cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}$

$1340.57=35^2+25^2-x^2$

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

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2 years ago

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kicyunya [14]

Answer:

Huh? otay....

Explanation:

Brainliest?????????

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3 years ago

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Back in their laboratory, the team of scientists carefully measure the changes in density of a freshwater sample as they decreas

BabaBlast [244]

Answer: As the temperature of the water decreases from 20 degrees celsius to 4 degrees celsius, the density increases.

Explanation:

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2 years ago

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in a softball game, a batter hits the ball at the velocity of 27m/s and angle of 40 shown below. What is the maximum range of th

Nata [24]

Answer:

R = 73.25 m

Explanation:

We have,

Initial speed of the ball is 27 m/s

It is projected at an angle of 40 degrees

The maximum range of the ball is given by :

$R=\dfrac{u^2\sin2\theta}{g}$

Plugging all the values we get :

$R=\dfrac{(27)^2\sin2(40)}{9.8}\\R=73.25\ m$

So, the maximum range of the ball is 73.25 m

8 0

3 years ago

A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin

gogolik [260]

Answer:

a) 23 m/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$p_{o} = p_{f} (1)$

The initial momentum p₀, can be written as follows:

$p_{o} = m_{1} * v_{1o} + m_{2}* v_{2o} = 6.0 kg * 5.0 m/s + 2.0 kg * v_{2o} (2)$

The final momentum pf, can be written as follows:

$p_{f} = (m_{1} + m_{2} )* v_{f} = 8.0 kg* (-2.0 m/s) (3)$

Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

$v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg} = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s (4)$

This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.

6 0

3 years ago