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adelina 88 [10]
4 years ago
5

Reduce, reuse, recycle! Soft drinks come in aluminum cans that are readily recycled. Storage of empty cans is never a problem, s

ince aluminum cans are easily crushed because they are very A) conductive. B) magnetic. C) malleable. D) soluble.
Physics
2 answers:
ryzh [129]4 years ago
8 0
Malleable is correct
natita [175]4 years ago
6 0
They are C, Malleable.

I hope this helps! (:
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A car is moving at a speed of 50 km/hour.the driver takes her foot off the accelerator and the car coasts to stop.why?
vodka [1.7K]
Because of the opposing forces on the car like friction because there is no more acceleration 
3 0
4 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!
Tasya [4]

Answer:

98 N

Explanation:

The weight of an object is given by:

W=mg

where

m is the mass of the object

g=9.8 m/s^2 is the acceleration due to gravity

In this problem, the mass of the lemur is m=10 kg, so its weight is

W=mg=(10 kg)(9.8 m/s^2)=98 N

3 0
3 years ago
Please answer this question
ElenaW [278]

Answer:

120

Explanation:

30 percent of 400 is 120

8 0
4 years ago
Read 2 more answers
Reactance Frequency Dependence: Sketch a graph of the frequency dependence of a resistor, capacitor, and inductor. RLC Circuit R
jolli1 [7]

Answer:

f=\frac{1}{2\pi \sqrt{LC}}

Explanation:

We know that impedance of a RLC circuit is given by Z=R+J(X_L-X_C)

So Z=\sqrt{R^2+(X_L-X_C)^2} here R is resistance X_L is inductive reactance and X_C is capacitive reactance

To minimize the impedance X_L-X_C should be zero we know that X_L=\omega L\ and \ X_C=\frac{1}{\omega C}

So \omega L-\frac{1}{\omega C}=0

\omega ^2=\frac{1}{LC}

\omega =\sqrt{\frac{1}{LC}}

We know that \omega =2\pi f

So \omega =2\pi f=\frac{1}{\sqrt{LC}}

f=\frac{1}{2\pi \sqrt{LC}}

Where f is resonance frequency  

8 0
3 years ago
A box of spherical jawbreaker candies is 23 g and its volume is 32.3 cm3. If the average mass of a single jawbreaker is 0.94 g,
Alex777 [14]

The radius of each jawbreaker is approximately 0.68 cm.

<h3>Volume of a sphere;</h3>
  • v = 4 /3 πr³

where

r = radius

Therefore,

23 g  = 32.3 cm³

0.94 g  = ?

cross multiply

volume of a single jawbreaker = 32.3 × 0.94 / 23 = 30.362 / 23 = 1.32 cm³

Therefore,

volume of each jawbreaker = 4 /3 πr³

1.32 = 4 / 3 × 3.14 × r³

r³ = 1.32 /4.18666666667

r³ = 0.31533683707

r = ∛0.31533683707

r = 0.680651651 = 0.68

Therefore, the radius of each jawbreaker is approximately 0.68 cm.

learn more on radius here: brainly.com/question/19172427

5 0
3 years ago
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