A 1.695 kg block of lead is 10 cm long, 5 cm high, and 30mm wide. Calculate the density of the lead

Which of the following statements describes a compound machine?

ASHA 777 [7]

From the word compound, the compound machine is already a combination of two or more types of simple machine. Thus, the answer is letter C. Because of its complexity, it is able to perform several other functions than a simple one.

6 0

2 years ago

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A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba

Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

The mass of the rubber ball is $m_r = 2 \ kg$

The initial speed of the rubber ball is $u = 3 \ m/s$

The final speed at which it bounces bank $v - 3 \ m/s$

The mass of the clay ball is $m_c = 2 \ kg$

The initial speed of the clay ball is $u = 3 \ m/s$

The final speed of the clay ball is $v = 0 \ m/s$

Generally Impulse is mathematically represented as

$I = \Delta p$

where $\Delta p$ is the change in the linear momentum so

$I = m(v-u)$

For the rubber is

$I_r = 2(-3 -3)$

$I_r = -12\ kg \cdot m/s$

=> $|I_r| = 12\ kg \cdot m/s$

For the clay ball

$I_c = 2(0-3)$

$I_c = -6 \ kg\cdot \ m/s$

=> $| I_c| = 6 \ kg\cdot \ m/s$

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

8 0

3 years ago

Justin Bieber is thrown horizontally at 10.0 m/s from the top of a cliff 122.5 m high.

sveticcg [70]

===> Distance fallen from rest in free fall =

   (1/2) (acceleration) (time²)

   (122.5 m) = (1/2) (9.8 m/s²) (time²)

Divide each side by (4.9 m/s²): (122.5 m / 4.9 m/s²) = time²

   (122.5/4.9) s² = time²

Take the square root of each side: 5.0 seconds

===> (Accelerating at 9.8 m/s², he will be dropping at
   (9.8 m/s²) x (5.0 s) = 49 m/s
when he goes 'splat'. We'll need this number for the last part.)

===> With no air resistance, the horizontal component of velocity
doesn't change.

Horizontal distance = (10 m/s) x (5.0 s) = 50 meters .

===> Impact velocity = (10 m/s horizontally) + (49 m/s vertically)

   = √(10² + 49²) = 50.01 m/s arctan(10/49)

   =    50.01 m/s at 11.5° from straight down,
      away from the base of the cliff.

7 0

3 years ago

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You want to produce three 2.00-mm-diametercylindrical wires, each with a resistance of 1.00 Ω at room temperature. One wire is g

Vlada [557]

Answer:

(a) L = 128.75 m

(b) L = 182.56 m

(c) L = 114.28 m

(d) Mass of Gold = 7.68 kg = 7680 gram

(e) Cost of Gold Wire = $ 307040

Explanation:

The resistance of the wire is given as:

R = ρL/A

where,

R = Resistance

ρ = resistivity

L = Length

A = cross-sectional area

(a)

For Gold Wire:

ρ = 2.44 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)

L = 128.75 m

(b)

For Copper Wire:

ρ = 1.72 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)

L = 182.56 m

(c)

For Aluminum Wire:

ρ = 2.75 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)

L = 114.28 m

(d)

Density = Mass/Volume

Mass = (Density)(Volume)

Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³

Therefore,

Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)

Mass of Gold = 7.68 kg = 7680 gram

(e)

Cost of Gold Wire = (Unit Price of Gold)(Mass of Gold)

Cost of Gold Wire = ($ 40/gram)(7680 grams)

Cost of Gold Wire = $ 307040

7 0

3 years ago

A box of mass 1 kg is hung from a spring scale. The reading on the spring scale is approximately 10N. What would be the reading

8090 [49]

The answer will be 50N.
This is because the spring reads weight and weight is mass times acceleration due to gravity.5kg*10m/s2=50N

8 0

2 years ago

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