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bogdanovich [222]
3 years ago
11

An object moving with a speed of 21 m/s and has a kinetic energy of 140 J, what is the mass of the object.

Physics
2 answers:
alexira [117]3 years ago
7 0

Answer:2940

Explanation:

Masteriza [31]3 years ago
7 0

Answer:

0.635 kg

Explanation:

Recall that Kinetic energy is defined as:

K.E. = (1/2) mv²,

where

K.E = Kinetic energy = given as 140J

m = mass (we are asked to find this)

v = velocity = given as 21 m/s

Substituting the above values into the equation:

K.E. = (1/2) mv²

140 = (1/2) m (21)²   (rearranging)

m = (140)(2) / (21)²

m = 0.635 kg

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horsena [70]

Answer:

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S = 1/2 g t^2      since initial speed is zero

S = 1/2 * 9.8 m/s^2 * 5.29 s^2 = 25.9 m

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2 years ago
Which structure reduces friction between joints?
Roman55 [17]

Answer:

The cartilage is the structure that reduces friction between joints.

Explanation:

Cartilage is the tissue between the bone joints. They are composed of several membranes and cellular groups. One of them called synovial membrane which provides lubricant to the edges of tissue that are touched by the bones. Reducing the friction and absorbing it so the body can move without obstacles.

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3 years ago
The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc
Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is 1.163 * 10^{-5} m

Explanation:

From the uncertainty Principle,

Δ_{y} Δ_{p} = \frac{h}{2\pi }

The momentum P_{y} = (mass of electron)(speed of electron)

                                = (9.109 * 10^{-31}kg)(995 * 10^{3} m/s)

                                = 9.0638 * 10^{-25}kgm/s

If the uncertainty is reduced to a 0.0010%, then momentum

                              = 9.068 * 10^{-30}kgm/s

Thus the uncertainty in the position would be:

                              Δ_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }

                              Δ_{y} \geq  1.163 * 10^{-5}m

5 0
3 years ago
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)

<u>L =  128.75 m</u>

<u></u>

(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)

<u>L =  182.56 m</u>

<u></u>

(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)

<u>L =  114.28 m</u>

<u></u>

(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)

<u>Mass of Gold = 7.68 kg = 7680 gram</u>

<u></u>

(e)

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

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

<u>Cost of Gold Wire = $ 307040</u>

7 0
3 years ago
How many degrees are there between the direction of motion and the force of
Aleks04 [339]

Answer:

180°

Explanation:

Friction, if it exists, ALWAYS opposes motion or attempted motion.

5 0
3 years ago
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