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3 years ago

How many millimeters are 120 meters ? Help ! Please <3

Physics

2 answers:

blsea [12.9K]3 years ago

8 0

Answer:

120,000

Explanation:

Millimeters to meters calculation-

Multiply by 1,000.

120 x 1,000 = 120,000.

This is the correct answer and formula.

Hope this helps!

natulia [17]3 years ago

4 0

The answer is 120000

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mel-nik [20]

Answer:

option C is correct

Explanation:

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4 years ago

Four rods that obey Hooke's law are each put under tension. (a) A rod 50.0 cm50.0 cm long with cross-sectional area 1.00 mm21.00

deff fn [24]

Answer:

c < a = b

Explanation:

The tensile stress = Force applied/(Cross sectional area)

(a) The applied force = 200 N

The cross sectional area = 1.00 mm² = 1 × 10⁻⁶ m²

The tensile stress = 200 N/(1 × 10⁻⁶ m²) = 200,000,000 Pa = 200 MPa

(b) The applied force = 200 N

The cross sectional area = 1.00 mm² = 1 × 10⁻⁶ m²

The tensile stress = 200 N/(1 × 10⁻⁶ m²) = 200,000,000 Pa = 200 MPa

The cross sectional area = 2.00 mm² = 2 × 10⁻⁶ m²

The tensile stress = 100 N/(2 × 10⁻⁶ m²) = 50,000,000 Pa = 50 MPa

Therefore, the tensile stress from smallest to largest are;

(a) 50 MPa, < (b) 200 MPa = (a) 200 MPa

Therefore, we have;

c < a = b.

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3 years ago

What is the movement of particle in a liquid

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3 years ago

A 2.81 μF capacitor is charged to 1220 V and a 6.61 μF capacitor is charged to 560 V. These capacitors are then disconnected fro

fenix001 [56]

Answer:

756.88 Volts will be the potential difference across each capacitor.

Explanation:

$Q=C\times V$

Q = Charge on capacitor

C = Capacitance

V = Voltage across capacitor

Capacitance of first capacitor = $C_1=2.81 \mu F=2.81\times 10^{-6} F$

Charge of first capacitor = $Q_1$

Voltage across first capacitor = $V_1=1220 V$

$Q_1=C_1V_1$

$Q_1=2.81\times 10^{-6} F\times 1220 V=0.0034282 C$

Capacitance of first capacitor = $C_2=6.61\mu F=6.61\times 10^{-6} F$

Charge of second capacitor = $Q_2$

Voltage across first capacitor = $V_2=560 V$

$Q_2=C_2V_2$

$Q_1=6.61\times 10^{-6} F\times 560 V=0.0037016 C$

Both the capacitors are disconnected and positive plates are now connected to each other and the negative plates are connected to each other. These capacitors are connected in parallel combination.

Total charge = Q

$Q =Q_1+Q_2=0.0034282 C+ 0.0037016 C=0.0071298 C$