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3 years ago

Question 1 of 20

Physics

2 answers:

jolli1 [7]3 years ago

5 0

Answer:

Explanation:

notka56 [123]3 years ago

4 0

Answer:

B. The magnet has magnetized the right side of the block.

Explanation:

a pe x

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Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of highe

NISA [10]

Answer:

Explanation:

24 - gauge wire , diameter = .51 mm .

Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m

R = ρ l / s

1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]

= 8.42 x 10⁻² ohm

= .084 ohm

B ) Current required through this wire

= 12 / .084 A

= 142.85 A

C )

Let required length be l

resistance = .084 l

2 = 12 / .084 l

l = 12 / (2 x .084)

= 71.42 m

6 0

3 years ago

Read 2 more answers

Which instrument should, ideally, have zero resistance?

Temka [501]

The answer is C voltmeter

6 0

3 years ago

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un litro de un gas es calentado a presión constante desde 20°C hasta 60°C que volumen final ocupará dicho gas?

WINSTONCH [101]

Answer:

Final volume, V2 = 3 Litres

Explanation:

Given the following data;

Initial volume, V1 = 1 litre

Initial temperature, T1 = 20°C

Final temperature, T2 = 60°C

To find the final volume, we would use Charles' law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

V1/T1 = V2/T2

Making V2 as the subject formula, we have;

V1T2 = V2T1

V2 = (V1T2)/T1

Substituting into the formula, we have;

V2 = (1 * 60)/20

V2 = 60/20

Final volume, V2 = 3 Litres

4 0

3 years ago

WILL MARK BRAINLIEST FOR CORRCET ANSWER!!!!!!!!!

jenyasd209 [6]

Answer:

It’s C

Explanation:

8 0

3 years ago

Read 2 more answers

A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

weqwewe [10]

Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

Pressure is defined as the force (F) applied per unit area (A)

P=F/A (N/m²)

P1=P2

$\frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }$

$F_{2} = \frac{F_{1}*A_{2} }{A_{1}}$ Equation (1)

Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

A₁ = 2.0 cm²

A₂ = 40 cm²

Calculation of the weight (W₂) can the large piston support

We replace data in the equation (1)

$F_{2} = \frac{(500)*(40) }{2}$

F₂ = 10000 N

W₂= F₂= 10000 N

6 0

3 years ago