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3 years ago

All of the following are kinds of forces except

Physics

1 answer:

dangina [55]3 years ago

3 0

Answer:

All of the following are kinds of forces except?

<h2><u><em>b) acceleration is the Answer</em></u></h2>

Explanation:

<h3><u><em>Acceleration is actually produced when a force acts upon a body therefore it is the result after we apply any force.</em></u></h3>

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An iron wire has a length of 1.50 m and a cross sectional area of 0.290 mm2. If the resistivity of iron is 10.0 ✕ 10−8 Ω · m and

lapo4ka [179]

Answer:

1.35 A

Explanation:

Applying,

V = IR

I = V/R..................... Equation 1

I = Current, V = Voltage, R = Resistance.

But,

R = Lρ/A............... Equation 2

Where L = Length of the wire, ρ = resistivity, A = Cross-sectional area of the wire.

Sustitute equation 2 into equation 1

V = AV/Lρ............... Equation 3

From the question,

Given: V = 0.7 V, A = 0.290 mm² = 2.9×10⁻⁷ m², L = 1.5 m, ρ = 10×10⁻⁸ Ω.m

Substitute these values into equation 3

I = (0.7× 2.9×10⁻⁷)/(1.5× 10×10⁻⁸ )

I = (2.03×10⁻⁷)/(15×10⁻⁸)

I = 1.35 A

5 0

3 years ago

An electric circuit through which no current flows is known as?

Afina-wow [57]

The answer is : Open circuit

4 0

4 years ago

Read 2 more answers

¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula

Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

$F = |q|vBsin(\theta)$ (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

$v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s$

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.

Espero que te sea de utilidad!

7 0

3 years ago

A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the s

leonid [27]

Complete question is;

A force stretches a wire by 0.60 mm. A second wire of the same material has the same cross section and twice the length.

a) How far will it be stretched by the same force?

b) A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force?

Answer:

0.15 mm

Explanation:

According to Hooke's Law,

E = Stress(σ)/Strain(ε)

Where E is youngs modulus

Formula for stress is;

Stress(σ) = Force(F)/Area(A)

Formula for strain is;

Strain(ε) = Change in length/original length = (Lf - Li)/Li

We are also told that a second wire of the same material has the same cross section and twice the length.

Thus;

Rearranging Hooke's Law to get the constants on one side, we have;

F/(AE) = ε

Thus from the conditions given;

ε1 = 0.6/Li

ε2 = (Change in length)/(2*Li)

And ε1 = ε2

Thus;

0.6/Li = Change in length/(2*Li)

Li will cancel out and we now have;

Change in length = 2 × 0.6 = 1.2 mm

Finally, we are told A third wire of the same material has the same length and twice the diameter as the first.

Area of a circle;A1 = πd²/4

Now, we are told d is doubled.

Thus, new area of the new circle is;

A2 = π(2d)²/4 = πd²

Rearranging Hooke's Law,we have;

F/A = εE

Since F and E are now constants, we have;

F/E = constant = Aε

Thus;

A1(ε1) = A2(ε2)

A1 = πd²/4

e1 = 0.60/Li

A2 = πd²

e2 = Change in length/Li

Thus;

((πd²/4) × 0.6)/Li = (πd² × Change in length)/ Li

Rearranging, Li and πd² will cancel out to give;

0.6/4 = Change in length

Change in length = 0.15 mm

4 0

3 years ago

How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?

aalyn [17]

There is too much information given, it's hard to understand exactly which variables are important in this problem.

7 0

3 years ago