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scoray [572]
3 years ago
7

On a spring scale that measures force, a reading shows the force of an object to be about half way between 0 N and 1 N. Which of

the following would be the most accurate measurement for the force?
49.0 N
0.49 N
0.049 N
4.90 N
Physics
1 answer:
deff fn [24]3 years ago
7 0
0.49 N
On a spring scale that measures force, a reading shows the force of an object to be about half way between 0 N and 1 N. 0. 49 Newton is the most accurate measurement for the force. If we cluster the number values of 0 N to 1 N It would be 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0 N. These numbers indicate the micro intervals of 0-1. It cannot be the other given because the other values are either higher or lower which exceeds the 0-1 scale. Observe.



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The coefficient of linear expansion of steel is 12 × 10-6 K-1 . What is the change in length of a 25-m steel bridge span when it
Westkost [7]

Answer:

0.012-m

Explanation:

∆L = α × Lo × (T-To)

α is the coefficient of linear expansion = 12 × 10-6 K-1

Lo = Initial length = 25-m

∆L = Change in length

(T-To) = 40 K

∆L = 12 × 10-6 × 25 × 40

∆L = 0.012-m

4 0
3 years ago
A planar electromagnetic wave is propagating in the +x direction. At a certain point P and at a given instant, the electric fiel
jeka94

Answer:

B=2.74\times 10^{-10}\ T

Explanation:

It is given that,

A planar electromagnetic wave is propagating in the +x direction.The electric field at a certain point is, E = 0.082 V/m

We need to find the magnetic vector of the wave at the point P at that instant.

The relation between electric field and magnetic field is given by :

c=\dfrac{E}{B}

c is speed of light

B is magnetic field

B=\dfrac{E}{c}\\\\B=\dfrac{0.082}{3\times 10^8}\\\\B=2.74\times 10^{-10}\ T

So, the magnetic vector at point P at that instant is 2.74\times 10^{-10}\ T.

3 0
2 years ago
A charge of 63.0 nC is located at a distance of 3.40 cm from a charge of -47.0 nC. What are the x- and y-components of the elect
Serjik [45]

Answer:

Ep_x = 288.97*10^3\frac{N}{C}

Ep_y = 2770.6*10^3\frac{N}{C}

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/r²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

r: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁: Total field at point P due to charge q₁. As the charge is positive ,the field leaves the charge.

Ep₂: Total field at point P due to charge q₂. As the charge is negative, the field enters the charge.

Known data

q₁ = 63 nC = 63×10⁻⁹ C

q₂ = -47 nC = -47×10⁻⁹ C

k = 8.99*10⁹ N×m²/C²

d₁ = 1.4cm = 1.4×10⁻² m

d₂ = 3.4cm = 3.4×10⁻² m

Calculation of r and β

r=\sqrt{d_1^2 + d_2^2} = \sqrt{(1.4*10^{-2})^2 + (3.4*10^{-2})^2} = 3.677*10^{-2}m

\beta = tan^{-1}(\frac{d_1}{d_2}) = tan^{-1}(\frac{1.4}{3.4}) = 22.38^o

Problem development

Ep: Total field at point P due to charges q₁ and q₂.

Ep = Ep_x i + Ep_y j

Ep₁ₓ = 0

Ep_{2x}=\frac{-k*q_2*Cos\beta}{r^2}=\frac{8.99*10^9*47*10^{-9}*Cos(22.38)}{(3.677*10^{-2})^2}=288.97*10^3\frac{N}{C}

Ep_{1y}=\frac{-k*q_1}{d_1^2}=\frac{8.99*10^9*63*10^{-9}}{(1.4*10^{-2})^2}=2889.6*10^3\frac{N}{C}

Ep_{2y}=\frac{-k*q_2*Sen\beta}{r^2}=\frac{-8.99*10^9*47*10^{-9}*Sen(22.38)}{(3.677*10^{-2})^2}=-119*10^3\frac{N}{C}

Calculation of the electric field components at point P

Ep_x = Ep_{1x} + Ep_{2x} = 0 + 288.97*10^3 = 288.97*10^3\frac{N}{C}

Ep_y = Ep_{1y} + Ep_{2y} = 2889.6*10^3 - 119*10^3 = 2770.6*10^3\frac{N}{C}

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