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kolbaska11 [484]

2 years ago

10

Which best describes a relationship among the vectors? Vectors P and Py are components of vector Px. Vectors Px and Py are compo

nents of vector P.

2 answers:

Semmy [17]2 years ago

4 0

Answer:

D - Vectors Px and Py are components of vector P.

Explanation:

mafiozo [28]2 years ago

3 0

If we have a vector P, then we name it's components by the axis we project it onto.

In 2D Cartesian coordinate system we have one x axis and one y axis.

Vector P can then be represented with two components: Px and Py.

Px is the component of vector P on the x axis and Py is the component of the vector P on the y axis.

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How many neutrons are contained in 2 kg? Mass of one neutron is 1.67x10-27 kg.

liraira [26]

Answer:

1.2 × 10^27 neutrons

Explanation:

If one neutron = 1.67 × 10^-27 kg

then in 2kg...the number of neutrons

; 2 ÷ 1.67 × 10^-27

There are.... 1.2 × 10^27 neutrons

4 0

2 years ago

Someone please help me!

kirill [66]

I think it’s the first one

4 0

2 years ago

Calculate the amount of momentum of an object with a mass of 10 kg travelling al a

irina [24]

Answer:

50kg.m/s

Explanation:

In order to find momentum you must use the formula P=mv

p= momentum

m=mass

v= velocity

so in other words, momentum= mass times velocity

or in this case, momentum= 10 times 5 :)

7 0

2 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

2 years ago

A vertical spring with a spring constant of 420 N/m is mounted on the floor. From directly above the spring, which is unstrained

mixas84 [53]

Answer:

19.53 cm

Explanation:

The computation of the height is as follows:

Here we applied the conservation of the energy formula

As we know that

P.E of the block = P.E of the spring

m g h = ( 1 ÷ 2) k x^2

where

m = 0.15

g = 9.81

k = 420

x = 0.037

So now put the values to the above formula

(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2

1.4715 (h) = 0.28749

h = 0.19537 m

= 19.53 cm

3 0

2 years ago