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3 years ago

The diagram below shows the velocity vectors for two cars that are moving relative to each other.

Physics

1 answer:

scoundrel [369]3 years ago

7 0

Answer:

The answer is " $5 \ \frac{m}{s} \ west$ "

Explanation:

$\to \vec{V_1} = (25 \frac{m}{s}) (\hat{-i})\\\\\to \vec{V_2} = (20 \frac{m}{s}) (\hat{-i})\\\\$

velocity of car | respect to car :

$\to \vec{V_{12}} = \vec{V_1} - \vec{V_2}\\\\$

$=\vec{-25} \hat{i}+ \vec{20} \hat{i}\\\\= 5 \ \frac{m}{s} \ west$

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Which of the following is most appropriate to follow motion in one dimension?

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The linear scale is applicable only as it moves in one dimension. From the word "linear" it means it deals with one equation only. Unlike the other options, the dimensions are many because it involves 2 or more variables for its equation.

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3 years ago

A 46.0-kg box is being pushed a distance of 8.80 m across the floor by a force P whose magnitude is 171 N. The force P is parall

dolphi86 [110]

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

$W_P=Px$

P: applied force = 171N

x: distance in which the for P is applied = 8.80m

you replace the values of P and x and obtain:

$W_P=(171N)(8.80m)=1504.8J$

(b) The work don by the friction force is:

$W_f=F_fx=\mu N x=\mu Mg x$

μ = coefficient of kinetic friction = 0.250

M: mass of the box = 46.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J$

$N=Mg=(46.0kg)(9,8m/s^2)=450.8N$

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

$W_N=0J$

(d) the same as before:

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2 years ago

Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum

Alina [70]

Answer:

Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
Uncertainty in momentum (∆P) = ?
Planck's constant (h) = 6.26 × 10⁻³⁴ Js

$\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi}$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} }$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} }$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192}$

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$\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21}$

$\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21}$

$\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}$

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fenix001 [56]

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