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3 years ago

6

Find the angle between forces of 41 pounds and 68 pounds given a magnitude of 87 pounds for the resultant force. (Hint: Write fo

rce one as a vector in the direction of the positive x-axis and force two as a vector at an angle θ with the positive x-axis.)

1 answer:

dedylja [7]3 years ago

7 0

Answer:

$\theta = 76.9 degree$

Explanation:

As we know that the resultant of two vectors is given as

$R = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 cos\theta$

here we know that

$R = 87 Lb$

$F_1 = 41 Lb$

$F_2 = 68 Lb$

now we have

$87 = \sqrt{41^2 + 68^2 + 2(41)(68)cos\theta$

$87^2 = 6305 + 5576 cos\theta$

$\theta = cos^{-1}(\frac{1264}{5576})$

$\theta = 76.9 degree$

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Leokris [45]

Answer:

1) $\Delta L= 0.612\ m$

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Explanation:

1)

given initial length, $L=1275\ m$

initial temperature, $T_i=-15^{\circ}C$

final temperature, $T_f=25^{\circ}C$

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<u>∴Change in temperature:</u>

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2)

Given relation:

$\Delta V=V.\beta.\Delta T$

where:

$\Delta V$ = change in volume

V= initial volume

$\Delta T$ =change in temperature

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initial volume of gasoline, $V_{Gi}=60\ L$

initial temperature of steel tank, $T_{Si}=15^{\circ}C$

initial temperature of gasoline, $T_{Gi}=15^{\circ}C$

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final temperature of gasoline, $T_{Gf}=25^{\circ}C$

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$\Delta T_G=10^{\circ}C$

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$\Delta V_G= V_G.\beta_G.\Delta T_G$

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$\Delta T_S=25-15$

$\Delta T_S=10^{\circ}C$

Now,

$\Delta V_S= V_S.\beta_S.\Delta T_S$

$\Delta V_S=60\times 35\times 10^{-6}\times 10$

$\Delta V_S=0.021\ L$

c)

Quantity of gasoline spilled after the given temperature change:

$V_0=\Delta V_G-\Delta V_S$

$V_0=0.57-0.021$

$V_0=0.549\ L$

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Karolina [17]

GPE = mgh Where m is the mass of the object (kg), g is acceleration due to gravity (which I will assume to be 9.81ms-2), and h is the height of the object above ground level. This is used in cases where the object is close to the earth, since any change in gravitational force is negligible.

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