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3 years ago

7

Suppose that a particular artillery piece has a range R = 9880 yards . Find its range in miles. Use the facts that 1mile=5280ft

and 3ft=1yard.

1 answer:

Solnce55 [7]3 years ago

6 0

Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

If you do it in steps

R = 9880 yd * 3 ft/yd = 29640 ft

R = 29640 ft / 5280 ft/mi = 5.61 mi

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Consider two uniform solid spheres where both have the same diameter, but one has twice the mass of the other. how much larger i

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<span>The moment of inertia of the large sphere will be twice that of the smaller sphere.
The formula for the moment of inertia for a solid sphere is:
I = (2/5)mr^2
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I = moment of inertia
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A sailboat moves north for a distance of 10.00 km when blown by a wind from the exact south with a force of 5.00 x 10^4 n. how m

Nataliya [291]

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What is the name of the scale used to measure hurricanes

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The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$

So we have:

$F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N$

So, the components of the resultant this time are:

$F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}$

Learn more about vector addition:

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3 years ago

Calculate Vector component in Y if the hypotenuse is 32 and angle is 45

Lerok [7]

Answer:

The correct option is;

c. 22.6

Explanation:

The given parameters are;

The hypotenuse of the vector = 32

The angle of the vector = 45°

Therefore, the vector component in the y-axis is given as follows;

$v_y = v \times sin(\theta)$

Substituting the values from the question gives;

$v_y = 32 \times sin(45^{\circ}) \approx 22.6$

The vector component in the y-axis, $v_y$ , is approximately 22.6.

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