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1 year ago

5

Hilda plans to display in a clear cube the needle betsy ross used to sew the first american flag. the needle will fit exactly as

the diagonal of the cube. if the side length of the clear cube is startroot 3 endroot inches, and the diagonal of a face of the cube is startroot 6 endroot inches, what is the length of the needle? startroot 6 endroot inches 3 inches 3 startroot 2 endroot inches 6 inches

2 answers:

inysia [295]1 year ago

8 0

Based on the length of side of the cube given, the length of the needle is 2.28 inches.

<h3>What is the length of the needle?</h3>

The length of the side of a cube and its diagonal are related by the formula below:

$Length \: of \: Diagonal = \sqrt{3s}$

where s is length of side

$Side \: Length \: of \: the \: Cube = \sqrt{3inches}$

$Diagonal \: of \: a \: face \: of \: the \: cube = \sqrt{6inches}$

$The \: length \: of \: the \: needle = \sqrt{3\sqrt{3}} = 2.28 \: inches \\$

Therefore, the length of the needle is 2.28 inches

Learn more about length of diagonal of cube at: brainly.com/question/800212

#SPJ4

Irina-Kira [14]1 year ago

7 0

Answer:

Its B 3in

Explanation:

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Can charging by friction only occur in solids

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The distance traveled, in feet, of a ball dropped from a tall building is modeled by the equation d(t) = 16t2 where d equals the

Ainat [17]

When t=2, the ball has fallen   d(2) = 16 (2²) = 64 feet .

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   (distance covered) / (time to cover the distance)

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That's what choice-C says.

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2 years ago

If a car increases its velocity from zero to 60 m/s in 10 seconds, its acceleration is

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2 years ago

A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t

Jobisdone [24]

Answer:

(a) E=0 : 0 cm from the center of the sphere

(b) E= 227.8*10³ N/C : 10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C : 40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C : 59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

$E=\frac{K*Q}{r^{2} }$ : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

$E=k*\frac{Q}{a^{3} } *r$ :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q: Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at 0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

$E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1$

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }$

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere

r>a , We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }$

E= 411.84 * 10³ N/C

4 0

2 years ago

3. Maverick and Goose are flying a training mission in their F-14. They are

Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of $y_0=1500m$ , and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

$y=y_0 - \frac{gt^2}{2}$ [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

$x = v.t$ [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

$t=\sqrt{\frac{2y_0}{g}}$

Since we have $y_0=1500m$

$t=\sqrt{\frac{2(1500)}{9.8}}$

$t=17.5 sec$

Replacing in [2]

$x = 688\ m/sec \ (17.5sec)$

$x = 12040\ m$

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0

2 years ago