Which vector is the sum of the vectors shown below?

A factory has a solid copper sphere that needs to be drawn into a wire. The mass of the copper sphere is 76.5 kg. The copper nee

baherus [9]

Answer:

120.125 m

Explanation:

Density = Mass/volume

D = m/v .............................. Equation 1.

Where D = Density of the solid copper sphere, m = mass of the solid copper sphere, v = volume of the solid copper sphere.

Making v the subject of the equation,

v = m/D............................... Equation 2

Given: m = 76.5 kg,

Constant: D = 8960 kg/m .

Substituting into equation 2

v = 76.5/8960

v = 0.0085379 m³

Since the copper sphere is to be drawn into wire,

Volume of the copper sphere = volume of the wire

v = volume of the wire

Volume of wire = πd²L/4

Where d = diameter of the wire, L = length of the wire.

Note: A wire takes the shape of a cylinder.

v = πd²L/4 ........................ equation 3.

making L the subject of the equation,

L = 4v/πd²..................... Equation 4

Given: v = 0.0085379 m³, d = 9.50 mm = 0.0095 and π = 3.14

Substitute into equation 4

L = 4×0.0085379/(3.15×0.0095²)

L = 0.0341516/0.0002843

L = 120.125 m.

L = 120.125 m

Thus the length of the wire produced = 120.125 m

4 0

3 years ago

The gauge pressure in the tires of your car is 210 kPa (30.5 psi) when the temperature is 25°C (77 °F). Several days later it is

DaniilM [7]

Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:

P/T = const.

P = pressure, T = temperature, the quotient of P/T must stay constant.

Initial P and T values:

P = 210kPa + 101.325kPa

P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)

T = 25°C = 298.15K

Final P and T values:

P = ?, T = 0°C = 273.15K

Set the initial and final P/T values equal to each other and solve for the final P:

311.325/298.15 = P/273.15

P = 285.220kPa

Subtract 101.325kPa to find the final gauge pressure:

285.220kPa - 101.325kPa = 183.895271kPa

The final gauge pressure is 184kPa or 26.7psi.

8 0

3 years ago

Artificial gravity is a must for any space station if humans are to live there for any extended length of time. Without artifici

wariber [46]

Answer:

Hi myself Shrushtee.

Explanation:

Artificial gravity is a must for any space station if humans are to live there for any extended length of time. Without artificial gravity, human growth is stunted and biological functions break down. An effective way to create artificial gravity is through the use of a rotating enclosed cylinder, as shown in the figure. Humans walk on the inside edge of the cylinder, which is sufficiently large (diameter of 2235 meters) that its curvature is not readably noticeable to the inhabitants. (The space station in the figure is not drawn to the scale of the human.) Once the space station is rotating at the necessary speed, how many minutes would it take the space station to make one revolution?

The distance traveled by the man in one revolution is simply the circumference of the space station, C = 2p R. From this result, you should be able to deduce the time it takes for the space station to sweep out a complete revolution.

<h2>Please mark me as brainleist</h2>

7 0

2 years ago

We are able to see an object when it:

xenn [34]

Answer:

C. reflects one or more visible light frequencies.

Explanation:

The color that we see is the color that is deflected from the RGBHSB.etc. This means that the light has all the colors within, and the object it hits absorbs all but one color, in which it deflects for our eyes to see.

~

4 0

3 years ago

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A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

2 years ago

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