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2 years ago

5

Can someone help me with this please

1 answer:

andrezito [222]2 years ago

3 0

Carbon: C, 12.011, 6, 12
Oxygen: O, 8, 8, 8, 16
Boron: B, 10.811, 5, 5, 11

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When something is hit harder how does the transverse wave change?

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When something is hit harder just like when sound is turned up the waves become higher and more frequent like a zig zag more so then wavy.

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2 years ago

Relationship between voltage, resistance and current in a circuit

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V = I×R

where -

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I = current flowing in the circuit

R = Equivalent Resistance in the circuit

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2 years ago

Explain the difference between speed and velocity and indicate if these are scalar or vector quantities

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3 years ago

A 3250-kg aircraft takes 12.5 min to achieve its cruising

scoundrel [369]

Answer:

effeciency n = = 49%

Explanation:

given data:

mass of aircraft 3250 kg

power P = 1500 hp = 1118549.81 watt

time = 12.5 min

h = 10 km = 10,000 m

v =85 km/h = 236.11 m/s

$n = \frac{P_0}{P}$

$P_o = \frac{total\ energy}{t} = \frac{ kinetic \energy + gravitational\ energy}{t}$

kinetic energy $= \frac{1}{2} mv^2 =\frac{1}{2} 3250* 236 = 90590389.66 kg m^2/s^2$

kinetic energy $= 90590389.66 kg m^2/s^2$

gravitational energy $= mgh = 3250*9.8*10000 = 315500000.00 kg m^2/s^2$

total energy $= 90590389.66 +315500000.00 = 409091242.28 kg m^2/s^2$

$P_o =\frac{409091242.28}{750} = 545454.99 j/s$

$effeciency\ n = \frac{P_o}{P} = \frac{545454.99}{1118549.81} = 0.49$

effeciency n = = 49%

8 0

3 years ago

A child\'s top is held in place, upright on a frictionless surface. The axle has a radius of r = 2.96 mm. Two strings are wrappe

sladkih [1.3K]

Answer:

Explanation:

Given

radius r=2.96 mm

Tension T=2.4 N

time taken=0.74 s

Let $\alpha$ be the angular acceleration

$2 T\times r=I\times \alpha$

$2\times 2.4\times 2.96\times 10^{-3}=0.5\times m\times (2.96\times 10^{-3})^2\times \alpha$

$\alpha =\frac{4\times 2.4}{m\times 2.96\times 10^{-3}}$

$\alpha =\frac{3.24\times 10^3}{m} rad/s^2$

$\omega =\omega _0+\alpha \cdot t$

$\omega =0+\frac{3.24\times 10^3}{m}\times 0.74$

$\omega =\frac{2.4\times 10^3}{m} rad/s$

Angular momentum

$L=I\omega$

$L=0.5\times mr^2\times \omega$

$L=0.5\times m\times (2.96\times 10^{-3})^2\times \frac{2.4\times 10^3}{m}$

$L=0.01051 kg-m^2/s$

4 0

3 years ago