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3 years ago

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if A and B are non zero vectors, is it possible for vector A×vector B and vector A.vector B both to be zero? Justify your answer

1 answer:

DedPeter [7]3 years ago

3 0

Answer:

not able to understand the question

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What is the launch speed of a projectile that rises vertically above the Earth to an altitude equal to 14 REarth before coming t

vivado [14]

Answer:

v = 1.078 10⁴ m / s

Explanation:

To solve this exercise we can use conservation of mechanical energy

Starting point. Just clearing

Em₀ = K + U = ½ m v² - G m M / $R_{e}$

Final point. At r = 14R_{e}

Em_{f} = U = -G m M / 14R_{e}

how energy is conserved

Em₀ = $Em_{f}$

½ m v² - G m M /R_{e} = -G m M / 14R_{e}

½ v² = G M / R_{e} (-1/14 + 1)

v² = 2 G M / R_{e} 13/14

we calculate

v² = 2 6.67 10⁻¹¹ 5.98 10²⁴ / 6.37 10⁶ 13/14

v = √ (1,16 10⁸)

v = 1.078 10⁴ m / s

5 0

3 years ago

a sound has a frequency of 230 hz. what is the velocity of this sound if it has a wavelength of 46 cm?

lapo4ka [179]

Answer:

V=f×λ

Where,

V is the velocity of the wave measure using m/s.

f is the frequency of the wave measured using Hz.

λ is the wavelength of the wave measured using m.

transform 46 cm in m = 0.46m and there u have it

Explanation:

6 0

3 years ago

an empty boat floats in water with 10% of its volume submerged. and when it is loaded with 1200kg , the volume submerged will in

Lostsunrise [7]

Let volume of empty boat be = 100% = 1V

and mass of boat be M

In water 10%, 0.1V of the volume is submerged.

Mass, m of 1200kg increases the submerging from 10%, 0.1V to 70%, 0.7V

M leads to 0.1V boat submerging

boat submerging.

M + 1200kg leads to 0.7V boat submerging.

This is 60%, 0.6 V increase

By comparison

(M+1200kg) * 0.1V = 0.7V * M

0.1M + 120kg = 0.7M

120kg = 0.7M - 0.1M

120kg = 0.6M

M = (120/0.6)kg

M = 200kg.

The mass of the boat is 200kg.

4 0

2 years ago

In scientific notation, (6.2 x 10^4) x (3.3 x 10^2 ) equals ____________________.

castortr0y [4]

You will get 20460000 as your answer which is broken down into, 2.046 x 10^7 as your number has to be between 1-10.

5 0

3 years ago

Read 2 more answers

A 25 kg circular disk has a diameter of 2.5 feet and a thickness of 2.5 cm. Find the density of the disk in kg/m3. Next, find th

Gre4nikov [31]

Answer:

Assume that $\rm g= 9.81\; N\cdot kg^{-1}$ ; $\rho(\text{Water}) = \rm 1000\;kg\cdot m^{-3}$ .

Density of the disk: approximately $\rm 2.19\times 10^{3}\; kg\cdot m^{-3}$ .

Weight of the disk: approximately $\rm 245\;N$ .

Buoyant force on the disk if it is submerged under water: approximately $\rm 112\; N$ .

The disk will sink when placed in water.

Explanation:

Convert the dimensions of this disk to SI units:

Diameter: $d = \rm 25\; inches = (25\times 0.3048)\; m = 0.762\;m$ .
Thickness $h = \rm 2.5\; cm = (2.5\times 0.01)\; m = 0.025\;m$ .

The radius of a circle is 1/2 its diameter:

$\displaystyle r = \rm \frac{1}{2}\times 0.762\;m = 0.381\; m$ .

Volume of this disk:

$V(\text{disk}) = \pi\cdot r^{2}\cdot h = \pi\times 0.381^{2}\times 0.025 \approx 0.0114009\; m^{3}$ .

Density of this disk:

$\displaystyle \rho(\text{disk}) = \frac{m}{V} = \rm \frac{25\; kg}{0.0114009\; m^{3}} = 2.19\times 10^{3}\;kg\cdot m^{-3}$ .

$\rho(\text{disk}) >\rho(\text{water})$ indicates that the disk will sink when placed in water.

Weight of the object:

$W(\text{disk}) = m\cdot g = \rm 25\times 9.81 = 245.25\; N$ .

The buoyant force on an object in water is equal to the weight of water that this object displaces. When this disk is submerged under water, it will displace approximately $\rm 0.0114009\; m^{3}$ of water. The buoyant force on the disk will be:

$\begin{aligned}F(\text{buoyant force}) &= W(\text{Water Displaced}) \\& = \rho\cdot V(\text{Water Displaced})\cdot g\\ & = \rm 1\times 10^{3}\; kg\cdot m^{-3}\times 0.0114009\; m^{3}\times 9.81\; N\cdot kg^{-1}\\ &\approx \rm 112\; N\end{aligned}$ .

The size of this disk's weight is greater than the size of the buoyant force on it when submerged under water. As a result, the disk will sink when placed in water.

3 0

3 years ago