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3 years ago

Please answer this question

Physics

1 answer:

sergij07 [2.7K]3 years ago

3 0

Explanation:

m = kg. v=m/s. g=m/s^2. h= m

>>1/2mv^2=mgh

>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m

>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m>>kg m^2/s^2=kg m^2/s^2 the fraction 1/2 won't be able to make any changes to to the dimensional expression of energy i.e half of energy is still energy therefore you can neglect the number .

>>kg m^2/s^2=kg m^2/s^2 

>>J= J

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A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may

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Answer:

The tension on the string is $T = 43.302 \ N$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 5.00 \ kg = 5000 \ g$

The density of the rock is $\rho = 4300 \ kg/m^3 = 4.3 g/dm^3$

Generally the volume of the rock is mathematically evaluated as

$V = \frac{m_r}{\rho}$

substituting values

$V = \frac{5000}{4.3}$

$V = 1162.7 \ dm^3$

The volume of the rock immersed in water is

$V_w = \frac{V}{2}$

substituting values

$V_w = \frac{1162.7 }{2}$

$V_w = 581.4 \ dm^3$

mass of water been displaced by the this volume is

$m_w = V_w$ According to Archimedes principle

=> $m_w = 581.4 \ g$

$m_w = 0.5814 \ kg$

The weight of the water displace is

$W _w = m_w * g$

$W _w = 0.5814 * 9.8$

$W _w = 5.698 \ N$

The actual weight of the rock is

$W_r = m_r * g$

$W_r = 5.0 * 9.8$

$W_r = 49.0 \ N$

The tension on the string is

$T = W_r - W_w$

substituting values

$T = 49.0 - 5.698$

$T = 43.302 \ N$

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2 years ago

All of the following are environmental factors that affect human development except _____.

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Answer:

The answer would be....

Explanation:

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3 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

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Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

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3 years ago

A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an

posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle $\theta=30°$

We need to calculate the angle in radian

$\theta=30\times\dfrac{\pi}{180}$

$\theta=0.523\ rad$

We need to calculate the path length

Using formula of path length

$Path\ length =angle\times radius$

$Path\ length=0.523\times5.9$

$Path\ length =3.09\ m$

(b). Angle = 30 rad

We need to calculate the path length

$Path\ length=30\times5.9$

$Path\ length=177\ m$

(c). Angle = 30 rev

We need to calculate the angle in rad

$\theta=30\times2\pi$

$\theta=188.4\ rad$

We need to calculate the path length

$Path\ length=188.4\times5.9$

$Path\ length =1111.56\ m$

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

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3 years ago

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3 years ago