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3 years ago

What's the minimum Out PUT WORK required to raise 14,0m3 of water 26.0m?

Physics

1 answer:

BartSMP [9]3 years ago

3 0

Answer:

3.57 MJ

Explanation:

ASSUMING it's fresh water with density of 1000 kg/m³

W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J

Salt water would require more.

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Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav

umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

$v_i = 4.0 \ m/s$

then,

$v_f=\frac{4.0}{2}$

$=2.0 \ m/s$

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ $a=\frac{v_f^2-v_i^2}{2s}$

On substituting the given values, we get

⇒ $=\frac{(2.0)^2-(4.0)^2}{2\times 18.2}$

⇒ $=\frac{4-8}{37}$

⇒ $=\frac{-4}{37}$

⇒ $=0.108 \ m/s^2$

As we know,

⇒ $f=ma$

and,

⇒ $\mu_rmg=ma$

⇒ $\mu_r=\frac{a}{g}$

On substituting the values, we get

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7 0

3 years ago

If two point masses 1kg & 4kg are seperated by a distance of 2m. Magnitude of gravitational force exerted by 1kg on 4kg is ?

Aliun [14]

Answer:

F = G Newtons

Explanation:

Given:

Mass of 1st body = $1\:kg$
Mass of 2nd body = $4\:kg$

To Find:

Magnitude of gravitational force

Solution:

Here, we have a formula

$F=\dfrac{G.M_{1}.M_{2}}{r^{2}}$

<u>Substituting the values</u>

$\implies\:\:F = \dfrac{G(1)(4)}{2^{2}}$

$\implies\:\:F = \dfrac{4G}{4}$

$\implies\:\:F = \dfrac{\cancel{4}G}{\cancel{4}}$

$\implies\:\:\red{F = G}$

Know More:

The applied formula for the above solution is

${\boxed{F_{G}=\dfrac{G.M_{1}.M_{2}}{r^{2}}}}$

where,

F $_{G}$ = Gravitational force
G = Gravitational constant
M $_{1}$ = mass of 1st body
M $_{2}$ = mass of 2nd body
r = distance between two bodies

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