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3 years ago

If a client's knees hurt after a one-week walking regimen, which one of the following suggestions is most appropriate?

Physics

1 answer:

Flura [38]3 years ago

5 0

A, it's better to take a break to let your body rejuvenate.

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Dimas [21]

Answer:

l and lll

Explanation:

4 0

3 years ago

Read 2 more answers

Consider three planets. All have the same mass as Earth, but with different radii (from largest to smallest: Planet 1, 2, 3). Fo

LuckyWell [14K]

Answer:

option C

Explanation:

given,

mass of the three planet is same

radius of the planets are

R₁ > R₂ > R₃

expression of escape velocity

$v = \sqrt{\dfrac{2GM}{R}}$

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

from the above expression we can clearly conclude that the escape velocity is inversely proportional to the radius of the Planet.

radius of planet increases escape velocity decreases.

Hence planet 3 has the smallest radius so the escape velocity of the third planet will be maximum.

The correct answer is option C

3 0

4 years ago

flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electr

Ksenya-84 [330]

Answer:

$6.29591\times 10^{-6}\ N/C^2$

Explanation:

Flux is given by

$\phi=EAcos\theta$

A = Area

$A=0.4\times 10^{-3}\times 0.6\times 10^{-3}$

E = Electric field = 76.7 N/C

Angle is given by

$\theta=90-20\\\Rightarrow \theta=70^{\circ}$

$\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2$

The flux through the sheet is $6.29591\times 10^{-6}\ N/C^2$

6 0

3 years ago

The number of planets, besides the earth, that are visible to the unaided eye is

sergejj [24]

The answer should be 5 it’s easy

3 0

3 years ago

Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters

Vesnalui [34]

Answer:

C. $\frac{3F}{8}$

Explanation:

Let initial charges on both spheres be, $q$

$F=\frac{Kq^2}{d^2} \ \ \ \ \ \ \ \ \ \ \_i$

When the sphere C is touched by A, the final charges on both will be, $\frac{q}{2}$

#Now, when C is touched by B, the final charges on both of them will be:

$q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\$

Now the force between A and B is calculated as:

$F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}$

Hence the electrostatic force becomes 3F/8

4 0

3 years ago