Which quote best reveals why sowers was successful in achieving her goal?

Precision measurements of the acceleration due to gravity show that the acceleration is slightly different in different location

DochEvi [55]

Gravity largely depends on the comparison of two objects; it's why you have the equation F= (GMm)/r^2. On Earth, you have different altitudes that, with the formula, will give different results for gravity because the radius is different everywhere. This difference on calculations, however, are seen to be miniscule. We know gravity as 9.81 m/s^2 but it might be different by thousandths or hundreds of thousandths of a decimal.

6 0

4 years ago

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Whats the definition of deposition

Luda [366]

Deposition:

- when a gas changes directly to a solid

- latent heat is released

- physical change, NOT a chemical change

7 0

3 years ago

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A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

8 0

3 years ago

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe

SVETLANKA909090 [29]

Answer:

$T_s = 6.8 degree C$

Explanation:

As per thermal radiation we know that rate is heat radiation is given as

$\frac{dQ}{dt} = \sigma eA (T^4 - T_s^4)$

here we know that

T = 34 degree C = 307 K

$A = 1.38 m^2$

e = 0.557

$\sigma = 5.67 \times 10^{-8} W/m^2K^4$

$\frac{dQ}{dt} = 120 J/s$

now we have

$120 = (5.67 \times 10^{-8})(0.557)(1.38)(307^4 - T_s^4)$

$120 = (4.36 \times 10^{-8})(307^4 - T_s^4)$

$T_s = 279.8 K$

$T_s = 6.8 degree C$

5 0

3 years ago

What is the frictional force between a box and the floor it is being pulled across if, the kinetic coefficient of friction is 0.

Artyom0805 [142]

If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that

<em>n</em> = 38 N

The friction force is proportional to the normal force by a factor of 0.27, so that

<em>f</em> = 0.27 (38 N) ≈ 10.3 N

and so the answer is D.

8 0

3 years ago