The pinion gear A rolls on the gear racks B and C.

Franklin needs to ship a box with FedEx. In order to calculate his shipping costs, he needs to measure the mass of the package.

balandron [24]

Answer:

meter

Explanation:

please mark as brainlist answer

8 0

3 years ago

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Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a ligh

guapka [62]

Answer:

This is because the acceleration of objects due to gravity is independent of the mass of the object and is constant for all objects, therefore, all objects fall with the same speed.

Explanation:

The weight of an object or force of gravity acting on an object on the surface of earth is a product of its mass and acceleration due to gravity.

Mathematically, w = mg

where, m=mass of the object; g = acceleration due to gravity

Also, from newton's law of gravitation, gravitational force on the object ,F = GMm/r²

where G is the gravitational constant; M is mass of Earth; m is mass of object; r is the distance of separation between the object and the center of mass of the earth which is approximately the radius of earth.

Since the weight of an object is equal to the force of gravitation acting on it

W = F

mg = GMm/r²

g = GM/r²

The expression above is that of the relationship between the force of gravity acting on a body on the earth's surface, the weight of that body and the acceleration due to gravity, g.

It can be seen that the acceleration due to gravity g is independent of the mass of the object. Therefore, the acceleration of objects due to gravity is constant for all objects and all objects fall with the same speed.

4 0

2 years ago

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The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

7 0

3 years ago

A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and

sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0

2 years ago

What can be contracted through contact with saliva?

m_a_m_a [10]

Tuberculosis mrsa are some things that are contracted by saliva where spores in the air from an infected person
sneezing as well

4 0

3 years ago