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3 years ago

According to Newton's second law of motion what is force equal to

2 answers:

8 0

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the netforce, and inversely proportional to the mass of the object

IrinaVladis [17]3 years ago

6 0

Explanation:

According to Newton's second law of motion, the force is equal to the product of mass and acceleration of the object. Mathematically, it is given by :

$F=m\times a$

It is also defined as the rate of change of linear momentum is directly proportional to the applied external force i.e.

$F=\dfrac{dp}{dt}$

If the momentum of the system is constant, the net force acting on it is equal to 0. Hence, this is the required solution.

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What is the total magnification for each lens setting on a microscope with 15x oculars and 4x, 10x, 45x, and 97x objectives lens

KengaRu [80]

Answer:

M = 60x

M = 150x

M = 675x

M = 1455 x

Explanation:

In a microscope the total magnification is found by multiplying the magnification of the ocular (eye piece) and objective lens.

$m_e$ =Ocular magnification = 15x

$m_o$ =Objective magnification = 4x

Total magnification

$M=m_e\times m_o\\\Rightarrow M=15\times 4\\\Rightarrow M=60x$

M = 60x

$M=m_e\times m_o\\\Rightarrow M=15\times 10\\\Rightarrow M=150x$

M = 150x

$M=m_e\times m_o\\\Rightarrow M=15\times 45\\\Rightarrow M=675x$

M = 675x

$M=m_e\times m_o\\\Rightarrow M=15\times 97\\\Rightarrow M=1455x$

M = 1455 x

7 0

3 years ago

An 89.2-kg person with a density 1025 kg/m3 stands on a scale while completely submerged in water. What does the scale read?

Luden [163]

Answer:

89.11kg

Explanation:

Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.

Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}

The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.

The weight of the fluid=0.087m3× 1kg/me = 0.087kg

Now the weight of the fluid displaced is referred to as the upthrust.

Now the real weight - the apparent weight = the upthrust.

Hence the apparent weight = real weight - upthrust

Apparent weight = 89.2-0.087 = 89.11kg

3 0

3 years ago

How to convert sin angles with values of 90 degrees, 60, 45, 30, 0 to cos degrees.

Inga [223]

Answer:

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6 0

3 years ago

Three forces act on an object. Two of the forces have the magnitudes 56 N and 27 N, and make angles 53° and 156°, respectively,

Ludmilka [50]

Answer:

Explanation:

Given

Two forces $F_1$ and $F_2$ at an angle of $\theta _1$ $\theta _2$

$F_1=56\ N$

$F_2=27\ N$

$\theta _1=53^{\circ}$

$\theta _2=156^{\circ}$

As resultant force is zero therefore horizontal component as well as vertical component of force is zero

$\sum F_x=F_1\cos \theta _1+F_2\cos \theta _2+F_3\cos \theta _3$

$\sum F_x=56\cos 53+27\cos 156+F_3\cos \theta$

$F_3\cos \theta_3=-9.035\ N----1$

$\sum F_y=F_1\sin \theta _1+F_2\sin \theta _2+F_3\sin \theta _3$

$F_3\sin \theta _3=55.704\ N----2$

squaring and adding 1 and 2

$F_3^2(\cos ^2\theta _3+\sin^2\theta _3)=86.583+3102.93$

$F_3=56.47\ N$

Divide 1 and 2 to get $\theta _3$

$\frac{\sin \theta_3 }{\cos \theta_3 }=\frac{-55.704}{9.305}$

$\tan \theta_3=-6.16$

$\theta _3=180-80.78$

$\theta _3=99.22^{\circ}$

6 0

3 years ago

Consider a block of mass m at rest on an inclined plane of angle theta. An acrobat of mass m_A is standing on the top corner of

shtirl [24]

Answer:

μ = tan θ

Explanation:

For this exercise let's use the translational equilibrium condition.

Let's set a datum with the x axis parallel to the plane and the y axis perpendicular to the plane.

Let's break down the weight of the block

sin θ = Wₓ / W

cos θ = W_y / W

Wₓ = W sin θ

W_y = W cos θ

The acrobat is vertically so his weight decomposition is

sin θ = = wₐₓ / wₐ

cos θ = wₐ_y / wₐ

wₐₓ = wₐ sin θ

wₐ_y = wₐ cos θ

let's write the equilibrium equations

Y axis

N- W_y - wₐ_y = 0

N = W cos θ + wₐ cos θ

X axis

Wₓ + wₐ_x - fr = 0

fr = W sin θ + wₐ sin θ

the friction force has the formula

fr = μ N

fr = μ (W cos θ + wₐ cos θ)

we substitute

μ (Mg cos θ + mg cos θ) = Mgsin θ + mg sin θ

μ = $\frac{(M +m) \ sin \ \theta }{(M +m) \ cos \ \theta }$

μ = tan θ

this is the minimum value of the coefficient of static friction for which the system is in equilibrium.

8 0

3 years ago