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3 years ago

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Physics

1 answer:

laila [671]3 years ago

4 0

Answer:

Explanation:

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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

On a cold day you can warm your hands by rubbing them together. (a) Assume that the coefficient of friction between your hands i

melomori [17]

To solve this problem it is necessary to apply the equations related to the Force of Friction and Energy.

By definition the friction force is defined as

$F= \mu * N$

Where,

$\mu =$ Frictional Constant

N = Normal Force -> mg

At the same time we have the definition of the Energy, which can be

E = F*v

Where,

Force

v = Velocity.

Then replacing with our values we have that,

$F = \mu N$

F = 0.5 *34

F = 17

The energy then would be,

E = f v

E = 17 * 0.30 = 5.1 W

Therefore the rate at which heat is generated is 5.1W

6 0

3 years ago

What is the relationship between Potential and Kinetic Energy of a falling object

artcher [175]

Object Falling from Rest. As an object falls from rest, its gravitational potential energy is converted to kinetic energy. Conservation of energy as a tool permits the calculation of the velocity just before it hits the surface. K.E. = J, which is of course equal to its initial potential energy.

8 0

3 years ago

A 15-turn circular wire loop with a radius of 3.0 cm is initially in a uniform magnetic field with a strength of 0.5 T. The fiel

miskamm [114]

Answer:

0.212V

Explanation:

The induced emf in circular loop is $\epsilon=-d\frac{\phi_B}{dt}=-NA\frac{dB}{dt}$