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3 years ago

How fast is the earth moving through the universe?

Physics

1 answer:

Dmitry_Shevchenko [17]3 years ago

8 0

The Earth moves 30 km/s in it's orbit around the sun.

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A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The sprin

Mariana [72]

Answer:

The velocity of block = 0.188 $\frac{m}{s}$

Explanation:

Mass m = 5.6 kg

k = 1040 $\frac{N}{m}$

$\mu$ = 0.26

$x_{1} =$ 0.035 m , $v_{1}$ = 0

$x_{2} =$ 0.02 m

From work energy theorem

$K_{1} + U_{1} + W_{other} = K_{2} + U_{2}$ --------- (1)

Kinetic energy

$K = \frac{1}{2} k x^{2}$ ------- (1)

Potential energy

$U = \frac{1}{2} k x^{2}$ ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

$R_{N}$ = mg

$R_{N}$ = 5.6 × 9.81 = 54.9 N

Friction force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

$W_{f}$ = - 21.9 × 0.015

$W_{f}$ = - 0.329 J

Now kinetic energy

At point 1

$K_{1} = \frac{1}{2} m v^{2} _{1}$

$K_{1} = 0$

$U_{1} = \frac{1}{2} k x^{2}$

$U_{1} = \frac{1}{2} (1040) 0.035^{2}$

$U_{1} =$ 0.637 J

At point 2

$K_{2} = \frac{1}{2} (5.6) v^{2} _{2}$

$K_{2} = 2.8 v_{2} ^{2}$

Potential energy

$U_{2} = \frac{1}{2} k x_2^{2}$

$U_{2} = \frac{1}{2} (1040) 0.02^{2}$

$U_{2} = 0.208$ J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 $v_{2} ^{2}$ + 0.208

2.8 $v_{2} ^{2}$ = 0.1

$v_{2} =$ 0.188 $\frac{m}{s}$

This is the velocity of block.

6 0

4 years ago

Beginning at time t = 0, a student exerts a horizontal force on a box of mass 30 kg, causing it to move at 1.2 m/s toward an ele

Alexandra [31]

Since the initial and the final velocity are just equal, it is implied that the acceleration is zero. This means that the net force acting on the body is also zero. The horizontal force should be equal to the force of friction. The force of friction is equal to the product of the coefficient and the normal force.

Ff = (coefficient of friction) x Fn

The normal force is equal to the object's weight if the surface is horizontal.

Ff = (0.20) x (30 kg) x (9.8 m/s²) = 58.8 N
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<span>Thus, the horizontal force exerted must be 58.8 N.

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6 0

3 years ago

Im board lolololololololololol

inysia [295]

Answer:

same bru

Explanation:

5 0

3 years ago

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LUCKY_DIMON [66]