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3 years ago

How does friction affect the energy in a closed system of moving parts?

Physics

1 answer:

Triss [41]3 years ago

7 0

It converts kinetic energy into heat that remains in the system.

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Bob the American has just driven his car across the Canadian border when he sees a

Ludmilka [50]

Answer:

it equals 53 miles per hour

4 0

3 years ago

The radar system of a navy cruiser transmits at a wavelength of 1.4 cm, from a circular antenna with a diameter of 2.7 m. At a r

Goshia [24]

Answer:

Distance will be 49.34 m

Explanation:

We have given wavelength $\lambda =1.4cm =0.014m$

Diameter of the antenna d = 2.7 m

Range L = 7.8 km = 7800 m

We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D

We know that distance is given by $D=\frac{L1.22\lambda }{d}=\frac{7800\times 1.22\times 0.014}{2.7}=49.3422m$

So distance D will be 49.34 m

4 0

3 years ago

What form of energy breaks glass

yan [13]

While breaking glass with sound is an interesting trick it really takes a lot of effort vibrations diminishes do you to energy being carried away by the sound waves.

3 0

4 years ago

A 0.25 kg mass is placed on a vertically oriented spring that is stretched 0.56 meters from its equilibrium position. If the spr

Vitek1552 [10]

Answer:

The speed of the ball when it reaches equilibrium position is 3.31 m/s

Explanation:

Given;

mass of the object, m = 0.25 kg

initial displacement of the object, h₁ = 0.56 m

spring constant, k = 105 N/m

displacement at equilibrium position, h₂ = 0

initial velocity of the object, v₁ = 0

velocity of the object at equilibrium position = v₂

The change in gravitational potential energy at the equilibrium position is given as;

ΔP.E = mg(h₂ - h₁)

The change in kinetic energy of the object at the equilibrium position is given as;

ΔK.E = ¹/₂m(v₂² - v₁²)

Apply the principle of conservation of mechanical energy;

ΔK.E + ΔP.E = 0

¹/₂m(v₂² - v₁²) + mg(h₂ - h₁) = 0

¹/₂m(v₂² - 0) + mg(0 - h₁) = 0

¹/₂mv₂² - mgh₁ = 0

¹/₂mv₂² = mgh

¹/₂v₂² = gh

v₂² = 2gh

v₂ = √2gh

v₂ = √(2 x 9.8 x 0.56)

v₂ = 3.31 m/s

Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s

8 0

3 years ago

An object attached to a spring is pulled across a horizontal frictionless surface. If the force constant (spring constant) of th

lawyer [7]

Answer:

28.3kg

Explanation:

45=0.88*ml1.4

36.6=0.88m

divide through by 0.88

is said that

m=36.6/0.88

m=28.2857kg

3 0

3 years ago