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zysi [14]
3 years ago
14

10. How far did the crate move horizontally?

Physics
1 answer:
kirill115 [55]3 years ago
4 0
The trámanos are broten to energy
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A 0.150-kg cart that is attached to an ideal spring with a force constant (spring constant) of 3.58 N/m undergoes simple harmoni
SVETLANKA909090 [29]

Answer:

E = 0.01 J

Explanation:

Given that,

The mass of the cart, m = 0.15 kg

The force constant of the spring, k = 3.58 N/m

The amplitude of the oscillations, A = 7.5 cm = 0.075 m

We need to find the total mechanical energy of the system. It can be given by the formula as follows :

E=\dfrac{1}{2}kA^2

Put all the values,

E=\dfrac{1}{2}\times 3.58\times (0.075)^2\\\\=0.01\ J

So, the value of total mechanical energy is equal to 0.01 J.

3 0
3 years ago
HELP!
Goryan [66]

Answer:

The solved problem is in the photo. Hope it helps.

3 0
3 years ago
True of false efficiency compared the output work to the output force
Lerok [7]
The statement about "<span>efficiency compared the output work to the output force" is false. Efficiency can be compared from the input work to the output work.</span>
8 0
4 years ago
Two cylindrical rods, one copper and the other iron, are identical in lengths and cross-sectional areas. They are joined, end to
Pie

Answer:

Vc = 2.41 v

Explanation:

voltage (v) = 16 v

find the voltage between the ends of the copper rods .

applying the voltage divider theorem

Vc = V x (\frac{Rc}{Rc + Ri})

where

  • Rc = resistance of copper = \frac{ρl}{a}  (l = length , a = area, ρ = resistivity of copper)
  • Ri = resistance of iron = \frac{ρ₀l}{a}  (l = length , a = area, ρ₀ = resistivity of copper)

Vc =  V x (\frac{\frac{ρl}{a}}{\frac{ρl}{a} + \frac{ρ₀l}{a}})

Vc = V x (\frac{ρ x (\frac{l}{a})}{(ρ + ρ₀) x (\frac{l}{a})})

Vc = V x (\frac{ρ}{ρ + ρ₀})

where

  • ρ = resistivity of copper = 1.72 x 10^{-8} ohm.meter
  • ρ₀ = resistivity of iron = 9.71 x 10^{-8} ohm.meter

Vc = 16 x (\frac{1.72 x 10^{-8}}{1.72 x 10^{-8} + 9.71 x 10^{-8}})

Vc = 2.41 v

5 0
3 years ago
If the PLATE SEPARATION of an isolated charged parallel-plate capacitor is doubled: A. the electric field is doubled
mash [69]

Explanation:

The electric field of an isolated charged parallel-plate capacitor is given by :

E=\dfrac{q}{A\epsilon_o}........(1)

Where

q is the electric charge

A is the area of cross section of parallel plate

It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.

So, the correct option is (E) i.e. "none of the above".

5 0
3 years ago
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