1 year ago

An object has a gravitational

Physics

1 answer:

kirill [66]1 year ago

4 0

Answer:

potential energy = (mass × height × acceleration due to gravity.)

750 = mass × 3 × 10

hence, mass = 750/30 = 25 kg.

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As air pressure decreases, what happens

Naya [18.7K]

Answer:

I believe it is B, not 100% sure though

Explanation:

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2 years ago

Sam's job at the amusement park is to slow down and bring to a stop the boats in the log ride. If a boat and its riders have a m

shutvik [7]

Answer:

Sam will do 1152 J of work to stop the boat

Explanation:

Work: This is defined as the product of force and distance, the S.I unit of work is Joules. At any point in science, during calculation Energy and worked can be interchange because they have the same unit.

E = W = 1/2mv²................ Equation 1

Where E = energy, W = work, m = mass, v = velocity.

Given: m = 900 kg, v = 1.6 m/s

Substituting these values into equation 1

W = 1/2(900)(1.6)²

W = 450×2.56

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Therefore Sam will do 1152 J of work to stop the boat

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3 years ago

The gas used to fill party ballons contain only helium atoms .This make helium a(n).

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a gas

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2 years ago

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Name three categories that are used to classify the elements in the periodic table?

nikitadnepr [17]

Answer:

metals,nonmetals, and inert gases

Explanation:

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2 years ago

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Two long parallel wires 40 cm apart are carrying currents of 10 A and 20 A in the opposite direction. What is the magnitude of t

Alex_Xolod [135]

Answer:

The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

Explanation:

Given;

distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m

current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively

The magnitude of the magnetic field halfway between the wires can be calculated as;

$B = \frac{\mu _oI_1}{2 \pi r} + \frac{\mu_oI_2}{2\pi r}$

where;

B is magnitude of the magnetic field halfway between the wires

I₁ is current in the first wire

I₂ is current the second wire

μ₀ is permeability of free space

r is distance half way between the wires

$B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r} (I_1 +I_2)\\\\B = \frac{4\pi *10^{-7} }{2\pi *0.2} (10 +20) = 3.0 *10^{-5}\ T$

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

5 0

2 years ago