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3 years ago

An orange of mass 3kg falls from a height 50m. When its height is H, its speed is

Physics

1 answer:

MArishka [77]3 years ago

8 0

Answer:

(10 m/s²) (50 m) = (10 m/s²) H + ½ (4 m/s)²

Explanation:

Initially, the orange has only gravitational potential energy.

As the orange falls, it has a combination of potential energy and kinetic energy.

Energy is conserved, so initial energy = final energy:

PE₀ = PE + KE

mgH₀ = mgH + ½ mv²

gH₀ = gH + ½ v²

Plugging in values:

(10 m/s²) (50 m) = (10 m/s²) H + ½ (4 m/s)²

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Answer:

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2 years ago

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A block of wood 3.0 cm on each side has a mass of 27g. what is the density of this block?

schepotkina [342]

The block of wood is 3cm on each side so it is a cube. The volume of a cube is given by s^3. So the volume of this block is 3cm x 3cm x3 cm = 27 cm^3. density = mass/volume =27 g / 27 cm^3 = 1 g/cm^3

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2 years ago

the refractive index of cooking oil is 1.47, and the refractive index of water is 1.33. A thick layer of cooking oil is floating

VARVARA [1.3K]

When light travels from a medium with higher refractive index to a medium with lower refractive index, there is a critical angle after which all the light is reflected (so, there is no refraction).

The value of this critical angle can be derived by Snell's law, and it is equal to
$\theta_C = \arcsin ( \frac{n_2}{n_1} )$
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.

In our problem, n1=1.47 and n2=1.33, so the critical angle is
$\theta_C = \arcsin( \frac{1.33}{1.47} )=\arcsin (0.91)=65^{\circ}$

4 0

2 years ago

An uncharged capacitor is connected to a 21-V battery until it is fully charged, after which it is disconnected from the battery

attashe74 [19]

Answer:

The voltage bewtween the plates will be 9.5V

Explanation:

Facts:

The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.

Where ϵ0 is the permittivity of free space.

A capacitor filled with dielectric slab of dielectric constant K, will have a new capacitance C1=ϵ0kA/d

C1=K(ϵ0A/d)

C1=KC ----------- 1

Where C is the capacitance with no dielectric and C1 is the capacitance with dielectric.

The new capacitance is k times the capacitance of the capacitor without dielectric slab.

This implies that the charge storing capacity of a capacitor increases k times that of the capacitor without dielectric slab.

Given points

• The terminal voltage of the battery to which the capacitor is connected to charge V=25V

• A dielectric slab of paraffin of dielectric constant K=2 is inserted in the space between the capacitor plates after the fully charged capacitor is disconnected

The charge stored in the original capacitor Q=CV

The charge stored in the original capacitor after inserting dielectric Q1=C1V1

The law of conservation of energy states that the energy stored is constant:

i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.

Q = Q1

CV = C1V1

 CV = C1V1 -------2

We derived C1=KC from equation 1. Inserting this into equation 2

CV = KCV1

V1 = (CV)/KC

V/K

= 21/2.2

= 9.5

4 0

3 years ago