What is the magnitude of the electric field on a +2 C charge if it experiences an electric force of 6 N?

A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is

AleksandrR [38]

(a) 198 g

When the rock is submerged into the water, there are two forces acting on the rock:

- its weight, equal to $W=mg$ (m=mass, g=acceleration of gravity), downward

- the buoyant force, equal to $B=m_w g$ ( $m_w$ =mass of water displaced), upward

So the resultant force, which is the apparent weight of the rock (W'), is

$W'=W-B$

which can be rewritten as

$m'g = mg-m_w g$

where m' is the apparent mass of the rock. Using:

m = 540 g

m' = 342 g

we find the mass of water displaced

$m_w = m-m'=540 g-342 g=198 g$

(b) $1.98\cdot 10^{-4} m^3$

If the rock is completely submerged, the volume of the rock corresponds to the volume of water displaced.

The volume of water displaced is given by

$V_w = \frac{m_w}{\rho_w}$

where

$m_w = 198 g = 0.198 kg$ is the mass of the water displaced

$\rho_w = 1000 kg/m^3$ is the density of the water

Substituting,

$V_w = \frac{0.198}{1000}=1.98\cdot 10^{-4} m^3$

And so this is also the volume of the rock.

(c) $2727 kg/m^3$

The average density of the rock is given by

$\rho = \frac{m}{V}$

where

m = 540 g = 0.540 kg is the mass of the rock

$V=1.98\cdot 10^{-4} m^3$ is its volume

Substituting into the equation, we find

$\rho = \frac{0.540 kg}{1.98\cdot 10^{-4}}=2727 kg/m^3$

3 0

3 years ago

The brakes of a 125 kg sled are applied while it is moving at 8.1 m/s, which exerts a force of 261 N to slow the sled down. How

sammy [17]

Answer:

15.7 m

Explanation:

m = mass of the sled = 125 kg

v₀ = initial speed of the sled = 8.1 m/s

v = final speed of sled = 0 m/s

F = force applied by the brakes in opposite direction of motion = 261

d = stopping distance for the sled

Using work-change in kinetic energy theorem

- F d = (0.5) m (v² - v₀²)

- (261) d = (0.5) (125) (0² - 8.1²)

d = 15.7 m

6 0

3 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

True or false your vocal cords act like a guitar string when it comes to sound

torisob [31]

Answer:

True

Explanation:

8 0

3 years ago

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5 meddi muttasıl örnek

alexandr1967 [171]

Answer:

WHAT'S YOUR LANGUAGE

I CAN'T UNDERSTAND

5 0

2 years ago

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