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3 years ago

PLEASE HELP! ILL GIVE BRAINLY!!

2 answers:

7 0

B

Because instead of expending energy into the atmosphere we are gaining it ultimately causing earths temperature to increase

LUCKY_DIMON [66]3 years ago

3 0

Answer:

Explanation:

if it is ganing more energy than it is giving off (which of cource it isint or we whould be baken) then it whould heat up

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Where would a boat produce the highest concentration of carbon monoxide?

mr_godi [17]

A boat would produce the highest concentration of carbon monoxide in the exhaust system.

Carbon monoxide (CO) is a colorless, odorless, and tasteless gas that is slightly less dense than air. It is toxic to hemoglobic animals (both invertebrate and vertebrate, including humans) when encountered in concentrations above about 35 ppm.

5 0

3 years ago

. Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connecte

Umnica [9.8K]

Given :

Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connected with wires as shown in the circuit given in figure. The current flowing in both the circuits is the same.

To Find :

Will the heat produced in both the cases be equal.

Solution :

Heat released is given by :

H = i²Rt

Here, R is resistance and is given by :

$R = \dfrac{\rho L}{A}$

So,

$H = i^2\times \dfrac{\rho L}{A} t\\\\H = \dfrac{i^2\rho Lt}{A}$

Now, in the question every thing is constant except for the length of the wire and from above equation heat is directly proportional to the length of the wire.

So, heat produced by Reem's wire is more than Nain one.

Hence, this is the required solution.

7 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