What planet has a visible mark referred to as the the Great Red Spot?

What is the potential difference across lamp 1

Alex73 [517]

Answer:

1+1=2

Explanation:

You just need to add and voila the answer

7 0

2 years ago

A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±

gladu [14]

Answer:

E(final)/E(initial)=2

Explanation:

Applying the law of gauss to two parallel plates with charge density equal σ:

$E=\sigma/\epsilon_{o}=Q/(L^{2}*\epsilon_{o})\\$

So, if the charge is doubled the Electric field is doubled too

E(final)/E(initial)=2

3 0

3 years ago

Which of the following statements about electric field lines associated with electric charges is false? Electric field lines can

Tcecarenko [31]

Answer:

The false statement is 'Electric field lines form closed loops'.

Explanation:

Electric field lines originate from positive end and terminates at negative end,i.e., field lines are inward in direction to the negative charges and outward from the positive charges.
These lines when close together represents high intensity and when far apart shows low intensity of the field.
These lines do not intersect, as the tangent drawn on these lines provides us with the field direction and intersection of these lines means two field directions which is not possible.
These lines unlike magnetic field lines do not form closed loops as they do not turn around but originate at positive end and terminates at negative end which ensures no loop formation.

8 0

3 years ago

A weight trainer lifts a 90.0-kg barbell from a stand 0.90 m high and raises it to a height of 1.75 m. What is the increase in t

nirvana33 [79]

Answer:

∆PE = 749.7 J

At 0.9 m high, PE = 793.8 J

At 1.75 m high, PE = 1543.5 J

7 0

2 years ago

A man standing 1.54 m in front of a shaving mirror produces a real, inverted image 15.2 cm from it. What is the focal length of

zavuch27 [327]

Answer:

The focal length is 16.86 cm and the distance of the man if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

Explanation:

Given that,

Object distance u=1.54 m =154 cm

Image distance v = 15.2 cm

Magnification = 2

We need to calculate the focal length

Using formula of mirror

$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{15.2}+\dfrac{1}{-154}$

$\dfrac{1}{f}=\dfrac{347}{5852}$

$f=16.86\ cm$

We need to calculate the focal length

Using formula of magnification

$m= \dfrac{-v}{u}$

Put the value into the formula

$2=\dfrac{v}{u}$

$v = -2u$

Using formula of for focal length

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{16.86}=\dfrac{1}{u}-\dfrac{1}{2u}$

$\dfrac{1}{16.86}=\dfrac{1}{2u}$

$2u=16.86$

$u=\dfrac{16.86}{2}$

$u=8.43\ cm$

Hence, The focal length is 16.86 cm and the distance of the man if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

3 0

3 years ago