If the speed and mass of an object are doubled, which of the following is true?

A large plate carries a uniform charge density σ = 8. 85 × 10-9 c/m2. a pattern showing equipotential surfaces with a 5 v potent

vfiekz [6]

The potential difference comes out to be

$10 \times 10 {}^{ - 3} m$

Given:

σ = 8. 85 × 10-9 c/m2

we know,

$E = \frac{σ}{2ε0}$

$E = \frac{8.85 \times 10 {}^{ - 9} }{2ε0}$

$E = \frac{v}{d}$

given the potential difference between two equipotential surface=5v

E=∆v

∆d=∆v/E

$= \frac{5 \times 8.85 \times 10 { }^{ - 12} \times 2 }{8.85 \times 10 {}^{ - 9} }$

$Δ = 10 \times 10 {}^{ - 3} m$

Thus the potential difference is

$10 \times 10 {}^{ - 3} m$

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5 0

2 years ago

A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object

Elina [12.6K]

Answer:

2f

Explanation:

The formula for the object - image relationship of thin lens is given as;

1/s + 1/s' = 1/f

Where;

s is object distance from lens

s' is the image distance from the lens

f is the focal length of the lens

Total distance of the object and image from the lens is given as;

d = s + s'

We earlier said that; 1/s + 1/s' = 1/f

Making s' the subject, we have;

s' = sf/(s - f)

Since d = s + s'

Thus;

d = s + (sf/(s - f))

Expanding this, we have;

d = s²/(s - f)

The derivative of this with respect to d gives;

d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²

Equating to zero, we have;

(2s/(s - f)) - s²/(s - f)² = 0

(2s/(s - f)) = s²/(s - f)²

Thus;

2s = s²/(s - f)

s² = 2s(s - f)

s² = 2s² - 2sf

2s² - s² = 2sf

s² = 2sf

s = 2f

8 0

3 years ago

If the tension in the rope is 160 n, how much work does the rope do on the skier during a forward displacement of 270 m?

Lunna [17]

If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

Given,

Tension force in the rope is (T) = 160 N

Displacement of the skier (S) = 270 m

The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.

Therefore, work done by the rope on the skier is,

$W=T.S$

⇒ $W=270*160*cos\pi \\W=-43200 J$

Hence work done by the rope is - 43200 J.

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8 0

2 years ago

Read 2 more answers

If the archerfish spits its water 30 degrees from the horizontal aiming at an insect 1.2 m above the surface of the water, how f

Burka [1]

Answer:

The speed is $v = 9.8 \ m/s$

Explanation:

From the question w are told that

The angle made is $\theta = 30^o$

The distance above the surface of the water is $h_{max} = 1.2 \ m$

The value of $g = 10 \ m/s^2$

The maximum height attained by the fish is mathematically evaluate as

$h_{max} = \frac{v^2 sin ^2 \theta }{2g }$

Making v which is the speed of the fish the subject of the formula

$v = \sqrt{ \frac{2gh_{max}}{ sin^2 \theta } }$

substituting values

$v = \sqrt{ \frac{2*10 *1.2 }{ [sin (30)]^2 } }$

$v = 9.8 \ m/s$

3 0

3 years ago

NASA scientists suggest using rotating cylindrical spacecraft to replicate gravity while in a weightless environment. Consider s

dusya [7]

Answer:

Explanation:

Given

diameter of spacecraft $d=148\ m$

radius $r=74\ m$

Force of gravity $F_g$ =mg

where m =mass of object

g=acceleration due to gravity on earth

Suppose v is the speed at which spacecraft is rotating so a net centripetal acceleration is acting on spacecraft which is given by

$F_c=\frac{mv^2}{r}$

$F_c=F_g$

$\frac{mv^2}{r}=mg$

$\frac{v^2}{r}=g$

$v=\sqrt{gr}$

$v=\sqrt{1450.4}$

$v=38.08\ m/s$

8 0

3 years ago