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3 years ago

Dierdre drew a diagram to compare the three types of mirrors.

Physics

1 answer:

Sholpan [36]3 years ago

8 0

<em>Labels that belong in the marked ares X, Y & Z include;</em>

X: Curves outward

Y: Image may be smaller than object

Z: Image is always virtual

<u>Since the rays never meet, the images formed by convex mirrors are always virtual and smaller than the object, and since they are smaller, the images appear to be further than they actually are.</u>

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Between which two points did they travel fastest?

marusya05 [52]

Answer:

During the section CD , the speed is fastest.

Explanation:

The rate of change of distance is called speed.

Speed = distance / time

Its SI unit ism/s. It is a scalar quantity.

The slope of the distance time graph is given by the speed of the object.

Here, the speed of AB is 30/3= 10 m/s .

The speed of BC is = 0 m/s

The speed of CD is (50 - 30)/(6 - 5) = 20 m/s

So, the speed is maximum during the section CD.

7 0

3 years ago

100 J OF HEAT IS PRODUCED EACH SECOND IN A 4 COULUMB RESISTER. THE POTENTIAL DIFFERANCE ACROSS THE RESISTER WILL BE

igomit [66]

Answer:

The correct answer is "20 Volts".

Explanation:

Given:

Heat,

H = 100 J

Resistance,

R = 4 Ω

As we know,

⇒ $P=\frac{v^2}{R}$

By putting the values, we get

⇒ $100=\frac{v^2}{4}$

⇒ $v^2=100\times 4$

⇒ $=400$

⇒ $v=\sqrt{400}$

⇒ $=20 \ volts$

6 0

3 years ago

A 66-kg fellow stands on a digital scale in an elevator that accelerates upwards from rest to 4.5 m/s in 2.00 s. show answer Inc

Nina [5.8K]

Answer:

Explanation:

mass of the fellow ( m ) = 66kg

acceleration of fellow a

v = u + at

4.5 = 0 + a x 2

a = 4.5 /2

= 2.25 m / s²

Net force acting on fellow in upward direction by the surface of elevator

R - mg where R is reaction force of the surface of the elevator

Applying Newton's law of motion

R - mg = ma

R = m (g +a )

= 66 x ( 9.8 + 2 )

= 778.8 N

This will be the scale reading .

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3 years ago

During the push-up, the hips should never hit the ground and should move 1 point

Monica [59]

Hahahahahahahahahha true

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4 years ago

The electric potential at the dot in the figure is 3160 V. What is charge q?

PSYCHO15rus [73]

Hi there!

Recall the equation for electric potential of a point charge:

$V = \frac{kQ}{r}$

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

$V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V$

Lower left charge's potential:

$V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V$

Add the two, and subtract from the total EP at the point:

$3160 + 1123.75 - 2247.5 = 2036.25$

The remaining charge must have a potential of 2036.25 V, so:

$2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}$

5 0

2 years ago