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3 years ago

Which type of mirror is most similar to a convex lens

Physics

2 answers:

denpristay [2]3 years ago

7 0

The exact answer is : Concave mirror

Mars2501 [29]3 years ago

6 0

There could be many different comparison features that would
lead to an observation of similarity or dis-similarity.

A mirror might be said to be similar to a convex lens if it is round,
or if it has a curved surface, or if it is made of glass.

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This diagram shows three horizontal forces acting on an object.

mixer [17]

Answer:

14N

Explanation:

Hope it will help

Brainlists please

5 0

3 years ago

A charge of 4 nc is placed uniformly on a square sheet of nonconducting material of side 17 cm in the yz plane. (a) what is the

Ratling [72]

The charge density of the sheet is 1.384×10⁻⁷C/m².

Charge density is defined as the charge per unit area.

The sheet is a square of length l=17 cm.

Calculate the area A of the sheet .

$A=l^2=(17 cm)^2= (17*10^-^2m)^2=0.0289 m^2$

The charge Q on the sheet is

$Q=4nC=4*10^-^9C$

The charge density σ is given by,

$\sigma=\frac{Q}{A}$

Substitute 4×10⁻⁹C for Q and 0.0289 m² for A.

$\sigma=\frac{Q}{A}\\ =\frac{4*10^-^9C}{0.0289 m^2} \\ =1.389*10^-^7C/m^2$

Thus, the charge density of the sheet is <u>1.384×10⁻⁷C/m².</u>

8 0

4 years ago

Explain it if u know please

Rudik [331]

Answer: B

Explanation: The straight line shows the theoretical average of the actual velocity and the curved line show that it is less than the average velocity of the object within 6 seconds.

straight line:

3. 3,80

6. 6,160

curved line:

1,2
2,17
3,32
4,60
5,100
6,160

A.K.A.-- the only place where the straight line and the curved line meet at is only at coordinate 6,160.

6 0

3 years ago

Read 2 more answers

. An 8-ton truck is initially going up a 20° ramp with a speed of 90 km/hr. The truck, then, brakes reaching a speed of 36 km/hr

marshall27 [118]

Answer:

a) $F = 13166.468\,N$ , b) $t = 2\,s.$

Explanation:

a) The equations of equilibrium for the truck are: (x' is the axis parallel to ramp, y' is the axis perpendicular to ramp)

During braking:

$\Sigma F_{x'} = - m\cdot g \cdot \sin \theta - F = m\cdot a$

$\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0$

The deceleration experimented by the truck is:

$a=\frac{(36\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} )-(90\,\frac{km}{h} )\cdot (\frac{1000\,m}{1\,km} )\cdot (\frac{1\,h}{3600\,s} )}{3\,s}$