Ask question

Ask question

All categories

Amiraneli [1.4K]

3 years ago

9

What type of image can a convex mirror produce?

2 answers:

Digiron [165]3 years ago

7 0

Convex mirrors only produce virtual images

nirvana33 [79]3 years ago

4 0

Answer:

Virtual,erect and diminished

Explanation:

A convex mirror is a spherical mirror which has a bulged reflecting surface. The bulge is towards the source of light. It also known as a diverging mirror.

A convex mirror always forms a diminished, virtual and an erect image. Such types of images cannot be obtained on a screen and appear to divert from a point behind the mirror.

You might be interested in

How does lifting electromagnet working

dmitriy555 [2]

Explanation:

Large electrical shifting magnets have concentrated retaining strength to lift dense, ferric objects and a deep-reaching magnetization. An immensely useful materials management technique is these electromagnetic rises.

3 0

2 years ago

3. Imagine a 10kg block moving with a speed of 20m/s<br> calculate the kinetic energy of this block

MatroZZZ [7]

The formula of the kinetic energy is:
$E_{k} = \frac{m v^{2} }{2}$
where m is a mass of the object, v is speed of the object at the moment of time. So we have:
$E_{k} = \frac{10* 20^{2} }{2} = 2000J$
The answer is 2000 Joules.

6 0

3 years ago

A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the

Marrrta [24]

Our values can be defined like this,

$m = 65kg$

$v = 3.5m / s$

$d = 0.55m$

The problem can be solved for part A, through the Work Theorem that says the following,

$W = \Delta KE$

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

$\frac {1} {2} mv ^ 2$

So,

$Fd = \frac {1} {2} m ^ 2$

Clearing F,

$F = \frac {mv ^ 2} {2d}$

Replacing the values

$F = \frac {(65) (3.5)} {2 * 0.55}$

$F = 723.9N$

B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

6 0

3 years ago

A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t

postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water $m=1.4\ kg$

at $200^{\circ}$ specific volume is $\nu =0.001157\ m^3\kg$ (From Table A-4,Saturated water Temperature table)

$V_1=m\nu _1$

$V_1=1.4\times 0.001157$

$V_1=1.6198\times 10^{-3}\ m^3$

Final Volume $V_2=4V_1$

$V_2=4\times (1.6198\times 10^{-3})$

$V_2=6.4792\times 10^{-3}\ m^3$

Specific volume at this stage

$\nu _2=\frac{V_2}{m}$

$\nu _2=\frac{6.4792\times 10^{-3}}{1.4}$

$\nu _2=0.004628\ m^3/kg$

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

$T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})$

$T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)$

$T_2=370.7^{\circ} C$

4 0

3 years ago

Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24C and leaves at 0.8 MPa and 60C. When the mass flow r

sergey [27]

Answer:

= 287kW

Explanation:

Knowing the enthalpy data, we have to

$h1=239.16kJ/kg\\h2=296.81kJ/kg$

So,

$E_i_n_p_u_t=mh_1$

Here,

m=mass flow rate

h= Enthalpy of refrigerant at the compressor

Replacing

= 1.2 × 239.16

= 287kW

4 0

3 years ago