1 point
30. A ray passing through the centre of curvature of a concave mirror.
after reflection is reflected back along the same path. Which is the correct
figure for the above statement.

Answers

Answer 1

Explanation:

when the object is between the pole and focus


Related Questions

which of the following are fundamental unit ?
I.candela
II.ampere
III.mole
IV.all of these​

Answers

Answer:

all of these very easy questions

The elevators in the John Hancock building in Chicago move 783 ft in 33 s. What is this speed in miles per hour?

Answers

Answer:

16.18 mph

Explanation:

the speed of elevators =

(783×3600)/(33×5280) =

2,818,800/174,240 = 16.18 mph

what is the density of a 75 g block of wood measuring 12 cm x 8cm x 9cm

Answers

Answer:

density is equal to mass divided by volume (m/v)

volume: 12×8×9=864cm³

75÷864=0.09

Answer: 0.09g/cm³

Round 0.010229 to four sig figs

Answers

It would be 0.01023.

0.010229 has 5 sig. fig.'s (leading zeros do not count)

So rounding to 4 sig. fig.'s is just a matter of rounding off the last digit, which leaves you with 0.01023.

Slove these problem physics ​

Answers

Answer:

element name = neon

no of proton = 10

no. if neutron = 9

atomic no. = 10

atomic weight = 20.17179u

it lies in group 18 and period 2

it is nonmetal because it is a noble gas .

hope this helped you ☺️

a radioactive element has a half-life of 30 days. calculate the mass of the element that remains after 20 days if 100g of the element decays. Also, calculate the decay constant.

Answers

Explanation:

the answer is in the image above

Two trucks left the warehouse at the same time and drove in
the same direction along a straight road. One truck drove at 68
km/h while the other drove at 57 km/h. Find theirs distance
apart after2 hours.

Answers

Truck a
V = 68km/h
t = 2hours

Distance = speed * time

D = 68 * 2

D = 136km




Truck B

Speed = 57km/h

t = 2 hours

Distance = speed * time

D = 57 * 2

D = 114km


Distance travelled by truck A = 136 km

Distance travelled by truck B = 114km

Distance apart = 136km - 114km = 22km

Hence both the trucks will be 22kms apart.

Hope this helps. Plz let me know if you have any doubts.

A train is moving with the velocity of 10m/s . It attains an acceleration of 4m/s after 5 sec . find the distance covered by train at that time.

Answers

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A shift inward of the production possibilities curve signifies that _

A. The demand for the products has increased, so the supply will increase
B. The available resources have increased, so potential production levels will increase
C. The available production resources have decreased, so potential production levels will decrease
D. The demand for the products has decreased so, the supply will decrease

Answers

Answer:

c

Explanation:

A shift inward of the production possibilities curve signifies that: C. The available production resources have decreased, so potential production levels will decrease.

A production possibilities curve (PPC) is also referred to as production possibilities frontier (PPF) and it can be defined as a curve that illustrates the maximum (best) combinations of two (2) products which can be produce in an economy if they both depend on these two (2) main factors;

Fixed technology. Fixed resources.

Basically, the production possibilities curve (PPC) illustrates the maximum combinations of goods that are available with fixed technology and resources.

Furthermore, an inward shift of the production possibilities curve (PPC) signifies that the available resources used for production have decreased, so potential production levels would also decrease.

Read more: https://brainly.com/question/17053992

seharitat -The distance of a panet from the earth is 2.5 x 10^7 km and the gravitational force between them them is 3.82 x 10^18 N. Mass of the planet and earth are equal, each being 5.98 x 10^24 Kg. Calculate the universal gravitation constant.​

Answers

F=3.82×10^18NMass of earth=M=5.98×10^24kgMass of planet=m=5.98×10^24kgDistance=r=2.5×10^7km

Universal gravitation constant=G

We know

[tex]\boxed{\sf F=G\dfrac{Mm}{r^2}}[/tex]

[tex]\\ \sf\longmapsto G=\dfrac{Fr^2}{Mm}[/tex]

[tex]\\ \sf\longmapsto G=\dfrac{3.82\times 10^{18}(2.5\times 10^7)^2}{(5.98\times 10^{24})^2}[/tex](M=m)

[tex]\\ \sf\longmapsto G=\dfrac{3.82\times 10^{18}(6.25\times 10^{14})}{35.7\times 10^{48}}[/tex]

[tex]\\ \sf\longmapsto G=\dfrac{23.87\times 10^{32}}{35.7\times 10^{48}}[/tex]

[tex]\\ \sf\longmapsto G=0.668\times10^{-16}[/tex]

[tex]\\ \sf\longmapsto G=6.68\times 10^{-15}[/tex]

If a ball is rolling at a velocity of 1.5 m/s and has a momentum of 10.0 kg times m/s, what is the mass of the ball?

Answers

Answer:

Mass of the ball, M = 6.667 kg

Explanation:

Given data:

Momentum, Mo = 10.0 kgm/s

Velocity of the rolling ball, V = 1.5 m/s

Mass of the body, M = ?

Momentum, Mo = Mass, M x Velocity, V

10.0 kgm/s = M x 1.5 m/s

Divide each side by 1.5 m/s

M = 10.0 kgm/s / 1.5 m/s

M = (6⅔) kg

:. Mass of the ball, M = 6.667 kg

A wombat runs south in a straight line with an average velocity of 5 m/s for 4
minutes and then with an average velocity of 4 m/s for 3 minutes in the same
direction.
What is the wombats total displacement?
What is the average velocity during that time?

Answers

We have that

The wombats total displacement is

[tex]d_t=2100m[/tex]

The average velocity during that time is

[tex]v_{avg}=5m/s[/tex]

From the question we are told that:

initial Velocity [tex]V_1=5m/s[/tex]

initial Time [tex]T_1=4min=>240sec[/tex]

Final Velocity [tex]V_2=4m/s[/tex]

Final time  [tex]T_2=3min=>180sec[/tex]

a)

Generally, the equation for displacement is mathematically given by

d_t=(Initial +final) displacement

Where

initial displacement

[tex]d_1=V_1*t_1\\\\d_1=5*240[/tex]

[tex]d_1=1200m[/tex]

Final displacement

[tex]d_2=V_2*t_2\\\\d_2=5*180[/tex]

[tex]d_2=900m[/tex]

Therefore

[tex]d_t=(Initial +final) displacement[/tex]

[tex]d_t=1200+900[/tex]

[tex]d_t=2100m[/tex]

b)

Generally, the equation for Average velocity is mathematically given by

[tex]v_{avg}=\frac{d_t}{t_1+t_2}[/tex]

[tex]v_{avg}=\frac{2100m}{240+180}[/tex]

[tex]v_{avg}=5m/s[/tex]

in conclusion

The wombats total displacement is

[tex]d_t=2100m[/tex]

The average velocity during that time is

[tex]v_{avg}=5m/s[/tex]

For more information on this visit

https://brainly.com/question/11934397

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