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2 years ago

Please help! The image produced by a concave mirror is ? .

Physics

2 answers:

Alexeev081 [22]2 years ago

6 0

Answer:

is a reflection.

The image is real light rays actually focus at the image location). As the object moves towards the mirror the image location moves further away from the mirror and the image size grows (but the image is still inverted). When the object is that the focal point, the image is at infinity.

Explanation:

SSSSS [86.1K]2 years ago

3 0

Answer:

virtual or real

Explanation:

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A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the

Mademuasel [1]

Answer:

(a) $a=2m/sec^2$

(b) 5220 j

(d) 3446.66 watt

Explanation:

We have given mass m = 290 kg

Initial velocity u = 0 m/sec

Final velocity v = 6 m/sec

Time t = 3 sec

From first equation of motion

v = u+at

So $a=\frac{v-u}{t}=\frac{6-0}{3}=2m/sec^2$

(a) We know that force is given by

F = ma

So force will be $F=290\times 2=580N$

(b) From second equation of motion we know that

$s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 2\times 3^2=9m$

We know that work done is given by

W = F s = 580×9 =5220 j

We know that power is given as

$P=\frac{W}{t}=\frac{5220}{3}=1740Watt$

(d) Time t = 1.5 sec

So $P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt$

5 0

3 years ago

A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the si

solmaris [256]

Answer:

4.384 * 10^13

Explanation:

Given the expression :

[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]

Applying the laws of indices

[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]

13.2733 * 10^19 ÷ 3.0276 * 10^6

(13.2733 / 3.0276) * 10^(19 - 6)

4.3840996 * 10^13

= 4.384 * 10^13

6 0

2 years ago

Scientists study how the continents move. Why might scientists use a model

tigry1 [53]

Answer:

B. It is too slow to observe directly

Explanation:

They move too slow to be able to observe how they move.

I hope it helps! Have a great day!
bren~

3 0

1 year ago

Determine the binding energy per nucleon of an mg-24 nucleus. the mg-24 nucleus has a mass of 24.30506. a proton has a mass of 1

My name is Ann [436]

The mass of Mg-24 is 24.30506 amu, it contains 12 protons and 12 neutrons.

Theoretical mass of Mg-24:

The theoretical mass of Mg-24 is:

Hydrogen atom mass = 12 × 1.00728 amu = 12.0874 amu

Neutron mass = 12 x 1.008665 amu = 12.104 amu

Theoretical mass = Hydrogen atom mass + Neutron mass = 24.1913 amu

Note that the mass defect is:

Mass defect = Actual mass - Theoretical mass : 24.30506 amu- 24.1913 amu= 0.11376 amu

Calculating the binding energy per nucleon:

$\frac{B.E.}{nucleon}=\frac{(0.11376amu)(931Mev/amu}{24nucleons} = 4.41294 Mev/nucleon$

So approximately 4.41294 Mev/necleon

3 0

2 years ago

A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns throu

anastassius [24]

Answer:

a. 3 s.

Explanation:

Given;

angular acceleration of the wheel, α = 4 rad/s²

time of wheel rotation, t = 4 s

angle of rotation, θ = 80 radians

Apply the kinematic equation below,

$\theta = \omega_1 t \ + \ \frac{1}{2} \alpha t^2\\\\80 = 4\omega_1 + \frac{1}{2}*4*4^2\\\\80 = 4\omega_1 + 32\\\\ 4\omega_1 = 48\\\\ \omega_1 = \frac{48}{4}\\\\ \omega_1 = 12 \ rad/s$

Given initial angular velocity, ω₀ = 0

Apply the kinematic equation below;

$\omega_1 = \omega_o + \alpha t_1\\\\12 = 0 + 4t\\\\4t = 12\\\\t = \frac{12}{4}\\\\t = 3 \ s$

Therefore, the wheel had been in motion for 3 seconds.

a. 3 s.

8 0

2 years ago