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3 years ago

Why do mirrors form inverted images?

2 answers:

7 0

Answer: Reflection from a Concave Mirror
When the object is far from the mirror, the image is inverted and at the focal point. The image is real light rays actually focus at the image location). ... As the object moves towards the mirror inside the focal point the image becomes virtual and upright behind the mirror.

marysya [2.9K]3 years ago

3 0

Answer:

Explanation:

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).

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1. A 1.30 kg ball strikes a wall with a velocity of -10.5 m/s. The ball bounces off with a velocity of 6.50 m/s. If the ball is

bekas [8.4K]

Let F be the magnitude of the force. The impulse of this force while the ball is in contact with the wall is

Ft = F (0.0210 s)

and this impulse is equal to the change in the ball's momentum,

m ∆v = (1.30 kg) (6.50 m/s - (-10.5 m/s)) = (1.30 kg) (17.0 m/s)

Solve for F :

F (0.0210 s) = (1.30 kg) (17.0 m/s)

F = (1.30 kg) (17.0 m/s) / (0.0210 s)

F ≈ 1050 N

4 0

3 years ago

Help!! What’s the answer

solmaris [256]

Nope the answer is condensation because because its going in, *wink* *wink*

4 0

3 years ago

An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a

Blababa [14]

Answer:

$\frac{\Delta L}{L} =5.37\times 10^{-4}$

Explanation:

Given:

cross sectional area of the wire, $A=5.75\times 10^{-6}\ m^2$

density of aluminium wire, $\rho=2.7\times 10^3\ kg.m^{-3}$

young's modulus of the material, $E=7\times 10^{10}\ N.m^{-2}$

wave speed, $v=118\ m.s^{-1}$

<u>We have mathematical expression for strain as:</u>

$\frac{\Delta L}{L} =\frac{\sigma}{E}$ ...............................(1)

and since, $\sigma =\frac{T}{A}$

where, T = tension force in the wire

equation (1) becomes:

$\frac{\Delta L}{L} =\frac{T}{A.E}$ ............................(2)

<u>Also velocity ofwave in tensed wire:</u>

$v=\sqrt{\frac{T}{\mu} }$ ...................................(3)

where: $\mu=$ linear mass density of the wire

$\therefore \mu=\rho \times A$

Now, equation (3) becomes

$v=\sqrt{\frac{T}{\rho \times A} }$

$T=v^2.\rho \times A$ ............................(4)

Using eq. (2) & (4) for tension T

$v^2.\rho \times A=A.E\times \frac{\Delta L}{L}$

$\frac{\Delta L}{L} =\frac{v^2.\rho}{E}$

putting the respective values

$\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}$

$\frac{\Delta L}{L} =5.37\times 10^{-4}$

6 0

3 years ago

7. A car is travelling along a road at 30 ms when a pedestrian steps into the road 55 m ahead. The

trasher [3.6K]

Answer: Car collide with man

Explanation:

Given

Speed of car is $u=30\ m/s$

Distance of the man from the car is $s=55\ m$

Reaction time $t_r=0.5\ s$

Rate of deceleration $a_d=-10\ m/s^2$

Distance traveled in the reaction time $d_o=30\times 0.5=15\ m$

Net effective distance to cover $d=55-15=40\ m$

Distance required to stop the car

$\Rightarrow v^2-30^2=2(-10)(s)\\\Rightarrow 0-900=-20s\\\Rightarrow s=45\ m$

Require distance is more than that of net effective distance. Hence, car collides with the man.

6 0

3 years ago

A student is preparing to take a bath when she realizes the hot water tap in the bathroom is not working the student go to the k

WITCHER [35]

Answer:

The final temperature of the bath water will be 55°C

Explanation:

A student is preparing to take a bath when she realizes that the hot water tap is not working.

The student goes to the kitchen and prepares 10 L of 100 degree Celsius water to mix with 10 L of 10 degree Celsius water. What will be the final temperature of the bath water?

Using the relation, Heat lost = Heat gained (assuming no heat is lost to the surroundings).

Heat lost by hot water = heat gained by cold water

Using the formula of heat, Q = mcΔT

where m = mass of water, c = specific heat capacity of water, ΔT = temperature difference

mass of water = density * volume (density of water = 1 kg/L)

mass of hot water, m₁ = 1 kg/L * 10 L = 10 kg;

mass of cold water, m₂ = 1 kg/L * 10 L = 10 kg;

specific heat capacity of water, c = 4200 j/kg

Let the final temperature of the mixture be T

-m₁cΔT₁ = m₂cΔT₂ ( since heat is lost by the hot water)

-m₁ΔT₁ = m₂ΔT₂

-10 * (T- 100) = 10 * (T - 10)

-10T + 1000 = 10T - 100

20T = 1100

T = 1100/20

T = 55°C

Therefore, the final temperature of the bath water will be 55°C

6 0

3 years ago