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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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Can anyone help? <br> Please and thank you

Juliette [100K]

Kinetic energy is calculated as such: $KE=\frac{1}{2} mv^{2}$ | therefore, $KE=0.913598 J$

7 0

3 years ago

How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0

1 year ago

3.1

Ilia_Sergeevich [38]

Answer:

The value is $C = 30729\ c$

Explanation:

From the question we are told that

The power rating of the stove is $P = 4.4 KW = 4.4 *10^{3} \ W$

The duration of its use everyday is $t_1 = 70 \ minutes = 1.167 \ hours$

The rating of the light bulbs is $P_2' = 150 W$

The number is $n = 7$

The power rating of the total bulb is $P_2 = 7 * 150 = 1050 \ W$

The duration of its use everyday is $t_2 = 7 hours$

The power rating of miscellaneous appliance $P_3 = 1.8 \ KW = 1.8 *10^{3} \ W$

The duration of its use everyday is $t_3 = 1 hour$

The power rating of hot water $P = 4 KW = 4 *10^{3} \ W$

The duration of its use everyday is $t_4 = 120 \ minutes = 2 hours$

Generally the total electrical energy used in 1 month is mathematically represented as

$E = P_1 * t_1 * 30 + P_2 * t_2 * 30 + P_3 * t_3 * 30+ P_4 * t_4 * 30$

=> $E = 4.4*10^{3} * 1.167 * 30 + 1050 * 7 * 30 + 1.8 *1 * 30+ 4*10^{3} * 2 * 30$

=> $E = 614598 \ W \cdot h$

=> $E = 614.6 \ K W \cdot h$

Generally the monthly electricity bill is mathematically represented as

$C = 614.6 * 50$

=> $C = 30729\ c$

8 0

2 years ago

Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft. Calculate the pressure at this elevation using three different

kramer

Answer:

a) P = 1240 lb/ft^2

b) P = 1040 lb/ft^2

c) P = 1270 lb/ft^2

Explanation:

Given:

- P_a = 2216.2 lb/ft^2

- β = 0.00357 R/ft

- g = 32.174 ft/s^2

- T_a = 518.7 R

- R = 1716 ft-lb / slug-R

- γ = 0.07647 lb/ft^3

- h = 14,110 ft

Find:

(a) Determine the pressure at this elevation using the standard atmosphere equation.

(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.

Solution:

- The standard atmospheric equation is expressed as:

P = P_a* ( 1 - βh/T_a)^(g / R*β)

(g / R*β) = 32.174 / 1716*0.0035 = 5.252

P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252

P = 1240 lb/ft^2

- The air density method which is expressed as:

P = P_a - γ*h

P = 2116.2 - 0.07647*14,110

P = 1040 lb/ft^2

- Using constant temperature ideal gas approximation:

P = P_a* e^ ( -g*h / R*T_a )

P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )

P = 1270 lb/ft^2

6 0

3 years ago

Read 2 more answers

A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (

kvv77 [185]

Answer:

v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

Lowest starting point.

Em₀ = K = ½ m v²

final point. Higher

$Em_{f}$ = U = m g h

Let's use trigonometry to lock her up

cos 60 = y / L

y = L cos 60

Height is the initial length minus the length at the maximum angle

h = L - L cos 60

h = L (1- cos 60)

energy is conserved

Em₀ = Em_{f}

½ m v² = mgL (1 - cos 60)

v = 2g L (1- cos 60)

let's calculate

v² = 2 9.8 3.0 (1- cos 60)

v = 29.4 m / s

6 0

3 years ago