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MariettaO [177]
3 years ago
12

Use the diagram below to answer the following question:

Physics
1 answer:
d1i1m1o1n [39]3 years ago
5 0

Answer:

3.0 cm

Explanation:

We can solve this problem by using the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length of the mirror

p is the distance of the object from the mirror

q is the distance of the image from the mirror

In this problem we have:

f = 1.5 cm is the focal length of the mirror (positive for a concave mirror)

p = 3.0 cm is the distance of the object from the mirror

Therefore, the distance of the image is:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{1.5}-\frac{1}{3.0}=\frac{1}{3.0}\\\rightarrow q=3.0 cm

And the positive sign means that the image is real.

(The second part of the exercise is just the description of the image of the first exercise).

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JUST PLZ HELP!!! Why does the lightbulb in the right electrical circuit turn on but not the one on the left?
jekas [21]

In the open circuit the current can not flow from one end of the power source to the other. Because of this there is no current flow, and therefore the light does not turn on.

8 0
3 years ago
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How much work is done by the force lifting a 0.1-kilogram hamburger vertically upward at a constant velocity 0.3 meter from a ta
pychu [463]

Answer:

0.2943 Nm

Explanation:

Work done is given a the product of force and diatance moved and expressed by the formula

W=Fd

Here W represent work, F is applied force and d is perpendicular distance

Also, we know that F=mg where m is the mass of an object and g is acceleration due to gravity. Substituting this back into the initial equation then

W=mgd

Taking acceleration due to gravity as 9.81 m/s2 and substituting mass with 0.1 kg and distance with 0.3 m then

W=0.1*9.81*0.3=0.2943 Nm

5 0
3 years ago
Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet
Valentin [98]

Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

       

starting point. Higher

         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

        v² = 215.6 -150.16

        v = √65.44

        v = 8.09   m/s

6 0
3 years ago
A construction crane, like the one shown, has a power output of 1,500 watts. If it takes 200 seconds to lift the roof to the top
Likurg_2 [28]

Answer:7.5j

Explanation:

P=w/s

4 0
3 years ago
Dos cargas Q1=2pc y Q2=4pc estan separadas por una distancia de 6cm ¿con que fuerza se atraen?
noname [10]

Here we can use coulomb's law to find the force between two charges

As per coulombs law

]tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we have

k = 9 * 10^9

q_1 = 2pC

q_2 = 4pC

r = 6cm = 0.06 m

now by using the above equation we have

F = \frac{9*10^9 * 2*10^{-12} * 4*10^{-12}}{0.06^2}

F = 2 * 10^{-11} N

so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.

3 0
3 years ago
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