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BlackZzzverrR [31]

2 years ago

6

A convex mirror has a focal length of -13.0 cm. When you place a 6.00 cm tall pencil 60.0 cm in front of that mirror, what is th

e location of the pencil's image?

1 answer:

wlad13 [49]2 years ago

6 0

The location of the pencil's image is =16.6cm

<h3>Calculation of image location</h3>

The focal length of the convex mirror (f)= -13.0 cm

The object distance (u) = 60cm

The image distance (v) = xcm

Using the formula,1/f= 1/v + 1/u

Make 1/v the subject of formula,

1/v = 1/f -1/u

1/v = 1/13 - 1/60

1/v = 60-13/780

1/v = 47/780

V = 780/47

V = 16.6 cm

Therefore, he location of the pencil's image is = 16.6cm

Learn more about mirror here:

brainly.com/question/13164847

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Can someone check my answers and tell me if their correct?

Otrada [13]

Seven

The magnitude is pointing towards the origin and is at - 20 degrees. The combination makes 160 with the x axis: C answer

Eight

They keep doing this. They use distance where they should use displacement but they use distance to try and fool you. It's a mighty poor practice.

The distance between the start and end points is the displacement. That "distance" is 180*sqrt(25) = 900 . The actual distance should be 180*4 + 180*3 = 720 + 540 = 1260. That's what a car's odometer or a bicycle odometer would read. the difference is 360.

I really do object to the wording, but what can I do?

Nine

Nine is the same thing as 8.

Displacement = sqrt(400^2 + 80^2)= sqrt(166400) = 408

The actual distance is 400 + 80 = 480

The difference is the answer = 480 - 408 = 72 <<<< Answer

Ten

This is just the displacement magnitude.

dis = sqrt(30^2 + 80^2)

dis = sqrt(900 + 6400)

dis = sqrt(7300)

dis = 85.44 <<<< Answer D

Twelve

Vi = 2.15*Sin(30) = 1.075 m/s

vf = 0

a = - 9.81

t = ?

<u>Formula</u>

a = (vf - vi)/t

<u>Solve</u>

-9.81 = (0 - 1.075)/t

- 9.81 * t = -1.075

t = 0.11 seconds

Thirteen

I'm leaving this last one to you. You need the initial height xo to answer it properly. Judging by the other questions, this one is right.

Edit

That is a surprise! Really quickly

d = 3.2 m

a = - 9.82

vf = 0

vi = ?

vf^2 = vi^2 - 2*a*d

0 = vi^2 - 2*9.81*3.2

vi = sqrt(19.62*3.2)

vi = 8.0 m/s But that is the vertical component of the speed

v = vi/sin(25)

v = 8.0/sin(25) = 11

6 0

3 years ago

A rocket achieves a lift-off velocity of 500.0 m/s from rest in 30.0 seconds. Calculate the average acceleration of the rocket.

Anettt [7]

Answer:

The average acceleration is 16.6 m/s² ⇒ 1st answer

Explanation:

A rocket achieves a lift-off velocity of 500.0 m/s from rest in

30.0 seconds

The given is:

→ The initial velocity = 0

→ The final velocity = 500 meters per seconds

→ The time is 30 seconds

Acceleration is the rate of change of velocity of the rocket

→ $a=\frac{v-u}{t}$

where a is the acceleration, v is the final velocity, u is the initial velocity

and t is the time

→ u = 0 , v = 500 m/s , t = 30 s

Substitute these values in the rule

→ $a=\frac{500-0}{30}=\frac{500}{30}=16.6$ m/s²

<em>The average acceleration is 16.6 m/s²</em>

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3 years ago

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2 years ago

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Suppose you could suddenly increase the speed of every molecule in a gas by a factor of 2. Would the gas pressure increase by a

Elena L [17]

<h2>The pressure will become double </h2>

Explanation:

The gas pressure is directly proportional to the mean root square velocity of the constituent molecules of gas .

P ∝ $\sqrt{\frac{C_1^2+C_2^2+C_3^2----C_n^2}{n} }$ I

Here C₁ , C₂ ------------ Cₙ is the velocities of molecules .

By making these velocities double

The pressure P₀ ∝ 2 $\sqrt{\frac{C_1^2+C_2^2+C_3^2----C_n^2}{n} }$ II

By dividing II by I

P₀ = 2 P

Thus pressure will become double than its previous value

5 0

3 years ago

Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on

Romashka [77]

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$\frac{1}{4\pi\spsilon_0}=9\times10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35\times10^{-4}$ N

So, from equation (i)

$1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}$

$\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$\frac{q_1+q_2}{2}$

d=20cm=0.2m, and $F_c=1.406\times10^{-4}$ N

So, from equation (i)

$1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}$

$\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}$

$\Rightarrow q_1+q_2=\pm5\times 10^{-8}$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)$

$\Rightarrow q_1=5\times 10^{-8}-q_2$

The equation (ii) become:

$(5\times 10^{-8}-q_2)q_2=6\times10^{-16}$

$\Rightarrow -(q_2)^2+5\times 10^{-8}q_2-6\times10^{-16}=0$

$\Rightarrow q_2=3\times10^{-8}, 2\times10^{-8}$

From equation (iii)

$q_1=2\times10^{-8}, 3\times10^{-8}$

So, the magnitude of initial charges on both the sphere are $3\times10^{-8}$ Coulombs $=0.03 \mu C$ and $2\times10^{-8}$ Colombs or $0.02 \mu C$ .

Considerion the nature of charges too,

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

4 0

3 years ago