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3 years ago

Scientists _____. Select all that apply.

1 answer:

5 0

The true answer is the choice A
Scientists can predict volcanic eruptions as long as adequate resources are available.Volcanic eruptions are one of the earthly activities which can be predicted by means of adequate resources (warming of the earth's crust, appearance of volcanic lava etc...). By contrast, earthquakes is a dangerous natural disaster that scientists cannot be predicted or mastering the time where it occurs.

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Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.

Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

We can solve this first part of the problem by applying the mirror equation:

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

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

$f=\frac{R}{2}$

For this mirror, R = 20 cm, so its focal length is

$f=\frac{20}{2}=+10 cm$ (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm$

We can solve this part by using the magnification equation:

$M=-\frac{y'}{y}=\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

$y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm$

and the negative sign means that the image is inverted.

#LearnwithBrainly

6 0

3 years ago

A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco

Sunny_sXe [5.5K]

Answer:

(a)

$x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km$

(b) $r=41.32km$

$\alpha =24.23^o$

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies $45^o$ south of east. The y-component will be negative and the x-component will be positive.

$x_1=25cos45^o=17.68km$

$y_1=-25sin45^o=-17.68km$

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

$x_2=40cos60^o=20km$

$y_2=40sin60^o=34.64km$

Therefore, total displacements:

$x=x_1+x_2=(17.68+20)km=37.68km$

$y=y_1+y_2=(-17.68+34.64)km=16.96km$

Magnitude of displacements,

$r=\sqrt{x^2+y^2}=41.32km$

Direction,

$\alpha =tan^{-1}(\frac{y}{x})=24.23^o$

6 0

3 years ago

Consider the cross section to the right with an overall height h 4.2 inch, bottom flange plate width b1-2.5 inch, top flange pla

ICE Princess25 [194]

Answer: a) 1.98inch

b) 2.737 2.44

c) 0.497inch

d) 1.58inch

Explanation:

Solution is attached in the pictures below

7 0

3 years ago

A box of mass 12 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.42. What is th

vladimir2022 [97]

As we know that friction force on box is given by

$F_s = \mu_s N$

here we know that

$N = mg$

here we have

m = 12 kg

$\mu_s = 0.42$

so now we have

$N = 12(9.8) = 117.6 N$

now we will have

$F_s = 0.42(12)(9.8)$

$F_s = 49.4 N$

so it required minimum 49 N(approx) force to move the block

5 0

3 years ago

Read 2 more answers

A brick falls to the ground. if the time for the collision of the brick and the ground is increased by a factor of 4, the force

melomori [17]

Answer:

By a factor of 1/4.

Explanation:

The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,

$\\\begin{aligned}\\\small F &=\small \frac{\Delta (mV)}{\Delta T}\end{aligned}$

in which $\small \Delta (mV)$ , $\small \Delta t$ represent the change in momentum and the time taken for that change.

If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by $\small F_1$ , the following manipulation confirms the answer to this question.

$\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}$

Here $\small F$ is the force that was applied to the object previously.

#SPJ4

4 0

2 years ago