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marissa [1.9K]
4 years ago
8

Why is a convex lens useful for starting fires

Physics
1 answer:
Anika [276]4 years ago
7 0

because if you were stuck on an island and you had a bag of of stuff and you also had a convex lens in there it could be good if u can't start a fire by rubbing to sticks instead you can use the convex lens to start a fire

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On the same spring day, a station near the equator has a surface temperature of 25°C, 15°C higher than the middle-latitude city
DedPeter [7]

Answer:

The air temperature at the tropopause is - 79 °C

Explanation:

We know that a station near the equator has a surface temperature of 25°C

Vertical soundings reveal an environmental lapse rate of 6.5 °C per kilometer.

The tropopause is encountered at 16 km.

In order to find the air temperature at the tropopause we are going to deduce a linear function for the temperature at the tropopause.

This linear function will have the following structure :

f(x)=ax+b

Where ''a'' and ''b'' are real numbers.

Let's write T(x) to denote the temperature '' T '' in function of the distance

'' x '' ⇒

T(x)=ax+b

We can reorder the function as :

T(x)=b+ax (I)

Now, at the surface the value of ''x'' is 0 km and the temperature is 25°C so in the function (I) we write :

T(0)=25=b+a(0) ⇒ b=25 ⇒

T(x)=25+ax (II)

In (II) the value of ''a'' represents the change in temperature per kilometer.

Because the temperature decrease with the height this number will be negative and also a data from the question ⇒

T(x)=25-(6.5)x (III)

In (III) we deduced the linear equation. The last step is to replace by x=16 in (III) ⇒

T(16)=25-(6.5)(16)=-79

The air temperature at the tropopause is - 79 °C

6 0
3 years ago
Describe what will happen to
Ostrovityanka [42]
It will be hotter and more like than 20celium degrees
4 0
3 years ago
A tennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height as the point of release. (a
Readme [11.4K]

Answer:

Part a)

a = -9.81 m/s/s

Part b)

v = 0

Part c)

v = 9.81 m/s

Part d)

H = 4.905 m

Explanation:

Part a)

During the motion of ball it will have only gravitational force on the ball

so here the acceleration of the ball is only due to gravity

so it is given as

a = g = 9.81 m/s^2

Part b)

As we know that ball is moving against the gravity

so here the velocity of ball will keep on decreasing as the ball moves upwards

so at the highest point of the motion of the ball the speed of ball reduce to zero

v_f = 0

Part c)

We know that the total time taken by the ball to come back to the initial position is T = 2 s

so in this time displacement of the ball will be zero

\Delta y = 0 = v_y t + \frac{1}{2} at^2

0 = v_y (2) - \frac{1}{2}(9.81)(2^2)

v_y = 9.81 m/s

Part d)

at the maximum height position we know that the final speed will be zero

so we will have

v_f^2 - v_i^2 = 2 a d

here we have

0 - (9.81^2) = 2(-9.81)H

H = 4.905 m

4 0
3 years ago
Two cars A and B are 100m apart moving towards each other with
maxonik [38]

Let car A's starting position be the origin, so that its position at time <em>t</em> is

A: <em>x</em> = (40 m/s) <em>t</em>

and car B has position at time <em>t</em> of

B: <em>x</em> = 100 m - (60 m/s) <em>t</em>

<em />

They meet when their positions are equal:

(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>

(100 m/s) <em>t</em> = 100 m

<em>t</em> = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) <em>t</em> = 200 m

<em>t</em> = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

3 0
3 years ago
How does a computer process data? *
elena55 [62]

The correct answer D: all of the above

5 0
3 years ago
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