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4 years ago

Burning oil and coal adds to the atmosphere.

Physics

1 answer:

True [87]4 years ago

8 0

Answer:

carbon dioxide

Explanation:

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A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above

erastova [34]

Answer:

t = 2 s

Explanation:

In order to find the time taken by the stone to fall from the top of the building to the ground we can use 2nd equation of motion. 2nd equation of motion is as follows:

s = Vit + (0.5)gt²

where,

t = time = ?

Vi = Initial Velocity = 20 m/s

s = height of building = 60 m

g = 9.8 m/s²

Therefore,

60 m = (20 m/s)t + (0.5)(9.8 m/s²)t²

4.9t² + 20t - 60 = 0

solving this quadratic equation we get:

t = -6.1 s (OR) t = 2 s

Since, the time cannot be negative in magnitude.

Therefore,

6 0

3 years ago

Which of the following is not a potential sign of chemical change? a. release of gas c. change of color b. evaporation of water

vitfil [10]

B is the answer. because it is a physical change

5 0

3 years ago

A mouse ran 25 meters in 5 seconds, stopped for 10 seconds to eat a piece of cheese, and finally ran another 5 seconds a distanc

Mkey [24]

Answer:

It ran at an average of 2 meters per second.

Explanation:

6 0

3 years ago

What are the two most important properties of a telescope?

tigry1 [53]

1.Light-collecting area
2.Angular resolution

8 0

4 years ago

A person 1.8m tall stands 0.75m from a reflecting globe in a garden.

Maru [420]

Answer:

1. The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2. The person's image is 3.38 m tall.

Explanation:

From the given question, object distance, u = 0.75 m, object height = 1.8 m, radius of curvature of the reflecting globe, r = 8 cm = 0.08 m.

f = $\frac{r}{2}$ = $\frac{0.08}{2}$ = 0.04 m

1. The image distance, v, can be determined by applying mirror formula:

$\frac{1}{f}$ = $\frac{1}{u}$ + $\frac{1}{v}$

$\frac{1}{0.04}$ = $\frac{1}{0.75}$ + $\frac{1}{v}$

$\frac{4}{100}$ - $\frac{75}{100}$ = $\frac{1}{v}$

$\frac{1}{v}$ = $\frac{4 - 75}{100}$

= - $\frac{71}{100}$

⇒ v = - $\frac{100}{71}$

= - 1.41 m

The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2. $\frac{image distance}{object distance}$ = $\frac{image height}{object height}$

$\frac{1.41}{0.75}$ = $\frac{v}{1.8}$

v = $\frac{2.538}{0.75}$

= 3.384

v = 3.38 m

The person's image is 3.38 m tall.

6 0

3 years ago