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3 years ago

An echo is a reflection of sound against another surface. Suppose you are standing on one side of a canyon and you

Physics

1 answer:

Morgarella [4.7K]3 years ago

4 0

Answer:

10-1 Temperature and Expansion. 135 ... text, the laboratory work that you do, or your physics teacher. ... Assume that the speed of sound in air is 340 m/s. How.

Explanation:

hope that helps

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Andrew kicks a ball along a straight path. The ball rolls straight forward for 13.2 meters. Then Andrew kicks the ball straight

tresset_1 [31]

Answer: 22.7 meters

Explanation: The distance traveled is how much the ball has rolled in total, this means the lenght of the path that it has followed from begining to end.

Since it first travels 13.2 meters and then 9.5 meters, if we sum this quantities:

13.2 + 9.5 = 22.7 meters

So 22.7 is the distance that the ball has traveled.

7 0

4 years ago

I need to talk to a girl my age

Greeley [361]

Answer:

that is my question

Explanation:

you want a girl to talk your age

6 0

3 years ago

What happens to a current when the electric field in the wire increase

Gnom [1K]

Answer:

If the voltage is increased then the electric field is higher, and electron velocity (average) is proportional to this field. Then you have an increase in speed. And current is total charge passing per time unit, so current is proportional to velocity value of charge (and to voltage in resistors and wire).

Explanation:

5 0

4 years ago

can someone help me answer this? i dont get it and I need help before the next 30 minutes. ill give a brainiest to whoever helps

Zinaida [17]

Answer:

S=5.83...

Explanation:

s=d/t

Speed=distance/time

s=35/10

s=5.83...

3 0

4 years ago

The gravitational attraction between a 20 kg cannonball and a 0.002 kg

Naya [18.7K]

Answer:

$2.966\times 10^{-11}\ N$

Explanation:

Given:

Mass of the cannonball (M) = 20 kg

Mass of the marble (m) = 0.002 kg

Distance between the cannonball and marble (d) = 0.30 m

Universal gravitational constant (G) = $6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2}$

Now, we know that, the gravitational force (F) acting between two bodies of masses (m) and (M) separated by a distance (d) is given as:

$F=\dfrac{GMm}{d^2}$

Plug in the given values and solve for 'F'. This gives,

$F=\frac{(6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2})\times (20\ kg)\times (0.002\ kg)}{(0.30\ m)^2}\\\\F=\frac{6.674\times 20\times 0.002\times 10^{-11}\ m^3 kg^{-1+2} s^{-2}}{0.09\ m^2}\\\\F=2.966\times 10^{-11}\ kg\cdot m\cdot s^{-2}\\\\F=2.966\times 10^{-11}\ N.........(1\ N = 1\ kg\cdot m\cdot s^{-2})$

The same force is experienced by both cannonball and marble.

Therefore, the gravitational force of the marble is $2.966\times 10^{-11}\ N$

3 0

3 years ago