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I am Lyosha [343]
2 years ago
15

if a copper coin with a volume of exactly 50 cubic cm is dropped into the volume of water 4cubic cm,how much water will be overf

lown (water is half way of the can)
Physics
1 answer:
Alenkinab [10]2 years ago
7 0

Answer:

50 cubic cm

Explanation:

As the water displaces 50 cubic centimeters the amount of water that must leave the container for it to be able to be dropped into it is also 50 cubic as the water which was in its place must have gone somewhere.

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Question 14 (2 points)
sveticcg [70]

Answer:

after 2 seconds its velocity is -20 m/s. after 3 seconds its velocity is -30 m/s. after 10 seconds its velocity is -100 m/s.

Explanation:

This is my answer.

7 0
3 years ago
If 20 beats are produced within a single second, which of the following frequencies could possibly be held by two sound waves tr
zhuklara [117]

The correct choice is

D. 22 Hz and 42 Hz.

In fact, the beat frequency is given by the difference between the frequencies of the two waves:

f_B = |f_1 -f_2|

In this problem, the beat frequency is f_B=20 Hz, therefore the only pair of frequencies that gives a difference equal to 20 Hz is

D. 22 Hz and 42 Hz.

4 0
3 years ago
Read 2 more answers
4)
Whitepunk [10]
A. 0 charge

15 . A
17. C
6 0
3 years ago
A river flows due south with a speed of 2.0 m/s. You steer a motorboat across the river; your velocity relative to the water is
neonofarm [45]

Answer:

a) 25.5°(south of east)

b) 119 s

c) 238 m

Explanation:

solution:

we have river speed v_{r}=2 m/s

velocity of motorboat relative to water is v_{m/r}=4.2 m/s

so speed will be:

a) v_{m}=v_{r}+v_{m/r}

solving graphically

v_{m}=\sqrt{v^2_{r}+v^2_{m/r}}

     =4.7 m/s

Ф=tan^{-1} (\frac{v_{r}}{v_{m/r}} )

  =25.5°(south of east)

b) time to cross the river: t=\frac{w}{v_{m/r}}=\frac{500}{4.2}=119 s

c) d=v_{r}t=(2)(119)=238 m

note :

pic is attached

6 0
3 years ago
g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts,
Stolb23 [73]

Answer:

523269.9\ \text{N/m}

Explanation:

q = Charge

r = Distance

q_1=25\ \text{C}

r_1=3000\ \text{m}

q_2=40\ \text{C}

r_2=850\ \text{m}

The electric field is given by

E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}

The electric field at the aircraft is 523269.9\ \text{N/m}

4 0
3 years ago
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