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kompoz [17]
2 years ago
14

Cho hai điện tích q1=q2=8.10^-7 C đặt cách nhau 5cm. Xác định cường độ điện trường tại điểm:

Physics
1 answer:
Aleksandr [31]2 years ago
5 0

Answer: b

Explanation:

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Problem #2: An apple is thrown upward with an initial velocity of +24.0 m/s. a. Sketch the apple's trip and label what you know.
bogdanovich [222]

Answer:

The answer is below

Explanation:

a) The initial velocity (u) = 24 m/s

We can solve this problem using the formula:

v² = u² - 2gh

where v = final velocity, g= acceleration due to gravity = 9.8 m/s², h = height.

At maximum height, the final velocity = 0 m/s

v² = u² - 2gh

0² = 24² - 2(9.8)h

2(9.8)h = 24²

2(9.8)h = 576

19.6h = 576

h = 29.4 m

b) The time taken to reach the maximum height is given as:

v = u - gt

0 = 24 - 9.8t

9.8t = 24

t = 2.45 s

The total time needed for the apple to return to its original position = 2t = 2 * 2.45 = 4.9 s

4 0
2 years ago
_____ are formed where bumps from two surfaces come into contact ?
Murrr4er [49]

Answer:

the answer would be microwelds.

3 0
3 years ago
. A person weighing 750 N gets on an elevator.
Kobotan [32]

 

F = 750 N  (Force)

d = 10 m  (displacement )

t = 25 s   (time)

L = ?   (Mechanical work )  =  (Energy)

P = ?   (Power)

Solve:

L = F × d = 750 × 10 = 7500 Joules

P = L / t = 7500 / 25 = 300 Watts

5 0
2 years ago
How many volts would it take to push 1 amp through a resistance of 1 ohm?
ELEN [110]
V=I x R so V= 1 x 1 =1V
5 0
2 years ago
Read 2 more answers
A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?
posledela

Answer:

118\; \rm Hz.

Explanation:

The frequency f of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of v= 295\; \rm m\cdot s^{-1}. In other words, the wave would have traveled 295\; \rm m in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that 295\; \rm m? The wavelength of this wave\lambda = 2.50\; \rm m gives the length of one wave cycle. Therefore:

\displaystyle \frac{295\;\rm m}{2.50\; \rm m} = 118.

That is: there are 118 wave cycles in 295\; \rm m of this wave.

On the other hand, Because that 295\; \rm m of this wave goes through that point in each second, that 118 wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

f = 118\; \rm s^{-1} = 118\; \rm Hz.

The calculations above can be expressed with the formula:

\displaystyle f = \frac{v}{\lambda},

where

  • v represents the speed of this wave, and
  • \lambda represents the wavelength of this wave.

6 0
3 years ago
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