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Elis [28]
2 years ago
15

Obtaine the resultant of the forces 20N-40N ​

Physics
1 answer:
ivann1987 [24]2 years ago
8 0

Answer:

- 20 N

Explanation:

20 N - 40 N = - 20 N

The resultant force is - 20 N.

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A ball is thrown upward from an initial height of 1.5m the ball reaches a height of 5m then falls to the ground . What Is the di
jeka94

Answer:

The distance traveled by the ball is 8.5 m

Explanation:

Initial height of the ball, h₁ = 1.5 m above the ground

final height of the ball, h₂ = 5m

Upward distance = distance traveled by the ball from a height of 1.5m to 5m = 5m - 1.5m = 3.5 m

Downward distance = distance traveled by the ball from 5m height to the ground =5m - 0 = 5m

Total distance traveled = upward distance + downward distance

Total distance traveled = 3.5 m + 5m = 8.5 m

Therefore, the distance traveled by the ball is 8.5 m

4 0
3 years ago
A tube with a cap on one end, but open at the other end, has a fundamental frequency of 130.8 Hz. The speed of sound is 343 m/s
sergey [27]

Answer:

Y = V / f      where Y equals wavelength

4 Y1 = V / f1       for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2   for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1      dividing equations

f2 = 2 f1  

the new fundamental frequency is 2 * 130.8 = 261.6

(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.

8 0
2 years ago
Which of the following is a scalar quantity?
neonofarm [45]

Answer:

55

Explanation:

I hope this answer help u

4 0
2 years ago
Read 2 more answers
A wildebeest runs with an average speed of 4.0\,\dfrac{\text m}{\text s}4.0 s m ​ 4, point, 0, start fraction, start text, m, en
devlian [24]

Answer:

60m

Explanation:

4 0
3 years ago
Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of
IRISSAK [1]

Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

6 0
3 years ago
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