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Nuetrik [128]
2 years ago
13

Could you help me with a science question really quick?

Physics
1 answer:
Rama09 [41]2 years ago
4 0
Yea I would love to help u
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Two loudspeakers are placed side by side and driven by the same source at 500 Hz. A listener is positioned in front of the two s
Oliga [24]

Answer:

0.68 m

Explanation:

We know that the speed of sound in air is a product of frequency and wavelength. Taking speed of sound in air as 340 m/s

V=frequency*wavelength

Then wavelength is given by 350/500=0.68 m

Therefore, to repeat constructive interference at the listener's ear, a distance of 0.68 m is needed

4 0
3 years ago
20 N effort is used in the 20 cm long spanner to unscrew a nut, then calculate momentum to unscrew the nut. [ Ans=4 Nm]
irga5000 [103]

Answer:

Refer to the attachment.

7 0
2 years ago
A 5kg object is moving at a height of 2 m. The potential energy of the object is closest to ___ j
elena55 [62]

Answer:

In this case, a body of mass 5 kg kept at a height of 10 m. So the potential energy is given as 5 * 10 *10 = 500 J.

6 0
3 years ago
Why do you think football players are usually large in size?
kiruha [24]
Football players in the NFL are large in size due to their eating habits since as players they have to work out constantly to stay and shape and eat to gain weight and be strong, so its important for them to be strong against their opponents in gameplay. Football players sometimes take steroids which make their muscles stronger and harder, and also gives them a growth spurt but the effects are worse and makes them sicker, high blood pressure, and even more. 
8 0
3 years ago
Read 2 more answers
An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Dete
dezoksy [38]

Answer:43.34 m

Explanation:

Given

acceleration(a)=2 m/s^2

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

v=0+2\times 6=12 m/s

After this object will start moving under gravity

height reached in first 6 s

s=ut+\frac{at^2}{2}

s=0+\frac{2\times 6^2}{2}

s=36 m

After fuel run out distance traveled in upward direction is

v^2-u^2=2as_0

here v=0

u=12 m/s

a=9.8 m/s^2

0-12^2=2(-9.8)(s)

s_0=\frac{144}{2\times 9.8}=7.34 m

s+s_0=36+7.34=43.34 m

7 0
3 years ago
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