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Natalija [7]
3 years ago
6

Is a bird sitting motionless on a perch, balanced or unbalanced?

Physics
1 answer:
4vir4ik [10]3 years ago
3 0
Balanced hope that helps
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A 4.0 kg block is resting on a rough horizontal table. The
Novosadov [1.4K]

Answer:

I have made a mistake in my answer

6 0
2 years ago
Sound travels through air at 343 m/s at 20 °C. A bat emits an ultrasonic squeak and hears the echo 0.05 second later . How far a
Rashid [163]

Answer:

Distance of the object is 8.6 m

Explanation:

As we know that the speed of sound at t degree C is given as

v = 332 + 0.6 t

here we know that the temperature is

t = 20 degree C

so we have

v = 332 + 0.6(20)

v = 344 m/s

now we know that bat heard the echo of sound in 0.05 s

so the to and fro distance of the object is d + d = 2d

so we have

2 d = v\times t

2d = 344(0.05)

d = 8.6 m

7 0
3 years ago
A 0.75-kg ball falls vertically downward from a height of 55.0 m and rebounds upward. if the ball reaches a height of 30.0 m and
Nikitich [7]
Impuls I is given by:
I = \delta mv = |F|t
where \delta mv is the change in momentum, |F| is the average force and t is the time.

Solve the equation for the force F:
|F| = \frac{\delta mv}{t} = \frac{m(v_f - v_i)}{t}

Energy should be conserved, so the velocities will be:
\frac{1}{2}mv^2 = mgh \\ v = \sqrt{2gh}

Combining both equations:
|F| =  \frac{m( \sqrt{2g(h_f + h_i)} )}{t}
where h_f = 30m, h_i = 55m, m = 0.75kg, t = 0.0025s
4 0
3 years ago
34.9x46x809 Please helpp
Fed [463]
<h2>34.9×46×809</h2><h3>=1605.4×809 </h3><h3>=1298768.6</h3>

please mark this answer as brainlist

6 0
2 years ago
Suppose you are planning a trip in which a spacecraft is totravel at a constant velocity for exactly six months, as measuredby a
Eva8 [605]

Answer: 0.999959 c

Explanation:  

According to the special relativity theory, time is measured differently by two observers moving one relative another, according to the Lorentz Transform Equation, as follows:

t = t’ / t=t^'/√(1-(v)2/c2 )

where t= time for the moving observer (relative to the spacecraft, fixed on Earth) = 110 years.

t’= time for the observer at rest respect from spacecraft = 1 year

v= spacecraft constant speed

c= speed of light

Solving for v, with a six decimals precision as a multiple of c, we get:

v = 0.999959 c

5 0
2 years ago
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