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2 years ago

. The Soviet Yuri Gagarin was the first human to orbit Earth true or false

Physics

1 answer:

Dmitrij [34]2 years ago

3 0

Answer:

true , I searched and got u the answer

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light of wavelength 485 nm passes through a single slit of width 8.32 *10^-6m. what is the single between the first (m=1) and se

Assoli18 [71]

Answer:

3.35

Explanation:

Got it on Acellus

3 0

2 years ago

An initially uncharged 3.47-μF capacitor and a 6.43-kΩ resistor are connected in series to a 1.50-V battery that has negligible

harkovskaia [24]

Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms

Explanation: In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:

I(t)=io*Exp(-t/τ)

and also we consider that io=V/R=(1.5/6.43*10^3)

=233.28 A

then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6

=22.31 ms

Finally the time to reduce the current to 2.57% of its initial value is obtained from:

I(t)=io*Exp(-t/τ) for I(t)/io=0.0257=Exp(-t/τ) then

ln(0.0257)*τ =-t

t=-ln(0.0257)*τ=81.68 ms

3 0

3 years ago

A bald eagle is flying to the left with a speed of 34 meters

Shtirlitz [24]

Answer:

the speed after 3 seconds is 10 m/s

Explanation:

The computation of the speed is shown below:

As we know that

V = U + at

Here,

U = 34 m/s

a = - 8 m/s²

t = 3 Sec

V = velocity after 3 sec

V = 34 + (-8)3

= 34 - 24

V = 10 m/s

Hence, the speed after 3 seconds is 10 m/s

4 0

2 years ago

A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi

Serggg [28]

Answer:

The spring force constant is $k=243\ \frac{N}{m}$ .

Explanation:

We are told the mass of the ball is $m=0.0555\ kg$ , the height above the spring where the ball is dropped is $h=0.536\ m$ , the length the ball compresses the spring is $d=0.04897\ m$ and the acceleration of gravity is $9.8\ \frac{m}{s^{2}}$ .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy $E_{mi}$ is equal to the final mechanical energy $E_{mf}$ :

$E_{mf}=E_{mi}$

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

$\frac{1}{2}kd^{2}=mgh$