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3 years ago

6

Hii please help i’ll give brainliest if you give a correct answer please

1 answer:

Diano4ka-milaya [45]3 years ago

6 0

Answer:

A

Explanation:

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

den301095 [7]

A: 0.7 -0.8 and 0hrs
B: 0.2 - 0.4 and 1.0 -1.1 hrs
C: 0-0.2 hrs and 0.8 - 1.0 hrs
D: 0.4 - 0.7 hrs

4 0

3 years ago

Read 2 more answers

A coil of conducting wire carries a current of i(t) = 14.0 sin(1.15 ✕ 103t), where i is in amperes and t is in seconds. A second

Advocard [28]

Answer:

The peak emf in second coil is 1.876 V

Explanation:

Given :

Inductance $L = 130 \times 10^{-6}$ H

The current $I(t) = 14 \sin(1.15\times 10^{3} t)$

We compare above equation with standard equation,

$I(t) = I_{o} \sin (\omega t + \phi)$

From above equation we have,

$\omega = 10^{3}$ and $\phi = 1.15$

Find the inductive resistance,

$X_{L} = \omega L$

$X_{L} = 10^{3} \times 130 \times 10^{-6}$

$X_{L} = 0.134$

The peak emf in second coil is,

$V = I_{o} X_{L}$

$V = 14 \times 0.134$

$V = 1.876$ V

Therefore, the peak emf in second coil is 1.876 V

8 0

4 years ago

Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out

PolarNik [594]

Answer:

The two value of the wavelength for the out of tune guitar is

$\lambda _2 = (6.48,6.52) \ cm$

Explanation:

From the question we are told that

The wavelength of the note is $\lambda = 6.50 \ cm = 0.065 \ m$

The difference in beat frequency is $\Delta f = 17.0 \ Hz$

Generally the frequency of the note played by the guitar that is in tune is

$f_1 = \frac{v_s}{\lambda}$

Where $v_s$ is the speed of sound with a constant value $v_s = 343 \ m/s$

$f_1 = \frac{343}{0.0065}$

$f_1 = 5276.9 \ Hz$

The difference in beat is mathematically represented as

$\Delta f = |f_1 - f_2|$

Where $f_2$ is the frequency of the sound from the out of tune guitar

$f_2 =f_1 \pm \Delta f$

substituting values

$f_2 =f_1 + \Delta f$

$f_2 = 5276.9 + 17.0$

$f_2 = 5293.9 \ Hz$

The wavelength for this frequency is

$\lambda_2 = \frac{343 }{5293.9}$

$\lambda_2 = 0.0648 \ m$

$\lambda_2 = 6.48 \ cm$

For the second value of the second frequency

$f_2 = f_1 - \Delta f$

$f_2 = 5276.9 -17$

$f_2 = 5259.9 Hz$

The wavelength for this frequency is

$\lambda _2 = \frac{343}{5259.9}$

$\lambda _2 = 0.0652 \ m$

$\lambda _2 = 6.52 \ cm$

8 0

3 years ago

How fast does a 500kg car need to drive to have 100,000 J of kinetic energy?

olchik [2.2K]

Answer:

The car must move at 2 m/s to have a Ke of 2,000 Joules.

Explanation:

Mark me pls

8 0

3 years ago

A hydrogen-filled balloon is used to lift a 120-kg stone off the ground. The basket holding the stone has a mass of 10.0 kg . Pa

Kaylis [27]

Answer:

Radius, r = 2.88 meters.

Explanation:

It is given that,

A hydrogen-filled balloon is used to lift a 120-kg stone off the ground, m₁ = 120 kg

The basket holding the stone has a mass of 10.0 kg, m₂ = 10 kg

Mass density of air, $d=1.29\ kg/m^3$

We know that buoyant force is equal to the weight of liquid displaced i.e.

$B=mg$

$B=(m_1+m_2)g$

$B=(120+10)\times 9.8$

B = 1274 N

Also, buoyant force is given by :

$B=dVg$

$B=\dfrac{4\pi r^3dg}{3}$

$r^3=\dfrac{3B}{4\pi dg}$

$r^3=\dfrac{3\times 1274}{4\pi \times 1.29\times 9.8}$

$r=2.88\ m$

So, the minimum radius R of the balloon be in order to lift the stone off the ground is 2.88 meters. Hence, this is the required solution.

3 0

4 years ago