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alexdok [17]
3 years ago
10

A 2-kg object is moving at 3 m/s. A 4-N force is applied in the direction of motion and if the object has a final velocity 7m/s.

calculate the distance that the object moves
Physics
1 answer:
aniked [119]3 years ago
7 0
Intial velocity u=3m/s
final velocity v
2
=u
2
+2as=3
2
+(2×2×5)=29 ⟹v=5.3m/s

KE=
2
1
​
m(v
2
−u
2
)=
2
1
​
×2×((5.3)
2
−3
2
)=20J
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The inductive reactance of the circuit is exactly twice the resistance: XL=2R. Adjust the phasor that represents the voltage acr
SVETLANKA909090 [29]

Answer:

∅=63.43^{0}

Explanation:

z=impedance

x_{l}=2R

R=R

The resultant of the resistances in the circuit is called impedance

x_{l} is inductive reactance of the circuit

R is the resistance of the resistor

z=\sqrt{xl^{2}+R^2 }

z=\sqrt{2R^{2}+R^2 }

Z=\sqrt{5R^2}

Z=R\sqrt{5}ohms

tan∅=2R/R

tan∅=2

∅=Tan^-1(2)

∅=63.43^{0}

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3 0
3 years ago
A wave traveling in the positive x-direction with a frequency of 50.0 Hz is shown in the figure below. Find the following values
Klio2033 [76]

Answer:

Explanation:

a. The amplitude is the measure of the height of the wave from the midline to the top of the wave or the midline to the bottom of the wave (called crests). The midline then divides the whole height in half. Thus, the amplitude of this wave is 9.0 cm.

b. Wavelength is measured from the highest point of one wave to the highest point of the next wave (or from the lowest point of one wave to the lowest point of the next wave, since they are the same). The wavelength of this wave then is 20.0 cm. or \lambda=20.0cm

c. The period, or T, of a wave is found in the equation

f=\frac{1}{T} were f is the frequency of the wave. We were given the frequency, so we plug that in and solve for T:

50.0=\frac{1}{T} so

T=\frac{1}{50.0} and

T = .0200 seconds to the correct number of sig fig's (50.0 has 3 sig fig's in it)

d. The speed of the wave is found in the equation

f=\frac{v}{\lambda} and since we already have the frequency and we solved for the wavelength already, filling in:

50.0=\frac{v}{20.0} and

v = 50.0(20.0) so

v = 1.00 × 10³ m/s

And there you go!

5 0
3 years ago
A skater rotates with her arms crossed at an angular speed of 8.0 rad/s and she has a moment of inertia of 100 kg/m2. She extend
solong [7]

Answer:

When her hands extends, her momen of inertia is 4.28\ kg-m^2.

Explanation:

Given that,

Initial angular speed, \omega_i=8\ rad/s

Initial moment of inertia, I_1=100\ kg-m^2

Final angular speed, \omega_f=7\ rad/s

Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :

I_1\omega_1=I_2\omega_2

I_2 is final moment of inertia

I_2=\dfrac{I_1\omega_1}{\omega_2}\\\\I_2=\dfrac{100\times 8}{7}\\\\I_2=114.28\ kg-m^2

So, when her hands extends, her momen of inertia is 4.28\ kg-m^2. Hence, this is the required solution.

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'A' is correct. B, C, and D are false statements.
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