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Kitty [74]
3 years ago
11

If the box moves at a constant velocity, what do you know about the forces acting on the object?

Physics
2 answers:
Anon25 [30]3 years ago
6 0
The answer is A , the forces are balanced
pantera1 [17]3 years ago
3 0
A, this is the answer hope this helps
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Kinetic energy is the answer to your question.
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3 years ago
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Calculate kinetic energy of a planet using 5.97 x 10^24 kg mass and v at 30.29 km s-1
kiruha [24]

Answer:

2.74 × 10^33 J

Explanation:

the formula to calculate kinetic energy is:

1/2mv²

m= mass (kg)

v= velocity (m/s)

given that,

m = 5.97 × 10^24

v = 30.29 km s-1

  = 30290 m s-1

1/2× 5.97 × 10^24 × 30290²

=2.74 × 10^33 J

4 0
2 years ago
What will the transistor look like in the future
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<u>TRANSISTOR:</u>

The future of transistors based on the ongoing researches suggests that, it would be based on nano technology. This technology is itself still under development and the case is similar in the case of transistors too.

Transistors due to their applications, especially its application as an amplifier has increased its usage and the dependability over them. The hard part often arises with the fabrication techniques which are being tested and developed as we speak in order to increase the efficiency of the transistors and also to reduce their physical size though, these would mean that, the design complications would be higher.

5 0
3 years ago
A flat screen is located 0.55 m away from a single slit. Light with a wavelength of 544 nm (in vacuum) shines through the slit a
kobusy [5.1K]

Answer:

Explanation:

The diffraction pattern is given as

Sinθ = mλ/ω

Where m=1,2,3,4....

Now, when m=1

Sinθ = λ/ω

Then,

ω = λ/Sinθ

The width of the central bright fringe is given as

y=2Ltanθ. From trigonometric

Then,

θ=arctan(y/2L)

Given that,

y=0.052m

L=0.55m

θ=arctan(0.052/2×0.55)

θ=arctan(0.0473)

θ=2.71°

Substituting this into

ω = λ/Sinθ

Since λ=544nm=544×10^-9m

Then,

ω = 544×10^-9/Si.2.71

ω = 1.15×10^-5m.

5 0
3 years ago
Consider a motor that exerts a constant torque of 25.0 N⋅m to a horizontal platform whose moment of inertia is 50.0 kg⋅m2 . Assu
Step2247 [10]

To solve this exercise it is necessary to apply the concepts related to Work and Kinetic Energy. Work from the rotational movement is described as

W=\tau \Delta\theta

In the case of rotational kinetic energy we know that

KE = \frac{1}{2}I\omega^2

PART A) \theta is given in revolutions and needs to be in radians therefore

\theta = 12rev(\frac{2\pi rad}{1rev})

\theta = 24\pi rad

Replacing in the work equation we have to

W=\tau \Delta\theta

W= (25)(24\pi)

W = 1884.95J

PART B) From the torque and moment of inertia it is possible to calculate the angular acceleration and the final speed, with which the kinetic energy can be determined.

\tau = I \alpha

Rearrange for the angular acceleration,

\alpha = \frac{\tau}{I}

\alpha = \frac{25}{50}

\alpha = 0.5rad/s

From the kinematic equations of angular motion we have,

\omega_f^2=\omega_i^2+2\alpha\theta

\omega_f^2=0+2*0.5*24\pi

\omega_f=\sqrt{0+2*0.5*24\pi}

\omega_f = 8.68rad/s

In this way the rotational kinetic energy would be given by

KE = \frac{1}{2}I\omega_f^2

KE = \frac{1}{2}(50)(8.68)^2

KE = 1883.56J

3 0
3 years ago
Read 2 more answers
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